standard,chapter_id,chapter_name,file_path,data class_6,1,Patterns in Mathematics,ncert_books/class_6/Ganita_Prakash/fegp101.pdf,"1 1.1 What is Mathematics? Mathematics is, in large part, the search for patterns, and for the explanations as to why those patterns exist. Such patterns indeed exist all around us—in nature, in our homes and schools, and in the motion of the sun, moon, and stars. They occur in everything that we do and see, from shopping and cooking, to throwing a ball and playing games, to understanding weather patterns and using technology. The search for patterns and their explanations can be a fun and creative endeavour. It is for this reason that mathematicians think of mathematics both as an art and as a science. This year, we hope that you will get a chance to see the creativity and artistry involved in discovering and understanding mathematical patterns. It is important to keep in mind that mathematics aims to not just find out what patterns exist, but also the explanations for why they exist. Such explanations can often then be used in applications well beyond the context in which they were discovered, which can then help to propel humanity forward. PATTERNS IN MATHEMATICS Chapter 1_Patterns in Mathematics.indd 1 13-08-2024 16:10:52 Reprint 2025-26 Ganita Prakash | Grade 6 For example, the understanding of patterns in the motion of stars, planets, and their satellites led humankind to develop the theory of gravitation, allowing us to launch our own satellites and send rockets to the Moon and to Mars; similarly, understanding patterns in genomes has helped in diagnosing and curing diseases—among thousands of other such examples. 1.2 Patterns in Numbers Among the most basic patterns that occur in mathematics are patterns of numbers, particularly patterns of whole numbers: Figure it Out 1. Can you think of other examples where mathematics helps 2. How has mathematics helped propel humanity forward? (You us in our everyday lives? might think of examples involving: carrying out scientific experiments; running our economy and democracy; building bridges, houses or other complex structures; making TVs, mobile phones, computers, bicycles, trains, cars, planes, calendars, clocks, etc.) Math Talk Chapter 1_Patterns in Mathematics.indd 2 13-08-2024 16:10:52 2 The branch of Mathematics that studies patterns in whole numbers is called number theory. Number sequences are the most basic and among the most fascinating types of patterns that mathematicians study. Table 1 shows some key number sequences that are studied in Mathematics. Reprint 2025-26 0, 1, 2, 3, 4, ... Figure it Out 1. Can you recognise the pattern in each of the sequences in 2. Rewrite each sequence of Table 1 in your notebook, along 1, 1, 1, 1, 1, 1, 1, ... (All 1’s) 1, 2, 3, 4, 5, 6, 7, ... (Counting numbers) 1, 3, 5, 7, 9, 11, 13, ... (Odd numbers) 2, 4, 6, 8, 10, 12, 14, ... (Even numbers) 1, 3, 6, 10, 15, 21, 28, ... (Triangular numbers) 1, 4, 9, 16, 25, 36, 49, ... (Squares) 1, 8, 27, 64, 125, 216, ... (Cubes) 1, 2, 3, 5, 8, 13, 21, ... (Virahānka numbers) 1, 2, 4, 8, 16, 32, 64, ... (Powers of 2) 1, 3, 9, 27, 81, 243, 729, ... (Powers of 3) Table 1? with the next three numbers in each sequence! After each sequence, write in your own words what is the rule for forming the numbers in the sequence. Table 1: Examples of number sequences Patterns in Mathematics Math Talk Chapter 1_Patterns in Mathematics.indd 3 13-08-2024 16:10:52 1.3 Visualising Number Sequences Many number sequences can be visualised using pictures. Visualising mathematical objects through pictures or diagrams can be a very fruitful way to understand mathematical patterns and concepts. Let us represent the first seven sequences in Table 1 using the following pictures. Reprint 2025-26 3 Ganita Prakash | Grade 6 Table 2: Pictorial representation of some number sequences 1 1 1 1 1 Counting 1 2 3 4 5 numbers 1 3 5 7 9 numbers 2 4 6 8 10 numbers 1 3 6 10 15 Odd Triangular numbers All 1’s Even Chapter 1_Patterns in Mathematics.indd 4 13-08-2024 16:10:52 4 1 4 9 16 25 81 27 64 125 Reprint 2025-26 Cubes Squares Figure it Out 1. Copy the pictorial representations of the number sequences 2. Why are 1, 3, 6, 10, 15, … called triangular numbers? Why 3. You will have noticed that 36 is both a triangular number and a 4. What would you call the following sequence of numbers? in Table 2 in your notebook, and draw the next picture for each sequence! are 1, 4, 9, 16, 25, … called square numbers or squares? Why are 1, 8, 27, 64, 125, … called cubes? square number! That is, 36 dots can be arranged perfectly both in a triangle and in a square. Make pictures in your notebook illustrating this! This shows that the same number can be represented differently, and play different roles, depending on the context. Try representing some other numbers pictorially in different ways! Patterns in Mathematics Math Talk Chapter 1_Patterns in Mathematics.indd 5 13-08-2024 16:10:53 5. Can you think of pictorial ways to visualise the sequence of Powers of 2? Powers of 3? That’s right, they are called hexagonal numbers! Draw these in your notebook. What is the next number in the sequence? 1 7 19 37 Reprint 2025-26 5 Ganita Prakash | Grade 6 1.4 Relations among Number Sequences Sometimes, number sequences can be related to each other in surprising ways. Example: What happens when we start adding up odd numbers? 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36 . . . This is a really beautiful pattern! Here is one possible way of thinking about Powers of 2: Why does this happen? Do you think it will happen forever? 1 2 4 8 16 Chapter 1_Patterns in Mathematics.indd 6 13-08-2024 16:10:53 6 The answer is that the pattern does happen forever. But why? As mentioned earlier, the reason why the pattern happens is just as important and exciting as the pattern itself. A picture can explain it Visualising with a picture can help explain the phenomenon. Recall that square numbers are made by counting the number of dots in a square grid. How can we partition the dots in a square grid into odd numbers of dots: 1, 3, 5, 7,... ? Think about it for a moment before reading further! Reprint 2025-26 Math Talk This picture now makes it evident that Because such a picture can be made for a square of any size, this explains why adding up odd numbers gives square numbers. By drawing a similar picture, can you say what is the sum of the first 10 odd numbers? Now by imagining a similar picture, or by drawing it partially, as needed, can you say what is the sum of the first 100 odd numbers? Another example of such a relation between sequences: Adding up and down Let us look at the following pattern: Here is how it can be done: 1 + 3 + 5 + 7 + 9 + 11 = 36. Patterns in Mathematics Chapter 1_Patterns in Mathematics.indd 7 13-08-2024 16:10:53 1 = 1 1 + 2 + 1 = 4 1 + 2 + 3 + 2 + 1 = 9 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36 This seems to be giving yet another way of getting the square numbers— by adding the counting numbers up and then down! . . . Reprint 2025-26 7 Ganita Prakash | Grade 6 Can you find a similar pictorial explanation? Figure it Out 1. Can you find a similar pictorial explanation for why adding 2. By imagining a large version of your picture, or drawing it partially, as needed, can you see what will be the value of 1 + 2 + 3 + ... + 99 + 100 + 99 + ... + 3 + 2 + 1? 3. Which sequence do you get when you start to add the All 1’s sequence up? What sequence do you get when you add the All 1’s sequence up and down? 4. Which sequence do you get when you start to add the counting numbers up? Can you give a smaller pictorial explanation? counting numbers up and down, i.e., 1, 1 + 2 + 1, 1 + 2 + 3 + 2 + 1, …, gives square numbers? Try This Chapter 1_Patterns in Mathematics.indd 8 13-08-2024 16:10:53 8 5. What happens when you add up pairs of consecutive triangular numbers? That is, take 1 + 3, 3 + 6, 6 + 10, 10 + 15, … Which sequence do you get? Why? Can you explain it with a picture? 6. What happens when you start to add up powers of 2 starting with 1, i.e., take 1, 1 + 2, 1 + 2 + 4, 1 + 2 + 4 + 8, … ? Now add 1 to each of these numbers—what numbers do you get? Why does this happen? Reprint 2025-26 1.5 Patterns in Shapes Other important and basic patterns that occur in Mathematics are patterns of shapes. These shapes may be in one, two, or three dimensions (1D, 2D, or 3D)—or in even more dimensions. The branch of Mathematics that studies patterns in shapes is called geometry. Shape sequences are one important type of shape pattern that mathematicians study. Table 3 shows a few key shape sequences that are studied in Mathematics. 7. What happens when you multiply the triangular numbers by 6 8. What happens when you start to add up hexagonal numbers, i.e., 9. Find your own patterns or relations in and among the sequences in Table 1. Can you explain why they happen with a picture or otherwise? and add 1? Which sequence do you get? Can you explain it with a picture? take 1, 1 + 7, 1 + 7 + 19, 1 + 7 + 19 + 37, … ? Which sequence do you get? Can you explain it using a picture of a cube? Patterns in Mathematics Chapter 1_Patterns in Mathematics.indd 9 13-08-2024 16:10:53 Reprint 2025-26 9 Ganita Prakash | Grade 6 Heptagon Triangle K2 K3 K4 K5 K6 Table 3: Examples of shape sequences Quadrilateral Regular Polygons Octagon Nonagon Pentagon Hexagon Decagon Complete Graphs Stacked Squares Chapter 1_Patterns in Mathematics.indd 10 13-08-2024 16:10:53 10 Reprint 2025-26 Koch Snowflake Stacked Triangles 1.6 Relation to Number Sequences Often, shape sequences are related to number sequences in surprising ways. Such relationships can be helpful in studying and understanding both the shape sequence and the related number sequence. Example: The number of sides in the shape sequence of Regular Polygons is given by the counting numbers starting at 3, i.e., 3, 4, 5, 6, 7, 8, 9, 10, .... That is why these shapes are called, respectively, regular triangle, quadrilateral (i.e., square), pentagon, hexagon, heptagon, octagon, nonagon, decagon, etc., respectively. The word ‘regular’ refers to the fact that these shapes have equal-length sides and also equal ‘angles’ (i.e., the sides look the same and the corners also look the same). We will discuss angles in more depth in the next chapter. The other shape sequences in Table 3 also have beautiful relationships with number sequences. Figure it Out 1. Can you recognise the pattern in each of the sequences in 2. Try and redraw each sequence in Table 3 in your notebook. Table 3? Can you draw the next shape in each sequence? Why or why not? After each sequence, describe in your own words what is the rule or pattern for forming the shapes in the sequence. Patterns in Mathematics Math Talk Chapter 1_Patterns in Mathematics.indd 11 13-08-2024 16:10:53 Figure it Out 1. Count the number of sides in each shape in the sequence 2. Count the number of lines in each shape in the sequence of Regular Polygons. Which number sequence do you get? What about the number of corners in each shape in the sequence of Regular Polygons? Do you get the same number sequence? Can you explain why this happens? of Complete Graphs. Which number sequence do you get? Can you explain why? Reprint 2025-26 Try This 11 Ganita Prakash | Grade 6 Mathematics may be viewed as the search for patterns and for the Among the most basic patterns that occur in mathematics are number Some important examples of number sequences include the counting explanations as to why those patterns exist. sequences. 3. How many little squares are there in each shape of the 4. How many little triangles are there in each shape of the sequence of Stacked Triangles? Which number sequence does this give? Can you explain why? (Hint: In each shape in the sequence, how many triangles are there in each row?) 5. To get from one shape to the next shape in the Koch Snowflake sequence, one replaces each line segment ‘—’ by a ‘speed bump’ . As one does this more and more times, the changes become tinier and tinier with very very small line segments. How many total line segments are there in each shape of the Koch Snowflake? What is the corresponding number sequence? (The answer is 3, 12, 48, ..., i.e., 3 times Powers of 4; this sequence is not shown in Table 1.) sequence of Stacked Squares? Which number sequence does this give? Can you explain why? Summary Try This Chapter 1_Patterns in Mathematics.indd 12 13-08-2024 16:10:54 12 Sometimes number sequences can be related to each other in beautiful Visualising number sequences using pictures can help to understand Shape sequences are another basic type of pattern in mathematics. numbers, odd numbers, even numbers, square numbers, triangular numbers, cube numbers, Virahānka numbers, and powers of 2. and remarkable ways. For example, adding up the sequence of odd numbers starting with 1 gives square numbers. sequences and the relationships between them. Some important examples of shape sequences include regular polygons, complete graphs, stacked triangles and squares, and Koch snowflake iterations. Shape sequences also exhibit many interesting relationships with number sequences. Reprint 2025-26 Page No. 2. Section 1.1 Page No. 3. Section 1.2 Q.1. Can you recognize the pattern in each of the sequences in Table 1? Ans. Yes, There is a pattern in each case. Powers of 2- 1,2,4 = 2 x 2, 8 = 2 x 2 x2 ,16 = 2 x 2 x2 x 2 x 2… Powers of 3 - 1, 3, 9 = 3 x 3 = 27 =3 x 3 x3,… Virahanka numbers- 1,2, 3, 5 = 2 + 3, 8 = 3 + 5, 13 = 5 + 8,…. Rest of the patterns have been shown on page 4 ,Table 2 Figure it Out Figure it Out Q1. Can you think of other examples where mathematics helps us in our everyday lives? Ans. Some examples are paying for fruits, vegetables, groceries etc. calculation of speed of vehicles, designs or patterns in different buildings, finding area of any plot or out own home. There could be many more such contexts in our everyday lives. Discuss for other examples also. Q2. How has mathematics helped propel humanity forward? (You might think of examples involving: carrying out scientific experiments; running our economy and democracy; building bridges, houses or other complex structures; making TVs, mobile phones, computers, bicycles, trains, cars, planes, calendars, clocks, etc.) Ans. Teacher student discussion is required. CHAPTER 1 — SOLUTIONS Patterns in Mathematics Page no. 5 Section 1.3 Figure it out Q.2. Why are 1, 3, 6, 10, 15, … called triangular numbers? Why are 1, 4, 9, 16, 25, … called square numbers or squares? Why are 1, 8, 27, 64, 125, … called cubes? Ans. Refer Table 2, page 4, and check for yourself. Q.3. You will have noticed that 36 is both a triangular number and a square number! That is, 36 dots can be arranged perfectly both in a triangle and in a square. Make pictures in your notebook illustrating this! Ans. Refer Table 2, page 4, and draw. [1] Q.4. What would you call the following sequence of numbers? Ans. 61. Q.5. Can you think of pictorial ways to visualise the sequence of powers of 2? powers of 3? Ans. Sequence of powers of 2—given on page 6 Sequence of powers of 3- One of the ways could be – 1 2 By drawing a similar picture, can you say what is the sum of the first 10 odd numbers? Ans. 100 3 9 Page No. 7 Section 1.4 [2] Q.1. Can you find a similar pictorial explanation for why adding counting numbers up and down, i.e., 1, 1 + 2 + 1, 1 + 2 + 3 + 2 + 1, …, gives square numbers? Ans. One of the ways isQ.2. By imagining a large version of your picture, or drawing it partially, as needed, can you see what will be the value of 1 + 2 + 3 + ... + 99 + 100 + 99 + ... + 3 + 2 + 1? Ans. 10,000 Q.3. Which sequence do you get when you start to add the All 1’s sequence up? What sequence do you get when you add the All 1’s sequence up and down? Now by imagining a similar picture, or by drawing it partially, as needed, can you say what is the sum of the first 100 odd numbers? Ans.10000 Figure it out 1, 1 + 2 + 1, 1 + 2 + 3 + 2 + 1, …, Ans. Q.4. Which sequence do you get when you start to add the Counting numbers up? Can you give a smaller pictorial explanation? Ans. 1, 1+2, 1+2+3, 1+2+3+4, ……… • 1 + (1 +1) + (1 +1 +1) + (1 + 1+1+1) … • 1, 1+(1+1) + 1, 1+(1+1) +(1+1+1) + (1+1) +1, 1+(1+1) +(1+1+1) + (1+1+1+1) + (1+1+1) + (1+1) +1, ………. Which is triangular number sequence. For pictorial representation refer Table 2 on page 4 (Try it for isosceles right triangle also.) [3] Q.5. What happens when you add up pairs of consecutive triangular numbers? That is, take 1 + 3, 3 + 6, 6 + 10, 10 + 15, … ? Which sequence do you get? Why? Can you explain it with a picture? Ans. We get: 4, 9, 16, 25, ……… (Square numbers). For the pictorial representation refer Table 2 on page 4 Q.6. What happens when you start to add up powers of 2 starting with 1, i.e., take 1, 1 + 2, 1 + 2 + 4, 1 + 2 + 4 + 8, …? Now add 1 to each of these numbers — what numbers do you get? Why does this happen? Ans. • We get, 1, 3, 7, 15, 31, ………… • After adding 1 to each number, we get:2, 4, 8, 16, 32, …… Refer the picture on page 6. Q.7. What happens when you multiply the triangular numbers by 6 and add 1? Which sequence do you get? Can you explain it with a picture? Ans. (1×6) +1, (3×6) +1, (6×6) +1, (10×6) +1, (15×6) +1, … Q.8. What happens when you start to add up hexagonal numbers, i.e., take 1, 1 + 7, 1 + 7 + 19, 1 + 7 + 19 + 37, … ? Which sequence do you get? Can you explain it using a picture of a cube? Ans. 1, 8, 27, 64, ………We get cube numbers. = 7, 19, 37, 61, 91, …. For picture refer picture in Q.4 page 5. For picture, refer Table 2, page 4 Page no. 10 Section 1.5 Q.1. Can you recognise the pattern in each of the sequences in Table 3? Ans. Yes. ➢ 3, 4, 5, 6, 7, 8, 9, 10 One of the ways to interpret this is that we get a sequence of number of sides of the pictures of the shapes ➢ 1, 3, 6, 10, 15 ➢ 1, 4, 9, 16, 25 ➢ 1, 4, 9, 16, 25 ➢ 3, 3×4, 3×4×4, 3×4×4×4, 3×4×4×4×4 Figure it out [4] Page no. 11 Section 1.6 Q.1. Count the number of sides in each shape in the sequence of Regular Polygons. Which number sequence do you get? What about the number of corners in each shape in the sequence of Regular Polygons? Do you get the same number sequence? Can you explain why this happens? Ans. • Number of sides = 3,4,5,6,7,8,9,10… We get counting number sequence, starting with 3. • Number of Corners = 3,4,5,6,7,8,9,10… Yes, we get the same number sequence • in any closed figure, number of sides = number of corners (vertices). Q.2. Count the number of lines in each shape in the sequence of Complete Graphs. Which number sequence do you get? Can you explain why? Ans. • 1, 3, 6, 10, 15. This is a Triangle number sequence Q.3. How many little squares are there in each shape of the sequence of Stacked Squares? Which number sequence does this give? Can you explain why? Ans. • 1, 4, 9, 16, 25. This is a Square number sequence. • Squares can be drawn using these number of dots. Q.4. How many little triangles are there in each shape of the sequence of Stacked Triangles? Which number sequence does this give? Can you explain why? (Hint: In each shape in the sequence, how many triangles are there in each row?) Figure it out Ans. 1, 4, 9, 16, 25 = 1, 1+2+1, 1+2+3+2+1, ………. Square number sequence (As adding up and down) gives us square number sequence. Q.5. To get from one shape to the next shape in the Koch Snowflake sequence, one replaces each line segment‘—’ by a ‘speed bump’ . As one does this more and more times, the changes become tinier and tinier with very very small line segments. How many total line segments are there in each shape of the Koch Snowflake? What is the corresponding number sequence? (The answer is 3, 12, 48, ..., i.e. 3 times Powers of 4; this sequence is not shown in Table 1) Ans. • Total line segments in each shape: 3, 12, 48, 192, 768 • Corresponding sequence: 3, 3×4, 3×4×4, 3×4×4×4×, 3×4×4×4×4,… [5]" class_6,2,Lines and Angles,ncert_books/class_6/Ganita_Prakash/fegp102.pdf,"2 In this chapter, we will explore some of the most basic ideas of geometry including points, lines, rays, line segments and angles. These ideas form the building blocks of ‘plane geometry’, and will help us in understanding more advanced topics in geometry such as the construction and analysis of different shapes. Mark a dot on the paper with a sharp tip of a pencil. The sharper the tip, the thinner will be the dot. This tiny dot will give you an idea of a point. A point determines a precise location, but it has no length, breadth or height. Some models for a point are given below. 2.1 Point Lines and Angles Chapter 2_Lines and Angles.indd 13 13-08-2024 16:14:25 If you mark three points on a piece of paper, you may be required to distinguish these three points. For this purpose, each of the three points may be denoted by a single capital letter such as The tip of a compass The sharpened end of a pencil The pointed end of a needle Reprint 2025-26 P Z T Ganita Prakash | Grade 6 Z, P and T. These points are read as ‘Point Z’, ‘Point P’ and ‘Point T’. Of course, the dots represent precise locations and must be imagined to be invisibly thin. Fold a piece of paper and unfold it. Do you see a crease? This gives the idea of a line segment. It has two end points, A and B. Mark any two points A and B on a sheet of paper. Try to connect A to B by various routes (Fig. 2.1). What is the shortest route from A to B? This shortest path from point A to Point B (including A and B) as shown here is called the line segment from A to B. It is denoted by either AB or BA. The points A and B are called the end points of the line segment AB. Imagine that the line segment from A to B (i.e., AB) is extended beyond A in one direction and beyond B in the other direction without any end (see Fig. 2.2). This is a model for a line. Do you think you can draw a complete picture of a line? No. Why? A line through two points A and B is written as AB. It extends forever in both directions. Sometimes a line is denoted by a letter like l or m. Observe that any two points determine a unique line that passes through both of them. 2.2 Line Segment 2.3 Line A A Fig. 2.1 B B B m Chapter 2_Lines and Angles.indd 14 13-08-2024 16:14:25 14 Reprint 2025-26 A Fig. 2.2 2.4 Ray A ray is a portion of a line that starts at one point (called the starting point or initial point of the ray) and goes on endlessly in a direction. The following are some models for a ray: Look at the diagram (Fig. 2.3) of a ray. Two points are marked on it. One is the starting point A and the other is a point P on the path of the ray. We then denote the ray by AP. Beam of light from a lighthouse Ray of light from a torch Sun rays  Figure it Out 1. Can you help Rihan and Sheetal find their answers? A Lines and Angles Fig. 2.3 P Chapter 2_Lines and Angles.indd 15 13-08-2024 16:14:25 Rihan marked a point on a piece of paper. How many lines can he draw that pass through the point? Reprint 2025-26 Sheetal marked two points on a piece of paper. How many different lines can she draw that pass through both of the points? 15 Ganita Prakash | Grade 6 2. Name the line segments in Fig. 2.4. Which of the five marked 3. Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays? 4. Draw a rough figure and write labels appropriately to illustrate each of the following: points are on exactly one of the line segments? Which are on two of the line segments? a. OP and OQ meet at O. L M T P Fig. 2.5 N B Fig. 2.4 A Q R Chapter 2_Lines and Angles.indd 16 13-08-2024 16:14:25 16 5. In Fig. 2.6, name: b. XY and PQ intersect at point M. c. Line l contains points E and F but not point D. d. Point P lies on AB. a. Five points b. A line c. Four rays d. Five line segments D E Reprint 2025-26 Fig. 2.6 O C B 2.5 Angle An angle is formed by two rays having a common starting point. Here is an angle formed by rays BD and BE where B is the common starting point (Fig. 2.8). The point B is called the vertex of the angle, and the rays BD and BE are called the arms of the angle. How can we name this angle? We can simply use the vertex and say that it is the Angle B. To be clearer, we use a point on each of the arms together with the vertex to name the angle. In this case, we name the angle as Angle DBE or Angle EBD. The word angle can be replaced by the symbol ‘∠’, i.e., ∠DBE or ∠EBD. Note that in specifying the angle, the vertex is always written as the middle letter. To indicate an angle, we use a small curve at the vertex (refer to Fig. 2.9). Vidya has just opened her book. Let us observe her opening the cover of the book in different scenarios. 6. Here is a ray OA (Fig. 2.7). It starts at O and passes through the point A. It also passes through the point B. a. Can you also name it as OB ? Why? b. Can we write OAas AO? Why or why not? vertex B O Fig. 2.8 arm arm Lines and Angles Fig. 2.7 D E B A Chapter 2_Lines and Angles.indd 17 13-08-2024 16:14:27 Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Reprint 2025-26 17 Ganita Prakash | Grade 6 Do you see angles being made in each of these cases? Can you mark their arms and vertex? Just as we talk about the size of a line based on its length, we also talk about the size of an angle based on its amount of rotation. So, the angle in Case 2 is greater as in this case she needs to rotate the cover more. Similarly, the angle in Case 3 is even larger than that of Case 2, because there is even more rotation, and Cases 4, 5, and 6 are successively larger angles with more and more rotation. The size of an angle is the amount of rotation or turn that is needed about the vertex to move the first ray to the second ray. Which angle is greater—the angle in Case 1 or the angle in Case 2? Vertex Initial position of ray Final position of ray Fig. 2.9 Amount of turn is the size of the angle Chapter 2_Lines and Angles.indd 18 13-08-2024 16:14:27 18 Let’s look at some other examples where angles arise in real life by rotation or turn: • In a compass or divider, we turn the arms to form an angle. The vertex is the point where the two arms are joined. Identify the arms and vertex of the angle. • A pair of scissors has two blades. When we open them (or ‘turn them’) to cut something, the blades form an angle. Identify the arms and the vertex of the angle. Reprint 2025-26 Do you see how these angles are formed by turning one arm with respect to the other? Teacher needs to organise various activities with the students to recognise the size of an angle as a measure of rotation. • Look at the pictures of spectacles, wallet and other common objects. Identify the angles in them by marking out their arms and vertices. Teacher’s Note Lines and Angles Chapter 2_Lines and Angles.indd 19 13-08-2024 16:14:28 Figure it Out 1. Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle. A B C D Reprint 2025-26 19 Ganita Prakash | Grade 6 2. Draw and label an angle with arms ST and SR. 3. Explain why ∠APC cannot be labelled as ∠P. 4. Name the angles marked in the given figure. P P B C Q A Math Talk Chapter 2_Lines and Angles.indd 20 13-08-2024 16:14:28 20 5. Mark any three points on your paper that are not on one line. Label them A, B, C. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C? Write them down, and mark each of them with a curve as in Fig. 2.9. T R Reprint 2025-26 2.6 Comparing Angles Look at these animals opening their mouths. Do you see any angles here? If yes, mark the arms and vertex of each one. Some mouths are open wider than others; the more the turning of the jaws, the larger the angle! Can you arrange the angles in this picture from smallest to largest? Is it always easy to compare two angles? 6. Now mark any four points on your paper so that no three of them are on one line. Label them A, B, C, D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C, D? Write them all down, and mark each of them with a curve as in Fig. 2.9. Lines and Angles Chapter 2_Lines and Angles.indd 21 13-08-2024 16:14:28 Here are some angles. Label each of the angles. How will you compare them? Draw a few more angles; label them and compare. Reprint 2025-26 Math Talk 21 O B Ganita Prakash | Grade 6 Comparing angles by superimposition Any two angles can be compared by placing them one over the other, i.e., by superimposition. While superimposing, the vertices of the angles must overlap. After superimposition, it becomes clear which angle is smaller and which is larger. The picture shows the two angles superimposed. It is now clear that ∠PQR is larger than ∠ABC. Equal angles. Now consider ∠AOB and ∠XOY in the figure. Which is greater? A B A C Q O Y X P R Q(B) O B Y P A X A CR Chapter 2_Lines and Angles.indd 22 13-08-2024 16:14:29 22 The corners of both of these angles match and the arms overlap with each other, i.e., OA ↔ OX and OB ↔ OY. So, the angles are equal in size. The reason these angles are considered to be equal in size is because when we visualise each of these angles as being formed out of rotation, we can see that there is an equal amount of rotation needed to move OB to OA and OY to OX. From the point of view of superimposition, when two angles are superimposed, and the common vertex and the two rays of both angles lie on top of each other, then the sizes of the angles are equal. Reprint 2025-26 Where else do we use superimposition to compare? Figure it Out 1. Fold a rectangular sheet of paper, then draw a line along the fold created. Name and compare the angles formed between the fold and the sides of the paper. Make different angles by folding a rectangular sheet of paper and compare the angles. Which is the largest and smallest angle you made? 2. In each case, determine which angle is greater and why. 3. Which angle is greater: ∠XOY or ∠AOB? Give reasons. c. ∠XOB or ∠XOC Discuss with your friends on how you decided which one is greater. a. ∠AOB or ∠XOY b. ∠AOB or ∠XOB X A Y O B C A X Lines and Angles Y Math Talk Math Talk Chapter 2_Lines and Angles.indd 23 13-08-2024 16:14:29 Comparing angles without superimposition Two cranes are arguing about who can open their mouth wider, i.e., who is making a bigger angle. Let us first draw their angles. How do we know which one is bigger? As seen O B Reprint 2025-26 Fig. 2.10 23 Ganita Prakash | Grade 6 before, one could trace these angles, superimpose them and then check. But can we do it without superimposition? Suppose we have a transparent circle which can be moved and placed on figures. Can we use this for comparison? Let us place the circular paper on the angle made by the first crane. The circle is placed in such a way that its centre is on the vertex of the angle. Let us mark the points A and B on the edge circle at the points where the arms of the angle pass through the circle. Can we use this to find out if this angle is greater than, or equal to or smaller than the angle made by the second crane? Let us place it on the angle made by the second crane so that the vertex coincides with the centre of the circle and one of the arms passes through OA. O O B B A A Chapter 2_Lines and Angles.indd 24 13-08-2024 16:14:29 24 Can you now tell which angle is bigger? Reprint 2025-26 O B Y A X Which crane was making the bigger angle? If you can make a circular piece of transparent paper, try this method to compare the angles in Fig. 2.10 with each other. 2.7 Making Rotating Arms Let us make ‘rotating arms’ using two paper straws and a paper clip by following these steps: Teacher’s Note A teacher needs to check the understanding of the students around the notion of an angle. Sometimes students might think that increasing the length of the arms of the angle increases the angle. For this, various situations should be posed to the students to check their understanding on the same. 1. Take two paper straws and a paper clip. 2. Insert the straws into the arms of the paper clip. Lines and Angles Chapter 2_Lines and Angles.indd 25 13-08-2024 16:14:30 Make several ‘rotating arms’ with different angles between the arms. Arrange the angles you have made from smallest to largest by comparing and using superimposition. Passing through a slit: Collect a number of rotating arms with different angles; do not rotate any of the rotating arms during this activity. 3. Your rotating arm is ready! Reprint 2025-26 25 Ganita Prakash | Grade 6 Take a cardboard and make an angle-shaped slit as shown below by tracing and cutting out the shape of one of the rotating arms. Now, shuffle and mix up all the rotating arms. Can you identify which of the rotating arms will pass through the slit? The correct one can be found by placing each of the rotating arms over the slit. Let us do this for some of the rotating arms: Chapter 2_Lines and Angles.indd 26 13-08-2024 16:14:31 26 Only the pair of rotating arms where the angle is equal to that of the slit passes through the slit. Note that the possibility of passing through the slit depends only on the angle between the rotating arms and not on their lengths (as long as they are shorter than the length of the slit). Slit angle is greater than the arms’ angle. The arms will not go through the slit. Slit angle is less than the arms’ angle. The arms will not go through the slit. Reprint 2025-26 Slit angle is equal to the arms’ angle. The arms will go through the slit. 2.8 Special Types of Angles Let us go back to Vidya’s notebook and observe her opening the cover of the book in different scenarios. She makes a full turn of the cover when she has to write while holding the book in her hand. She makes a half turn of the cover when she has to open it on her table. In this case, observe the arms of the angle formed. They lie in a straight line. Such an angle is called a straight angle. Challenge: Reduce this angle. Angle reduced. The angle is still the same! Lines and Angles Chapter 2_Lines and Angles.indd 27 12-12-2024 10:36:03 Let us consider a straight angle ∠AOB. Observe that any ray OC divides it into two angles, ∠AOC and ∠COB. A O B Fig. 2.11 Reprint 2025-26 27 Ganita Prakash | Grade 6 Is it possible to draw OC such that the two angles are equal to each other in size? Let’s Explore We can try to solve this problem using a piece of paper. Recall that when a fold is made, it creates a crease which is straight. Take a rectangular piece of paper and on one of its sides, mark the straight angle AOB. By folding, try to get a line (crease) passing through O that divides ∠AOB into two equal angles. How can it be done? Math Talk Chapter 2_Lines and Angles.indd 28 13-08-2024 16:14:33 28 Fold the paper such that OB overlaps with OA. Observe the crease and the two angles formed. Reprint 2025-26 Justify why the two angles are equal. Is there a way to superimpose and check? Can this superimposition be done by folding? Each of these equal angles formed are called right angles. So, a straight angle contains two right angles. Why shouldn't you argue with a 90̊ angle? Because they're always right. Lines and Angles Chapter 2_Lines and Angles.indd 29 13-08-2024 16:14:33 If a straight angle is formed by half of a full turn, how much of a full turn will form a right angle? Observe that a right angle resembles the shape of an ‘L’. An angle is a right angle only if it is exactly half of a straight angle. Two lines that meet at right angles are called perpendicular lines. Figure it Out 1. How many right angles do the windows of your classroom contain? Do you see other right angles in your classroom? Reprint 2025-26 29 Ganita Prakash | Grade 6 2. Join A to other grid points in the figure by a straight line to get a straight angle. What are all the different ways of doing it? 3. Now join A to other grid points in the figure by a straight line to get a right angle. What are all the different ways of doing it? A B A A B A B B Chapter 2_Lines and Angles.indd 30 13-08-2024 16:14:33 30 Hint: Extend the line further as shown in the figure below. To get a right angle at A, we need to draw a line through it that divides the straight angle CAB into two equal parts. C Reprint 2025-26 A B Classifying Angles Angles are classified in three groups as shown below. Right angles are shown in the second group. What could be the common feature of the other two groups? 4. Get a slanting crease on the paper. Now, try to get another crease that is perpendicular to the slanting crease. a. How many right angles do you have now? Justify why the angles are exact right angles. b. Describe how you folded the paper so that any other person who doesn’t know the process can simply follow your description to get the right angle. Lines and Angles Chapter 2_Lines and Angles.indd 31 13-08-2024 16:14:34 In the first group, all angles are less than a right angle or in other words, less than a quarter turn. Such angles are called acute angles. In the third group, all angles are greater than a right angle but less than a straight angle. The turning is more than a quarter turn and less than a half turn. Such angles are called obtuse angles. Figure it Out 1. Identify acute, right, obtuse and straight angles in the previous figures. 2. Make a few acute angles and a few obtuse angles. Draw them in different orientations. Reprint 2025-26 31 Ganita Prakash | Grade 6 What will be the next figure and how many acute angles will it have? Do you notice any pattern in the numbers? 2.9 Measuring Angles We have seen how to compare two angles. But can we actually quantify how big an angle is using a number without having to compare it to another angle? We saw how various angles can be compared using a circle. Perhaps a circle could be used to assign measures for angles? 3. Do you know what the words acute and obtuse mean? Acute means sharp and obtuse means blunt. Why do you think these words have been chosen? 4. Find out the number of acute angles in each of the figures below. Chapter 2_Lines and Angles.indd 32 13-08-2024 16:14:34 32 To assign precise measures to angles, mathematicians came up with an idea. They divided the angle in the centre of the circle into Reprint 2025-26 Fig. 2.12 360 equal angles or parts. The angle measure of each of these unit parts is 1 degree, which is written as 1°. This unit part is used to assign measure to any angle: the measure of an angle is the number of 1° unit parts it contains inside it. For example, see this figure: It contains 30 units of 1° angle and so we say that its angle measure is 30°. Measures of different angles: What is the measure of a full turn in degrees? As we have taken it to contain 360 degrees, its measure is 360°. What is the measure of a straight angle in degrees? A straight angle is half of a full turn. As a full-turn is 360°, a half turn is 180°. What is the measure of a right angle in degrees? Two right angles together form a straight angle. As a straight angle measures 180°, a right angle measures 90°. 30 units 180 units Lines and Angles Chapter 2_Lines and Angles.indd 33 14-08-2024 14:20:43 A pinch of history A full turn has been divided into 360°. Why 360? The reason why we use 360° today is not fully known. The division of a circle into A O B A O B Reprint 2025-26 A O A O B 90 units B 33 Ganita Prakash | Grade 6 360 parts goes back to ancient times. The Rigveda, one of the very oldest texts of humanity going back thousands of years, speaks of a wheel with 360 spokes (Verse 1.164.48). Many ancient calendars, also going back over 3000 years—such as calendars of India, Persia, Babylonia and Egypt—were based on having 360 days in a year. In addition, Babylonian mathematicians frequently used divisions of 60 and 360 due to their use of sexagesimal numbers and counting by 60s. Perhaps the most important and practical answer for why mathematicians over the years have liked and continued to use 360 degrees is that 360 is the smallest number that can be evenly divided by all numbers up to 10, aside from 7. Thus, one can break up the circle into 1, 2, 3, 4, 5, 6, 8, 9 or 10 equal parts, and still have a whole number of degrees in each part! Note that 360 is also evenly divisible by 12, the number of months in a year, and by 24, the number of hours in a day. These facts all make the number 360 very useful. The circle has been divided into 1, 2, 3, 4, 5, 6, 8, 9 10 and 12 parts below. What are the degree measures of the resulting angles? Write the degree measures down near the indicated angles. Chapter 2_Lines and Angles.indd 34 14-08-2024 14:20:57 34 Degree measures of different angles How can we measure other angles in degrees? It is for this purpose that we have a tool called a protractor that is either a circle divided into 360 equal parts as shown in Fig. 2.12 (on page 32), or a half circle divided into 180 equal parts. Reprint 2025-26 Unlabelled protractor Here is a protractor. Do you see the straight angle at the center divided into 180 units of 1 degree? Only part of the lines dividing the straight angle are visible, though! Starting from the marking on the rightmost point of the base, there is a long mark for every 10°. From every such long mark, there is a medium sized mark after 5°.  Figure it out K L 1. Write the measures of the following angles: a. ∠ KAL Notice that the vertex of this angle coincides with the centre of the protractor. So the number of units of 1 degree angle between KA and AL gives the measure of ∠KAL. By counting, we get— ∠KAL = 30° Making use of the medium sized and large sized marks, is it possible to count the number of units in 5s or 10s? Lines and Angles Chapter 2_Lines and Angles.indd 35 13-08-2024 16:14:34 A T W Reprint 2025-26 b. ∠WAL c. ∠TAK 35 Ganita Prakash | Grade 6 Labelled protractor This is a protractor that you find in your geometry box. It would appear similar to the protractor above except that there are numbers written on it. Will these make it easier to read the angles? There are two sets of numbers on the protractor: one increasing from right to left and the other increasing from left to right. Why does it include two sets of numbers? Name the different angles in the figure and write their measures. 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 S 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 R 90 90 Chapter 2_Lines and Angles.indd 36 12-12-2024 10:36:04 36 U T 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 Reprint 2025-26 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 90 90 O Q P Can you use the numbers marked to find the angle without counting the number of markings? Here, OT and OS pass through the numbers 20 and 55 on the outer scale. How many units of 1 degree are contained between these two arms? Can subtraction be used here? How can we measure angles directly without having to subtract? Place the protractor so the center is on the vertex of the angle. Align the protractor so that one the arms passes through the 0º mark as in the picture below. Did you include angles such as ∠TOQ? Which set of markings did you use—inner or outer? What is the measure of ∠TOS? 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 90 90 A Lines and Angles Chapter 2_Lines and Angles.indd 37 13-08-2024 16:14:35 What is the degree measure of ∠AOB? Make your own Protractor! You may have wondered how the different equally spaced markings are made on a protractor. We will now see how we can make some of them! 1. Draw a circle of a convenient radius on a sheet of paper. Cut out the circle (Fig. 2.13). A circle or one full turn is 360°. 2. Fold the circle to get two equal halves and cut it through the crease to get a semicircle. Write ‘0°’ in the bottom right corner of the semicircle. Reprint 2025-26 O B 37 Ganita Prakash | Grade 6 3. Fold the semi-circular sheet in half as shown in Fig. 2.15 to form a quarter circle. Fig. 2.14 A O The measure of half a circle is 1 2 of a full turn. (Fig. 2.14) So, the measure of half a turn = 1 2 of ____ = 180°. Thus, write 180° in the left bottom corner of the semicircle. The measure of a quarter circle is 1 4 of a full turn. The measure of a 1 4 turn = 1 4 of 360° = ________. Or, the measure of a 1 4 turn = 1 2 of a half turn = 1 2 of 180° = ______. Thus, mark 90° at the top of the semicircle. Fig. 2.13 180 units B Chapter 2_Lines and Angles.indd 38 13-08-2024 16:14:35 38 Fig. 2.15 Reprint 2025-26 A O B 90 units 4. Fold the sheet again as shown in Figs. 2.16 and 2.17: 5. Continuing with another half fold as shown in Fig. 2.18, we get an angle of measure ________________________. 6. Unfold and mark the creases as OB, OC, ..., etc., as shown in Fig. 2.19 and Fig. 2.20. When folded, this is 1 8 of the circle, or 1 8 of a turn, or 1 8 of 360°, or 1 4 of 180° or 1 2 of 90° = ________________________. The new creases formed give us measures of 45° and 180°−  45° = 135° as shown. Write 45° and 135° at the correct places on the new creases along the edge of the semicircle. Fig. 2.16 Fig. 2.17 180O 0O Fig. 2.18 135O 90O 45O Lines and Angles Chapter 2_Lines and Angles.indd 39 13-08-2024 16:14:35 I H G C D E F Fig. 2.19 O A B Reprint 2025-26 Fig. 2.20 180O 0O 157.5O 45O 67.5O 112.5O 135O 90O 22.5O 39 Ganita Prakash | Grade 6 At each step, we folded in halves. This process of getting half of a given angle is called bisecting the angle. The line that bisects a given angle is called the angle bisector of the angle. Identify the angle bisectors in your handmade protractor. Try to make different angles using the concept of angle bisector through paper folding. Think! In Fig. 2.19, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = ∠GOH = ∠HOI=_____. Why? Figure it Out 1. Find the degree measures of the following angles using your protractor. I J Angle Bisector H H I J Chapter 2_Lines and Angles.indd 40 12-12-2024 10:36:04 40 Teacher’s Note It is important that students make their own protractor and use it to measure different angles before using the standard protractor so that they know the concept behind the marking of the standard protractor. 2. Find the degree measures of different angles in your classroom using your protractor. H Reprint 2025-26 I J G K 3. Find the degree measures for the angles given below. Check if your paper protractor can be used here! 4. How can you find the degree measure of the angle given below using a protractor? c H I J I H Lines and Angles J Chapter 2_Lines and Angles.indd 41 13-08-2024 16:14:36 5. Measure and write the degree measures for each of the following angles: a. b. Reprint 2025-26 41 Ganita Prakash | Grade 6 6. Find the degree measures of ∠BXE, ∠CXE, ∠AXB and ∠BXC. e. f. 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 d. c. B 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 90 90 C Chapter 2_Lines and Angles.indd 42 12-12-2024 10:36:04 42 7. Find the degree measures of ∠PQR, ∠PQS and ∠PQT. A X E P R S Reprint 2025-26 Q T 8. Make the paper craft as per the given instructions. Then, unfold and open the paper fully. Draw lines on the creases made and measure the angles formed. 9. Measure all three angles of the triangle shown in Fig. 2.21 (a), and write the measures down near the respective angles. Now add up the three measures. What do you get? Do the same for the triangles in Fig. 2.21 (b) and (c). Try it for other triangles as well, and then make a conjecture for what happens in general! We will come back to why this happens in a later year. 1 2 3 7 5 6 Lines and Angles 4 8 Chapter 2_Lines and Angles.indd 43 13-08-2024 16:14:36 A (a) C B B C Fig. 2.21 Reprint 2025-26 (b) A A (c) C B 43 Ganita Prakash | Grade 6 Mind the Mistake, Mend the Mistake! A student used a protractor to measure the angles as shown below. In each figure, identify the incorrect usage(s) of the protractor and discuss how the reading could have been made and think how it can be corrected. 90 ∠W = 70⁰ 90 90 ∠U = 35⁰ ∠V = 80⁰ 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 U V 90 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 90 ∠X = 150⁰ 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 90 90 90 Chapter 2_Lines and Angles.indd 44 13-08-2024 16:14:37 44 ∠Y = 120⁰ 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 100 80 90 90 Y 130 50 140 40 51 0 03 160 20 170 10 180 0 120 60 110 70 80 100 Reprint 2025-26 60 120 70 110 80 100 100 80 110 70 21 0 06 130 140 50 40 150 30 160 20 170 10 180 0 ∠Z = 85⁰ 90 90 50 130 40 140 30 150 20 160 0180 10 170 Figure it Out Where are the angles? 1. Angles in a clock: 2. The angle of a door: a. The hands of a clock make different angles at different times. At 1 o’clock, the angle between the hands is 30°. Why? b. What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock? c. Explore other angles made by the hands of a clock. Is it possible to express the amount by which a door is opened using an angle? What will be the vertex of the angle and what will be the arms of the angle? 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 5 6 7 8 9 10 11 2 4 3 12 1 Lines and Angles 2 4 3 Chapter 2_Lines and Angles.indd 45 13-08-2024 16:14:38 3. Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle? Reprint 2025-26 45 Ganita Prakash | Grade 6 It is important that students see the application of each mathematical concept in their daily lives. Teacher can organise some activities where students can appreciate the practical applications of angles in real-life situations, e.g., clocks, doors, swings, concepts of uphill and downhill, location of the sun, the giving of directions, etc. 4. Here is a toy with slanting slabs attached to its sides; the greater the angles or slopes of the slabs, the faster the balls roll. Can angles be used to describe the slopes of the slabs? What are the arms of each angle? Which arm is visible and which is not? 5. Observe the images below where there is an insect and its rotated version. Can angles be used to describe the amount of rotation? How? What will be the arms of the angle and the vertex? Hint: Observe the horizontal line touching the insects. Teacher’s Note Chapter 2_Lines and Angles.indd 46 13-08-2024 16:14:39 46 2.10 Drawing Angles Vidya wants to draw a 30° angle and name it ∠TIN using a protractor. In will be the vertex, IT and IN will be the arms of the angle. Keeping one arm, say IN, as the reference (base), the other arm IT should take a turn of 30°. Reprint 2025-26 Step 1: We begin with the base and draw : Step 2: We will place the centre point of the protractor on I and align IN to the 0 line. Step 3: Now, starting from 0, count your degrees (0, 10, 30) up to 30 on the protractor. Mark point T at the label 30°. 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 0 180 10 170 20 160 30 150 40 140 50 130 60 120 70 110 80 100 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 180 0 I N 90 90 90 90 I N Lines and Angles Chapter 2_Lines and Angles.indd 47 13-08-2024 16:14:40 Step 4: Using a ruler join the point I and T. ∠TIN = 30° is the required angle. I N Reprint 2025-26 T 47 Ganita Prakash | Grade 6 This is an angle guessing game! Play this game with your classmates by making two teams, Team 1 and Team 2. Here are the instructions and rules for the game: We now change the rules of the game a bit. Play this game with your classmates by again making two teams, Team 1 and Team 2. Here are the instructions and rules: Fig. 2.22  Let’s Play a Game #1 • Team 1 secretly choose an angle measure, for example, 49° and makes an angle with that measure using a protractor without Team 2 being able to see it. • Team 2 now gets to look at the angle. They have to quickly discuss and guess the number of degrees in the angle (without using a protractor!). • Team 1 now demonstrates the true measure of the angle with a protractor. I N 30º T Chapter 2_Lines and Angles.indd 48 12-12-2024 10:36:04 48  Let’s Play a Game #2 • Team 2 scores the number of points that is the absolute difference in degrees between their guess and the correct measure. For example, if Team 2 guesses 39°, then they score 10 points (49°–39°). • Each team gets five turns. The winner is the team with the lowest score! Reprint 2025-26 • Team 1 announces to all, an angle measure, e.g., 34°. • A player from Team 2 must draw that angle on the board without using a protractor. Other members of Team 2 can help the player by speaking words like ‘Make it bigger!’ or ‘Make it smaller!’. • A player from Team 1 measures the angle with a protractor for all to see. • Team 2 scores the number of points that is the absolute difference in degrees between Team 2’s angle size and the intended angle size. For example, if player’s angle from Team 2 is measured to be 25°, then Team 2 scores 9 points (34°–25°). • Each team gets five turns. The winner is again the team with the lowest score. Figure it Out Teacher’s Note These games are important to play to build intuition about angles and their measures. Return to this game at least once or twice on different days to build practice in estimating angles. Note that these games can also be played between pairs of students. 1. In Fig. 2.23, list all the angles possible. Did you find them all? Now, guess the measures of all the angles. Then, measure the angles with a protractor. Record all your numbers in a table. See how close your guesses are to the actual measures. Lines and Angles Chapter 2_Lines and Angles.indd 49 13-08-2024 16:14:40 C A D L S Fig. 2.23 P R Reprint 2025-26 B 49 Ganita Prakash | Grade 6 2.11 Types of Angles and their Measures We have read about different types of angles in this chapter. We have seen that a straight angle is 180° and a right angle is 90°. How can other types of angles—acute and obtuse—be described in terms of their degree measures? 2. Use a protractor to draw angles having the following degree measures: a. 110° b. 40° c. 75° d. 112° e. 134° 3. Draw an angle whose degree measure is the same as the angle given below: Also, write down the steps you followed to draw the angle. I H J Chapter 2_Lines and Angles.indd 50 13-08-2024 16:14:40 50 Acute Angle: Angles that are smaller than the right angle, i.e., less than 90° and are greater than 0°, are called acute angles. 40⁰ P R Q Examples of acute angles Reprint 2025-26 S 50⁰ T R 75⁰ E F Q Obtuse Angle: Angles that are greater than the right angle and less than the straight angle, i.e., greater than 90° and less than 180°, are called obtuse angles. Have we covered all the possible measures that an angle can take? Here is another type of angle. Reflex angle: Angles that are greater than the straight angle and less than the whole angle, i.e., greater than 180° and less than 360°, are called reflex angles. Figure it Out P A S 130º S I X 110º C S Examples of obtuse angles Examples of reflex angles B T W T M Lines and Angles Chapter 2_Lines and Angles.indd 51 13-08-2024 16:14:40 1. In each of the below grids, join A to other grid points in the figure by a straight line to get: a. An acute angle Reprint 2025-26 51 Ganita Prakash | Grade 6 b. An obtuse angle c. A reflex angle Mark the intended angles with curves to specify the angles. One has been done for you. Chapter 2_Lines and Angles.indd 52 13-08-2024 16:14:40 52 2. Use a protractor to find the measure of each angle. Then classify each angle as acute, obtuse, right, or reflex. a. ∠PTR b. ∠PTQ c. ∠PTW d. ∠WTP Q R P T Reprint 2025-26 W In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of ∠SET? 80o 90o Hint: Observe that ∠REB is a straight angle. Hence, the degree measure of ∠REB = 180° of which 80° is covered by ∠TER. A similar argument can be applied to find the measure of ∠SET. Let’s Explore Figure it Out 1. Draw angles with the following degree measures: 2. Estimate the size of each angle and then measure it with a protractor: a. b. c. a. 140° b. 82° c. 195° d. 70° e. 35° B E S T Lines and Angles R Chapter 2_Lines and Angles.indd 53 13-08-2024 16:14:40 3. Make any figure with three acute angles, one right angle and two obtuse angles. 4. Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 60°. 5. Draw the letter ‘Y’ such that the three angles formed are 150°, 60° and 150°. d. e. f. Classify these angles as acute, right, obtuse or reflex angles. Reprint 2025-26 53 Ganita Prakash | Grade 6 A point determines a location. It is denoted by a capital letter. A line segment corresponds to the shortest distance between two points. The line segment joining points S and T is denoted by ST. A line is obtained when a line segment like ST is extended on both sides indefinitely; it is denoted by ST or sometimes by a single small letter like m. A ray is a portion of a line starting at a point D and going in one direction indefinitely. It is denoted by DP where P is another point on the ray. An angle can be visualised as two rays starting from a common starting 6. The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes? 7. Puzzle: I am an acute angle. If you double my measure, you get an acute angle. If you triple my measure, you will get an acute angle again. If you quadruple (four times) my measure, you will get an acute angle yet again! But if you multiply my measure by 5, you will get an obtuse angle measure. What are the possibilities for my measure? Summary Chapter 2_Lines and Angles.indd 54 13-08-2024 16:14:43 54 The size of an angle is the amount of rotation or turn needed about the The sizes of angles can be measured in degrees. One full rotation or Degree measures of angles can be measured using a protractor. Angles can be straight (180°), right (90°), acute (more than 0° and less than 90°), obtuse (more than 90° and less than 180°), and reflex (more than 180° and less than 360°). point. Two rays OP and OM form the angle ∠POM (also called ∠MOP); here, O is called the vertex of the angle, and the rays OP and OM are called the arms of the angle. vertex to rotate one ray of the angle onto the other ray of the angle. turn is considered as 360 degrees and denoted as 360°. Reprint 2025-26 Section 2.4 Page No. 15 Figure it Out Q.1. Can you help Rihan and Sheetal find their answers? Ans. Rihan can draw many/uncountable number of lines through the given point. Q.2. Name the line segments in Fig. 2.4. Which of the five marked points are on exactly one of the line segments? Which are on two of the line segments? Ans. LM, MP, PQ, QR Q.3. Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays? Ans. TA, TB, TN and NB Rihan marked a point on a piece of paper. How many lines can he draw that pass through the point? Sheetal can draw only one line through the two given points. Points L and R are exactly on one line segment. Points M, P and Q are on two line segments. No, T is the starting point of TB, TN and TA but not of NB. CHAPTER 2 — SOLUTIONS Lines and Angles Sheetal marked two points on a piece of paper. How many different lines can she draw that pass through both of the points? Q.4. Draw a rough figure and write labels appropriately to illustrate each of the following: Ans. a. OP and OQ meet at O. b. XY and PQ intersect at point M. c. Line l contains points E and F but not point D. d. Point P lies on AB. [1] Q.5. In Fig. 2.6, name: a. Five points b. A line c. Four rays d. Five line segments Ans. a) D, E, O, B and C b) DE or DO or DB or EO or EB or OB c) OC, OB, OE, OD (Try for other rays) Q.6. Here is a ray OA (Fig. 2.7). It starts at O and passes through the point A. It also passes Ans. a) Yes, O is the starting point and point B lies on the rays that goes endlessly in the d) DE, DO, DB, EO, EB (OB; OC are also possible) through the point B. a. Can you also name it as OB? Why? b. Can we write OA as AO? Why or why not? b) No, OA is a ray with starting point O whereas AO is a ray with starting point A. direction of A. OA is the extension of OB. [2] Section 2.5 Page – 19 Figure it Out Q.1. Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle. Ans. Yes, one of the angles is ∠BDC. It’s vertex is D. One ray is DC and the other rays DB. Try for other pictures. Q.2. Draw and label an angle with arms ST and SR. Ans. Q.4. Name the angles marked in the given figure. Ans. ∠RTQ, ∠RTP Q.5. Mark any three points on your paper that are not on one line. Label them A, B, C. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C? Write them down, and mark each of them with a curve as in Fig. 2.9. Ans. We get three lines AB, BC, CA. Using A, B & C we can name three angles: ∠ABC or ∠CBA, ∠BCA or ∠ACB & ∠CAB or ∠BAC. [3] Q.6. Now mark any four points on your paper so that no three of them are on one line. Label them A, B, C, D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C, D? Write them all down, and mark each of them with a curve as in Fig. 2.9. Ans. We get six lines AB, BC, CD, DA, AC & BD Section 2.6 Page 20 Is it always easy to compare two angles? Page 23 No. it is not always easy to compare two angles. For eg. 89º & 91º angles cannot be compared without measuring or overlapping. But for the given figures, comparison is easy. Using A, B, C & D we can name the following angles ∠BAC, ∠CAD, ∠BAD, ∠ADB, ∠BDC, ∠ADC, ∠DCA, ∠ACB, ∠DCB, ∠CBD, ∠DBA & ∠CBA Q.1. Fold a rectangular sheet of paper, then draw a line along the fold created. Name and compare the angles formed between the fold and the sides of the paper. Make different angles by folding a rectangular sheet of paper and compare the angles. Which is the largest and smallest angle you made? Where else do we use superimposition to compare? A few examples are- line segments, squares and circles. Think of more. Figure it out [4] Ans. Q.2. In each case, determine which angle is greater and why. Ans. (a)- ∠AOB; ∠XOY is an acute angle and ∠AOB = ∠AOX + ∠XOY+∠YOB (b)- ∠AOB (c) – None. ∠XOB = ∠XOC Q.3. Which angle is greater: ∠XOY or ∠AOB? Give reasons. Ans. By looking at the figure we cannot say. Superimposition or measurement is necessary here. a. ∠AOB or ∠XOY b. ∠AOB or ∠XOB c. ∠XOB or ∠XOC Discuss with your friends on how you decided which one is greater. Angles Formed: ∠AEF, ∠BEF, ∠DFE, ∠CFE Here ∠AEF & ∠CFE are larger than ∠BEF & ∠DFE Try more cases by folding rectangular sheets in different ways. Section 2.8 Page 28 Q. Is it possible to draw OC such that the two angles are equal to each other in size? Ans. Yes, when OA and OB overlap each other on folding the Vidya’s notebook, the crease OC will divide ∠AOB in two equal sized angles. Page 29 Q. If a straight angle is formed by half of a full turn, how much of a full turn will form a right angle? Ans. 1 4 of a full turn. [5] Section 2.8 Page No. 29 Figure it Out Q.4. Get a slanting crease on the paper. Now, try to get another crease that is perpendicular to the slanting crease. Ans. a. Four right angle. Each angle is ¼ the of the complete angle. Page 31 Figure it Out Q.2. Make a few acute angles and a few obtuse angles. Draw them in different orientations. Ans. Acute angle a. How many right angles do you have now? Justify why the angles are exact right angles. b. Describe how you folded the paper so that any other person who doesn’t know the process can simply follow your description to get the right angle. b. Explore different ways of doing it. Obtuse angle Q.3. Do you know what the words acute and obtuse mean? Acute means sharp and obtuse means blunt. Why do you think these words have been chosen? Ans. In acute angles the opening of the edges is lesser than the obtuse angle which have larger openings. Q.4. Find out the number of acute angles in each of the figures below. What will be the next figure and how many acute angles will it have? Do you notice any pattern in the numbers? Ans. (i) three (ii) Twelve (iii) Twenty-one The next figure will have thirty acute angles. Yes, the pattern is 3 × 0 + 1, 3 × 1 + 1, 3 × 2 + 1, 3 × 3 + 1,… The number 0,1,2,3,4,….. are number of inner triangles. [6] Section 2.9 Page No. 35 Figure it out Q.1. Write the measures of the following angles: Ans. a. ∠KAL = 30° Yes, it is possible to count the number of units in 5s or 10s. Page No. 36 Q. Name the different angles in the figure and write their measures. Ans. ∠POQ = 35° a. ∠ KAL b. ∠WAL c. ∠TAK b. ∠WAL = 50° c. ∠TAK = 120° ∠POR = 95° ∠POS = 125° ∠POT = 160° ∠QOR = 60° ∠QOS = 90° ∠QOT = 125° ∠QOU = 145° ∠ROS = 30° ∠ROT = 65° ∠ROU = 85° ∠SOT = 35° ∠SOU = 55° ∠TOU = 20° [7] Page No. 40 Think! Q. In Fig. 2.20, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = Ans. Each of the angles = 22.5º Figure it Out Q.1. Find the degree measures of the following angles using your protractor. Ans. ∠IHJ = ∠JHI = 47º Q.3. Find the degree measures for the angles given below. Check if your paper protractor can be used here! Ans. ∠IHJ = 42º, ∠IHJ = 116º Q.4. How can you find the degree measure of the angle given below using a protractor? Ans. Measure of marked angle = 360º – Measure of unmarked angle ∠GOH = ∠HOI=_____. Why? As the straight angle of 180º is divided into eight equal parts so each of the right angles will be of measure = 180º 8 = 22.5º ∠GHK = ∠IHJ = 23º ∠IHJ = ∠JHI = 108º No, paper protractor cannot work here. = 360º - 100º = 260º Try other ways to find the marked angle Q.5. Measure and write the degree measures for each of the following angles: Ans. Q.6. Find the degree measures of ∠BXE, ∠CXE, ∠AXB and ∠BXC. Ans. ∠BXE = 115º, ∠CXE = 85º, ∠AXB = 65º ∠BXC = 30º Q.7. Find the degree measures of ∠PQR, ∠PQS and ∠PQT. Ans. ∠PQR =45º ∠PQS=100º ∠PQT=150º. a. 80º b. 120º c. 60º d. 130º e. 130º f. 60º [8] Page 45 Figure it Out Q.1. Angles in a clock: Ans. (a) The angles at the centre of the clock is 360º which is divided into 12 equal parts. So. Q.2. The angle of a door: Ans. Yes, the vertex of the angle will be the point where door meets the wall. The arms will be the edges of the door and the wall. Q.3. Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle? Is it possible to express the amount by which a door is opened using an angle? What will be the vertex of the angle and what will be the arms of the angle? a. The hands of a clock make different angles at different times. At 1 o’clock, the angle between the hands is 30°. Why? b. What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock? c. Explore other angles made by the hands of a clock. The angle between two successive numbers = 360 12 = 30º (b) At 2’O clock = 60º = 2×30 At 4’O clock = 120º = 4×30 At 6’O clock = 180º = 6×30 (c) At 3’O clock = 90º At 9’O clock = 270º Try for other angles made by the hands of a clock. Q.4. Here is a toy with slanting slabs attached to its sides; the greater the angles or slopes of the slabs, the faster the balls roll. Can angles be used to describe the slopes of the slabs? What are the arms of each angle? Which arm is visible and which is not? Ans. Student may not see the angle, but when the starting arm is fixed as the position where she starts swinging. The angle can be thought to be between positions where she starts the swinging (the initial position) and the position where she attains the greatest position of the swing at any one side. Ans. Yes, angles can be used directly to describe slopes of the slab, larger the angle, greater the slope of the slab. For each angle, one arm is a side and one arm is the slope. The vertical arm is not visible, whereas the other arm is visible. Here in this toy, edges of the slabs are the arms of the angles. Top horizontal ray is not visible, other arms in the form of edges of the slab are visible. Teacher should motivate students to get other possible answers. [9] Page 49 Section 2.10 Figure it Out Q.1. In Fig. 2.23, list all the angles possible. Did you find them all? Now, guess the measures of all the angles. Then, measure the angles with a protractor. Record all your numbers in a table. See how close your guesses are to the actual measures. Ans. ∠CAP, ∠ACD, ∠APL, ∠DLP, ∠RPL, ∠SLP, ∠PRS, ∠LSR, ∠BRS, ∠CLP Try more! Page 52 Section 2.11 Figure it Out Q.2. Use a protractor to find the measure of each angle. Then classify each angle as acute, obtuse, right or reflex. Ans. Let’s Explore: Q. In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of ∠SET? Ans. ∠BET = 100º, ∠SET = 10º a. ∠PTR = 30º (Acute angle) b. ∠PTQ = 60º (Acute angle) c. ∠PTW = 102º (Obtuse angle) d. ∠WTP = 258º (Reflex angle) Page – 53 Figure it out Q.3. Make any figure with three acute angles, one right angle and two obtuse angles. Ans. ∠A, ∠B & ∠C are three acute angles ∠D, ∠F are obtuse angle ∠E is right angle [10] Q.4. Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 60°. Ans. Ans. Q.5. Draw the letter ‘Y’ such that the three angles formed are 150°, 60° and 150°. Q.6. The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes Ans. The angle between two spokes next to each other is 15º. Largest acute angle between spokes is 75º. Q.7. Puzzle: I am an acute angle. If you double my measure, you get an acute angle. If you triple my measure, you will get an acute angle again. If you quadruple (four times) my measure, you will get an acute angle yet again! But if you multiply my measure by 5, you will get an obtuse angle measure. What are the possibilities for my measure? Ans. The acute angle can be 19º, 20º, 21º & 22º. next to each other? What is the largest acute angle formed between two spokes? [11]" class_6,3,Number Play,ncert_books/class_6/Ganita_Prakash/fegp103.pdf,"3 NUMBER PLAY Numbers are used in different contexts and in many different ways to organise our lives. We have used numbers to count, and have applied the basic operations of addition, subtraction, multiplication and division on them, to solve problems related to our daily lives. In this chapter, we will continue this journey, by playing with numbers, seeing numbers around us, noticing patterns, and learning to use numbers and operations in new ways. Think about various situations where we use numbers. List five different situations in which numbers are used. See what your classmates have listed, share, and discuss. 3.1 Numbers can Tell us Things What are these numbers telling us? Math Talk Chapter 3_Number Play.indd 55 14-08-2024 14:43:33 Some children in a park are standing in a line. Each one says a number. The children now rearrange themselves, and again each one says a number based on the arrangement. What do you think these numbers mean? Reprint 2025-26 Ganita Prakash | Grade 6 Did you figure out what these numbers represent? Hint: Could their heights be playing a role? A child says ‘1’ if there is only one taller child standing next to them. A child says ‘2’ if both the children standing next to them are taller. A child says ‘0’, if neither of the children standing next to them are taller. That is each person says the number of taller neighbours they have. Try answering the questions below and share your reasoning. 1. Can the children rearrange themselves so that the children standing at the ends say ‘2’? Math Talk Chapter 3_Number Play.indd 56 14-08-2024 14:43:38 56 2. Can we arrange the children in a line so that all would say 3. Can two children standing next to each other say the same 4. There are 5 children in a group, all of different heights. Can 5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible? 6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not? 7. How would you rearrange the five children so that the only 0s? number? they stand such that four of them say ‘1’ and the last one says ‘0’? Why or why not? maximum number of children say ‘2’? Reprint 2025-26 3.2 Supercells Observe the numbers written in the table below. Why are some numbers coloured? Discuss. A cell is coloured if the number in it is larger than its adjacent cells. The number 626 is coloured as it is larger than 577 and 345, whereas 200 is not coloured as it is smaller than 577. The number 198 is coloured as it has only one adjacent cell with 109 in it, and 198 is larger than 109. 1. Colour or mark the supercells in the table below. 2. Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells. Figure it Out 200 577 626 345 790 694 109 43 79 75 63 10 29 28 34 6828 670 9435 3780 3708 7308 8000 5583 52 5346 1258 9635 Number Play 198 Chapter 3_Number Play.indd 57 14-08-2024 14:43:38 3. Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions. 4. Out of the 9 numbers, how many supercells are there in the table above? ___________ 5. Find out how many supercells are possible for different numbers of cells. Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy. Reprint 2025-26 Math Talk 57 Ganita Prakash | Grade 6 Let’s do the supercells activity with more rows. Here the neighbouring cells are those that are immediately to the left, right, top and bottom. The rule remains the same: a cell becomes a supercell if the number in it is greater than all the numbers in its neighbouring cells. In Table 1, 8632 is greater than all its neighbours 4580, 8280, 4795 and 1944. 6. Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not? 7. Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not? 8. Fill a table such that the cell having the second largest number is not a supercell. 9. Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible? 10. Make other variations of this puzzle and challenge your classmates. 2430 7500 7350 9870 3115 4795 9124 9230 4580 8632 8280 3446 Table 1 Try This Chapter 3_Number Play.indd 58 14-08-2024 14:43:38 58 Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours. The biggest number in the table is ____________ . Reprint 2025-26 5785 1944 5805 6034 13,609 60,319 19,306 10,963 96,301 36,109 Table 2 60,193 The smallest even number in the table is ____________. The smallest number greater than 50,000 in the table is ____________. Once you have filled the table above, put commas appropriately after the thousands digit. 3.3 Patterns of Numbers on the Number Line We are quite familiar with number lines now. Let’s see if we can place some numbers in their appropriate positions on the number line. Here are the numbers: 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400. Identify the numbers marked on the number lines below, and label the remaining positions. Figure it Out a. 1000 2000 2180 2754 3000 4000 5000 6000 7000 8000 9000 10,000 Number Play Chapter 3_Number Play.indd 59 14-08-2024 14:43:39 Put a circle around the smallest number and a box around the largest number in each of the sequences above. b. 15,077 15,078 15,083 c. 86,705 87,705 d. 9996 9997 2010 2020 Reprint 2025-26 59 Ganita Prakash | Grade 6 We start writing numbers from 1, 2, 3 … and so on. There are nine 1-digit numbers. Find out how many numbers have two digits, three digits, four digits, and five digits. Digit sums of numbers Komal observes that when she adds up digits of certain numbers the sum is the same. For example, adding the digits of the number 68 will be same as adding the digits of 176 or 545. 3.4 Playing with Digits 1-digit numbers From 1–9  Figure it Out 9 2-digit numbers 3-digit numbers 4-digit numbers 5-digit numbers Chapter 3_Number Play.indd 60 14-08-2024 14:43:39 60 1. Digit sum 14 a. Write other numbers whose digits add up to 14. b. What is the smallest number whose digit sum is 14? c. What is the largest 5-digit whose digit sum is 14? d. How big a number can you form having the digit sum of 14? Can you make an even bigger number? 2. Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class. 3. Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue? Reprint 2025-26 . Math Talk Digit Detectives After writing numbers from 1 to 100, Dinesh wondered how many times he would have written the digit ‘7’! Among the numbers 1–100, how many times will the digit ‘7’ occur? Among the numbers 1–1000, how many times will the digit ‘7’ occur? 3.5 Pretty Palindromic Patterns What pattern do you see in these numbers: 66, 848, 575, 797, 1111? These numbers read the same from left to right and from right to left. Try and see. Such numbers are called palindromes or palindromic numbers. The numbers 121, 313, 222 are some examples of palindromes using the digits ‘1’, ‘2’, 3’. Reverse-and-add palindromes Write all possible 3-digit palindromes using these digits. All palindromes using 1, 2, 3 Number Play Chapter 3_Number Play.indd 61 14-08-2024 14:43:40 Now, look at these additions. Try to figure out what is happening. Steps to follow: Start with a 2-digit number. Add this number to its reverse. Stop if you get a palindrome or else repeat the steps of reversing the digits and adding. Try the same procedure for some other numbers, and perform the same steps. Stop if Reprint 2025-26 61 Ganita Prakash | Grade 6 you get a palindrome. There are numbers for which you have to repeat this a large number of times. Are there numbers for which you do not reach a palindrome at all? Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.* Explore Puzzle time I am a 5-digit palindrome. I am an odd number. My ‘t’ digit is double of my ‘u’ digit. My ‘h’ digit is double of my ‘t’ digit. Who am I? _________________ tth Write the number in words: th h ut Math Talk Chapter 3_Number Play.indd 62 14-08-2024 14:43:40 62 3.6 The Magic Number of Kaprekar D.R. Kaprekar was a mathematics teacher in a government school in Devlali, Maharashtra. He liked playing with numbers very much and found many beautiful patterns in numbers that were previously unknown. In 1949, he discovered a fascinating and magical phenomenon when playing with 4-digit numbers. *The answer is yes! For 3-digit numbers the answer is unknown. It is suspected that starting with 196 never yields a palindrome! Reprint 2025-26 A = 8632 A = 6642 A = 7641 A = B = 2368 B = 2466 B = 1467 B = C = 8632–2368 C = 6642–2466 C = 7641–1467 C = = 6264 = 4176 = 6174 Follow these steps and experience the magic for yourselves! Pick any 4-digit number having at least two different digits, say 6382. Make the smallest number from these Make the largest number from these Subtract B from A. Call it C. What happens if we continue doing this? Take a 4-digit number. digits. Call it A. digits. Call it B. C = A – B Use digits of C Number Play Chapter 3_Number Play.indd 63 14-08-2024 14:43:40 Take different 4-digit numbers and try carrying out these steps. Find out what happens. Check with your friends what they got. You will always reach the magic number ‘6174’! The number ‘6174’ is now called the ‘Kaprekar constant’. Carry out these same steps with a few 3-digit numbers. What number will start repeating? Explore Reprint 2025-26 63 Ganita Prakash | Grade 6 3.7 Clock and Calendar Numbers On the usual 12-hour clock, there are timings with different patterns. For example, 4:44, 10:10, 12:21. Try and find out all possible times on a 12-hour clock of each of these types. Manish has his birthday on 20/12/2012 where the digits ‘2’, ‘0’, ‘1’, and ‘2’ repeat in that order. Find some other dates of this form from the past. His sister, Meghana, has her birthday on 11/02/2011 where the digits read the same from left to right and from right to left. Jeevan was looking at this year’s calendar. He started wondering, “Why should we change the calendar every year? Can we not reuse a calendar?”. What do you think? You might have noticed that last year’s calendar was different from this year’s. Also, next year’s calendar will also be different from the previous years. Find all possible dates of this form from the past. Chapter 3_Number Play.indd 64 14-08-2024 14:43:45 64 But, will any year’s calendar repeat again after some years? Will all dates and days in a year match exactly with that of another year? 1. Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4-digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4 -digits to make: Figure it Out a. the difference between the largest and smallest numbers greater than 5085. Reprint 2025-26 Try This 3.8 Mental Math Observe the figure below. What can you say about the numbers and the lines drawn? 2. What is the sum of the smallest and largest 5-digit palindrome? What is their difference? 3. The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that? 4. How many rounds does the number 5683 take to reach the Kaprekar constant? 38,800 3,400 28,000 63,000 61,600 19,500 b. the difference between the largest and smallest numbers less than 5085. c. the sum of the largest and smallest numbers greater than 9779. d. the sum of the largest and smallest numbers less than 9779. 25,000 13,000 400 Number Play Chapter 3_Number Play.indd 65 14-08-2024 14:43:45 Numbers in the middle column are added in different ways to get the numbers on the sides (1500 + 1500 + 400 = 3400). The numbers in the middle can be used as many times as needed to get the desired sum. Draw arrows from the middle to the numbers on the sides to obtain the desired sums. Two examples are given. It is simpler to do it mentally! 38,800 = 25,000 + 400 × 2 + 13,000 3400 = 1500 + 1500 + 400 31,000 20,900 60,000 1,500 Reprint 2025-26 65 Ganita Prakash | Grade 6 Can we make 1,000 using the numbers in the middle? Why not? What about 14,000, 15,000 and 16,000? Yes, it is possible. Explore how. What thousands cannot be made? Adding and Subtracting Here, using the numbers in the boxes, we are allowed to use both addition and subtraction to get the required number. An example is shown. Digits and Operations An example of adding two 5-digit numbers to get another 5-digit number is 12,350 + 24,545 = 36,895. An example of subtracting two 5-digit numbers to get another 5-digit number is 48,952 – 24,547 = 24,405. 12,000 800 40,000 7,000 Figure it Out 300 1,500 39,800 = 40,000 – 800 + 300 + 300 45,000 = 5,900 = 17,500 = 21,400 = Math Talk Chapter 3_Number Play.indd 66 14-08-2024 14:43:45 66 1. Write an example for each of the below scenarios whenever possible. 5-digit – 5-digit to give a difference less than 56,503 5-digit + 5-digit to give a 5-digit sum more than 90,250 5-digit + 3-digit to give a 6-digit sum 5-digit – 3-digit to give a 4-digit difference Reprint 2025-26 4-digit + 4-digit to give a 6-digit sum 5-digit − 4-digit to give a 4-digit difference 5-digit + 5-digit to give a 6-digit sum 5-digit − 5-digit to give a 3-digit difference 5-digit + 5-digit to give 18,500 5-digit − 5-digit to give 91,500 3.9 Playing with Number Patterns Here are some numbers arranged in some patterns. Find out the sum of the numbers in each of the below figures. Should we add them one by one or can we use a quicker way? Share and discuss in class the different methods each one of you used to solve these questions. 2. Always, Sometimes, Never? Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates. Below are some statements. Think, explore and find out if each of the statement is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning and discuss this with the class. a. 5-digit number + 5-digit number gives a 5-digit number b. 4-digit number + 2-digit number gives a 4-digit number c. 4-digit number + 2-digit number gives a 6-digit number d. 5-digit number – 5-digit number gives a 5-digit number e. 5-digit number – 2-digit number gives a 3-digit number Number Play Math Talk Chapter 3_Number Play.indd 67 14-08-2024 14:43:45 a. b. 50 50 40 40 40 50 50 40 40 40 50 50 40 40 40 50 50 Reprint 2025-26 40 40 40 50 50 67 Ganita Prakash | Grade 6 c. d. 15 e. f. 35 35 35 35 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 35 35 25 1515 15 35 1515 15 15 25 25 25 25 25 125 125 125 125 125 125 125 15 15 25 25 2525 25 35 35 35 35 35 35 35 15 25 25 35 35 35 15 35 35 15 15 2525 35 25 35 15 25 1515 15 15 15 15 25 25 25 25 25 15 35 125 125 125 125 125 250 125 125 125 125 500 1000 500 250 250 125 500 125 250 500 250 250 250 250 Chapter 3_Number Play.indd 68 14-08-2024 14:43:46 68 3.10 An Unsolved Mystery — the Collatz Conjecture! Look at the sequences below—the same rule is applied in all the sequences: Do you see how these sequences were formed? a. 12, 6, 3, 10, 5, 16, 8, 4, 2, 1 b. 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 c. 21, 64, 32, 16, 8, 4, 2, 1 d. 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 Reprint 2025-26 The rule is: one starts with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat. Notice that all four sequences above eventually reached the number 1. In 1937, the German mathematician, Lothar Collatz conjectured that the sequence will always reach 1, regardless of the whole number you start with. Even today—despite many mathematicians working on it — it remains an unsolved problem as to whether Collatz’s conjecture is true! Collatz’s conjecture is one of the most famous unsolved problems in mathematics. Make some more Collatz sequences like those above, starting with your favourite whole numbers. Do you always reach 1? Do you believe the conjecture of Collatz that all such sequences will eventually reach 1? Why or why not? 3.11 Simple Estimation At times, we may not know or need an exact count of things and an estimate is sufficient for the purpose at hand. For example, your school headmaster might know the exact number of students enrolled in your school, but you may only know an estimated count. How many students are in your school? About 150? 400? A thousand? Paromita’s class section has 32 children. The other 2 sections of her class have 29 and 35 children. So, she estimated the number of children in her class to be about 100. Along with Class 6, her school also has Classes 7–10 and each class has 3 sections each. She assumed a similar number in each class and estimated the number of students in her school to be around 500. Number Play Chapter 3_Number Play.indd 69 14-08-2024 14:43:46 We shall do some simple estimates. It is a fun exercise, and you may find it amusing to know the various numbers around us. Remember, Figure it Out Reprint 2025-26 69 Ganita Prakash | Grade 6 Estimate the answer Try to guess within 30 seconds. Check your guess with your friends. we are not interested in the exact numbers for the following questions. Share your methods of estimation with the class. 1. Steps you would take to walk: a. From the place you are sitting to the classroom door b. Across the school ground from start to end c. From your classroom door to the school gate d. From your school to your home 2. Number of times you blink your eyes or number of breaths you take: a. In a minute b. In an hour c. In a day 3. Name some objects around you that are: a. a few thousand in number b. more than ten thousand in number 1. Number of words in your maths textbook: a. More than 5000 2. Number of students in your school who travel to school by bus: a. More than 200 b. Less than 5000 Chapter 3_Number Play.indd 70 14-08-2024 14:43:46 70 3. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹100. Do you agree with him? Why or why not? 4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). b. Less than 200 Hint: Look at the map of India to locate these cities. Reprint 2025-26 3.12 Games and Winning Strategies Numbers can also be used to play games and develop winning strategies. Here is a famous game called 21. Play it with a classmate. Then try it at home with your family!  Rules for Game #1: The first player says 1, 2 or 3. Then the two players take turns adding 1, 2, or 3 to the previous number said. The first player to reach 21 wins! Play this game several times with your classmate. Are you starting to see the winning strategy? Which player can always win if they play correctly? What is the pattern of numbers that the winning player should say? There are many variations of this game. Here is another common variation: 5. Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. Do you agree with her? Why or why not? 6. Earlier, people used to walk long distances as they had no other means of transport. Suppose you walk at your normal pace. Approximately, how long would it take you to go from: 7. Make some estimation questions and challenge your classmates! a. Your current location to one of your favourite places nearby. b. Your current location to any neighbouring state’s capital city. c. The southernmost point in India to the northernmost point in India. Number Play Chapter 3_Number Play.indd 71 14-08-2024 14:43:46  Rules for Game #2: The first player says a number between 1 and 10. Then the two players take turns adding a number between 1 and 10 to the previous number said. The first player to reach 99 wins! Play this game several times with your classmate. See if you can figure out the corresponding winning strategy in this case! Which Reprint 2025-26 71 Ganita Prakash | Grade 6 player can always win? What is the pattern of numbers that the winning player should say this time? Make your own variations of this game — decide how much one can add at each turn, and what number is the winning number. Then play your game several times, and figure out the winning strategy and which player can always win! 1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap. 2. How many rounds does your year of birth take to reach the Kaprekar constant? 3. We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000? 4. Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then, try to get an exact number and see how close your estimate is. Figure it Out 19,381 50,319 38,408 16,200 39,344 29,765 23,609 45,306 62,871 Try This Chapter 3_Number Play.indd 72 14-08-2024 14:43:46 72 5. Estimate the number of liters a mug, a bucket and an overhead tank can hold. 6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670. 7. Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number. Reprint 2025-26 Numbers can be used for many different purposes including, to convey information, make and discover patterns, estimate magnitudes, pose and solve puzzles, and play and win games. Thinking about and formulating set procedures to use numbers for these purposes is a useful skill and capacity (called ‘computational thinking’). Many problems about numbers can be very easy to pose, but very difficult to solve. Indeed, numerous such problems are still unsolved (for example, Collatz’s Conjecture). 8. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence? 9. Check if the Collatz Conjecture holds for the starting number 100. 10. Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now? Summary Number Play Chapter 3_Number Play.indd 73 14-08-2024 14:43:46 Reprint 2025-26 73 Q. Think about various situations where we use numbers. List five different situations in which numbers are used. See what your classmates have listed, share, and discuss. Ans. Five different possible situations in which numbers are used - Section 3.1 Page No. 55 Q. What do you think these numbers mean? Ans. Refer page 56. Q1. Can the children rearrange themselves so that the children standing at the ends say ‘2’? Ans. No; There will be no one standing on the other side of the child standing at the end. Q2. Can we arrange the children in a line so that all would say only 0s? Ans. Yes; All the children in the line should be of same height. Page No. 56 1. Time 2. Calendar 3. Counting objects/Marks 4. Measurement of height & weight 5. Money There could many more. CHAPTER 3 — SOLUTIONS Number Play Q3. Can two children standing next to each other say the same number? Ans. Yes; Refer picture on page 55. Q4. There are 5 children in a group, all of different heights. Can they stand such that four of them say ‘1’ and the last one says ‘0’? Why or why not? Ans. Yes, they can, if they are standing in ascending order of height. [1] Q5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible? Ans. No; the tallest child at the end cannot say1. Q6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not? Ans. Yes, it is possible. Q7. How would you rearrange the five children so that the maximum number of children say ‘2’? Ans. At the most only 2 children can say 2 as given is the following arrangement. Section 3.2 Page No. 57 Figure it out Q1. Colour or mark the supercells in the table below. Ans. 6828 670 9435 3780 3708 7308 8000 5583 52 6828 670 9435 3780 3708 7308 8000 5583 52 [2] Q2. Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells. Ans. One of the ways could be-5346; 9636.Try more Q3. Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions. Ans. Q4. Out of the 9 numbers, how many supercells are there in the table above? ___________ Ans. 5 Q5 Find out how many supercells are possible for different numbers of cells. Ans. For even number of cells say,2,4,6,… the number of supercells would be respectively, 2/2 =1,4/2 =2,6/2=3,… For odd number of cells , say 1,3,5,7,… the number of supercells would be respectively (1+1)/2= 1, (3+1)/2 = 2, (5+1)/2= 3,(7+1)/2 = 4,… To get the maximum number ofsupercells, we have to start by filling the first cell as super cell & then fill alternately. Q6. Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not? 5346 1258 9635 5346 5347 1000 1258 1100 1200 1300 9635 9636 110 100 150 130 280 200 230 210 270 Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy. Ans. No; the cell which is filled by the greatest number among the given numbers chosen, will become super cell irrespective of its position in the table. Q7. Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not? Ans. Yes, the largest number in a table will always be a supercell. Q8. Fill a table such that the cell having the second largest number is not a supercell. Ans. One of the ways could beNo, the smallest number in a table can never be a supercell as the number in all the adjacent cells will be greater than it. 1 2 3 4 5 6 7 9 8 [3] Q9. Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible? Ans. One of the ways isQ10. Make other variations of this puzzle and challenge your classmates. Ans. Some of these could beCan you fill the table with 9 cells such that there are more than 5 super cells? Can you fill the table with 9 cells such that there are exactly 4 super cells? Page No. 58 Q. Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours. Ans. Table 2 (One of the ways) – Q. The biggest number in the table is ____________. Ans. The biggest number in the table is 96,310 Q. The smallest even number in the table is ____________. Ans. The smallest even number in the table is 10,396 Second smallest number a super cell 2 1 3 4 5 6 7 9 8 Second largest number 8 is not a supercell. 96,310 96,301 36,109 39,160 96,103 13,609 60,319 19,306 13,906 10,396 60,193 60,931 10,369 10,963 10,936 69,031 Q. The smallest number greater than 50,000 in the table is ____________. Ans. The smallest number greater than 50,000 in the table is 60,193. Section 3.3 Page no.59 Ans. 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Q. We are quite familiar with number lines now. Let’s see if we can place some numbers in their appropriate positions on the number line. Here are the numbers: 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400. 1050 1500 2180 2754 3600 5030 5300 [4] 8400 9590 9950 (a). (c). (d). Put a circle around the smallest number and a box around the largest number in each of the sequences above. Section 3.4 Page no. 60 (b). Q. Identify the numbers marked on the number lines below, and label the remaining positions. 9993 9994 9995 9996 9997 9998 9999 10000 10001 10002 1990 1995 2000 2005 2010 2015 2020 2025 2030 2035 15077 15078 15079 15080 15081 15082 15083 15084 15085 15086 83705 84705 85705 86705 87705 88705 89705 90705 91705 92705 Q. Find out how many numbers have two digits, three digits, four digits, and five digits: Ans. 1 Digit numbers 2 Digit numbers 3 Digit numbers 4 Digit numbers 5 Digit numbers 9 1 Digit numbers 2 Digit numbers 3 Digit numbers 4 Digit numbers 5 Digit numbers 9 90 900 9,000 90,000 [5] Q.1. Digit sum 14 Ans. a. Some such numbers are:248, 653, 356, 815, 833, 12335, 23351. b. 59 c. 95000 d. 95, 9005, 900005, 90000005, 9000000005, 90000000000005 … Q.3. Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue? Ans. 123 → 1+2+3 = 6 234 → 2+3+4 = 9 345 → 3+4+5 = 12 456 → 4+5+6 = 15 567 → 5+6+7 = 18 678 → 6+7+8 = 21 789 → 7+8+9 = 24 • Yes, there is a pattern, all the sums are multiples of 3. • No. Page no. 61 Figure it out a. Write other numbers whose digits add up to 14. b. What is the smallest number whose digit sum is 14? c. What is the largest 5-digit number whose digit sum is 14? d. How big a number can you form having the digit sum 14? Can you make an even bigger number? Ans. • 20 times. • 300 times. Q. Among the numbers 1–100, how many times will the digit ‘7’ occur? Among the numbers 1–1000, how many times will the digit ‘7’ occur? [6] Section 3.5 Page no. 61 Ans. Palindromes: 111, 121, 131 222, 212, 232 313, 323, 333 Explore Page no. 62 Q. Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.* Try more Yes, it will always give a palindrome. Q. I am a 5-digit palindrome. I am an odd number. My ‘t’ digit is double of my ‘u’ digit. My ‘h’ digit is double of my ‘t’ digit. Who am I? _________________ Ans. tth th h t u Q. Write all possible 3-digit palindromes using these digits. Puzzle time Some of these are- 12 +21 33 1 2 4 2 1 47 +74 121 Section 3.6 Page no. 63 Q. Carry out these same steps with a few 3-digit numbers. What number will start repeating? Ans. Take a 3- Digit number say, 321. 321 -123 198 Try for other 3-digit numbers. 954 -459 495 The number 495 starts repeating. 981 -189 792 972 -279 693 Twelve thousand four hundred twenty one. [7] 963 -369 594 954 -459 495 Section 3.7 Page no. 64 Q. Try and find out all possible times on a 12-hour clock of each of these types. Ans. 4:44 2:22 3:33 10:10 11:11 12:12 09:09 12:21 05:50 10:01 Think of some more! Q. Find some other dates of this form from the past. Ans. 20/04/2004, 20/06/2006, Try for yourself! Q. Find all possible dates of this form from the past. Ans. 01/02/2001, 02/02/2002, Think of some more! Q. Will any year’s calendar repeat again after some years? Will all dates and days in a year match exactly with that of another year? Ans. Yes, The calendar repeats itself after 6 years if only one leap year is included in these 6 years. If 2 leap years are included, then it will repeat after 5 years. Page no. 64 Figure it out Q.1. Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4- digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4– digits to make: a. the difference between the largest and smallest numbers greater than 5085. b. the difference between the largest and smallest numbers less than 5085. c. the sum of the largest and smallest numbers greater than 9779. d. the sum of the largest and smallest numbers less than 9779. Ans. Some of the possibilities are– a. 7431 – 1347 = 6084 b. 7433 – 3347 = 4086 c. 7433 + 3347 = 10780 d. 7431 + 1347 = 8778 Q.2. What is the sum of the smallest and largest 5-digit palindrome? What is their difference? Ans. Smallest 5 digit palindrome = 10001 largest 5 digit palindrome = 99999 Sum = 10001 + 99999 = 110,000 Difference = 99999 – 10001 = 89,998 [8] Q.3. The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that? Ans. Time Now → 10:01 Next palindrome time → 11:11 After 1 hr. 10 min = 70 min the clock will show next palindrome time. Next palindrome time = 12:21 which will occur after 2 hr. 20 min = 140 min from 10:01. Q.4. How many rounds does the number 5683 take to reach the Kaprekar constant? Ans. 5683 8653 -3568 5085 Page no. 66 Section 3.8 Q. Can we make 1,000 using the numbers in the middle? Why not? What about 14,000, 15,000 and 16,000? Yes, it is possible. Explore how. What thousands cannot be made? Ans. No; the only number which is smaller than 1000 is 400 and 1000 is not a multiple of 400. 14000 = 1500 × 8 + 400 × 5 = 12000 + 2000 = 14000 15000 = 13000 + 400 × 5 = 13000 + 2000 = 15000 16000 = 1500 × 8 + 400 × 10 = 12000 + 4000 = 16000 Only one thousand (1000) cannot be made. 7641 -1467 6174 1 2 3 4 5 6 7 8 It will take 8 rounds to reach the Kaprekar constant. 8550 -5058 3492 9432 -2349 7083 8730 -3078 5652 6552 -2556 3996 9963 -3699 6264 6642 -2466 4176 Figure it out Q.1. Write an example for each of the below scenarios whenever possible. Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates. Ans. • 5 digit + 5 digit > 90,250 45,000 + 45,400 = 90,400 > 90,250 • 5 digit + 3 digit = 6 digit sum 99,999 + 999 = 100,998 • 4 digit + 4 digit = 6 digit sum Not possible as even the sum of the greatest 4 digit numbers will not give a six digit sum. (9999 + 9999 = 19,998) [9] • 5 digit + 5 digit = 6 digit sum 60,000 + 40,000 = 1,00,000 • 5 digit + 5 digit = 18,500 Not possible as smallest 5-digit number is 10,000. If both the numbers are 10,000 then the sum is 20,000, which is more than 18,500. • 5 digit – 5 digit < 56,503 80,000 – 50,000 < 56,503 < 56,503 • 5 digit – 3 digit = 4 digit difference 10,000 – 999 = 9001 • 5 digit – 4 digit = 4 digit difference 12,000 – 2,500 = 9,500 • 5 digit – 5 digit = 3 digit difference 50,999 – 50,000 = 999 • 5 digit – 5 digit = 91,500 Not possible as the difference of the greatest and the smallest 5 digit numbers, the maximum difference, can be 99,999 – 10,000 = 89,999 Some examples of other such questions are1. 5 digit + 5 digit = 7 digit sum 2. 4 digit + 4 digit = 2900 More such examples can be made. Q.2. Always, Sometimes, Never? Below are some statements. Think, explore and find out if each of the statement is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning; discuss this with the class. a. 5-digit number + 5-digit number gives a 5-digit number b. 4-digit number + 2-digit number gives a 4-digit number c. 4-digit number + 2-digit number gives a 6-digit number d. 5-digit number – 5-digit number gives a 5-digit number e. 5-digit number – 2-digit number gives a 3-digit number Ans. a. Only sometimes true. 20,000 + 80,000 = 1,00,000 not a 5 digit number b. Only sometimes true 9,999 + 99 = 10,098 not a 4 digit number c. Never true 9,999 + 99 = 10,098 On adding the greatest 4 digit and the greatest 2 digit numbers, we can reach only 5 digit number 10,098. So there is no possibility of getting a 6 digit number. d. Only sometimes true Ex. 12,000 – 10,000 = 2,000 5 digit – 5 digit = 4 digit e. Never true Ex. 10,000 – 99 = 9901 [10] Page no. 69 Section 3.10 Q. Make some more Collatz sequences like those above, starting with your favourite whole numbers. Do you always reach 1? Do you believe the conjecture of Collatz that all such sequences will eventually reach 1? Why or why not? Ans. a) 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 b) 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 Yes we always reach 1. The even numbers are halved and when we have an odd number we convert it into an even number by multiplying by 3 and adding 1 so that eventually it can be halved again. The smallest even number is 2 so we will reach 1 for sure. Q.3. Name some objects around you that are: a. a few thousand in number b. more than ten thousand in number Ans. A few thousands: car numbers, 4 digit pin More than ten thousands: Salary, Mobile numbers. Q.3. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹ 100. Do you agree with him? Why or why not? Ans. Yes, it is possible with less quantity of serving and with purchase of 1-1-1 fruit of each type. However, it is not possible with costly fruits and more quantity of serving. Q.4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). Ans. 2500 kilometer Even if the greatest 2 digit numbers is subtracted from the smallest 5 digit number, 4 digit number will be obtained. Q. 5. Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. Do you agree with her? Why or why not? Ans. No, I do not agree with her. There are 6 school hours in a day and around 200 working days in a year. , × = 10.8 years (Nursery, KG, 1,2,3,4,5,6) 8 years She is in school for 8 years, 13,000 hours is way too high. Q.7. Make some estimation questions and challenge your classmates! Ans. Some such are • How many students are there in your school? • How many hours does a person sleep in his lifetime on an average? (More such questions can be made) [11] Section 3.12 Page No. 72 Figure it Out Q.1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap. 16,200 39,344 29,765 Ans. If I exchange the digits 1 and 6 in the number 62,871 then there will be 4 Super cells. Q.2. How many rounds does your year of birth take to reach the Kaprekar constant? Ans. Suppose the birth year is 1980, then, - 9810 -1089 8721 Q.3. We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000? Ans. With repeating digit With non repeating digit Largest number → 73,999 73,951 It takes 6 rounds. (Try now for your year of birth.) 23,609 62,871 45,306 19,381 50,319 38,408 16,200 39,344 29,765 23,609 12,876 45,306 19,381 50,319 38,408 8721 -1278 7443 7443 -3447 3996 9963 -3699 6264 6642 -2466 4176 7641 -1467 6174 Q.6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670. Ans. 18000 + 300 + 370 = 18670. Try for more. Smallest number → 35,111 35,179 Closest to 50000 51,111 51,379 With repeating digit With non repeating digit [12] Q.7. Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number. Ans. Number Chosen: 250 50 50 50 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 Q.8. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence? Ans. when we divide 28 = 2x2x2x2x2x2x2x2 by 2 it become 27 and every time you divide by 2 the same will continue happening, until you are left with 2 which when divided by 2 will leave 1. Q.9. Check if the Collatz Conjecture holds for the starting number 100. Ans. 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Q.10 Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now? Ans. Winning strategy is to be the first player. 25 25 25 25 Or [13]" class_6,4,Data Handling and Presentation,ncert_books/class_6/Ganita_Prakash/fegp104.pdf,"4 If you ask your classmates about their favourite colours, you will get a list of colours. This list is an example of data. Similarly, if you measure the weight of each student in your class, you would get a collection of measures of weight—again data. Any collection of facts, numbers, measures, observations or other descriptions of things that convey information about those things is called data. We live in an age of information. We constantly see large amounts of data presented to us in new and interesting ways. In this chapter, we will explore some of the ways that data is presented, and how we can use some of those ways to correctly display, interpret and make inferences from such data! Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 74 13/08/2024 15:33:11 4.1 Collecting and Organising Data Navya and Naresh are discussing their favourite games. most popular game in I think cricket is the favourite game! I play cricket sometimes but hockey is the game I like the most. Cricket is my our class. Reprint 2025-26 I am not sure. How can we find the most popular game in our class? Naresh and Navya decided to go to each student in the class and ask what their favourite game is. Then they prepared a list. Navya is showing the list: She says (happily), “I have collected the data. I can figure out the most popular game now!”. A few other children are looking at the list and wondering, “We can’t yet see the most popular game. How can we get it from this list?”. Mehnoor – Kabaddi Pushkal – Satoliya (Pittu) Anaya – Kabaddi Jubimon – Hockey Densy – Badminton Jivisha – Satoliya (Pittu) Simran – Kabaddi Jivika – Satoliya (Pittu) Rajesh – Football Nand – Satoliya (Pittu) Leela – Hockey Thara – Football Ankita – Kabaddi Afshan – Hockey Soumya – Cricket Imon – Hockey Keerat – Cricket Navjot – Hockey Yuvraj – Cricket Gurpreet – Hockey Hemal – Satoliya (Pittu) Rehana – Hockey Arsh – Kabaddi Debabrata – Football Aarna – Badminton Bhavya – Cricket Ananya – Hockey Kompal – Football Sarah – Kabaddi Hardik – Cricket Tahira – Cricket To figure out the most popular game in their class, what should Navya and Naresh do? Can you help them? Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 75 13/08/2024 15:33:11  Figure it Out 1. What would you do to find the most popular game among Naresh’s and Navya’s classmates? 2. What is the most popular game in their class? 3. Try to find out the most popular game among your classmates. 4. Pari wants to respond to the questions given below. Put a tick () for the questions where she needs to carry out data collection and Reprint 2025-26 75 75 Ganita Prakash | Grade 6 Shri Nilesh is a teacher. He decided to bring sweets to the class to celebrate the new year. The sweets shop nearby has jalebi, gulab jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of the children. He wrote the names of the sweets on the board and asked each child to tell him their preference. He put a tally mark ‘|’ for each student and when the count reached 5, he put a line through the previous four and marked it as ||||. Gulab jamun |||| |||| 9 put a cross () for the questions where she doesn’t need to collect data. Discuss your answers in the classroom. a. What is the most popular TV show among her classmates? b. When did India get independence? c. How much water is getting wasted in her locality? d. What is the capital of India? Sweets Tally Marks No. of Students Gujiya |||| |||| ||| ____________ Jalebi |||| | 6 Barfi ||| ____________ Chapter 4_Data Handling and Presentation.indd 76 13/08/2024 15:33:11 76  Figure it Out 1. Complete the table to help Shri Nilesh to purchase the correct numbers of sweets: Rasgulla |||| || ____________ a. How many students chose jalebi? b. Barfi was chosen by students? c. How many students chose gujiya? d. Rasgulla was chosen by students? e. How many students chose gulab jamun? Reprint 2025-26 Shri Nilesh requested one of the staff members to bring the sweets as given in the table. The above table helped him to purchase the correct numbers of sweets. Sushri Sandhya asked her students about the sizes of the shoes they wear. She noted the data on the board. To organise the data, we can write the name of each sweet in one column and using tally signs, note the number of students who prefer that sweet. The numbers 6, 9, … are the frequencies of the sweet preferences for jalebi, gulab jamun … respectively.  Figure it Out 2. Is the above table sufficient to distribute each type of sweet to the correct student? Explain. If it is not sufficient, what is the alternative? She then arranged the shoe sizes of the students in ascending order — 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7 4 5 3 4 3 4 5 5 4 5 5 4 5 6 4 3 5 6 4 6 4 5 7 5 6 4 5 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 77 13/08/2024 15:33:11 1. Help her to figure out the following: 2. How did arranging the data in ascending order help to answer these questions? 3. Are there other ways to arrange the data? a. The largest shoe size in the class is _________. b. The smallest shoe size in the class is _________. c. There are _________ students who wear shoe size 5. d. There are _________ students who wear shoe sizes larger than 4. Reprint 2025-26 Math Talk 77 Ganita Prakash | Grade 6 4. Write the names of a few trees you see around you. When you observe a tree on the way from your home to school (or while walking from one place to another place), record the data and fill in the following table: 5. Take a blank piece of paper and paste any small news item from a newspaper. Each student may use a different article. Now, prepare a table on the piece of paper as given below. Count the number of each of the letters ‘c’, ‘e’, ‘i’, ‘r’, and ‘x’ in the words of the news article, and fill in the table. a. Which tree was found in the greatest number? b. Which tree was found in the smallest number? c. Were there any two trees found in the same numbers? Letter c e i r x Any other letter of your choice Peepal Neem Tree No. of Trees …. … Chapter 4_Data Handling and Presentation.indd 78 13/08/2024 15:33:12 78 a. The letter found the most number of times is ________. b. The letter found the least number of times is ________. c. List the five letters ‘c’, ‘e’, ‘i’, ‘r’, ‘x’ in ascending order of frequency. Now, compare the order of your list with that of your classmates. Is your order the same or nearly the same as theirs? (Almost everyone is likely to get the order ‘x, c, r, i, e’.) Why do you think this is the case? Number of times found in the news item Reprint 2025-26 4.2 Pictographs Pictographs are one visual and suggestive way to represent data without writing any numbers. Look at this picture — you may be familiar with it from previous classes. Provide more opportunities to collect and organise data. Ask students to guess what is the most popular colour, game, toy, school subject, etc., amongst the students in their classroom and then collect the data for it. It can be a fun activity in which they also learn about their classmates. Discuss how they can organise the data in different ways, each way having its own advantages and limitations. For all these tasks and the tasks under ‘Figure it Out’, discuss the tasks with the children and let them understand the tasks, and then let them plan and present their research processes and conclusions in the class. d. Write the process you followed to complete this task. e. Discuss with your friends the processes they followed. f. If you do this task with another news item, what process would you follow? Teacher’s Note Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 79 13/08/2024 15:33:12 This picture helps you understand at a glance the different modes of travel used by students. Based on this picture, answer the following questions: Modes of Travelling Number of Students = 1 Student Private car School bus Public bus Walking Cycle Reprint 2025-26 79 Ganita Prakash | Grade 6 • Which mode of travel is used by the most number of students? • Which mode of travel is used by the least number of students? A pictograph represents data through pictures of objects. It helps answer questions about data with just a quick glance. In the above pictograph, one unit or symbol ( ) is used to represent one student. There are also other pictographs where one unit or symbol stands for many people or objects.  Example: Nand Kishor collected responses from the children of his middle school in Berasia regarding how often they slept at least 9 hours during the night. He prepared a pictograph from the data: Answer the following questions using the pictograph: 1. What is the number of children who always slept at least 9 hours at night? Sometimes Response Number of Children ( = 10 Children) Always Never Chapter 4_Data Handling and Presentation.indd 80 13/08/2024 15:33:12 80 Solutions 2. How many children sometimes slept at least 9 hours at night? 3. How many children always slept less than 9 hours each night? Explain how you got your answer. 1. In the table, there are 5 pictures for ‘Always’. Each picture represents 10 children. Therefore, 5 pictures indicate 5 × 10 = 50 children. 2. There are 2 complete pictures (2 × 10 = 20) and a half picture (half of 10 = 5). Therefore, the number of children who sleep at least 9 hours only sometimes is 20 + 5 = 25. Reprint 2025-26 Drawing a Pictograph One day, Lakhanpal collected data on how many students were absent in each class: He created a pictograph to present this data and decided to show 1 student as in the pictograph— 3. There are 4 complete pictures for ‘Never’. Hence, 4 × 10 = 40 children never sleep at least 9 hours in a night, i.e., they always sleep less than 9 hours. No. of students 3 5 4 2 0 1 5 7 Class I II III IV V VI VII VIII Class es VIII VII III VI IV II V Data Handling and Presentation = 1 student Chapter 4_Data Handling and Presentation.indd 81 12-12-2024 10:48:23 Meanwhile, his friends Jarina and Sangita collected data on how many students were present in each class: No. of students 30 35 20 25 30 25 30 20 Class I II III IV V VI VII VIII I No. of students absent Reprint 2025-26 81 Ganita Prakash | Grade 6  If they want to show their data through a pictograph, where they also use one symbol for each student, as Lakhanpal did, what are the challenges they might face? Jarina made a plan to address this — since there were many students, she decided to use to represent 5 students. She figured that would save time and space too. Sangita decided to use one to represent 10 students instead. Since she used one to show 10 students, she had a problem in showing 25 students and 35 students in the pictograph. Then, she realised she could use to show 5 students. Class es VIII VII VI IV III V II I No. of students present = 5 students Chapter 4_Data Handling and Presentation.indd 82 13/08/2024 15:33:12 82 Class es VIII VII III VI IV II V I No. of students present Reprint 2025-26 = 10 students What could be the problems faced in preparing such a pictograph, if the total number of students present in a class is 33 or 27? • Pictographs are a nice visual and suggestive way to represent data. They represent data through pictures of objects. • Pictographs can help answer questions and make inferences about data with just a quick glance (in the examples above— about favourite games, favourite colours, most common modes of conveyance, number of students absent, etc.). • By reading a pictograph, we can quickly understand the frequencies of the different categories (for example, cricket, hockey, etc.) and the comparisons of these frequencies. • In a pictograph, the categories can be arranged horizontlly or vertically. For each category, simple pictures and symbols are then drawn in the designated columns or rows according to the frequency of that category. • A scale or key (for example, : 1 student or : 5 students) is added to show what each symbol or picture represents. Each symbol or picture can represent one unit or multiple units. • It can be more challenging to prepare a pictograph when the amount of data is large or when the frequencies are not exact multiples of the scale or key. Data Handling and Presentation Math Talk Chapter 4_Data Handling and Presentation.indd 83 14-08-2024 14:46:59  Figure it Out 1. The following pictograph shows the number of books borrowed by students, in a week, from the library of Middle School, Ginnori: Reprint 2025-26 83 Ganita Prakash | Grade 6 Wednesday Thursday Saturday Tuesday Monday Friday Day Number of Books Borrowed ( =1 Book ) 2. Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below — a. On which day were the minimum number of books borrowed? b. What was the total number of books borrowed during the week? c. On which day were the maximum number of books borrowed? What may be the possible reason? Shopkeeper Number of Kites Sold Chaman 250 Chapter 4_Data Handling and Presentation.indd 84 13/08/2024 15:33:13 84 Poonam Ben 700 Rukhsana 100 Jetha Lal 250 Jasmeet 450 Rani 300 Reprint 2025-26 4.3 Bar Graphs Have you seen graphs like this on TV or in a newspaper? Like pictographs, such bar graphs can help us to quickly understand and interpret information, such as the highest value, the comparison of values of different categories, etc. However, when the amount of data is large, presenting it by a pictograph is not only time consuming but at times difficult too. Let us see how data can be presented instead using a bar graph. Let’s take the data collected by Lakhanpal earlier, regarding the number of students absent on one day in each class: Prepare a pictograph using the symbol to represent 100 kites. Answer the following questions: a. How many symbols represent the kites that Rani purchased? b. Who purchased the maximum number of kites? c. Who purchased more kites, Jasmeet or Chaman? d. Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why? Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 85 12-12-2024 10:48:24 No. of students 3 5 4 2 0 1 5 7 Class I II III IV V VI VII VIII Source: https://www.statista.com/chart/17122/ number-of-threatened-species-red-list/ Reprint 2025-26 85 Ganita Prakash | Grade 6 If the students have not noticed, please point out the equally spaced horizontal lines. Explain that this means that each pair of consecutive numbers on the left has the same gap. He presented the same data using a bar graph: Number o f students 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 No. of students absent in each class 1 unit length = 1 student Teacher’s Note Class Chapter 4_Data Handling and Presentation.indd 86 13/08/2024 15:33:14 86 1. In Class 2, ___________ students were absent that day. 2. In which class were the maximum number of students absent? ___________ 3. Which class had full attendance that day? ___________ When making bar graphs, bars of uniform width can be drawn horizontally or vertically with equal spacing between them; then the Answer the following questions using the bar graph: Reprint 2025-26 length or height of each bar represents the given number. As we saw in pictographs, we can use a scale or key when the frequencies are larger. Let us look at an example of vehicular traffic at a busy road crossing in Delhi, which was studied by the traffic police on a particular day. The number of vehicles passing through the crossing each hour from 6 a.m. to 12:00 noon is shown in the bar graph. One unit of length stands for 100 vehicles. We can see that the maximum traffic at the crossing is shown by the longest bar, i.e., for the time interval 7–8 a.m. The bar graph shows that 1200 vehicles passed through the crossing at that time. The second longest bar is for 8–9 a.m. During that time, 1000 vehicles passed through the crossing. Similarly, the minimum traffic is shown by the smallest bar, i.e., the bar for the time interval 6–7 a.m. During that time, only about 150 vehicles passed through the crossing. The second smallest bar is that for the time interval 11 a.m.–12 noon, when about 600 vehicles passed through the crossing. The total number of cars passing through the crossing during the two-hour interval 8.00–10.00 a.m. as shown by the bar graph is about 1000 + 800 = 1800 vehicles. Tim e in terv als 11–12 10–11 9–10 7–8 8–9 6–7 100 200 300 400 600 700 800 900 110 0 500 100 0 120 0 Number of vehicles Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 87 13/08/2024 15:33:14 Reprint 2025-26 87 Ganita Prakash | Grade 6 Example:  Figure it Out 1. How many total cars passed through the crossing between 6 a.m. and noon? 2. Why do you think so little traffic occurred during the hour of 6–7 a.m., as compared to the other hours from 7 a.m.–noon? 3. Why do you think the traffic was the heaviest between 7–8 a.m.? 4. Why do you think the traffic was lesser and lesser each hour after 8 a.m. all the way until noon? Pop ulat ion of Ind ia in cror es 110 100 90 80 70 60 50 40 30 20 36 44 54 68 84 102 Chapter 4_Data Handling and Presentation.indd 88 13/08/2024 15:33:14 88 This bar graph shows the population of India in each decade over a period of 50 years. The numbers are expressed in crores. If you were to take 1 unit length to represent one person, drawing the bars will 10 1951 1961 1971 1981 1991 2001 Population of India in crores Reprint 2025-26 Years be difficult! Therefore, we choose the scale so that 1 unit represents 10 crores. The bar graph for this choice is shown in the figure. So a bar of length 5 units represents 50 crores and of 8 units represents 80 crores. • On the basis of this bar graph, what may be a few questions you may ask your friends? • How much did the population of India increase over 50 years? How much did the population increase in each decade? 4.4 Drawing a Bar Graph In a previous example, Shri Nilesh prepared a frequency table representing the sweet preferences of the students in his class. Let’s try to prepare a bar graph to present his data — 1. First, we draw a horizontal line and a vertical line. On the horizontal line, we will write the name of each of the sweets, equally spaced, from which the bars will rise in accordance with their frequencies; and on the vertical line we will write the frequencies representing the number of students. Gulab jamun 9 Rasgulla 7 Sweet No. of Students Gujiya 13 Jalebi 6 Barfi 3 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 89 13/08/2024 15:33:14 2. We must choose a scale. That means we must decide how many students will be represented by a unit length of a bar so that it fits nicely on our paper. Here, we will take 1 unit length to represent 1 student. 3. For jalebi, we therefore need to draw a bar having a height of 6 units (which is the frequency of the sweet jalebi), and similarly for the other sweets we have to draw bars as high as their frequencies. Reprint 2025-26 89 Ganita Prakash | Grade 6 4. We, therefore, get a bar graph as shown below — When the frequencies are larger and we cannot use the scale of 1 unit length = 1 number (frequency), we need to choose a different scale like we did in the case of pictographs. Example: The number of runs scored by Smriti in each of the 8 matches are given in the table below: Match Match 1 Match 2 Match 3 Match 4 Match 5 Match 6 Match 7 Match 8 Nu mb e r o f st u d e n ts 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Jalebi Gulab jamun Gujia Barfi Rasgulla Sweet preferences of students Sweets Chapter 4_Data Handling and Presentation.indd 90 13/08/2024 15:33:14 90 In this example, the minimum score is 0 and the maximum score is 100. Using a scale of 1 unit length = l run would mean that we have to go all the way from 0 to 100 runs in steps of l. This would be unnecessarily tedious. Instead, let us use a scale where 1 unit length = 10 runs. We mark this scale on the vertical line and draw the bars according to the scores in each match. We get the following bar graph representing the above data. Runs 80 50 10 100 90 0 90 50 Reprint 2025-26 Example: The following table shows the monthly expenditure of Imran’s family on various items: Runs 100 0 10 20 30 40 50 60 70 80 90 Match 1 Match 2 Match 3 Match 4 Match 5 Match 6 Match 7 Match 8 House rent Food Education Items Expenditure (in `) Runs scored by Smriti Matches 3000 3400 800 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 91 12-12-2024 10:48:24 To represent this data in the form of a bar graph, here are the steps— • Draw two perpendicular lines, one horizontal and one vertical. • Along the horizontal line, mark the ‘items’ with equal spacing between them and mark the corresponding expenditures along the vertical line. Electricity Transport Miscellaneous Reprint 2025-26 1200 400 600 91 Ganita Prakash | Grade 6 • Take bars of the same width, keeping a uniform gap between them. • Choose a suitable scale along the vertical line. Let, 1 unit length = ` 200, and then mark and write the corresponding values (` 200, ` 400, etc.) representing each unit length. Finally, calculate the heights of the bars for various items as shown below — Here is the bar graph that we obtain based on the above steps: Miscellaneous House rent Education Electricity Transport Food Expen ditur e ( in `) 3600 3400 3200 3000 2800 2600 2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 1200 ÷ 200 3000 ÷ 200 3400 ÷ 200 800 ÷ 200 400 ÷ 200 600 ÷ 200 15 units 17 units 4 units 2 units 3 units 6 units Chapter 4_Data Handling and Presentation.indd 92 13/08/2024 15:33:14 92 House rent Reprint 2025-26 Miscell aneous Food Item Educati on Electri city Transp ort Figure it Out 1. On which item does Imran’s family spend the most and the second most? 2. Is the cost of electricity about one-half the cost of education? 3. Is the cost of education less than one-fourth the cost of food? 1. Samantha visited a tea garden, and collected data of the insects and critters she saw there. Here is the data she collected: 2. Pooja collected data on the number of tickets sold at the Bhopal railway station for a few different cities of Madhya Pradesh over a two-hour period. Use the bar graph to answer the following questions: Help her prepare a bar graph representing this data. Mites Caterpillars Beetles Butterflies Grasshoppers 6 10 5 3 2 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 93 13/08/2024 15:33:16 She used this data and prepared a bar graph on the board to discuss the data with her students, but someone erased a portion of the graph. Number of tickets 24 20 16 28 16 City Vidisha Jabalpur Seoni Indore Sagar Reprint 2025-26 93 Ganita Prakash | Grade 6 3. Chinu listed the various means of transport that passed across the road in front of his house from 9 a.m. to 10 a.m.: No. o f Tic kets a. Write the number of tickets sold for Vidisha above the bar. b. Write the number of tickets sold for Jabalpur above the bar. c. The bar for Vidisha is 6 unit lengths and the bar for Jabalpur d. Draw the correct bar for Sagar. e. Add the scale of the bar graph by placing the correct numbers f. Are the bars for Seoni and Indore correct in this graph? If not, draw the correct bar(s). Vidisha Jabalpur Seoni Indore Sagar is 5 unit lengths. What is the scale for this graph? on the vertical axis. City Chapter 4_Data Handling and Presentation.indd 94 13/08/2024 15:33:16 94 auto rickshaw auto rickshaw bike bicycle bus bike scooter bicycle bike bullock cart auto rickshaw scooter bicycle scooter bicycle scooter bike bus bicycle scooter bus scooter auto rickshaw bike bike car bike bus bike bike bike auto rickshaw bicycle bullock cart bicycle auto rickshaw car scooter car auto rickshaw bicycle bike car auto rickshaw bike scooter bike car car scooter Reprint 2025-26 4. Roll a die 30 times and record the number you obtain each time. Prepare a frequency distribution table using tally marks. Find the number that appeared: 5. Faiz prepared a frequency distribution table of data on the number of wickets taken by Jaspreet Bumrah in his last 30 matches: a. Prepare a frequency distribution table for the data. b. Which means of transport was used the most? c. If you were there to collect this data, how could you do it? Write the steps or process. a. The minimum number of times. b. The maximum number of times. c. Find numbers that appeared an equal number of times. Wickets Taken Number of Matches 0 2 1 4 2 6 3 8 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 95 13/08/2024 15:33:16 a. What information is this table giving? b. What may be the title of this table? c. What caught your attention in this table? d. In how many matches has Bumrah taken 4 wickets? 4 3 5 5 6 1 7 1 Reprint 2025-26 95 Ganita Prakash | Grade 6 6. The following pictograph shows the number of tractors in five different villages. Village A Village B Village C Village D Villages Number of Tractors ( = 1 Tractor ) e. Mayank says, “If we want to know the total number of wickets he has taken in his last 30 matches, we have to add the numbers 0, 1, 2, 3 …, up to 7.” Can Mayank get the total number of wickets taken in this way? Why? f. How would you correctly figure out the total number of wickets taken by Bumrah in his last 30 matches, using this table? Chapter 4_Data Handling and Presentation.indd 96 13/08/2024 15:33:18 96 Observe the pictograph and answer the following questions— Village E a. Which village has the smallest number of tractors? b. Which village has the most tractors? c. How many more tractors does Village C have than Village B? d. Komal says, “Village D has half the number of tractors as Village E.” Is she right? Reprint 2025-26 7. The number of girl students in each class of a school is depicted by the pictograph: Classes Number of Girl Students ( = 4 Girls ) 1 2 3 4 5 6 7 Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 97 13/08/2024 15:33:27 Observe this pictograph and answer the following questions: a. Which class has the least number of girl students? b. What is the difference between the number of girls in Classs 5 and 6? c. If two more girls were admitted in Class 2, how would the graph change? d. How many girls are there in Class 7? 8 Reprint 2025-26 97 Ganita Prakash | Grade 6 Village A : 18, Village B : 36, Village C : 12, Village D : 48, Village E : 18, Village F : 24 Prepare a pictograph and answer the following questions: 8. Mudhol Hounds (a type of breed of Indian dogs) are largely found in North Karnataka’s Bagalkote and Vijaypura districts. The government took an initiative to protect this breed by providing support to those who adopted these dogs. Due to this initiative, the number of these dogs increased. The number of Mudhol dogs in six villages of Karnataka are as follows— 9. A survey of 120 school students was conducted to find out which activity they preferred to do in their free time: a. What will be a useful scale or key to draw this pictograph? b. How many symbols will you use to represent the dogs in Village B? c. Kamini said that the number of these dogs in Village B and Village D together will be more than the number of these dogs in the other 4 villages. Is she right? Give reasons for your response. Playing Preferred Activity Number of Students 45 Chapter 4_Data Handling and Presentation.indd 98 12-12-2024 10:48:25 98 Draw a bar graph to illustrate the above data taking the scale of 1 unit length = 5 students. Which activity is preferred by most students other than playing? Reading story books Watching TV Listening to music Painting Reprint 2025-26 30 20 10 15 10. Students and teachers of a primary school decided to plant tree saplings in the school campus and in the surrounding village during the first week of July. Details of the saplings they planted are as follows— a. The total number of saplings planted on Wednesday and Thursday is ___________. b. The total number of saplings planted during the whole week is ___________. c. The greatest number of saplings were planted on ___________ and the least number of saplings were planted on ___________. Why do you think that is the case? Why were more saplings planted on certain days of the week and less on others? Can you think of possible explanations or reasons? How could you try and figure out whether your explanations are correct? Day Thursday Friday Saturday Sunday Nu mb er of s apl ing s p lan ted 70 60 50 40 30 20 10 0 Monday Tuesday Wednesday Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 99 12-12-2024 10:48:25 11. The number of tigers in India went down drastically between 1900 and 1970. Project Tiger was launched in 1973 to track and protect the tigers in India. Starting in 2006, the exact number of tigers in India was tracked. Shagufta and Divya looked up information about the number of tigers in India between 2006 and 2022 in fouryear intervals. They prepared a frequency table for this data and a bar graph to present this data, but there are a few mistakes in the graph. Can you find those mistakes and fix them? Reprint 2025-26 99 Ganita Prakash | Grade 6 • Like pictographs, bar graphs give a nice visual way to represent different categories. • Each category is represented by a bar where the length or height depicts the corresponding frequency (for example, cost) or quantity (for example, runs). • The bars have uniform spaces between them to indicate that they are free standing and represent equal categories. • The bars help in interpreting data much faster than a frequency table. By reading a bar graph, we can compare frequencies of different categories at a glance. • We must decide the scale (for example, 1 unit length = 1 student or 1 unit length = ` 200) for a bar graph on the basis of the data including the minimum and maximum frequencies, so that the resulting bar graph fits nicely and looks visually appealing on the paper or poster we are preparing. The markings of the unit lengths as per the scale must start from zero. data. They represent data through equally-spaced bars, each of equal width, where the lengths or heights give frequencies of the Year Number of Tigers (approx.) 2006 1400 2010 1700 2014 2200 2018 3000 2022 3700 Yea r 2022 2018 2014 2010 2006 0 1000 2000 3000 4000 Number of Tigers in India Number of Tigers Chapter 4_Data Handling and Presentation.indd 100 13/08/2024 15:33:28 100 Teacher’s Note The main focus of this chapter is to learn how to handle data to find answers to specific questions or inquiries, to test hypotheses or to take specific decisions. This should be kept in mind when providing practice opportunities to collect, organise and analyse data. Reprint 2025-26 4.5 Artistic and Aesthetic Considerations In addition to the steps described in previous sections, there are also some other more artistic and aesthetic aspects one can consider when preparing visual presentations of data to make them more interesting and effective. First, when making a visual presentation of data such as a pictograph or bar graph, it is important to make it fit in the intended space; this can be controlled, for example, by choosing the scale appropriately, as we have seen earlier. It is also desirable to make the data presentation visually appealing and easy-to-understand, so that the intended audience appreciates the information being conveyed. Let us consider an example. Here is a table naming the tallest mountain on each continent, along with the height of each mountain in meters. Continent Asia South America North America Africa Europe Antarctica Australia Mountain Everest Aconcagua Denali Kilimanjaro Elbrus Vinson Massif Koscuiszko Tallest Height 8848m 6962m 6194m 5895m 5642m 4892m 2228m Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 101 13/08/2024 15:33:28 How much taller is Mount Everest than Mount Koscuiszko? Are Mount Denali and Mount Kilimanjaro very different in height? This is not so easy to quickly discern from a large table of numbers. As we have seen earlier, we can convert the table of numbers into a bar graph, as shown on the right. Here, each value is drawn as a horizontal box. These are longer or shorter depending on the number they represent. This makes it easier to compare the heights of all these mountains at a glance. Reprint 2025-26 101 Ganita Prakash | Grade 6 South America — Aconcagua However, since the boxes represent heights, it is better and more visually appealing to rotate the picture, so that the boxes grow upward, vertically from the ground like mountains. A bar graph with vertical bars is also called a column graph. Columns are the pillars you find in a building that hold up the roof. Below is a column graph for our table of tallest mountains. From this column graph, it becomes easier to compare and visualise the heights of the mountains. Antarctica — Vinson Massif North America — Denali Australia — Koscuiszko Africa — Kilimanjaro Europe — Elbrus Asia — Everest 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Chapter 4_Data Handling and Presentation.indd 102 13/08/2024 15:33:28 102 10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0 Asia Everest South America Aconcagua North America Denali Africa Reprint 2025-26 Kilimanjaro Europe Elbrus Antarctica Vinson Massif Australia Koscuiszko In general, it is more intuitive, suggestive and visually appealing to represent heights, that are measured upwards from the ground, using bar graphs that have vertical bars or columns. Similarly, lengths that are parallel to the ground (for example, distances between location on Earth) are usually best represented using bar graphs with horizontal arcs. Infographics When data visualisations such as bar graphs are further beautified with more extensive artistic and visual imagery, they are called information graphics or infographics for short. The aim of infographics is to make use of attention-attracting and engaging visuals to communicate information even more clearly and quickly, in a visually pleasing way. As an example of how infographics can be used to communicate data even more suggestively, let us go back to the table above listing the tallest mountain on each continent. We drew a bar graph with vertical bars (columns) rather than horizontal bars, to be more indicative of mountains. But instead of rectangles, we could use triangles, which look a bit more like mountains. And, we can add a splash of colour as well. Here is the result.  Figure it Out 1. If you wanted to visually represent the data of the heights of the tallest persons in each class in your school, would you use a graph with vertical bars or horizontal bars? Why? 2. If you were making a table of the longest rivers on each continent and their lengths, would you prefer to use a bar graph with vertical bars or with horizontal bars? Why? Try finding out this information, and then make the corresponding table and bar graph! Which continents have the longest rivers? Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 103 13/08/2024 15:33:28 Reprint 2025-26 103 Ganita Prakash | Grade 6 While this infographic might look more appealing and suggestive at first glance, it does have some issues. The goal of our bar graph earlier was to represent the heights of various mountains — using bars of the appropriate heights but the same widths. The purpose of using the same widths was to make it clear that we are only comparing heights. However, in this infographic, the taller triangles are also wider! Are taller mountains always wider? The infographic is implying additional information that may be misleading and may or may not be correct. Sometimes going for more appealing pictures can also accidentally mislead. Taking this idea further, and to make the picture even more visually stimulating and suggestive, we can further change the shapes of the mountains to make them look even more like mountains, and add other details, while attempting to preserve the heights. For example, we can create an imaginary mountain range that contains all these mountains. Is the infographic below better than the column graph with rectangular columns of equal width? The mountains look more realistic, but is the picture accurate? For example, Everest appears to be twice as tall as Elbrus. 8000m 7000m 6000m 5000m 4000m 3000m 2000m 1000m Everest 8848m Asia Aconcagua S America 6962m North America Denali 6194m Kilimanjaro 5895m Africa Europe Elbrus 5642m Vinson Massif Antarctica 4892m Koscuiszko Australia 2228m Chapter 4_Data Handling and Presentation.indd 104 13/08/2024 15:33:28 104 Reprint 2025-26 What is 5642 × 2? While preparing visually-appealing presentations of data, we also need to be careful that the pictures we draw do not mislead us about the facts. In general, it is important to be careful when making or reading infographics, so that we do not mislead our intended audiences and we, ourselves, are not misled. The seven highest peaks on the seven continents. Data Handling and Presentation Chapter 4_Data Handling and Presentation.indd 105 13/08/2024 15:33:39 Facts, numbers, measures, observations and other descriptions of things that convey information about those things is called data. Data can be organised in a tabular form using tally marks for easy analysis and interpretation. Frequencies are the counts of the occurrences of values, measures or observations. Summary Reprint 2025-26 105 Ganita Prakash | Grade 6 Pictographs represent data in the form of pictures, or objects or parts of objects. Each picture represents a frequency which can be 1 or more than 1 — this is called the scale and it must be specified. Bar graphs have bars of uniform width; the length or height that indicates the total frequency of occurrence. The scale that is used to convert length or height to frequency again, must be specified. Choosing the appropriate scale for a pictograph or bar graph is important to accurately and effectively convey the desired information or data and to also make it visually appealing. Other aspects of a graph also contribute to its effectiveness and visual appeal such as how colours are used, what accompanying pictures are drawn, and whether the bars are horizontal or vertical. These aspects correspond to the artistic and aesthetic side of data handling and presentation. However, making visual representations of data ‘fancy’ can also sometimes be misleading. By reading pictographs and bar graphs accurately, we can quickly understand and make inferences about the data presented. Chapter 4_Data Handling and Presentation.indd 106 13/08/2024 15:33:39 106 Reprint 2025-26 Section 4.1 Page No. 75 Figure it Out Q.1. What would you do to find the most popular game among Naresh’s and Navya’s classmates? Ans. One of the ways could be to arrange and organize the given data in a table. Think of some other ways. Q.2. What is the most popular game in their class? Ans. Hockey (frequency – 8). Q.4. Pari wants to respond to the questions given below. Put a tick () for the questions where she needs to carry out data collection and put a cross (⨉) for the questions where she doesn’t need to collect data. Discuss your answers in the classroom. Ans. a. a. What is the most popular TV show among her classmates? b. When did India get independence? c. How much water is getting wasted in her locality? d. What is the capital of India? b. c. d. ⨉ ⨉ CHAPTER 4 — SOLUTIONS Data Handling and Presentation Page No. 76 Figure it Out Q.1. Complete the table to help Shri Nilesh to purchase the correct numbers of sweets: • How many students chose jalebi? • Barfi was chosen by students? • How many students chose gujiya? • Rasgulla was chosen by students? Ans. • 6 • 3 • 13 • 7 • 9 Q.2. Is the above table sufficient to distribute each type of sweet to the correct student? Explain. If it is not sufficient, what is the alternative? Ans. No. Section 4.1 Page No. 77 Figure it Out Q.1. Help her to figure out the following – Shri Nilesh requested one of the staff members to bring the sweets as given in the table. The above table helped him to purchase the correct numbers of sweets. • How many students chose gulab jamun? • The largest shoe size in the class is _________ It shows only the number of students who chose a particular sweet, and not about the sweet chosen by a particular student. One of the alternatives could be, students should be categorized according to their choice of sweet. Ans. • 7 • 3 • 10 • 15 Q.2. How did arranging the data in ascending order help to answer these questions? Ans. Ordered data is helpful because in this form, frequency of any data is easy to count and the given information can be used easily. • The smallest shoe size in the class is _________ • There are _________ students who wear shoe size 5. • There are _________ students who wear shoe sizes larger than 4. Q.3. Are there other ways to arrange the data? Ans. Yes, the data can be arranged in a frequency table. Section 4.2 Page No. 83 Q. What could be the problems faced in preparing such a pictograph, if the total number of students present in a class is 33 or 27? Ans. represents 10 students and represent 5 students. But to represent 3 or 7 students accurately by dividing these symbols is not possible. Figure it Out Q.1. The following pictograph shows the number of books borrowed by students, in a week, from the library of Middle School, Ginnori — Ans. a. Thursday b. 24 c. Saturday. One of the reasons could be that Sunday being a holiday students can read the borrowed books on Sunday. Q.2. Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below — a. On which day were the minimum number of books borrowed? b. What was the total number of books borrowed during the week? c. On which day were the maximum number of books borrowed? What may be the possible reason? Prepare a pictograph using the symbol to represent 100 kites. Shopkeeper No. of Kites sold ( =100 Kits) Chaman Rani Answer the following questions: d. Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why? Ans. a. 3 symbols b. Poonam Ben c. Jasmeet d. Yes. Number. of kites purchased by Poonam Ben = 700 = 2 x 300 +100 Section 4.3 Page no. 86 Rukhsana Jasmeet Jetha Lal Poonam Ben a. How many symbols represent the kites that Rani purchased? b. Who purchased the maximum number of kites? c. Who purchased more kites, Jasmeet or Chaman? Q1. In Class 2, ___________ students were absent that day. Ans. 5 Q2. In which class were the maximum number of students absent? ___________ Ans. Class 8 Q3. Which class had full attendance that day? ___________ Ans. Class 5 Page no. 88 Figure it Out Q.1. How many total cars passed through the crossing between 6 am and noon? Ans. 4450 cars Section 4.4 Page No. 93 Q. On which item does Imran’s family spend the most and the second most? Ans. Imran’s family spends the most on food and the second most on house rent. Q.2. Is the cost of electricity about one-half the cost of education? Ans. Yes, Q.3. Is the cost of education less than one-fourth the cost of food? Ans. Yes. Cost of education = Rs. 800 and cost of food = Rs 3400 =4 x Rs 850 So, cost of education = less than one-fourth of cost of food. Figure it Out Q.1. Samantha visited a tea garden and collected data of the insects and critters she saw there. Here is the data she collected — Help her prepare a bar graph representing this data. Ans. Q.2. Pooja collected data on the number of tickets sold at the Bhopal railway station for a few different cities of Madhya Pradesh over a 2-hour period. Ans. a. 24 b. 20 c. 1unit length = 4 tickets d. She used this data and prepared a bar graph on the board to discuss the data with her students, but someone erased a portion of the graph. a. Write the number of tickets sold for Vidisha above the bar. b. Write the number of tickets sold for Jabalpur above the bar. c. The bar for Vidisha is 6 unit lengths and the bar for Jabalpur is 5 unit lengths. What is the scale for this graph? d. Draw the correct bar for Sagar. e. Add the scale of the bar graph placing the correct numbers on the vertical axis. f. Are the bars for Seoni and Indore correct in this graph? If not, draw the correct bar(s). e. Q.3. Chinu listed the various means of transport that passed across the road in front of his house from 9 am to 10 am: a. Prepare a frequency distribution table for the data. b. Which means of transport was used the most? c. If you were there to collect this data, how could you do it? Write the steps or process. Ans. a. f. The bar for Seoni is correct but for Indore is incorrect. Means of Transport Number Bike 13 Car 6 Bicycle 8 Auto Rickshaw 8 b. Bike c. To collect the data of various means of transport, one of the ways could be – (i) Prepare a table with 2 columns. First for means of transport and second for its frequency. (ii) List the various means of transport as they pass across the road. (iii) Frequency should be presented using tally marks. Scooter 9 Bus 4 Bullock Cart 2 Q.5. Faiz prepared a frequency distribution table of data on the number of wickets taken by Jaspreet Bumrah in his last 30 matches: a. What information is this table giving? b. What may be the title of this table? c. What caught your attention in this table? d. In how many matches has Bumrah taken 4 wickets? e. Mayank says “If we want to know the total number of wickets he has taken in his last 30 matches, we have to add the numbers 0, 1, 2, 3 …, up to 7.” Can Mayank get the total number of wickets taken in this way? Why? f. How would you correctly figure out the total number of wickets taken by Bumrah in his last 30 matches, using this table? Ans. a. This table is giving the information about different number of wickets taken by Jaspreet Bumrah in different number of matches. b. The title of this table may be: Wickets Taken by Jaspreet Bumrah in Last 30 Matches.(think of more!) c. In this table, attention seeking point is that in 1 match 7 wickets were taken by Jaspreet Bumrah only.(think of more!) d. In 3 matches Bumrah has taken 4 wickets. e. No, Mayank can’t get the total number of wickets taken in this way. For example, 3 wickets per match were taken in 8 matches, which means total 24 wickets were taken. f. To know the total number of wickets, prepare a next column of total wickets .It will have products of corresponding numbers of the two given columns. Then the sum of the numbers in the 3 rd column, will be the total number of wickets. Total number of wickets =90 Q.6. The following pictograph shows the number of tractors in five different villages. Observe the pictograph and answer the following questions— a. Which village has the smallest number of tractors? b. Which village has the most tractors? c. How many more tractors does Village C have than Village B? d. Komal says, “Village D has half the number of tractors as Village E.” Is she right? Ans. a. Village D. b. Village C. c. 3. d. Yes Q.7. The number of girl students in each class of a school is depicted by a pictograph: a. Which class has the least number of girl students? b. What is the difference between the number of girls in Classs 5 and 6? c. If 2 more girls were admitted in Class 2, how would the graph change? d. How many girls are there in Class 7? Ans. a. Class 8. b. 6. Observe this pictograph and answer the following questions: Q.8. Mudhol Hounds (a type of breed of Indian dogs) are largely found in North Karnataka’s Bagalkote and Vijaypura districts. The government took an initiative to protect this breed by providing support to those who adopted these dogs. Due to this initiative, the number of these dogs increased. The number of Mudhol dogs in six villages of Karnataka are as follows — Village A : 18, Village B : 36, Village C : 12, Village D : 48, Village E : 18, Village F : 24 Prepare a pictograph and answer the following questions: c. Kamini said that the number of dogs in Village B and Village D together will be more than the number of dogs in the other 4 villages. Is she right? Give reasons for your response. Ans. Village Number of Mudhol Hounds ) = 6 dogs) Village A Village B Village C Village D a. What will be a useful scale or key to draw this pictograph? b. How many symbols will you use to represent the dogs in Village B? c. If 2 more girls were admitted in class 2, the last half symbol will be converted into full symbol. d. 12. Village E Village F a. Scale is = 6 dogs, as numbers of dogs are multiples of 6. b. 6 symbols. c. Yes, Kamini is right as the number of dogs in village B and village D together is 84 which is more than the number of dogs in the other 4 villages (72). Q.9 A survey of 120 school students was conducted to find out which activity they preferred to do in their free time. Draw a bar graph to illustrate the above data taking the scale of 1 unit length = 5 students. Which activity is preferred by most students other than playing? Ans. Q.10 Students and teachers of a primary school decided to plant tree saplings in the school campus and in the surrounding village during the first week of July. Details of the saplings they planted are as follows — a. The total number of saplings planted on Wednesday and Thursday is ___________. b. The total number of saplings planted during the whole week is ___________. c. The greatest number of saplings were planted on ___________, and the least number of saplings were planted on ___________. Why do you think that is the case? Why were more saplings planted on certain days of the week and less on others? Can you think of possible explanations or reasons? How could you try and figure out whether your explanations are correct? Other than playing, reading story books is preferred by most students. Playing Reading Story Books Watching TV Listening to music Painting N u m be r o f S tu de nts 0 5 10 15 20 25 30 35 40 45 50 Preferred Activity Ans. a. 70 b. 310 c. Saturday, Wednesday The variation in the number of saplings planted on different days may be because of rainy season or different number of students present on different days. (Discuss for other reasons.) Q.11. The number of tigers in India went down drastically between 1900 and 1970. Project Tiger was launched in 1973 to track and protect tigers in India. Starting in 2006, the exact number of tigers in India was tracked. Shagufta and Divya looked up information about the number of tigers in India between 2006 and 2022 in 4-year intervals. They prepared a frequency table for this data and a bar graph to present this data, but there are a few mistakes in the graph. Can you find those mistakes and fix them? Ans. There are mistakes in the bars of 2006, 2010, 2014, 2018. Section 4.5 Page no. 103 Figure it out Q.1. If you wanted to visually represent the data of the heights of the tallest persons in each class in your school, would you use a graph with vertical bars or horizontal bars? Why? Ans. To present the data of the heights of the tallest persons in each class in a school, vertical bar graph is useful as heights of persons can be compared by heights of bars easily. (Although both bar graphs can be used.) (Think of more reasons!) Q.2. If you were making a table of the longest rivers on each continent and their lengths, would you prefer to use a bar graph with vertical bars or with horizontal bars? Why? Try finding out this information, and then make the corresponding table and bar graph! Which continents have the longest rivers? Ans. In this case, bar graph with horizontal bars is useful as length of a river is a horizontal feature, so lengths of rivers can be compared with lengths of horizontal bars easily. (Although both bar graphs can be used.) (can you think of any other reasons?)" class_6,5,Prime Time,ncert_books/class_6/Ganita_Prakash/fegp105.pdf,"5 PRIME TIME 5.1 Common Multiples and Common Factors Children sit in a circle and play a game of numbers. One of the children starts by saying ‘1’. The second player says ‘2’, and so on. But when it is the turn of 3, 6, 9, … (multiples of 3), the player should say ‘idli’ instead of the number. When it is the turn of 5, 10, … (multiples of 5), the player should say ‘vada’ instead of the number. When a number is both a multiple of 3 and a multiple of 5, the player should say ‘idli-vada’! If a player makes any mistake, they are out. The game continues in rounds till only one person remains. For which numbers should the players say ‘idli’ instead of saying the number? These would be 3, 6, 9, 12, 18, … and so on. For which numbers should the players say ‘vada’? These would be 5, 10, 15, 20, … and so on. Which is the first number for which the players should say, ‘idli-vada’? It is 15, which is a multiple of 3, and also a multiple of 5. Find out other such numbers that are multiples of both 3 and 5. These numbers are called _____________________________. Idli-Vada Game Chapter 5_Prime Time.indd 107 13-12-2024 15:59:16 Reprint 2025-26 Ganita Prakash | Grade 6 Let us now play the ‘idli-vada’ game with different pairs of numbers: a. 2 and 5, b. 3 and 7, c. 4 and 6. We will say ‘idli’ for multiples of the smaller number, ‘vada’ for multiples of the larger number and ‘idli-vada’ for common multiples. Draw a figure similar to Fig. 5.1 if the game is played up to 60.  Figure it Out 1. At what number is ‘idli-vada’ said for the 10th time? 2. If the game is played for the numbers 1 to 90, find out: a. How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)? b. How many times would the children say ‘vada’ (including the times they say ‘idli-vada’)? c. How many times would the children say ‘idli-vada’? 3. What if the game was played till 900? How would your answers change? 4. Is this figure somehow related to the ‘idli-vada’ game? Hint: Imagine playing the game till 30. Draw the figure if the game is played till 60. 27 9 3 Multiples of 3 Multiples of 5 21 25 12 Common multiples of 3 and 5 24 18 Fig. 5.1 30 15 20 10 5 Chapter 5_Prime Time.indd 108 12-12-2024 10:52:07 108 Yesterday, we played this game with two numbers. We ended up saying just ‘idli’ or ‘idli-vada’ and nobody said just ‘vada’! Oh, what could those numbers be!? One of the numbers was 4. Reprint 2025-26 Jump Jackpot Jumpy and Grumpy play a game. • Grumpy places a treasure on some number. For example, he may place it on 24. • Jumpy chooses a jump size. If he chooses 4, then he has to jump only on multiples of 4, starting at 0. • Jumpy gets the treasure if he lands on the number where Grumpy placed it. What about the jump of sizes 1 and 24? Yes, they also will land on 24. The numbers 1, 2, 3, 4, 6, 8, 12, 24 all divide 24 exactly. Recall that such numbers are called factors or divisors of 24. Grumpy increases the level of the game. Two treasures are kept on two different numbers. Jumpy has to choose a jump size and stick to it. Jumpy gets the treasures only if he lands on both the numbers with the chosen jump size. As before, Jumpy starts at 0. Grumpy has kept the treasures on 14 and 36. And, Jumpy chooses a jump size of 7. Will Jumpy land on both the treasures? Starting from 0, he jumps to 7 → 14 → 21 → 28 → 35 → 42 … We see that he landed on 14 but 9876543210 242322212019181716151413121110 Which jump sizes will get Jumpy to land on 24? If he chooses 4: Jumpy lands on 4 → 8 → 12 → 16 → 20 → 24 → 28 → ... Other successful jump sizes are 2, 3, 6, 8 and 12. Which of the following could be the other number: 2, 3, 5, 8, 10? Prime Time Chapter 5_Prime Time.indd 109 13-08-2024 17:20:36 Reprint 2025-26 109 Ganita Prakash | Grade 6 did not land on 36, so he does not get the treasure. What jump size should he have chosen? The factors of 14 are: 1, 2, 7, 14. So, these jump sizes will land on 14. The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36. These jump sizes will land on 36. So, the jump sizes of 1 or 2 will land on both 14 and 36. Notice that 1 and 2 are the common factors of 14 and 36. The jump sizes using which both the treasures can be reached are the common factors of the two numbers where the treasures are placed. What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. Look at the table below. What do you notice? 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 Chapter 5_Prime Time.indd 110 13-08-2024 17:20:36 110  Figure it Out 1. Find all multiples of 40 that lie between 310 and 410. In the table, 1. Is there anything common among the shaded numbers? 2. Is there anything common among the circled numbers? 3. Which numbers are both shaded and circled? What are these numbers called? Reprint 2025-26 Math Talk 2. Who am I? a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8. b. I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other. 3. A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10. 4. Find the common factors of: a. 20 and 28 b. 35 and 50 c. 4, 8 and 12 d. 5, 15 and 25 5. Find any three numbers that are multiples of 25 but not multiples of 50. 6. Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both smaller than 10. The first time anybody says ‘idlivada’ is after the number 50. What could the two numbers be which are assigned ‘idli’ and ‘vada’? 7. In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both the numbers? 8. In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions. Prime Time Math Talk Chapter 5_Prime Time.indd 111 14-08-2024 14:51:28 from 1 to 10, except for 7. 10. Find the smallest number that is a multiple of all the numbers from 1 to 10. 9. Find the smallest number that is a multiple of all the numbers Multiples of ____ Common multiples 24 Multiples of ____ 72 Reprint 2025-26 48 Try This 111 Ganita Prakash | Grade 6 5.2 Prime Numbers Guna and Anshu want to pack figs (anjeer) that grow in their farm. Guna wants to put 12 figs in each box and Anshu wants to put 7 figs in each box. How many arrangements are possible? Think and find out the different ways how— 1. Guna can arrange 12 figs in a rectangular manner. 2. Anshu can arrange 7 figs in a rectangular manner. Guna has listed out these possibilities. Observe the number of rows and columns in each of the arrangements. How are they related to 12? In the second arrangement, for example, 12 figs are arranged in two columns of 6 each or 12 = 2 × 6. Anshu could make only one arrangement: 7 × 1 or 1 × 7. There are no other rectangular arrangements possible. In each of Guna’s arrangements, multiplying the number of rows by the number of columns gives the number 12. So, the number of rows or columns are factors of 12. We saw that the number 12 can be arranged in a rectangle in more than one way as 12 has more than two factors. The number 7 can be arranged in only one way, as it has only two factors—1 and 7. Numbers that have only two factors are called prime numbers or primes. Here are the first few primes—2, 3, 5, 7, 11, 13, 17, 19. Notice that the factors of a prime number are 1 and the number itself. What about numbers that have more than two factors? They are called composite numbers. The first few composite numbers are—4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20. Chapter 5_Prime Time.indd 112 13-08-2024 17:20:36 112 Reprint 2025-26 What about 1, which has only one factor? The number 1 is neither a prime nor a composite number. How many prime numbers are there from 21 to 30? How many composite numbers are there from 21 to 30? Can we list all the prime numbers from 1 to 100? Here is an interesting way to find prime numbers. Just follow the steps given below and see what happens. Step 1: Cross out 1 because it is neither prime nor composite. Step 2: Circle 2, and then cross out all multiples of 2 after that, i.e., 4, 6, 8, and so on. Step 3: You will find that the next uncrossed number is 3. Circle 3 and then cross out all the multiples of 3 after that, i.e., 6, 9, 12, and so on. Step 4: The next uncrossed number is 5. Circle 5 and then cross out all the multiples of 5 after that, i.e., 10, 15, 20, and so on. Step 5: Continue this process till all the numbers in the list are either circled or crossed out. 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 1 2 3 4 5 6 7 8 9 10 Prime Time Chapter 5_Prime Time.indd 113 13-08-2024 17:20:36 All the circled numbers are prime numbers. All the crossed out numbers, other than 1, are composite numbers. This method is called the Sieve of Eratosthenes. This procedure can be carried on for numbers greater than 100 also. Eratosthenes was a Greek mathematician who lived around 2200 years ago and developed this method of listing primes. Reprint 2025-26 It is definitely not some magic; there should be a reason why it works. 113 Ganita Prakash | Grade 6 Guna and Anshu started wondering how this simple method is able to find prime numbers! Think how this method works. Read the steps given above again and see what happens after each step is carried out. Primes through the Ages Prime numbers are the building blocks of all whole numbers. Starting from the time of the Greek civilisation (more than 2000 years ago) to this day, mathematicians are still struggling to uncover their secrets! Food for thought: is there a largest prime number? Or does the list of prime numbers go on without an end? A mathematician named Euclid found the answer and so will you in a later class! Fun fact: The largest prime number that anyone has ‘written down’ is so large that it would take around 6500 pages to write it! So they could only write it on a computer!  Figure it Out 1. We see that 2 is a prime and also an even number. Is there any other even prime? 2. Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference? 3. Are there an equal number of primes occurring in every row in the table on the previous page? Which decades have the least number of primes? Which have the most number of primes? Chapter 5_Prime Time.indd 114 13-08-2024 17:20:37 114 4. Which of the following numbers are prime: 23, 51, 37, 26? 5. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5. 6. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100. 7. Find seven consecutive composite numbers between 1 and 100. 8. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100. Reprint 2025-26 5.3 Co-prime numbers for safekeeping treasures Which pairs are safe? Let us go back to the treasure finding game. This time, treasures are kept on two numbers. Jumpy gets the treasures only if he is able to reach both the numbers with the same jump size. There is also a new rule—a jump size of 1 is not allowed. 9. Identify whether each statement is true or false. Explain. a. There is no prime number whose units digit is 4. b. A product of primes can also be prime. c. Prime numbers do not have any factors. d. All even numbers are composite numbers. e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite. 10. Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330? 11. How many three-digit prime numbers can you make using each of 2, 4 and 5 once? 12. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples. Prime Time Chapter 5_Prime Time.indd 115 14-08-2024 14:53:54 Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures? Will placing the treasure on 12 and 26 work? No! If the jump size is chosen to be 2, then Jumpy will reach both 12 and 26. What about 4 and 9? Jumpy cannot reach both using any jump size other than 1. So, Grumpy knows that the pair 4 and 9 is safe. Check if these pairs are safe: a. 15 and 39 b. 4 and 15 c. 18 and 29 d. 20 and 55 Reprint 2025-26 115 Ganita Prakash | Grade 6 What is special about safe pairs? They don’t have any common factor other than 1. Two numbers are said to be co-prime to each other if they have no common factor other than 1. Example: As 15 and 39 have 3 as a common factor, they are not co-prime. But 4 and 9 are co-prime. While playing the ‘idli-vada’ game with different number pairs, Anshu observed something interesting! 1. Sometimes the first common multiple was the same as the product of the two numbers. 2. At other times the first common multiple was less than the product of the two numbers. Find examples for each of the above. How is it related to the number pair being co-prime? Co-prime art Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread-gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class. Which of the following pairs of numbers are co-prime? a. 18 and 35 b. 15 and 37 c. 30 and 415 d. 17 and 69 e. 81 and 18 Math Talk Math Talk Chapter 5_Prime Time.indd 116 13-08-2024 17:20:37 116 9 In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to the two numbers (the number of pegs and the thread-gap) being co-prime? 12 11 10 8 7 6 5 4 3 2 1 13 1 2 5 67 8 9 10 11 12 16 1 2 Reprint 2025-26 4 3 6 7 8 9 10 11 12 13 14 15 24 1 2 3 4 5 6 7 8 9 10 13 12 11 14 15 16 17 18 19 20 21 22 23 3 5 4 5.4 Prime Factorisation Checking if two numbers are co-prime Teacher: Are 56 and 63 co-prime? Anshu and Guna: If they have a common factor other than 1, then they are not co-prime. Let us check. Anshu: I can write 56 = 14 × 4 and 63 = 21 × 3. So, 14 and 4 are factors, of 56. Further, 21 and 3 are factors of 63. So, there are no common factors. The numbers are co-prime. Guna: Hold on. I can also write 56 = 7 × 8 and 63 = 9 × 7. We see that 7 is a factor of both numbers, so, they are not co-prime. Writing 56 = 14 × 4 tells us that 14 and 4 are both factors of 56, but it does not tell all the factors of 56. The same holds for the factors of 63. Try another example: 80 and 63. There are many ways to factorise both numbers. 80 = 40 × 2 = 20 × 4 = 10 × 8 = 16 × 5 = ??? 63 = 9 × 7 = 3 × 21 = ??? We have written ‘???’ to say that there may be more ways to factorise these numbers. But if we take any of the given factorisations, for example, 80 = 16 × 5 and 63 = 9 × 7, then there are no common factors. Can we conclude that 80 and 63 are co-prime? As Anshu’s mistake above shows, we cannot conclude that as there may be other ways to factorise the numbers. What this means is that we need a more systematic approach to check if two numbers are co-prime. Clearly Guna is right, as 7 is a common factor. But where did Anshu go wrong? Make such pictures for the following: a. 15 pegs, thread-gap of 10 b. 10 pegs, thread-gap of 7 c. 14 pegs, thread-gap of 6 d. 8 pegs, thread-gap of 3 Prime Time Chapter 5_Prime Time.indd 117 13-08-2024 17:20:37 Reprint 2025-26 117 Ganita Prakash | Grade 6 Prime factorisation Take a number such as 56. It is composite, as we saw that it can be written as 56 = 4 × 14 . So, both 4 and 14 are factors of 56. Now take one of these, say 14. It is also composite and can be written as 14 = 2 × 7. Therefore, 56 = 4 × 2 × 7. Now, 4 is composite and can be written as 4 = 2 × 2. Therefore, 56 = 2 × 2 × 2 × 7. All the factors appearing here, 2 and 7, are prime numbers. So, we cannot divide them further. In conclusion, we have written 56 as a product of prime numbers. This is called prime factorisation of 56. The individual factors are called prime factors. For example, the prime factors of 56 are 2 and 7. Every number greater than 1 has a prime factorisation. The idea is the same: keep breaking the composite numbers into factors till only primes are left. The number 1 does not have any prime factorisation. It is not divisible by any prime number. What is the prime factorisation of a prime number like 7? It is just 7 (we cannot break it down any further). Let us see a few more examples. By going through different ways of breaking down the number, we wrote 63 as 3 × 3 × 7 and as 3 × 7 × 3. Are they different? Not really! The same prime numbers 3 and 7 occur in both cases. Further, 3 appears two times in both and 7 appears once. Here, you see four different ways to get prime factorisation of 36. Observe that in all four cases, we get two 2s and two 3s. Multiply back to see that you get 36 in all four cases. For any number, it is a remarkable fact that there is only one prime factorisation, except that the prime factors may come in different 36 Chapter 5_Prime Time.indd 118 13-08-2024 17:20:37 118 Reprint 2025-26 2×2×3×3 3×3×2×2 2×2×3×3 2×3×2×3 2×2×9 3×3×4 2×2×9 2×3×6 2×18 3×12 4×9 6×6 2×2×3×3 2×3×2×3 orders. As we explain below, the order is not important. However, as we saw in these examples, there are many ways to arrive at the prime factorisation! Does the order matter? Using this diagram, can you explain why 30 = 2 × 3 × 5, no matter which way you multiply 2, 3, and 5? When multiplying numbers, we can do so in any order. The end result is the same. That is why, when two 2s and two 3s are multiplied in any order, we get 36. In a later class, we shall study this under the names of commutativity and associativity of multiplication. Thus, the order does not matter. Usually we write the prime numbers in increasing order. For example, 225 = 3 × 3 × 5 × 5 or 30 = 2 × 3 × 5. Prime Time Chapter 5_Prime Time.indd 119 13-08-2024 17:20:37 Prime factorisation of a product of two numbers When we find the prime factorisation of a number, we first write it as a product of two factors. For example, 72 = 12 × 6. Then, we find the prime factorisation of each of the factors. In the above example, 12 = 2 × 2 × 3 and 6 = 2 × 3. Now, can you say what the prime factorisation of 72 is? The prime factorisation of the original number is obtained by putting these together. 72 = 2 × 2 × 3 × 2 × 3 Reprint 2025-26 119 Ganita Prakash | Grade 6 We can also write this as 2 × 2 × 2 × 3 × 3. Multiply and check that you get 72 back! Observe how many times each prime factor occurs in the factorisation of 72. Compare it with how many times it occurs in the factorisations of 12 and 6 put together. Prime factorisation is of fundamental importance in the study of numbers. Let us discuss two ways in which it can be useful. Using prime factorisation to check if two numbers are co-prime Let us again take the numbers 56 and 63. How can we check if they are co-prime? We can use the prime factorisation of both numbers— Figure it Out 1. Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000. 2. The prime factorisation of a number has one 2, two 3s, and one 11. What is the number? 3. Find three prime numbers, all less than 30, whose product is 1955. 4. Find the prime factorisation of these numbers without multiplying first a. 56 × 25 b. 108 × 75 c. 1000 × 81 5. What is the smallest number whose prime factorisation has: a. three different prime numbers? b. four different prime numbers? Chapter 5_Prime Time.indd 120 13-08-2024 17:20:37 120 Now, we see that 7 is a prime factor of 56 as well as 63. Therefore, 56 and 63 are not co-prime. What about 80 and 63? Their prime factorisations are as follows: There are no common prime factors. Can we conclude that they are co-prime? Suppose they have a common factor that is composite. Would the prime factors of this composite common factor appear in the prime factorisation of 80 and 63? 80 = 2 × 2 × 2 × 2 × 5 and 63 = 3 × 3 × 7 56 = 2 × 2 × 2 × 7 and 63 = 3 × 3 × 7 Reprint 2025-26 Therefore, we can say that if there are no common prime factors, then the two numbers are co-prime. Let us see some examples. Example: Consider 40 and 231. Their prime factorisations are as follows: 40 = 2 × 2 × 2 × 5 and 231 = 3 × 7 × 11 We see that there are no common primes that divide both 40 and 231. Indeed, the prime factors of 40 are 2 and 5 while, the prime factors of 231 are 3, 7, and 11. Therefore, 40 and 231 are co-prime! Example: Consider 242 and 195. Their prime factorisations are as follows: The prime factors of 242 are 2 and 11. The prime factors of 195 are 3, 5, and 13. There are no common prime factors. Therefore, 242 and 195 are co-prime. Using prime factorisation to check if one number is divisible by another We can say that if one number is divisible by another, the prime factorisation of the second number is included in the prime factorisation of the first number. We say that 48 is divisible by 12 because when we divide 48 by 12, the remainder is zero. How can we check if one number is divisible by another without carrying out long division? 242 = 2 × 11 × 11 and 195 = 3 × 5 × 13 Prime Time Chapter 5_Prime Time.indd 121 13-08-2024 17:20:37 Example: Is 168 divisible by 12? Find the prime factorisations of both: Example: Is 75 divisible by 21? Find the prime factorisations of both: Since we can multiply in any order, now it is clear that, Therefore, 168 is divisible by 12. 168 = 2 × 2 × 2 × 3 × 7 and 12 = 2 × 2 × 3 168 = 2 × 2 × 3 × 2 × 7 = 12 × 14 75 = 3 × 5 × 5 and 21 = 3 × 7 Reprint 2025-26 121 Ganita Prakash | Grade 6 As we saw in the discussion above, if 75 was a multiple of 21, then all prime factors of 21 would also be prime factors of 75. However, 7 is a prime factor of 21 but not a prime factor of 75. Therefore, 75 is not divisible by 21. Example: Is 42 divisible by 12? Find the prime factorisations of both: All prime factors of 12 are also prime factors of 42. But the prime factorisation of 12 is not included in the prime factorisation of 42. This is because 2 occurs twice in the prime factorisation of 12 but only once in the prime factorisation of 42. This means that 42 is not divisible by 12. We can say that if one number is divisible by another, then the prime factorisation of the second number is included in the prime factorisation of the first number.  Figure it Out 1. Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer. a. 30 and 45 b. 57 and 85 c. 121 and 1331 d. 343 and 216 2. Is the first number divisible by the second? Use prime factorisation. a. 225 and 27 b. 96 and 24 c. 343 and 17 d. 999 and 99 3. The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other? 4. Guna says, “Any two prime numbers are co-prime?”. Is he right? 42 = 2 × 3 × 7 and 12 = 2 × 2 × 3 Chapter 5_Prime Time.indd 122 13-08-2024 17:20:37 122 5.5 Divisibility Tests So far, we have been finding factors of numbers in different contexts, including to determine if a number is prime or not, or if a given pair of numbers is co-prime or not. Reprint 2025-26 It is easy to find factors of small numbers. How do we find factors of a large number? Let us take 8560. Does it have any factors from 2 to 10 (2, 3, 4, 5, ..., 9, 10)? It is easy to check if some of these numbers are factors or not without doing long division. Can you find them? Divisibility by 10 Let us take 10. Is 8560 divisible by 10? This is another way of asking if 10 is a factor of 8560. For this, we can look at the pattern in the multiples of 10. The first few multiples of 10 are: 10, 20, 30, 40, … Continue this sequence and observe the pattern. Is 125 a multiple of 10? Will this number appear in the previous sequence? Why or why not? Can you now answer if 8560 is divisible by 10? Consider this statement: Numbers that are divisible by 10 are those that end with ‘0’. Do you agree? Divisibility by 5 The number 5 is another number whose divisibility can easily be checked. How do we do it? Explore by listing down the multiples: 5, 10, 15, 20, 25, ... What do you observe about these numbers? Do you see a pattern in the last digit? What is the largest number less than 399 that is divisible by 5? Is 8560 divisible by 5? Prime Time Math Talk Chapter 5_Prime Time.indd 123 13-08-2024 17:20:37 Consider this statement: Numbers that are divisible by 5 are those that end with either a ‘0’ or a ‘5’. Do you agree? Divisibility by 2 The first few multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... What do you observe? Do you see a pattern in the last digit? Reprint 2025-26 Math Talk 123 Ganita Prakash | Grade 6 Is 682 divisible by 2? Can we answer this without doing the long division? Is 8560 divisible by 2? Why or why not? Consider this statement: Numbers that are divisible by 2 are those that end with ‘0’, ‘2’, ‘4, ‘6’ or ‘8’. Do you agree? What are all the multiples of 2 between 399 and 411? Divisibility by 4 Checking if a number is divisible by 4 can also be done easily! Look at its multiples: 4, 8, 12, 16, 20, 24, 28, 32, … Are you able to observe any patterns that can be used? The multiples of 10, 5 and 2 have a pattern in their last digits which we are able to use to check for divisibility. Similarly, can we check if a number is divisible by 4 by looking at the last digit? It does not work! Look at 12 and 22. They have the same last digit, but 12 is a multiple of 4 while 22 is not. Similarly 14 and 24 have the same last digit, but 14 is not a multiple of 4 while 24 is. Similarly, 16 and 26 or 18 and 28. What this means is that by looking at the last digit, we cannot tell whether a number is a multiple of 4. Can we answer the question by looking at more digits? Make a list of multiples of 4 between 1 and 200 and search for a pattern. Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe? Math Talk Chapter 5_Prime Time.indd 124 14-08-2024 14:54:36 124 Is 8536 divisible by 4? Consider these statements: 1. Only the last two digits matter when deciding if a given number is divisible by 4. 2. If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4. 3. If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4. Do you agree? Why or why not? Reprint 2025-26 Divisibility by 8 Interestingly, even checking for divisibility by 8 can be simplified. Can the last two digits be used for this? Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe? Change the last two digits of 8560 so that the resulting number is a multiple of 8. We have seen that long division is not always needed to check if a number is a factor or not. We have made use of certain observations to come up with simple methods for 10, 5, 2, 4, 8. Do we have such simple methods for other numbers as well? We will discuss simple methods to test divisibility by 3, 6, 7, and 9 in later classes! Consider these statements: 1. Only the last three digits matter when deciding if a given number is divisible by 8. 2. If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8. 3. If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8. Do you agree? Why or why not? Prime Time Math Talk Chapter 5_Prime Time.indd 125 13-08-2024 17:20:37  Figure it Out 1. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. 2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes. 3. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning. a. From the year you were born till now, which years were leap years? b. From the year 2024 till 2099, how many leap years are there? Reprint 2025-26 125 Ganita Prakash | Grade 6 5.6 Fun with numbers Special numbers There are four numbers in this box. Which number looks special to you? Why do you say so? 4. Find the remainders obtained when each of the following numbers are divided by (a) 10, (b) 5, (c) 2. 5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be? 6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160. 7. Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit. a. Sum of two even numbers gives a multiple of 4. b. Sum of two odd numbers gives a multiple of 4. 78, 99, 173, 572, 980, 1111, 2345 25 43 9 16 Math Talk Chapter 5_Prime Time.indd 126 12-12-2024 10:52:07 126 Look at the what Guna’s classmates have to share: • Karnawati says, “9 is special because it is a single-digit number whereas all the other numbers are 2-digit numbers”. • Gurupreet says, “9 is special because it is the only number that is a multiple of 3”. • Murugan says, “16 is special because it is the only even number and also the only multiple of 4”. • Gopika says, “25 is special as it is the only multiple of 5”. • Yadnyikee says, “43 is special because it is the only prime number”. • Radha says, “43 is special because it is the only number that is not a square”. Reprint 2025-26 5.6 Fun with numbers Special numbers There are four numbers in this box. Which number looks special to you? Why do you say so? 4. Find the remainders obtained when each of the following numbers are divided by a) 10, b) 5, c) 2. 5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be? 6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160. 7. Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit. a. Sum of two even numbers gives a multiple of 4. b. Sum of two odd numbers gives a multiple of 4. 78, 99, 173, 572, 980, 1111, 2345 25 43 9 16 A prime puzzle The figure on the left shows the puzzle. The figure on the right shows the solution of the puzzle. Think what the rules can be to solve the puzzle. Math Talk Below are some boxes with four numbers in each box. Within each box try to say how each number is special compared to the rest. Share with your classmates and find out who else gave the same reasons as you did. Did anyone give different reasons that may not have occurred to you?! 170 30 63 12 35 5 7 11 24 3 8 102 75 42 170 30 63 123 31 17 2 3 102 27 3 5 5 3 75 2 3 7 42 17 27 44 65 Prime Time Math Talk Look at the what Guna’s classmates have to share: • Karnawati says, “9 is special because it is a single-digit number whereas all the other numbers are 2-digit numbers”. • Gurupreet says, “9 is special because it is the only number that is a multiple of 3”. • Murugan says, “16 is special because it is the only even number and also the only multiple of 4”. • Gopika says, “25 is special as it is the only multiple of 5”. • Yadnyikee says, “43 is special because it is the only prime number”. • Radha says, “43 is special because it is the only number that is not a square”. Chapter 5_Prime Time.indd 127 13-08-2024 17:20:37 Rules Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column. 28 125 18 105 20 30 Reprint 2025-26 30 70 28 105 70 8 127 Ganita Prakash | Grade 6 If a number is divisible by another, the second number is called a factor of the first. For example, 4 is a factor of 12 because 12 is divisible by 4 (12 ÷ 4 = 3). Prime numbers are numbers like 2, 3, 5, 7, 11, … that have only two factors, namely 1 and themselves. Composite numbers are numbers like 4, 6, 8, 9, … that have more than 2 factors, i.e., at least one factor other than 1 and themselves. For example, 8 has the factor 4 and 9 has the factor 3, so 8 and 9 are both composite. Every number greater than 1 can be written as a product of prime numbers. This is called the number’s prime factorisation. For example, 84 = 2 × 2 × 3 × 7. 45 42 171 190 63 27 28 154 231 343 66 44 Chapter 5_Prime Time.indd 128 12-12-2024 10:52:09 128 There is only one way to factorise a number into primes, except for the ordering of the factors. Two numbers that do not have a common factor other than 1 are said to be co-prime. To check if two numbers are co-prime, we can first find their prime factorisations and check if there is a common prime factor. If there is no common prime factor, they are co-prime, and otherwise they are not. A number is a factor of another number if the prime factorisation of the first number is included in the prime factorisation of the second number. Reprint 2025-26 Section 5.1 Page no. 108 Q.1. At what number is ‘idli-vada’ said for the 10th time? Ans. 150. Q.2. If the game is played for the numbers from 1 till 90, find out: Ans. 30 times. Ans. 18 times. Ans. 6 times. Q.3. What if the game was played till 900? How would your answers change? Ans. If the game was played till 900, idli-vada would be said 60 times by the children. ‘Vada’ will be said 180 time and ‘idli’ will be said 300 times. Q.4. Is this figure somehow related to the ‘idli-vada’ game? Hint: Imagine playing the game till 30. Draw the figure if the game is played till 60. Figure it out a. How many times would the children say ‘idli’ (including the times they say ‘idlivada’)?b. How many times would the children say ‘vada’ (including the times they say ‘idlivada’)?c. How many times would the children say ‘idli-vada’? CHAPTER 5 — SOLUTIONS Prime Time Ans. Yes, the common numbers represent the numbers when to say ‘idli-vada’. 1 Page no. 109 Ans. The other number will be 8. Page no. 110 Q. What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. Ans. Jump size of 3 or 5 will take us to 15 & 30. Ans. All shaded numbers are multiples of 3. Ans. All circled numbers are multiples of 4. Ans. 36, 48, 60. These numbers are called common multiples of 3 and 4. Section 5.1 Page no. 110 Q.1. Find all multiples of 40 that lie between 310 and 410. Figure it out Q.1. Is there anything common among the shaded numbers? Q.2. Is there anything common among the circled numbers? Q.3. Which numbers are both shaded and circled? What are these numbers called? Which of the following could be the other number: 2, 3, 5, 8, 10? Other possible jump sizes are 1, 15. Ans. Multiples of 40 that lie between 310 and 410 are: Q.2. Who am I? Ans. a. 35 a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8. b. I am a number less than 100. Two of my factors are 3 and 5.One of my digits is 1 more than the other. 320, 360, 400. b. 45 2 Q.3. A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10. Ans. 6. Q.4. Find the common factors of: Ans. a. Common factors of 20 and 28 = 1, 2, 4. b. Common factors of 35 and 50 = 1, 5. c. Common factors of 4, 8 and 12 = 1, 2, 4. d. Common factors of 5, 15 and 25 = 1, 5 Q.5. Find any three numbers that are multiples of 25 but not multiples of 50. Ans. Some such numbers are: 25, 75, 125, 175. Q.6. Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both smaller than 10. The first time anybody says ‘idlivada’ is after the number 50. What could the two numbers be which are assigned ‘idli’ and ‘vada’? Ans. 7, 8 Q.7. In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both the numbers? Ans. The jump sizes are 1, 2, 7 or 14 Q.8. In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions. a. 20 and 28 b. 35 and 50 c. 4, 8 and 12 d. 5, 15 and 25 8, 9 (Try for other possibility) Ans. Multiples of 3 Multiples of 4 Q.9. Find the smallest number that is a multiple of all the numbers from 1 to 10 except for 7. Ans. 360 Q.10. Find the smallest number that is a multiple of all the number from 1 to 10. Ans. 2520 Think of some more. 3 Section 5.2 Page No. 113 Ans. Prime numbers from 21 to 30: 2 (23,29) Page No. 114 Q.1. We see that 2 is a prime and also an even number. Is there any other even prime? Ans. No. Q.2. Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference? Ans. The smallest difference between two successive primes is one (3-2=1) the largest difference is 8 (97-89). Q.3. Are there an equal number of primes occurring in every row in the table on the previous page? Which decades have the least number of primes? Which have the most number of primes? Ans. No. The least number of primes occur in the decades 91 to 100 and most number of primes and occur in the decades 1 to 10, 11 to 20. Q.4. Which of the following numbers are prime: 23, 51, 37, 26? Ans. 23 and 37. Q. How many prime numbers are there from 21 to 30? How many composite numbers are there from 21 to 30? Figure it out Composite numbers from 21 to 30: 8 Q.5. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5. Ans. Pairs of prime numbers less than 20 whose sum is a multiple of 5 are (Try for other possibilities). Q.6. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100. Ans. 17 and 71, 37 and 73, 79 and 97. Q.7. Find seven consecutive composite numbers between 1 and 100. Ans. 90, 91, 92, 93, 94, 95, 96. (2,3), (3,7), (2,13) 4 Q.8. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100. Ans. 3 and 5, 5 and 7, 11 and 13, 17 and 19, 29 and 31, 41 and 43, 59 and 61, 71 and 73. Q.9 Identify whether each statement is true or false. Explain. Ans. a. True, unit digit 4 means even number. We know that 2 is the only even prime number. Q.10. Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330? Ans. 105 = 3 × 5 × 7 is the only number which is the product of three distinct prime numbers. Q.11. How many three-digit prime numbers can you make using each of 2, 4 and 5 once? Ans. None Q.12. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples. a. There is no prime number whose units digit is 4. b. A product of primes can also be prime. c. Prime numbers do not have any factors. d. All even numbers are composite numbers. e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite. b. False, product of primes becomes a composite number. c. False, prime numbers have exactly two factors: 1 and the number itself. d. False, all even numbers except 2 are composite numbers e. True, as every pair, other than (2, 3), contains an even number which is a composite number. Ans. Yes, 11 = 2 × 5 + 1, 47 = 2 × 23 + 1, 83 = 2 × 41 + 1 23 = 2 × 11 +1, 59 = 2 × 29 +1 (Try other possibilities.) 5 Section 5.3 Page No. 115 Ans. Safe pairs are: 4 and 15, 18 and 29. Page no. 116 Ans. The pairs of co-prime numbers are: Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures? Q. Which of the following pairs of numbers are co-prime? Q. While playing the ‘idli-vada’ game with different number pairs, Anshu observed something interesting! Check if these pairs are safe: a. 15 and 39 b. 4 and 15 c. 18 and 29 d. 20 and 55 1. Sometimes the first common multiple was the same as the product of the two numbers. 2. At other times the first common multiple was less than the product of the two numbers. a. 18 and 35 b. 15 and 37 c. 30 and 415 d. 17 and 69 e. 81 and 18 a. 18 and 35 b. 15 and 37 d. 17 and 69 Ans. 1. Examples when the first common multiple is the product of the two numbers: numbers: Find examples for each of the above. How is it related to the number pair being coprime?3, 5; 3, 7 and 4, 9. 2. Examples when the first common multiple is less than the product of the two 3, 6; 3, 12 and 6, 15 Whenever the number pair is co-prime the first common multiple is the product of the two numbers. 6 Section 5.4 Page no. 120 Q.1. Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000. Ans. 64 = 2 × 2 × 2 × 2 × 2 × 2 Q.2. The prime factorisation of a number has one 2, two 3s, and one 11. What is the number? Ans. 198 = 2 × 3 × 3 × 11 Q.3. Find three prime numbers, all less than 30, whose product is 1955. Ans. 1955 = 5 × 17 × 23 Figure it out 104 = 2 × 2 × 2 × 13 105 = 3 × 5 × 7 243 = 3 ×3 ×3 ×3 ×3 320 = 2 × 2 × 2 × 2 × 2 × 2 × 5 141 = 3 × 47 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 729 = 3 ×3 ×3 × 3 ×3 ×3 1024 = 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 1331 = 11 × 11 × 11 1000 = 2 × 2 × 2 × 5 × 5 × 5 Q.4. Find the prime factorisation of these numbers without multiplying first Ans. Q.5. What is the smallest number whose prime factorisation has: a. three different prime numbers? b. four different prime numbers? a. 56 × 25 b. 108 × 75 c. 1000 × 81 a. 56 × 25 = 2 × 2 × 2 × 7 × 5 × 5 b. 108 × 75 = 2 × 2 × 3 × 3 × 3 ×3 × 5 × 5 c. 1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 ×3 × 5 × 5 × 5 7 Ans. a. 2 × 3 × 5 = 30 Page no. 122 Q.1. Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer. Ans. a. No Figure it out a. 30 and 45 b. 57 and 85 c. 121 and 1331 d. 343 and 216 b. 2 × 3 × 5 × 7 = 210 b. Yes c. No c. Yes 30 = 2 × 3 × 5 45 = 3 × 3 × 5 57 = 19 × 3 85 = 17 × 5 121 = 11 × 11 1331 = 11 × 11 × 11 343 = 7 × 7 × 7 216 = 2 × 2 × 2 × 3 × 3 × 3. Q.2. Is the first number divisible by the second? Use prime factorisation. Ans. a. No ( 3 × 3 × 5 × 5 3 × 3 × 3 ) a. 225 and 27 b. 96 and 24 c. 343 and 17 d. 999 and 99 b. Yes ( 2 × 2 × 2 × 2 × 2 × 3 2 × 2 ×2 ×3 ) c. No ( 7 × 7 × 7 17 ) d. No ( 3 × 3 × 3 × 3 × 37 3 × 3 × 11 ) 8 Q.3. The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other? Ans. No, they are not co-prime. Q.4. Guna says, “Any two prime numbers are co-prime”. Is he right? Ans. Yes, for example 2, 3; 3,11 (Try other examples) Section 5.5 Page no. 124 Ans. Yes, as 36 is divisible by 4. Q. Consider these statements: Ans. 1. Yes Q. Is 8536 divisible by 4? 1. Only the last two digits matter when deciding if a given number is divisible by 4. 2. If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4. 3. If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4. Do you agree? Why or why not? No, one of them is not dividing the other. 2. Yes 3. Yes Page no. 125 Ans. The numbers between 120 & 140 divisible by 8 are = 128, 136. Q. Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe? Yes,we agree. Consider numbers 124,364,4028 etc. These are divisible by 4.Check for more numbers. The numbers between 1120 & 1140 divisible by 8 are = 1128, 1136. The numbers between 3120 & 3140 divisible by 8 are = 3128, 3136. 9 Ans. 8552 is a multiple of 8. Ans. a. Yes Q.1. 2024 is a leap year (as February has 29 days). Leap years occurs in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. Q. Change the last two digits of 8560 so that the resulting number is a multiple of 8. Q. Consider this statement: Figure it out 1. Only the last three digits matter when deciding if a given number is divisible by 8. 2. If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8. 3. If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8. Do you agree? Why or why not? a. From the year you were born till now, which years were leap years? b. From the year 2024 till 2099, how many leap years are there? if the number formed by one’s, ten’s and hundred’s digit is divisible by 8, then the number is divisible by 8. b. Yes c. Yes Yes. Some examples are: 8576,7648,5024. Try some more. Ans. b. From the year 2024 till 2099, the number of leap years is 19 Q.2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes. Ans. The largest 4-digit number divisible by 4 and also a palindrome: 9999 Q.3. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning. a. Sum of two even numbers gives a multiple of 4. b. Sum of two odd numbers gives a multiple of 4. The smallest 4-digit number divisible by 4 and also a palindrome: 1001. 10 Ans. a. Sometimes true Q.4. Find the remainders obtained when each of the following numbers are divided by i) 10, ii) 5, iii) 2. Ans. Q.5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be? Ans. The two numbers that are sufficient to declare the divisibility of 14560 by 2, 4, 5, 8 and 10 are 5 and 8. Numbers a) Remainder when divided by 10 b) Remainder when divided by 5 c) Remainder when divided by 2 78 99 173 572 980 1111 2345 78, 99, 173, 572, 980, 1111, 2345 b. Sometimes true Note: Think of more such examples. Example 2+6 = 8, a multiple of 4. but 2+4 = 6, not a multiple of 4. Example 1+3 = 4, a multiple of 4. but 1+5 = 6, not a multiple of 4. 8 9 3 2 0 1 5 3 4 3 2 0 1 0 0 1 1 0 0 1 1 Q.6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160. Ans. The numbers divisible by 2, 4, 5, 8 and 10 are: Q.7. Write two numbers whose product is 10000. The two numbers should not have 0 as their unit digit. Ans. 10000 = 16 × 625. 5600, 6000 and 77622160. 11" class_6,6,Perimeter and Area,ncert_books/class_6/Ganita_Prakash/fegp106.pdf,"6 PERIMETER AND AREA 6.1 Perimeter Do you remember what the perimeter of a closed plane figure is? Let us refresh our understanding! The perimeter of any closed plane figure is the distance covered along its boundary when you go around it once. For a polygon, i.e., a closed plane figure made up of line segments, the perimeter is simply the sum of the lengths of its all sides, i.e., the total distance along its outer boundary. Let us revise the formulas for the perimeter of rectangles, squares, and triangles. The perimeter of a polygon = the sum of the lengths of its all sides. Chapter 6_Perimeter and Area.indd 129 13-08-2024 16:16:29 Perimeter of a rectangle Consider a rectangle ABCD whose length and breadth are 12 cm and 8 cm, respectively. What is its perimeter? Perimeter of the rectangle = Sum of the lengths of its four sides 8 c m A 12 cm B D C = AB + BC + CD + DA Reprint 2025-26 Ganita Prakash | Grade 6 = 2 × AB + 2 × BC = 2 × (AB + BC) = 2 × (12 cm + 8 cm) = 2 × (20 cm) = 40 cm. From this example, we see that— Perimeter of a rectangle = length + breadth + length + breadth. Perimeter of a rectangle = 2 × (length + breadth). The perimeter of a rectangle is twice the sum of its length and breadth. Perimeter of a square Debojeet wants to put coloured tape all around a square photo frame of side 1m as shown. What will be the length of the coloured tape he requires? Since Debojeet wants to put the coloured tape all around the square photo frame, he needs to find the perimeter of the photo frame. Thus, the length of the tape required = perimeter of the square = sum of the lengths of all four sides of the square = 1 m + 1 m + 1 m + 1 m = 4 m. Now, we know that all four sides of a square are equal in length. Therefore, in place of adding the lengths of each side, we can simply multiply the length of one side by 4. Thus, the length of the tape required = 4 × 1 m = 4 m. From this example, we see that = AB + BC + AB + BC Opposite sides of a rectangle are always equal. So, AB = CD and AD = BC 1 m Chapter 6_Perimeter and Area.indd 130 13-08-2024 16:16:29 130 The perimeter of a square is quadruple the length of its side. Perimeter of a square = 4 × length of a side. Reprint 2025-26 Perimeter of a triangle Consider a triangle having three given sides of lengths 4 cm, 5 cm and 7 cm. Find its perimeter. Perimeter of the triangle = 4 cm + 5 cm + 7 cm = 16 cm. Perimeter of a triangle = sum of the lengths of its three sides. Example: Akshi wants to put lace all around a rectangular tablecloth that is 3 m long and 2 m wide. Find the length of the lace required. Solution = 2 × (3 m + 2 m) Length of the rectangular table cover = 3 m. Breadth of the rectangular table cover = 2 m. Akshi wants to put lace all around the Therefore, the length of the lace required will be the perimeter of the Now, the perimeter of the rectangular tablecloth = 2 × (length + breadth) = 2 × 5 m = 10 m. Hence, the length of the lace required is 10 m. tablecloth. rectangular tablecloth. 5 cm 4 cm Perimeter and Area 7 cm Chapter 6_Perimeter and Area.indd 131 12-12-2024 11:04:16 Example: Find the distance travelled by Usha if she takes three rounds of a square park of side 75 m. Solution Perimeter of the square park = 4 × length Distance covered by Usha in one Therefore, the total distance travelled by of a side = 4 × 75 m = 300 m. round = 300 m. Usha in three rounds = 3 × 300 m = 900 m. Reprint 2025-26 131 Ganita Prakash | Grade 6  Figure it Out 1. Find the missing terms: 2. A rectangle having sidelengths 5 cm and 3 cm is made using a 3. Find the length of the third side of a triangle having a perimeter 4. What would be the cost of fencing a rectangular park whose length 5. A piece of string is 36 cm long. What will be the length of each 6. A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed? piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square? of 55 cm and having two sides of length 20 cm and 14 cm, respectively. is 150 m and breadth is 120 m, if the fence costs `40 per metre? side, if it is used to form: a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?. b. Perimeter of a square = 20 cm; side of a length = ?. c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?. a. A square, b. A triangle with all sides of equal length, and c. A hexagon (a six sided closed figure) with sides of equal length? Chapter 6_Perimeter and Area.indd 132 13-08-2024 16:16:30 132 Reprint 2025-26 30 m 40 m Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round. Matha Pachchi! Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance. 60 m 70 m Starting Point for Toshi Perimeter and Area Starting Point for Akshi Chapter 6_Perimeter and Area.indd 133 13-08-2024 16:16:30  Figure it Out 1. Find out the total distance Akshi has covered in 5 rounds. 2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? 3. Think and mark the positions as directed— a. Mark ‘A’ at the point where Akshi will be after she ran 250 m. b. Mark ‘B’ at the point where Akshi will be after she ran 500 m. c. Now, Akshi ran 1000 m. How many full rounds has she finished d. Mark ‘X’ at the point where Toshi will be after she ran 250 m. e. Mark ‘Y’ at the point where Toshi will be after she ran 500 m. running around her track? Mark her position as ‘C’. Reprint 2025-26 133 Ganita Prakash | Grade 6 Deep Dive: In races, usually there is a common finish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A’ and the runner on the outer track as ‘B’. Take a rough sheet of paper or a sheet of newspaper. Make a few random shapes by cutting the paper in different ways. Estimate the total length of the boundaries of each shape then use a scale or measuring tape to measure and verify the perimeter for each shape.  Estimate and Verify f. Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’. Common Finishing Line 150 m 100 m Chapter 6_Perimeter and Area.indd 134 13-08-2024 16:16:31 134 Akshi says that the perimeter of this triangle shape is 9 units. Toshi says it can’t be 9 units and the perimeter will be more than 9 units. What do you think? The perimeter is 9 units. Reprint 2025-26 No, it will be more than 9 units. This figure has lines of two different unit lengths. Measure the lengths of a red line and a blue line; are they same? We will call the red lines—straight lines and the blue lines—diagonal lines. So, the perimeter of this triangle is 6 straight units + 3 diagonal units. We can write this in a short form as: 6s + 3d units. Write the perimeters of the figures below in terms of straight and diagonal units. Perimeter of a regular polygon Like squares, closed figures that have all sides and all angles equal are called regular polygons. We studied the sequence of regular polygons as ‘Shape Sequence’ #1 in Chapter 1. Examples of regular polygons are the equilateral triangle (where all three sides and all three angles are equal), regular pentagon (where all five sides and all five angles are equal), etc. Perimeter and Area Chapter 6_Perimeter and Area.indd 135 14-08-2024 14:56:30 Perimeter of an equilateral triangle We know that for any triangle its perimeter is sum of all three sides. Using this understanding, we can find the perimeter of an equilateral triangle. Perimeter of an equilateral triangle = AB + BC + AC = AB + AB + AB = 3 times length of one side. Perimeter of an equilateral triangle = 3 × length of a side. What is a similarity between a square and an equilateral triangle? Reprint 2025-26 B C A 135 Ganita Prakash | Grade 6 Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons. Split and rejoin A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown into two equal pieces. These two pieces are joined in different ways. Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below. For example, the arrangement a. has a perimeter of 28 cm. Discuss more about regular polygons and encourage students to come up with a general formula for the perimeter of a regular polygon. a. 6 cm 6 cm 2 cm Teacher’s Note 6 cm 4 cm Chapter 6_Perimeter and Area.indd 136 13-08-2024 16:16:31 136 Arrange the two pieces to form a figure with a perimeter of 22 cm. b. c. d. 2 cm Reprint 2025-26 2 cm 2 cm 3 cm 6.2 Area We have studied the areas of closed figures (regular and irregular) in previous grades. Let us recall some key points. The amount of region enclosed by a closed figure is called its area. In previous grades, we arrived at the formula for the area of a rectangle and a square using square grid paper. Do you remember? Example: A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted. Solution Length of the floor = 5 m. Width of the floor = 4 m. Area of the floor = length × width = 5 m × 4 m = 20 sq m. Area of a square = ________ Area of a rectangle = _______ Let’s see some real-life problems related to these ideas. Help students in recalling the method of finding the area of a rectangle and a square using grid papers. Provide square grid papers to students and let them come up with the formula. Teacher’s Note Perimeter and Area Chapter 6_Perimeter and Area.indd 137 13-08-2024 16:16:31 Therefore, the area of the floor that is not carpeted is: area of the floor minus the area of the floor laid with carpet = 20 sq m – 9 sq m = 11 sq m. Example: Four square flower beds each of side 4 m are in four corners on a piece of land 12 m long and 10 m wide. Find the area of the remaining part of the land. Length of the square carpet = 3 m. Area of the carpet = length × length = 3 m × 3 m = 9 sq m. Hence, the area of the floor laid with carpet is 9 sq m. Reprint 2025-26 137 Ganita Prakash | Grade 6 Solution Length of the land (l) = 12 m. Therefore, the area of the remaining part of the land is: area of the complete land minus the area of all four flower beds = 120 sq m – 64 sq m = 56 sq m. Width of land (w) = 10 m. Area of the whole land = l × w = 12 m × 10 m = 120 sq m. The sidelength of each of the four square flower beds is (s) = 4 m. Area of one flower bed = s × s = 4 m × 4 m = 16 sq m. Hence, the area of the four flower beds = 4 × 16 sq m = 64 sq m.  Figure it Out 1. The area of a rectangular garden 25 m long is 300 sq m. What is 2. What is the cost of tiling a rectangular plot of land 500 m long and 3. A rectangular coconut grove is 100 m long and 50 m wide. If each 4. By splitting the following figures into rectangles, find their areas the width of the garden? 200 m wide at the rate of ` 8 per hundred sq m? coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove? (all measures are given in metres). Chapter 6_Perimeter and Area.indd 138 13-08-2024 16:16:31 138 a. b. 4 4 3 2 3 2 3 1 Reprint 2025-26 3 1 1 5 3 2 Hint: In the tangram pieces, by placing the shapes over each other, we can find out that Shapes A and B have the same area, Shapes C and E have the same area. You would have also figured out that Shape D can be exactly covered using Shapes C and E, which means Shape D has twice the area of Shape C or shape E, etc.  Figure it Out Cut out the tangram pieces given at the end of your textbook. 1. Explore and figure out how many pieces have the same area. 2. How many times bigger is Shape D as compared to Shape C? What 3. Which shape has more area: Shape D or F? Give reasons for your 4. Which shape has more area: Shape F or G? Give reasons for your 5. What is the area of Shape A as compared to Shape G? Is it twice as is the relationship between Shapes C, D and E? answer. answer. big? Four times as big? A G F E C B D Perimeter and Area Chapter 6_Perimeter and Area.indd 139 13-08-2024 16:16:31 6. Can you now figure out the area of the big square formed with all 7. Arrange these 7 pieces to form a rectangle. What will be the area 8. Are the perimeters of the square and the rectangle formed from seven pieces in terms of the area of Shape C? of this rectangle in terms of the area of Shape C now? Give reasons for your answer. these 7 pieces different or the same? Give an explanation for your answer. Reprint 2025-26 139 Ganita Prakash | Grade 6 Look at the figures below and guess which one of them has a larger area. We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit × 1 unit or 1 square unit. To estimate the area, we can trace the shape onto a piece of transparent paper and place the same on a piece of squared or graph paper and then follow the below conventions— 1. The area of one full small square of the squared or graph paper is taken as 1 sq unit. 2. Ignore portions of the area that are less than half a square. 3. If more than half of a square is in a region, just count it as 1 sq unit. 4. If exactly half the square is counted, take its area as 1 2 sq unit. a. b. Chapter 6_Perimeter and Area.indd 140 13-08-2024 16:16:31 140   Find the area of the following figures. Reprint 2025-26 Let’s Explore! Why is area generally measured using squares? Draw a circle on a graph sheet with diameter (breadth) of length 3. Count the squares and use them to estimate the area of the circular region. As you can see, circles can’t be packed tightly without gaps in between. So, it is difficult to get an accurate measurement of area using circles as units. Here, the same rectangle is packed in two different ways with circles—the first one has 42 circles and the second one has 44 circles. Try using different shapes (triangle and rectangle) to fill the given space (without overlaps and gaps) and find out the merits associated with using a square shape to find the area rather than another shape. List out the points that make a square the best shape to use to measure area. 1. Find the area (in square metres) of the floor outside of the corridor. Perimeter and Area Why can’t we use circles instead of squares to find the area? Chapter 6_Perimeter and Area.indd 141 13-08-2024 16:16:32 Let’s Explore! On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 24 square units. a. Which rectangle has the greatest perimeter? b. Which rectangle has the least perimeter? 2. Find the area (in square metres) occupied by your school playground. Reprint 2025-26 Math Talk 141 Ganita Prakash | Grade 6 c. If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer. 6.3 Area of a Triangle Draw a rectangle on a piece of paper and draw one of its diagonals. Cut the rectangle along that diagonal and get two triangles. Check! whether the two triangles overlap each other exactly. Do they have the same area? Try this with more rectangles having different dimensions. You can check this for a square as well. Can you draw any inferences from this exercise? Please write it here. Now, see the figures below. Is the area of the blue rectangle more or less than the area of the yellow triangle? Or is it the same? Why? Chapter 6_Perimeter and Area.indd 142 13-08-2024 16:16:32 142 Can you see some relationship between the blue rectangle and the yellow triangle and their areas? Write the relationship here. Draw suitable triangles on grid paper to verify your inferences and relationships observed in the above exercises. Help students in articulating their inferences and in defining the relationships they have observed in their own words, gradually leading to a common statement for whole classroom. Recall the definition of a diagonal in the classroom. Reprint 2025-26 Teacher’s Note Use your understanding from previous grades to calculate the area of any closed figure using grid paper and— 1. Find the area of blue triangle BAD. __________ 2. Find the area of red triangle ABE. ___________ Area of rectangle ABCD = ________________ So, the area of triangle BAD is half of the area of the rectangle ABCD. Both the red and blue triangles have the same area but they look very different. D E C A F B Perimeter and Area Chapter 6_Perimeter and Area.indd 143 13-08-2024 16:16:33 Here, the area of triangle AEF = half of the area of rectangle AFED. Similarly, the area of triangle BEF = half of the area of rectangle BFEC. Thus, the area of triangle ABE = half of the area of rectangle AFED +half of the area of rectangle BFEC Area of triangle ABE = Area of triangle AEF + Area of triangle BEF. What about triangle ABE? Reprint 2025-26 There are two halves of two different rectangles. 143 Ganita Prakash | Grade 6 = half of the sum of the areas of the rectangles AFED and BFEC = half of the area of rectangle ABCD. Conclusion ____________________________________________________ ____________________________________________________ Figure it Out 1. Find the areas of the figures below by dividing them into rectangles and triangles. a b c Chapter 6_Perimeter and Area.indd 144 13-08-2024 16:16:33 144 e d Reprint 2025-26 Making it ‘More’ or ‘Less’ Observe these two figures. Is there any similarity or difference between the two? Using 9 unit squares (having an area of 9 sq units), we have made figures with two different perimeters—the first figure has a perimeter of 12 units and the second has a perimeter of 20 units. Arrange or draw different figures with 9 sq units to get other perimeters. Each square should align with at least one other square on at least one side completely and together all squares should form a single connected figure with no holes. Using 9 unit squares, solve the following. 1. What is the smallest perimeter possible? 2. What is the largest perimeter possible? 3. Make a figure with a perimeter of 18 units. Perimeter and Area Chapter 6_Perimeter and Area.indd 145 13-08-2024 16:16:33 Let’s do something tricky now! We have a figure below having perimeter 24 units. Without calculating all over again, observe, think and find out what will be the change in the perimeter if a new square is attached as shown on the right. 4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning? Reprint 2025-26 145 Ganita Prakash | Grade 6 Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: a) increases; b) decreases; c) stays the same? Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What do you notice? Hall Area = ____ Small Bedroom (15 ft × ____ ft) Area = 180 sq ft Master Bedroom (15 ft × 15 ft) Area=225 sq ft (5 ft × 1 0 f t) Kitchen (15 ft × 12 ft) Area = 180 sq ft To ile t Utility (____ ft × ____ ft) Area = ____ 30 ft Chapter 6_Perimeter and Area.indd 146 13-08-2024 16:16:33 146 Some of the measurements are given. a. Find the missing measurements. b. Find out the area of his house. Garden (____ ft × ____ ft) Area = ____ Reprint 2025-26 Parking (____ ft × ____ ft) Area = ____ Now, find out the missing dimensions and area of Sharan’s home. Below is the plan: Some of the measurements are given. Master Bedroom (12 ft × 15 ft) Area=180 sq ft Ar ea = _ __ _ Small Bedroom (12 ft × 10 ft) Area = ____ a. Find the missing measurements. Toilet (____ ft × ____ ft) Area = ____ 42 ft Hall (23 ft × ____ ft) Area = ____ Kitchen (18 ft × 10 ft) Area = 180 sq ft Perimeter and Area Ar ea = 7 0 s q f t (__ __ ft × _ __ _ f t) (__ __ ft × _ __ _ f t) Ut ili ty En tr an ce Chapter 6_Perimeter and Area.indd 147 13-08-2024 16:16:33 What are the dimensions of all the different rooms in Sharan’s house? Compare the areas and perimeters of Sharan’s house and Charan’s house. b. Find out the area of his house. Reprint 2025-26 147 Ganita Prakash | Grade 6 In each figure, find the missing value of either the length of a side or the area of a region. a. c. Area Maze Puzzles 13 sq cm 15 sq cm ? sq cm 3 cm 26 sq cm ? sq cm b. d. 2 cm 3 cm ? cm 10 sq cm 10 sq cm ? sq cm 2 cm 3 cm Chapter 6_Perimeter and Area.indd 148 13-08-2024 16:16:33 148 15 cm 42 sq cm 60 sq cm 5 cm 6 cm 38 sq cm Reprint 2025-26 5 cm 4 cm 18 sq cm Figure it Out 1. Give the dimensions of a rectangle whose area is the sum of the 2. The area of a rectangular garden that is 50 m long is 1000 sq m. 3. The floor of a room is 5 m long and 4 m wide. A square carpet 4. Four flower beds having sides 2 m long and 1 m wide are dug at 5. Shape A has an area of 18 square units and Shape B has an area 6. On a page in your book, draw a rectangular border that is 1 cm 7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle 8. A square piece of paper is folded in half. The square is then cut the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn? areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m. Find the width of the garden. whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted. of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions. from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border? inside it, without touching the outer rectangle that occupies exactly half the area. Perimeter and Area Chapter 6_Perimeter and Area.indd 149 14-08-2024 14:57:42 a. The area of each rectangle is larger than the area of the square. b. The perimeter of the square is greater than the perimeters of c. The perimeters of both the rectangles added together is always 11 2 times the perimeter of the square. d. The area of the square is always three times as large as the areas of both rectangles added together. into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here? both the rectangles added together. Reprint 2025-26 149 Ganita Prakash | Grade 6 The perimeter of a polygon is the sum of the lengths of all its sides. a. The perimeter of a rectangle is twice the sum of its length and width. b. The perimeter of a square is four times the length of any one of its sides. The area of a closed figure is the measure of the region enclosed by the figure. Area is generally measured in square units. The area of a rectangle is its length times its width. The area of a square is the length of any one of its sides multiplied by itself. Two closed figures can have the same area with different perimeters, or the same perimeter with different areas. Areas of regions can be estimated (or even determined exactly) by breaking up such regions into unit squares, or into more generalshaped rectangles and triangles whose areas can be calculated. Summary Chapter 6_Perimeter and Area.indd 150 13-08-2024 16:16:34 150 Reprint 2025-26 Section 6.1 Page no. 132 Q.1. Find the missing terms: Ans. (a) length = 5 cm Q.2. A rectangle having side lengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square? Ans. Length of a side of square = 4 cm Q.3. Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm, respectively? Ans. Length of the third side of triangle = 21 cm Q.4. What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs Rs.40 per metre? Figure it Out a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ? b. Perimeter of a square = 20 cm; side of a length = ? c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ? (b) length of a side = 5 cm (c) breadth = 3cm CHAPTER 6 — SOLUTIONS Perimeter and Area Ans. P = 2 × (150+120) = 540; So, 540 × 40 = Rs. 21,600 Q.5. A piece of string is 36 cm long. What will be the length of each side, if it is used to form: Ans. (a) Length of each side of square = 9 cm a. A square, b. A triangle with all sides of equal length, and c. A hexagon (a six sided closed figure) with sides of equal length? (b) Length of each side of regular triangle = 12 cm (c) Length of each side of regular hexagon = 6 cm [1] Q.6. A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed? Ans. Perimeter of the field = 780 m Section 6.1 Page No. 133 Q.1. Find out the total distance Akshi has covered in 5 rounds. Ans. Perimeter of the outer track = 220 m Q.2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? Ans. Perimeter of the inner track = 180 m Deep Dive Ans. From point A to flag: 100 + 100 + 100 + 50 Figure it out Page No. 134 Total length of the rope needed = 2340 m Total distance Akshi has covered in 5 rounds = 1100 m Total distance Toshi has covered in 7 rounds = 1260 m Toshi ran longer distance From point B to flag: 125 + 150 + 75 = 350 m Section 6.1 Page no. 134 Ans. The perimeter will be more than 9 units as the length of a diagonal of a square is always greater than the sides of the square. Q. Akshi says that the perimeter of this triangle shape is 9 units. Toshi says it can’t be 9 units and the perimeter will be more than 9 units. What do you think? = 350 m [2] P.135 Ans. The perimeters of the figures: 8s + 2d, 4s + 6d, 12s + 6d, 18s + 6d. Ans. In general perimeter of regular polygons = Number of sides × Length of a side. Section 6.2 Page No. 138 Q.1. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden? Ans. Width of the garden = 300 25 = 12 m. Q.2. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs.8 per hundred sq. m? Ans. Cost of tiling = Rs. 8000 Q.3. A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove? Ans. 200. Q.4. By splitting the following figures into rectangles, find their areas (all measures are given in meters): Q. Write the perimeters of the figures below in terms of straight and diagonal units. Figure it out Q. Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons. Ans. (a) 28 sq. m. Page No. 140 Q. Find the area of the following figures. Ans. 4 sq. units, 9 sq. units, 10 sq. units, 11 sq. units. (b) 9 sq. m. [3] Let’s Explore! Page No. 141 Section 6.3 Figure it Out Page No. 144 Q.1. Find the areas of the figures below by dividing them into rectangles. Ans. a. 24 sq. units. b. 30 sq. units. c. 48 sq. units d. 16 sq. units e. 12 sq. units. Page No. 145 Q. Using 9 unit squares, solve the following. Ans. 1. The smallest possible perimeter is 12 cm obtained from a square of side 3 cm 1. What is the smallest perimeter possible? 2. What is the largest perimeter possible? 3. Make a figure with a perimeter of 18 units. 4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning? 2. The largest possible perimeter is 20 cm. obtained from a rectangle of length 9 cm and breadth 1 cm. 3. 4. Yes, except for the smallest perimeter (12 sq. units). [4] Page No. 146 Q. Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What do you notice? Ans. a. Page No. 147 Q. Now, find out the missing dimensions and area of Sharan’s home. Below is the plan: Some of the measurements are given. a. Find the missing measurements. b. Find out the area of his house. Some of the measurements are given. a. Find the missing measurements. b. Find out the area of his house. b. Area of his house = 35ft × 30ft =1050 sq ft i. Small bedroom 15ft × 12ft Area = 180 sq ft ii. Utility 15ft × 3ft Area = 45 sq ft iii. Hall 20ft × 12ft Area = 240 sq ft iv. Parking 15ft × 3ft Area = 45 sq ft v. Garden 20ft × 3ft Area = 60 sq ft Ans. a. Dimensions of Sharan’s house. What are the dimensions of all the different rooms in Sharan’s house? Compare the areas and perimeters of Sharan’s house and Charan’s house. i. Utility 7ft × 10 ft Area = 70 sq ft ii. Hall 23ft × 15 ft Area = 345 sq ft iii. Entrance 7 ft × 15ft Area = 105 sq ft iv. Small bedroom 12ft × 10ft Area = 120 sq ft v. Toilet 5ft × 10 ft [5] Page No. 148 Area Maze Puzzles Ans. Section 6.3 Area of Charan’s house = 35ft × 30 ft = 1050 sq ft Area of Sharan’s house = 42 ft × 25ft = 1050 sq ft Area of both the houses are equal. Now, Perimeter of Charan’s house = 130 ft Perimeter of Sharan’s house = 134 ft Perimeter of Sharan’s house is greater than the Perimeter of Charan’s house. In each fig. find the missing value of either the length of the side or the area of the region. a. 30 sq cm. b. 9 sq cm. c. 16 sq cm. d. 5 cm. b. Area of his house = 42 ft × 25 ft = 1050 sq ft Area = 50 sq ft Page No. 149 Figure it Out Q.1. Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m. Ans. Possible dimensions are 16 m and 4 m Or 32 m and 2m Or 8 m and 8 m Q.2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden? Ans. 20 m. Q.3. The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted. Ans. Area of floor not carpeted = 11 sq. m. [6] Q.4. Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn? Ans. Available area for lawn = 172 sq. m. Q.5. Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions. Ans. Possible dimensions of shape A are 6m and 3m; 2m and 9m; 18m and 1m Q.7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area. Ans. Area of outer rectangle = 96 sq. units. Q.8. A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here? Corresponding Perimeters =18 m, 22m, 38m respectively Possible dimensions of shape B are 5m and 4m; 10m and 2m; 20m and 1m Corresponding Perimeters =18m, 24m, 42m respectively Given P(A) > P(B) We may take P(A) = 22m or 38m and P(B) = 18 m and draw the figures accordingly. Area of inner rectangle = 48 sq. units. Draw the rectangles accordingly. Ans. Only C is true. a. The area of each rectangle is larger than the area of the square. b. The perimeter of the square is greater than the perimeters of both the rectangles added together. c. The perimeters of both the rectangles added together is always times the d. The area of the square is always three times as large as the areas of both rectangles added together. perimeter of the square. [7]" class_6,7,Fractions,ncert_books/class_6/Ganita_Prakash/fegp107.pdf,"7 Recall that when some whole number of things are shared equally among some number of people, fractions tell us how much each share is. Shabnam: Do you remember, if one roti is divided equally Mukta: Each child will get half a roti. Shabnam: The fraction ‘one half’ is written as Mukta: If one roti is equally shared among Fractions between two children, how much roti will each child get? 1 2 . We also sometimes read this as ‘one upon two.’ Chapter 7_Fractions.indd 151 13-08-2024 15:39:38 Shabnam: Each child’s share is 1 4 roti. Mukta: And which is more 1 2 roti or 1 4 roti? Shabnam: When 2 children share 1 roti equally, each child gets 1 2 roti. When 4 children share 1 roti equally, each child gets 1 4 roti. Since, in the second group more children share the 4 children, how much roti will one child get? Reprint 2025-26 Ganita Prakash | Grade 6 7.1 Fractional Units and Equal Shares Beni: Which fraction is greater— 1 5 or 1 9 ? Arvin: 9 is bigger than 5. So I would guess that 1 9 is greater than 1 5 . Am I right? Beni: No! That is a common mistake. Think of these fractions as shares. Arvin: If one roti is shared among 5 children, each one gets a share of 1 5 roti. If one roti is shared among 9 children, each one gets a share of 1 9 roti? Beni: Exactly! Now think again - which share is higher? Arvin: If I share with more people, I will get less. So, 1 9 < 1 5 . Beni: You got it! When one unit is divided into several equal parts, each part is called a fractional unit. These are all fractional units: 1 2 , 1 3 , 1 4 , 1 5 , 1 6 , …, 1 10 , …, 1 50 , …, 1 100 , etc. same one roti, each child gets a smaller share. So, 1 2 roti is more than 1 4 roti. Oh, so 1 100 is bigger than 1 200! 1 2 > 1 4 Chapter 7_Fractions.indd 152 13-08-2024 15:39:38 152 We also sometimes refer to fractional units as ‘unit fractions.’  Figure it Out 1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____ kg. 2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is ___ kg. 3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ____ glass of sugarcane juice. Fill in the blanks with fractions. Reprint 2025-26 Math Talk Find out and discuss the words for fractions that are used in the different languages spoken in your home, city, or state. Ask your grandparents, parents, teachers, and classmates what words they use for different fractions, such as for one and a half, three quarters, one and a quarter, half, quarter, and two and a half, and write them here: ___________ ___________ ___________ ___________ ___________ ___________ Knowledge from the past! Fractions have been used and named in India since ancient times. In the Rig Veda, the fraction 3 4 is referred to as tri-pada. This has the same meaning as the words for 3 4 in many Indian languages today, e.g., ʻteen paavʼ in colloquial Hindi and ‘mukkaal’ in Tamil. Indeed, words for fractions used today in many Indian languages go back to ancient times. 5. Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: One and a half, three quarters, one and a quarter, half, quarter, two and a half. 4. The big fish weighs 1 2 kg. The small one weighs 1 4 kg. Together they weigh ____ kg. Fractions Math Talk Chapter 7_Fractions.indd 153 13-08-2024 15:39:38 Write your answer here. Reprint 2025-26 153 Ganita Prakash | Grade 6 7.2 Fractional Units as Parts of a Whole The picture shows a whole chikki. A whole chikki A picture of the chikki broken into 2 pieces is shown below. How much of the original chikki is each piece? We can see that the bigger piece has 3 pieces of 1 4 chikki in it. So, we can measure the bigger piece using the fractional unit 1 4. We see that the bigger piece is 3 4 chikki. 1 4 Chapter 7_Fractions.indd 154 12-12-2024 11:07:23 154 A whole chikki cut into 6 equal pieces. By dividing the whole chikki into 6 equal parts in different ways, we get 1 6 chikki pieces of different shapes. Are they of the same size? 1 6 Reprint 2025-26 A whole chikki cut into 6 equal pieces in a different way. Math Talk 1 6 What is the fractional unit of chikki shown below? The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?  Figure it Out a. A whole chikki 1 3 b. c. d. We get this piece by breaking the chikki into 3 equal pieces. So this is 1 3 chikki. Fractions Chapter 7_Fractions.indd 155 13-08-2024 15:39:39 e. f. g. h. Reprint 2025-26 155 Ganita Prakash | Grade 6 7.3 Measuring Using Fractional Units Take a strip of paper. We consider this paper strip to be one unit long. Fold the strip into two equal parts and then open up the strip again. Taking the strip to be one unit in length, what are the lengths of the two new parts of the strip created by the crease? What will you get if you fold the previously-folded strip again into two equal parts? You will now get four equal parts. 1 2 1 2 1 Strip Paper Chapter 7_Fractions.indd 156 13-08-2024 15:39:39 156 Do it once more! Fill in the blank boxes. 1 4 2 times 1 4 = 2 4 3 times 1 4 = 3 4 Reprint 2025-26 4 times 1 4 = 4 4 Fractional quantities can be measured using fractional units. 2 times 1 8 4 times 1 8 = 6 times 1 8 8 times 1 8 8 8 = = 1 = = = = Fractions Chapter 7_Fractions.indd 157 13-08-2024 15:39:39 We can describe how much the quantity is by collecting together the fractional units. 1 2 = 1 times half Let us look at another example, 1 2 + 1 2 = 2 times half 1 2 + 1 2 + 1 2 = 3 times half Reprint 2025-26 1 2 + 1 2 + 1 2 + 1 2 = 4 times half Represents a full roti (whole) 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 5 times half 157 Ganita Prakash | Grade 6 a. 5 times 1 4 of a roti b. 9 times 1 4 of a roti 5. Match each fractional unit with the correct picture: Reading Fractions We usually read the fraction 3 4 as ‘three quarters’ or ‘three upon four’, but reading it as ‘3 times 1 4’ helps us to understand the size of the fraction because it clearly shows what the fractional unit is (1 4) and how many such fractional units (3) there are.  Figure it Out 1. Continue this table of 1 2 for 2 more steps. 2. Can you create a similar table for 1 4? 3. Make 1 3 using a paper strip. Can you use this to also make 1 6 ? 4. Draw a picture and write an addition statement as above to show: 1 3 1 5 1 8 1 6 Chapter 7_Fractions.indd 158 13-08-2024 15:39:40 158 Give several opportunities to the children to explore the idea of fractional units with different shapes like circles, squares, rectangles, triangles, etc. Recall what we call the top number and the bottom number of fractions. In the fraction 5 6 , 5 is the numerator and 6 is the denominator. Teacher’s Note Reprint 2025-26 7.4 Marking Fraction Lengths on the Number Line We have marked lengths equal to 1, 2, 3, … units on the number line. Now, let us try to mark lengths equal to fractions on the number line. What is the length of the blue line? Write the fraction that gives the length of the blue line in the box. The distance between 0 and 1 is one unit long. It is divided into two equal parts. So, the length of each part is 1 2 unit. So, this blue line is 1 2 unit long. Now, can you find the lengths of the various blue lines shown below? Fill in the boxes as well. 1. Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the blue line in the box or in your notebook. 0 1 2 0 1 2 1 3 Fractions Chapter 7_Fractions.indd 159 13-08-2024 15:39:40 2. Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the blue lines in the respective boxes or in your notebook. 3. Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook. 0 1 2 1 5 3 5 Reprint 2025-26 159 Ganita Prakash | Grade 6 1 2 0 1 2  Figure it Out 1. On a number line, draw lines of lengths 1 10 , 3 10 , and 4 5 . 2. Write five more fractions of your choice and mark them on the 3. How many fractions lie between 0 and 1? Think, discuss with 4. What is the length of the blue line and black line shown below? The 5. Write the fraction that gives the lengths of the black lines in the number line. your classmates, and write your answer. distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is 1 2 . So the blue line is 1 2 units long. Write the fraction that gives the length of the black line in the box. respective boxes. Math Talk Chapter 7_Fractions.indd 160 13-08-2024 15:39:40 160 0 1 1 2 5 2 5 3 5 4 5 Draw these lines on the board and ask the students to write the answers in their notebooks. Reprint 2025-26 Teacher’s Note 7.5 Mixed Fractions Fractions greater than one You marked some fractions on the number line earlier. Did you notice that the lengths of all the blue lines were less than one and the lengths of all the black lines were more than 1? Write down all the fractions you marked on the number line earlier. Now, let us classify these in two groups: Did you notice something common between the fractions that are greater than 1? smaller than the denominator, while in the fractions that are more than 1 unit, the numerator is larger than the denominator. We know that 3 2 , 5 2 and 7 2 are all greater than 1 unit. But can we see how many whole units they contain? In all the fractions that are less than 1 unit, the numerator is Lengths less than 1 unit Lengths more than 1 unit Fractions Chapter 7_Fractions.indd 161 13-08-2024 15:39:40 3 2 = 1 2 + 1 2 + 1 2 = 1 + 1 2 5 2 = 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 2 + 1 2 I know that 1 3 + 1 3 + 1 3 = 3 3 = 1. If I add one more 1 3, I will get more than 1 unit! So, 4 3 > 1. Reprint 2025-26 161 Ganita Prakash | Grade 6 Writing fractions greater than one as mixed numbers We saw that: 3 2 = 1 + 1 2 . We can write other fractions in a similar way. For example, 3 × 1 3 = 1  Figure it Out  Figure it Out 1. How many whole units are there in 7 2 ? 2. How many whole units are there in 4 3 and in 7 3 ? 1. Figure out the number of whole units in each of the following fractions: 4 3 = 1 3 + 1 3 + 1 3 + 1 3 = 1 + 1 3 . a. 8 3 b. 11 5 c. 9 4 We saw that Fraction Mixed number 8 3 = 2 + 2 3 This number is thus also called ‘two and two thirds’. We also write it as 2 2 3 . Math Talk Chapter 7_Fractions.indd 162 13-08-2024 15:39:40 162 A mixed number or mixed fraction contains a whole number (called the whole part) and a fraction that is less than 1 (called the fractional part). 2. Can all fractions greater than 1 be written as such mixed numbers? 3. Write the following fractions as mixed fractions (e.g., 9 2 = 4 1 2 ): a. 9 2 b. 9 5 c. 21 19 d. 47 9 e. 12 11 f. 19 6 Reprint 2025-26 Jaya: When I have 3 + 3 4 , this means 1 + 1 + 1 + 3 4 . I know So I get ( 1 4 + 1 4 + 1 4 + 1 4 ) + ( 1 4 + 1 4 + 1 4 + 1 4 ) + ( 1 4 + 1 4 + 1 4 + 1 4 ) + ( 1 4 + 1 4 + 1 4 ) = 15 4 . Therefore, (4 × 1 4 ) + (4 × 1 4 ) + (4 × 1 4 ) + (3 × 1 4 ) = 15 4 .  Figure it Out Write the following mixed numbers as fractions: a. 3 1 4 b. 7 2 3 c. 9 4 9 d. 3 1 6 e. 2 3 11 f. 3 9 10 1 = 1 4 + 1 4 + 1 4 + 1 4 . Can we write a mixed number (mixed fraction) as a regular fraction? Yes! I figured out a way to write a mixed number as a regular fraction! Fractions Math Talk Chapter 7_Fractions.indd 163 13-08-2024 15:39:40 7.6 Equivalent Fractions Using a fraction wall to find equal fractional lengths! In the previous section, you used paper folding to represent various fractions using fractional units. Let us do some more activities with the same paper strips. Reprint 2025-26 163 Ganita Prakash | Grade 6 These are ‘equivalent fractions’ that denote the same length, but they are expressed in terms of different fractional units. Now, check whether 1 3 and 2 6 are equivalent fractions or not, using paper strips. Make your own fraction wall using such strips as given in the picture below! 1 2 2 4 4 8 1 • Are the lengths 1 2 and 2 4 equal? • Are the lengths 2 4 and 4 8 equal? We can say that 1 2 = 2 4 = 4 8 . What do you observe? Chapter 7_Fractions.indd 164 13-08-2024 15:39:41 164 1. Are the lengths 1 2 and 3 6 equal? 2. Are 2 3 and 4 6 equivalent fractions? Why? 3. How many pieces of 4. How many pieces of length 1 6 will make a length of 1 3 ? Answer the following questions after looking at the fraction wall: length 1 6 will make a length of 1 2 ? Reprint 2025-26 1 6 2 6 3 6 4 6 5 6 6 6 1 5 2 5 3 5 4 5 5 5 1 4 2 4 3 4 4 4 1 3 2 3 3 3 1 UNIT 1 2 2 2 We can extend this idea to make a fraction wall up to the fractional unit 1 10 . (This fraction wall is given at the end of the book.)  Figure it Out 1. Are 3 6 , 4 8 , 5 10 equivalent fractions? Why? 2. Write two equivalent fractions for 2 6 . 3. 4 6 = = = = ............ (Write as many as you can) 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 9 2 8 3 8 4 8 5 8 6 8 7 8 8 8 2 7 3 7 4 7 5 7 6 7 1 7 2 6 3 6 4 6 5 6 6 6 2 5 3 5 4 5 5 5 1 UNIT 2 4 3 4 4 4 2 3 3 3 2 2 Fractions Chapter 7_Fractions.indd 165 13-08-2024 15:39:41 Understanding equivalent fractions using equal shares One roti was shared equally by four children. What fraction of the whole did each child get? The adjoining picture shows the division of a roti among four children. Fraction of roti each child got is 1 4 . Reprint 2025-26 The four shares must be equal to each other! 165 Ganita Prakash | Grade 6 You can also express this event through division facts, addition facts, and multiplication facts.  Figure it Out The division fact is 1 ÷ 4 = 1 4 . The addition fact is 1 = 1 4 + 1 4 + 1 4 + 1 4 . The multiplication fact is 1 = 4 × 1 4 . 1. Three rotis are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts. 2. Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts. Fraction of roti each child gets is ______. Division fact: Addition fact: Multiplication fact: Compare your picture and answers with your classmates! Chapter 7_Fractions.indd 166 13-08-2024 15:39:42 166 Now, if there are 10 children cakes will I need so that they get same amount of cake as 3. Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get? in my group, how many Anil? Reprint 2025-26 between 5 children, and another What if we put two such groups group again with 4 cakes and together? one group where 2 cakes are divided equally 10 children. Let us examine the shares of each child in the following situations. • 1 roti is divided equally between 2 children. • 2 rotis are divided equally among 4 children. • 3 rotis are divided equally among 6 children. Let us draw and share! Did you notice that in each situation the share of every child is the same? So, we can say that 1 2 = 2 4 = 3 6 . 1 roti is divided So, the share of each child is the same in both these situations! Group 1 Group 2 2 rotis are divided equally among 4 3 rotis are divided equally among 6 So, 2 5 = 4 10! Fractions Chapter 7_Fractions.indd 167 13-08-2024 15:39:44 equally between 2 Fractions where the shares are equal are called ‘equivalent fractions’. 1 2 1 2 2 4 2 4 2 4 2 4 3 6 3 6 3 6 3 6 3 6 3 6 Reprint 2025-26 167 Ganita Prakash | Grade 6 boxes here: Equally divide the rotis in the situations shown below and write down the share of each child. Are the shares in each of these cases the same? Why? equally among 2 rotis divided 3 children 2 3 2 3 2 3 So, 1 2 , 2 4 , and 3 6 are all equivalent fractions. Find some more fractions equivalent to 1 2 . Write them in the 2 3 is also called the simplest form of 4 6 . It is also the simplest form of 6 9 as well. equally among 4 rotis divided 6 children equally among 6 rotis divided 9 children Chapter 7_Fractions.indd 168 13-08-2024 15:39:45 168  Figure it Out Find the missing numbers: a. 5 glasses of juice shared equally among 4 friends is the same as ____ glasses of juice shared equally among 8 friends. Do you notice anything about the relationship between the numerator and denominator in each of these fractions? So, 5 4 = 8 . b. 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ___ bags. So, 4 3 = 12 Reprint 2025-26 Math Talk boxes here: Equally divide the rotis in the situations shown below and write down the share of each child. Are the shares in each of these cases the same? Why? equally among 2 rotis divided 3 children So, 1 2 , 2 4 , and 3 6 are all equivalent fractions. Find some more fractions equivalent to 1 2 . Write them in the 2 3 2 3 2 3 2 3 is also called the simplest form of 4 6 . It is also the simplest form of 6 9 as well. equally among 4 rotis divided 6 children equally among 6 rotis divided 9 children 1 chikki divided between 2 children or 5 chikkis divided among 8 children. Mukta: So, we must compare 1 2 and 5 8 . Which is more? Shabnam: Well, we have seen that 1 2 = 4 8 ; and clearly 4 8 < 5 8 . So, the children for whom 5 chikkis is divided equally among 8 will get more than those children for whom 1 chikki is divided equally among 2. The children of the second group will get more chikki each. What about the following groups? In which group will each child get more? 1 chikki divided between 2 children or 4 chikkis divided among 7 children. Shabnam: The children of which group will get more chikki this time? Mukta: We must compare 1 7 and 4 7 . In which group will each child get more chikki? c. 7 rotis divided among 5 children is the same as ____ rotis divided among _____ children. So, 7 5 = . Fractions  Figure it Out Find the missing numbers: a. 5 glasses of juice shared equally among 4 friends is the same as ____ glasses of juice shared equally among 8 friends. Do you notice anything about the relationship between the So, 5 4 = 8 . b. 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ___ bags. numerator and denominator in each of these fractions? So, 4 3 = 12 Math Talk Chapter 7_Fractions.indd 169 13-08-2024 15:39:45 Now 1 × 4 2 × 4 = 4 8 so, 1 2 = 4 8 . Shabnam: But why did you multiply the numerator and denominator by 4 again? Mukta: You will see! When 4 chikkis are divided equally among 7 children, each one will get 4 7 chikki. When 4 chikkis are divided equally among 8 children, each one will get 4 8 chikki. So 4 7 > 4 8. Reprint 2025-26 169 Ganita Prakash | Grade 6 Suppose the number of children is kept the same, but the number of units that are being shared is increased? What can you say about each child’s share now? Why? Discuss how your reasoning explains Now, decide in which of the two groups will each child get a larger share: 1. Group 1 : 3 glasses of sugarcane juice divided equally among 4 children. Group 2: 7 glasses of sugarcane juice divided equally among 10 children. 1 5 < 2 5 , 3 7 < 4 7 , and 1 2 < 5 8 . Therefore, 4 7 > 4 8 and 4 8 = 1 2 , so 4 7 > 1 2 . Now I understood why you multiplied the numerator and denominator by 4. If the number of units that are shared is the same, but the number of children among whom the units are shared is more, then the share is less. Math Talk Chapter 7_Fractions.indd 170 13-08-2024 15:39:46 170 2. Group 1 : 4 glasses of sugarcane juice divided equally among 7 children. Group 2: 5 glasses of sugarcane juice divided equally among 7 children. Which groups were easier to compare? Why? Shabnam: To compare the first two groups, we have to find fractions equivalent to the fractions 3 4 and 7 10. Mukta: How about 6 8 = 3 4 and 21 30 = 7 10? Reprint 2025-26 children is same, it is easier When the number of to compare, isn’t it? Shabnam: There is a condition. The fractional unit used for the two Mukta: Okay, so let us start making equivalent fractions then: 3 4 = 6 8 = 9 12 = 12 16 = 15 20 … But when do I stop? Shabnam: Got it! How about we go on till 4 × 10 = 40. Mukta: You mean the product of the two denominators? Sounds good! We have 3 4 and 7 10. The product of the two denominators (4 and 10) is 40. 3 4 = 6 8 = 9 12 = 12 16 = 15 20 = 18 24 = … = 27 36 = 30 40 . But notice that 15 20 and 14 20 also had the same denominator! 7 10 = 14 20 = 21 30 = 28 40. fractions have to be the same! Like 2 6 and 3 6 both use the same fractional unit 1 6 (i.e., the denominators are the same). But 6 8 and 21 30 do not use the same fractional units (they have different denominators). same fractional units for each Yes! We just needed to get the fraction. Go till we reach the denominator 40. Fractions Chapter 7_Fractions.indd 171 13-08-2024 15:39:46 Shabnam: So, fractions equivalent to 3 4 and 7 10 with the same fractional Since clearly 30 40 > 28 40, we conclude that 3 4 > 7 10. unit (same denominators) are 30 40 and 28 40, or 15 20 and 14 20. Reprint 2025-26 171 Ganita Prakash | Grade 6 fractional units are the same. Expressing a fraction in lowest terms (or in its simplest form) In any fraction, if its numerator and denominator have no common factor except 1, then the fraction is said to be in lowest terms or in its simplest form. In other words, a fraction is said to be in lowest terms if its numerator and denominator are as small as possible. Any fraction can be expressed in lowest terms by finding an equivalent fraction whose numerator and denominator are as small as possible. Let’s see how to express fractions in lowest terms. Example: Is the fraction 16 20 in lowest terms? No, 4 is a common factor of 16 and 20. Let us reduce 16 20 to lowest terms. We know that both 16 (numerator) and 20 (denominator) are divisible by 4. So, 16 ÷ 4 20 ÷ 4 = 4 5. Now, there is no common factor between 4 and 5. Hence, 16 20 expressed in lowest terms is 4 5. So, 4 5 is called the simplest form of 16 20, since 4 and 5 have no common factor other than 1. Find equivalent fractions for the given pairs of fractions such that the a. 7 2 and 3 5 b. 8 3 and 5 6 c. 3 4 and 3 5 d. 6 7 and 8 5 e. 9 4 and 5 2 f. 1 10 and 2 9 g. 8 3 and 11 4 h. 13 6 and 1 9 Chapter 7_Fractions.indd 172 13-08-2024 15:39:46 172 numerator and denominator by the lowest terms by dividing both the Any fraction can be converted to highest common factor between them. Reprint 2025-26 Expressing a fraction in lowest terms can also be done in steps. Suppose we want to express 36 60 in lowest terms. First, we notice that both the numerator and denominator are even. So, we divide both by 2, and see that 36 60 = 18 30. Both the numerator and denominator are even again, so we can divide them each by 2 again; we get 18 30 = 9 15. We now notice that 9 and 15 are both multiples of 3, so we divide both by 3 to get 9 15 = 3 5 . Now, 3 and 5 have no common factor other than 1, so, 36 60 in lowest terms is 3 5 . Alternatively, we could have noticed that in 36 60 , both the numerator and denominator are multiples of 12 : we see that 36 = 3 × 12 and 60 = 5 × 12. Therefore, we could have concluded that 36 60 = 3 5 straight away. Either method works and will give the same answer! But sometimes it can be easier to go in steps. Figure it Out Express the following fractions in lowest terms: a. 17 51 b. 64 144 e. 126 147 d. 525 112 Fractions Chapter 7_Fractions.indd 173 12-12-2024 11:07:23 7.7 Comparing Fractions Which is greater, 4 5 or 7 9? It can be difficult to compare two such fractions directly. However, we know how to find fractions equivalent to two fractions with the same denominator. Let us see how we can use it: 4 5 = 4×9 5×9 = 36 45 7 9 = 7×5 9×5 = 35 45. 45 is a common multiple of 5 and 9, so we can use 45 as a common denominator. Reprint 2025-26 173 Ganita Prakash | Grade 6 Clearly, 36 45 > 35 45 So, 4 5 > 7 9 ! 63 is a common multiple of 9 and 21. We can then write: Clearly, 49 63 < 51 63 . So, 7 9 < 17 21! Let’s Summarise! Steps to compare the sizes of two or more given fractions: Step 1: Change the given fractions to equivalent fractions so that they all are expressed with the same denominator or same fractional unit. Step 2: Now, compare the equivalent fractions by simply comparing the numerators, i.e., the number of fractional units each has. 7 9 = 7×7 9×7 = 49 63 , 17 21 = 17×3 21×3 = 51 63.  Figure it Out Let us try this for another pair: 7 9 and 17 21. 1. Compare the following fractions and justify your answers: Chapter 7_Fractions.indd 174 13-08-2024 15:39:46 174 a. 7 10 , 11 15 , 2 5 b. 19 24 , 5 6 , 7 12 a. 25 16 , 7 8, 13 4 , 17 32 b. 3 4 , 12 5 , 7 12 , 5 4 2. Write the following fractions in ascending order. 3. Write the following fractions in descending order. a. 8 3 , 5 2 b. 4 9 , 3 7 c. 7 10 , 9 14 d. 12 5 , 8 5 e. 9 4 , 5 2 Reprint 2025-26 7.8 Addition and Subtraction of Fractions Meena’s father made some chikki. Meena ate 1 2 of it and her younger brother ate 1 4 of it. How much of the total chikki did Meena and her brother eat together? We can arrive at the answer by visualising it. Let us take a piece of chikki and divide it into two halves first like this. Meena ate 1 2 of it as shown in the picture. Let us now divide the remaining half into two further halves as shown. Each of these pieces is 1 4 of the whole chikki. Meena’s brother ate 1 4 of the whole chikki, as is shown in the picture. Meena ate Meena ate Fractions Chapter 7_Fractions.indd 175 13-08-2024 15:39:47 The total chikki eaten is 1 2 (by Meena) and 1 4 (by her brother) The total chikki eaten = 1 2 + 1 4 How much of the total chikki is remaining? = 1 4 + 1 4 + 1 4 = 3 × 1 4 = 3 4. Brother ate Reprint 2025-26 Total chikki eaten 175 Ganita Prakash | Grade 6 Adding fractions with the same fractional unit or denominator Example: Find the sum of 2 5 and 1 5. Let us represent both using the rectangular strips. In both fractions, the fractional unit is the same 1 5 , so, each strip will be divided into 5 equal parts. So 2 5 will be represented as— And 1 5 will be represented as— Adding the two given fractions is the same as finding out the total number of shaded parts, each of which represent the same fractional unit 1 5 . In this case, the total number of shaded parts is 3. Since, each shaded part represents the fractional unit 1 5 , we see that the 3 shaded parts together represent the fraction 3 5. Chapter 7_Fractions.indd 176 13-08-2024 15:39:47 176 Therefore, 2 5 + 1 5 = 3 5? Example: Find the sum of 4 7 and 6 7. Let us represent both again using the rectangular strip model. Here in both fractions, the fractional unit is the same, i.e., 1 7 , so each strip will be divided into 7 equal parts. Then 4 7 will be represented as— Reprint 2025-26 and 6 7 will be represented as— shaded parts is 10, and each shaded part represents the fractional unit 1 7, so, the 10 shaded parts together represent the fraction 10 7 as seen here. Therefore, 4 7 + 6 7 = 10 7 = 1 + 3 7 = 1 3 7. In this case, the total number of unit, just add the number of fractional units from each with the same fractional While adding fractions fraction. Fractions Chapter 7_Fractions.indd 177 13-08-2024 15:39:47 Adding fractions with different fractional units or denominators Example: Find the sum of 1 4 and 1 3. To add fractions with different fractional units, first convert the fractions into equivalent fractions with the same denominator or Try adding 4 7 + 6 7 using a number line. Do you get the same answer? Reprint 2025-26 177 Ganita Prakash | Grade 6 fractional unit. In this case, the common denominator can be made 3 × 4 = 12, i.e., we can find equivalent fractions with fractional unit 1 12 . Let us write the equivalent fraction for each given fraction. Therefore, 1 4 + 1 3 = 3 12 + 4 12 = 7 12 . This method of addition, which works for adding any number of fractions, was first explicitly described in general by Brahmagupta in the year 628 CE! We will describe the history of the development of fractions in more detail later in the chapter. For now, we simply summarise the steps in Brahmagupta’s method for addition of fractions. Brahmagupta’s method for adding fractions 1. Find equivalent fractions so that the fractional unit is common for all fractions. This can be done by finding a common multiple of the denominators (e.g., the product of the denominators, or the smallest common multiple of the denominators). 2. Add these equivalent fractions with the same fractional units. This can be done by adding the numerators and keeping the same denominator. 1 4 = 1 × 3 4 × 3 = 3 12 , 1 3 = 1 × 4 3 × 4 = 4 12 . Now, 3 12 and 4 12 have the same fractional unit, i.e., 1 12 . Chapter 7_Fractions.indd 178 13-08-2024 15:39:47 178 Let us carry out another example of Brahmagupta’s method. Example: Find the sum of 2 3 and 1 5. The denominators of the given fractions are 3 and 5. The lowest common multiple of 3 and 5 is 15. Then we see that 3. Express the result in lowest terms if needed. 2 3 = 2 × 5 3 × 5 = 10 15 , 1 5 = 1 × 3 5 × 3 = 3 15 . Reprint 2025-26 Therefore, 2 3 + 1 5 = 10 15 + 3 15 = 13 15 . Example: Find the sum of 1 6 and 1 3 . The smallest common multiple of 6 and 3 is 6. 1 6 will remain 1 6 . 1 3 = 1 × 2 3 × 2 = 2 6 Therefore, 1 6 + 1 3 = 1 6 + 2 6 = 3 6 . The fraction 3 6 can now be re-expressed in lowest terms, if desired. This can be done by dividing both the numerator and denominator by 3 (the biggest common factor of 3 and 6): Therefore, 1 6 + 1 3 = 1 2. 3 6 = 3 ÷ 3 6 ÷ 3 = 1 2 .  Figure it Out 1. Add the following fractions using Brahmagupta’s method: a. 2 7 + 5 7 + 6 7 b. 3 4 + 1 3 c. 2 3 + 5 6 d. 2 3 + 2 7 e. 3 4 + 1 3 + 1 5 f. 2 3 + 4 5 g. 4 5 + 2 3 h. 3 5 + 5 8 i. 9 2 + 5 4 j. 8 3 + 2 7 Fractions Chapter 7_Fractions.indd 179 13-08-2024 15:39:47 2. Rahim mixes 2 3 litres of yellow paint with 3 4 litres of blue paint to make green paint. What is the volume of green paint he has made? 3. Geeta bought 2 5 meter of lace and Shamim bought 3 4 meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border? k. 3 4 + 1 3 + 1 5 l. 2 3 + 4 5 + 3 7 m. 9 2 + 5 4 + 7 6 Reprint 2025-26 179 Ganita Prakash | Grade 6 Subtraction of fractions with the same fractional unit or denominator Brahmagupta’s method also applies when subtracting fractions! Let us start with the problem of subtracting 4 7 from 6 7, i.e., what is 6 7 – 4 7? To solve this problem, we can again use the rectangular strips. In both fractions, the fractional unit is the same, i.e., 1 7. Let us first represent the bigger fraction using a rectangular strip model as shown: Each shaded part represents 1 7. Now, we need to subtract 4 7 . To do this let us remove 4 of the shaded parts: 6 7 Fractional parts to be removed. Chapter 7_Fractions.indd 180 13-08-2024 15:39:47 180 So, we are left with 2 shaded parts, i.e., 6 7 – 4 7 = 2 7. Try doing this same exercise using the number line. Reprint 2025-26 We can do this here directly because both fractions have the same fractional units. Subtraction of fractions with different fractional units or denominators Example: What is 3 4 – 2 3? As we already know the procedure for subtraction of fractions with the same fractional units, let us convert each of the given fractions into equivalent fractions with the same fractional units. and similarly, 2 3 = (2×4) (3×4) = 8 12 .  Figure it Out 3 4 = (3×3) (4×3) = 9 12 1. 5 8 – 3 8 2. 7 9 – 5 9 3. 10 27 – 1 27 multiply both the numerator and Yes! By doing this we can easily Think! Why did we choose to subtract the two fractions. Again! Why did we choose to multiply both the numerator and denominator denominator by 3? Fractions Chapter 7_Fractions.indd 181 13-08-2024 15:39:48 Therefore, 3 4 – 2 3 = 9 12 – 8 12 = 1 12 . Brahmagupta’s method for subtracting two fractions — 1. Convert the given fractions into equivalent fractions with the same fractional unit, i.e., the same denominator. 2. Carry out the subtraction of fractions having the same fractional units. This can be done by subtracting the numerators and keeping the same denominator. 3. Simplify the result into lowest terms if needed. Reprint 2025-26 here by 4? 181 Ganita Prakash | Grade 6 a. Jaya’s school is 7 10 km from her home. She takes an auto for 1 2 km from her home daily, and then walks the remaining b. Jeevika takes 10 3 minutes to take a complete round of the 7.9 A Pinch of History Do you know what a fraction was called in ancient India? It was called bhinna in Sanskrit, which means ‘broken’. It was also called bhaga or ansha meaning ‘part’ or ‘piece’. The way we write fractions today, globally, originated in India. In ancient Indian mathematical texts, such as the Bakshali manuscript (from around the year 300 CE), when they wanted to write 1 2, they wrote it as 1 2 which is indeed very similar to the way we write it today! This method of writing and working with fractions continued to be used in India for the next several centuries, including by Aryabhata (499 CE), Brahmagupta (628 CE), Sridharacharya (c. 750 CE), and Mahaviracharya (c. 850 CE), among others. The line segment between the numerator and denominator in ‘1 2’ and in other Figure it Out 1. Carry out the following subtractions using Brahmagupta’s method: 2. Subtract as indicated: 3. Solve the following problems: a. 8 15 – 3 15 b. 2 5 – 4 15 c. 5 6 – 4 9 d. 2 3 – 1 2 a. 13 4 from 10 3 b. 18 5 from 23 3 c. 29 7 from 45 7 distance to reach her school. How much does she walk daily to reach the school? park and her friend Namit takes 13 4 minutes to do the same. Who takes less time and by how much? Chapter 7_Fractions.indd 182 13-08-2024 15:39:48 182 Reprint 2025-26 fractions was later introduced by the Moroccan mathematician Al-Hassar (in the 12th century). Over the next few centuries the notation then spread to Europe and around the world. Fractions had also been used in other cultures such as the ancient Egyptian and Babylonian civilisations, but they primarily used only fractional units, that is, fractions with a 1 in the numerator. More general fractions were expressed as sums of fractional units, now called ‘Egyptian fractions’. Writing numbers as the sum of fractional units, e.g., 19 24 = 1 2 + 1 6 + 1 8, can be quite an art and leads to beautiful puzzles. We will consider one such puzzle below. General fractions (where the numerator is not necessarily 1) were first introduced in India, along with their rules of arithmetic operations like addition, subtraction, multiplication, and even division of fractions. The ancient Indian treatises called the ‘Sulbasutras’ shows that even during Vedic times, Indians had discovered the rules for operations with fractions. General rules and procedures for working with and computing with fractions were first codified formally and in a modern form by Brahmagupta. Brahmagupta’s methods for working with and computing with fractions are still what we use today. For example, Brahmagupta described how to add and subtract fractions as follows: “By the multiplication of the numerator and the denominator of each of the fractions by the other denominators, the fractions are reduced to a common denominator. Then, in case of addition, the numerators (obtained after the above reduction) are added. In case of subtraction, their difference is taken.’’ (Brahmagupta, Brahmasphuṭasiddhānta, Verse 12.2, 628 CE) The Indian concepts and methods involving fractions were transmitted to Europe via the Arabs over the next few centuries and they came into general use in Europe in around the 17th century and then spread worldwide. Fractions Chapter 7_Fractions.indd 183 13-08-2024 15:39:48 Reprint 2025-26 183 Ganita Prakash | Grade 6 Puzzle! It is easy to add up fractional units to obtain the sum 1, if one uses the same fractional unit, for example, However, can you think of a way to add fractional units that are all different to get 1? It is not possible to add two different fractional units to get 1. The reason is that 1 2 is the largest fractional unit, and 1 2 + 1 2 = 1. To get different fractional units, we would have to replace at We can try to look instead for a way to write 1 as the sum of three different fractional units. 1.  Can you find three different fractional units that add up to 1? It turns out there is only one solution to this problem (up to changing the order of the 3 fractions)! Can you find it? Try to find it before reading further. least one of the 1 2’s with some smaller fractional unit - but then the sum would be less than 1! Therefore, it is not possible for two different fractional units to add up to 1. 1 2 + 1 2 = 1, 1 3 + 1 3 + 1 3 = 1, 1 4 + 1 4 + 1 4 + 1 4 = 1, etc. Try This Chapter 7_Fractions.indd 184 14-08-2024 15:06:07 184 1 3 + 1 3 + 1 3 = 1. To get the fractional units to be different, we will have to increase at least one of the 1 3’s, and decrease at least one of the other 1 3’s to compensate for that increase. The only way to increase 1 3 to another fractional unit is to replace it by 1 2. So 1 2 must be one of the fractional units. Now 1 2 + 1 4 + 1 4 = 1. To get the fractional units to be different, we will have to increase one of the 1 4’s and decrease the other 1 4 to compensate for that increase. Now the only way to increase 1 4 to Here is a systematic way to find the solution. We know that Reprint 2025-26 another fractional unit, that is different from 1 2, is to replace it by 1 3. So two of the fractions must be 1 2 and 1 3! What must be third fraction then, so that the three fractions add up to 1? 2.  Can you find four different fractional units that add up to 1? It turns out that this problem has six solutions! Can you find at least one of them? Can you find them all? You can try using similar reasoning as in the cases of two and three fractional units—or find your own method! Once you find one solution, try to divide a circle into parts like in the figure above to visualise it! This explains why there is only one solution to the above problem. What if we look for four different fractional units that add up to 1? 1 2 + 1 3 + 1 6 = 1 Fractions Try This Chapter 7_Fractions.indd 185 14-08-2024 15:06:18 Reprint 2025-26 185 Ganita Prakash | Grade 6 Fraction as equal share: When a whole number of units is divided Fractional Units: When one whole basic unit is divided into equal Reading Fractions: In a fraction such as 5 6, 5 is called the numerator and 6 is called the denominator. Mixed fractions contain a whole number part and a fractional part. Number line: Fractions can be shown on a number line. Every fraction has a point associated with it on the number line. Equivalent Fractions: When two or more fractions represent the same share or number, they are called equivalent fractions. Lowest terms: A fraction whose numerator and denominator have no common factor other than 1 is said to be in lowest terms or in its simplest form. Brahmagupta’s method for adding fractions: When adding fractions, convert them into equivalent fractions with the same fractional unit (i.e., the same denominator), and then add the number of fractional units in each fraction to obtain the sum. This is accomplished by adding the numerators while keeping the same denominator. Brahmagupta’s method for subtracting fractions: When subtracting fractions, convert them into equivalent fractions with the same fractional unit (i.e., the same denominator), and then subtract the number of fractional units. This is accomplished by subtracting the numerators while keeping the same denominator. into equal parts and shared equally, a fraction results. parts, then each part is called a fractional unit. Chapter 7_Fractions.indd 186 13-08-2024 15:39:49 186 Reprint 2025-26 Page no. 152 Figure it out Q1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____kg. Ans. 1 3 Q2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is_______ kg. Ans. 1 4 Q3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ____ glass of sugarcane juice Ans. 3 4 Q4. The big fish weighs kg. The small one weighs kg. Together they weigh ____ kg. Ans. 3 4 Q5. Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: Section 7.1 One and a half, three quarters, one and a quarter, half, quarter, two and a half. CHAPTER 7 — SOLUTIONS Fractions Ans. 1 4 , 1 2 , 3 4 , 1 1 4 , 1 1 2 , 2 1 2 Page No. 154 Q. By dividing the whole chikki into 6 equal parts in different ways, we get 1/6 chikki pieces of different shapes. Are they of the same size? Ans. Yes, they are of the same size. Section 7.2 Page No. 155 Figure it out Q. The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece? Ans. a. 1 12 b. 1 4 c. 1 8 d. 1 6 e. 1 8 f. 1 6 g. 1 24 h. 1 24 Figure it out Page No. 158 Q1. continue this table of for 2 more steps. Ans. 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 6 times 1 2 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 7 times 1 2 Q2. Can you create a similar table for ? Ans. 1 4 = 1-time quarter Section 7.3 1 4 + 1 4 = 2 times quarter 1 4 + 1 4 + 1 4 = 3 times quarter 1 4 + 1 4 + 1 4 + 1 4 = 4 times quarter Try Further! Q4. Draw a picture and write an addition statement as above to show: a. 5 times of a roti b. 9 times of a roti Ans. a. 1 4 + 1 4 + 1 4 + 1 4 + 1 4 = 5 4 = 5 times quarter b. Q5. Match each fractional unit with the correct picture: Ans. 1 3 1 4 + 1 4 + 1 4 + 1 4 + 1 4 + 1 4 + 1 4 + 1 4 + 1 4 = 9 4 = 9 times quarter = 2 + 1 4 1 5 1 6 1 8 Page no. 159 Q1. Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the blue line in the box or in your notebook. Ans. 2 3 Q2. Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the blue lines in the respective boxed or in your notebook. Ans. 2 5 , 4 5 Section 7.4 Q3. Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook. Ans. 1 8 , 2 8 , 3 8 ,… Page no. 160 Figure it out Q1. On a number line, draw lines of lengths , , and . Ans. Q3. How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer. Ans. Uncountable number of fractions Q4. What is the length of the blue line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part Ans. 3 2 Q5. Write the fraction that gives the lengths of the black lines in the respective boxes. Ans. 6 5 , 7 5 , 8 5 , 9 5 0 is . So the blue line is units long. Write the fraction that gives the length of the black line in the box. 4 5 1 Page No. 162 Figure it out Q1. How many whole units are there in ? Ans. There are 3 whole units in 7 2 Q2. How many whole units are there in and in ? Ans. There is 1 whole unit in 4 3 and 2 whole units in 7 3 . Section 7.5 Figure it out Q1. Figure out the number of whole units in each of the following fractions: a. b. c. Ans. a. 2 Q2. Can all fractions greater than 1 be written as such mixed numbers? Ans. No. For example: 8 4 = 2 cannot be written as a mixed number. Q3. Write the following fractions as mixed fractions (e.g., = ): a. b. c. d. e. f. Ans. a. 9 2 = 4 1 2 b. 2 c. 2 b. 9 5 = 1 4 5 c. 21 19 = 1 2 19 d. 47 9 = 52 9 e. 12 11 = 1 1 11 f. 19 6 = 31 6 Page No. 163 Figure it out Q1. Write the following mixed numbers as fractions: a. 3 b. 7 c. 9 d. 3 e. 2 f. 3 Ans. a. 3 1 4 = 3 + 1 4 = 1 + 1 + 1 + 1 4 = (1 4 + 1 4 + 1 4 + 1 4 ) + (1 4 + 1 4 + 1 4 + 1 4 ) + (1 4 + 1 4 + 1 4 + 1 4 ) + 1 4 Page no. 164 Q1. Are the lengths and equal? Ans. Yes. Q2. Are and equivalent fractions? Why? Ans. Yes, since they are of equal length which can be seen in fractional wall also. Q3. How many pieces of length will make a length of ? Ans. 3 pieces. Section 7.6 = (4 × 1 4 ) + (4 × 1 4 ) + (4 × 1 4 ) + 1 4 = 13 4 b. 23 3 c. 85 9 d. 19 6 e. 25 11 f. 39 10 Q4. How many pieces of length will make a length of ? Ans. 2 pieces. Page no. 165 Figure it out Q1. Are , , and are equivalent fractions? Why? Ans. Yes. In the fraction wall their lengths can be seen to be equal. Q2. Write two equivalent fractions for . Ans. Two equivalent fractions for 2 6 are 1 3 and 3 9 . (Try for other equivalent fractions also) Q3 . = = = = = ……. (Write as many as you can) Ans. 4 6 = 2 3 = 6 9 = 8 12 = 10 15 = ……. Page no. 166 Figure it out Q1. Three rotis are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts. Fraction of roti each child gets is ______. Division fact: Addition fact: Multiplication fact: Compare your picture and answers with your classmates! Ans. Each child gets 3 4 roti. Q2. Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts. Division fact: 3 4 = 3 4 Addition fact: 3 = 3 4 + 3 4 + 3 4 + 3 4 Multiplication fact: 3 = 4 × 3 4 gets 1 2 Ans. 2 Rotis equally shared by 4 children. Each child roti. Q3. Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get? Ans. Anil would get 2 5 cake. Division fact :2 4 = 2 4 = 1 2 Addition fact: 2 = 1 2 + 1 2 + 1 2 + 1 2 Multiplication fact: 2 = 4 × 1 2 Page no. 168 Section 7.6 Figure it out Q1. Find the missing numbers: Ans. 10 Ans. 9 Ans. One of the choices is 14, 10 (Try for other options also.) Page no. 170 Q. Suppose the number of children is kept the same, but the number of units that are being shared is increased? What can you say about each child’s share now? Why? Discuss how your reasoning explains Section 7.6 a. 5 glasses of juice shared equally among 4 friends is the same as ____ glasses of juice shared equally among 8 friends. b. 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in __ bags. c. 7 rotis divided among 5 children is the same as____rotis divided among _____ children. So, = So, = So, = Ans. Each child will have larger share now. If the number of units to be shared increase with same number of children then each child will have larger share. Q. Now, decide in which of the two groups will each child get a larger share: 1. Group 1: 3 glasses of sugarcane juice divided equally among 4 children. 2. Group 1: 4 glasses of sugarcane juice divided equally among 7 children. < , < and < . Group 2: 7 glasses of sugarcane juice divided equally among 10 children. Group 2: 5 glasses of sugarcane juice divided equally among 7 children. Which groups were easier to compare? Why? Ans. Each child’s share in group 1 = 4 7 Each child’s share in group 2 = 5 7 And 5 7 > 4 7 The groups in second part were easier to compare because the number of children is same. Page no. 172 Q. Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. Ans. Section 7.6 a. 35 10 and 6 10 b. 16 6 and 5 6 . 15 20 and 12 20 . 30 35 and 56 35 . 9 4 and 10 4 a. and b. and c. and d. and e. and f. and g. and h. and . 9 90 and 20 90 . 32 12 and 33 12 ℎ. 39 18 and 2 18 Page no. 173 Figure it out Q. Express the following fractions in lowest terms: a. b. c. d. Ans. a. 1 3 . 4 9 Page no. 174 Figure it out Q1. Compare the following fractions and justify your answers: Ans. Section 7.6 Section 7.7 a. , b. , c. , d. , e. , c. 6 7 d. 75 16 a. 8 3 > 5 2 b. 4 9 > 3 7 . 7 10 > 9 14 d. 12 5 > 8 5 . 9 4 < 5 2 Q2. Write the following fractions in ascending order. Ans. a. 2 5 < 7 10 < 11 15 Q3. Write the following fractions in descending order. Ans. a. 13 4 > 25 16 > 7 8 > 17 32 Page no. 177 Q. Try adding + using a number line. Do you get the same answer? Ans. 4 7 + 6 7 4 7 + 6 7 = 10 7 = 1 + 3 7 Yes, using a number line, the answer is same. a. , , b. , , Section 7.8 b. 7 12 < 19 24 < 5 6 a. , , , b. , , , b. 12 5 > 5 4 > 3 4 > 7 12 Page no. 179 Figure it out Q1. Add the following fractions using Brahmagupta’s method: Section 7.8 Ans. Q2. Rahim mixes litres of yellow paint with litres of blue paint to make green paint. What is the volume of green paint he has made? a. 13 7 b. 13 12 c. 9 6 = 93 6 3 = 3 2 d. 20 21 e. 77 60 f. 22 15 g. 22 15 h. 49 40 i. 23 4 j. 62 21 k. 77 60 l. 199 105 m. 83 12 Ans. 1 5 12 litres Q3. Geeta bought meter of lace and Shamim bought meter of the same lace to put a Ans. Total length of the lace = 1 3 20 m complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border? Yes. Figure it out (Page no. 181) Q1. 5 8 - 3 8 Ans. 2 8 (= 1 4 , in lowest terms) Q2. - Ans. 2 9 Q3. - Ans. 9 27 (= 1 3 , in lowest terms) Page no. 181 Figure it out Ans. 1. 1 4 2. 2 9 3 1 3 Figure it out Q1. Carry out the following subtractions using Brahmagupta’s method: Section 7.8 Section 7.8 a. 8 15 - 3 15 b. 2 5 - 4 15 c. 5 6 - 4 9 d. 2 3 - 1 2 Ans. a. 1 3 b. 2 15 c. 7 18 d. 1 6 Q2. Subtract as indicated: Ans. a. 1 12 Q3. Solve the following problems: Ans. a. 1 5 Km b. 61 15 c. 16 7 a. from b. from c. from a. Jaya’s school is km from her home. She takes an auto for km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school? b. Jeevika takes minutes to take a complete round of the park and her friend b. Namit takes less time than Jeevika, by 1 12 minutes Namit takes minutes to do the same. Who takes less time and by how much?" class_6,8,Playing with Constructions,ncert_books/class_6/Ganita_Prakash/fegp108.pdf,"8 8.1 Artwork Observe the following figures and try drawing them freehand. Playing with Constructions Chapter 8_Playing with Constructions.indd 187 13-08-2024 16:38:44 Reprint 2025-26 Fig. 8.1 Ganita Prakash | Grade 6 Now, arm yourself with a ruler and a compass. Let us explore if we can draw these figures with these tools and get familiar with a compass. Observe the way a compass is made. What can one draw with the compass? Explore! Do you know what curves are? They are any shapes that can be drawn on paper with a pencil, and include straight lines, circles and other figures as shown below: Mark a point ‘P’ in your notebook. Then, mark as many points as possible, in different directions, that are 4 cm away from P. Think: Imagine marking all the points of 4 cm distance from the point P. How would they look? Chapter 8_Playing with Constructions.indd 188 13-08-2024 16:38:44 188 Try to draw it and verify if it is correct by taking some points on the curve and checking if their distances from P are indeed 4 cm. Explore, if you have not already done so, and see if a compass can be used for this purpose. You can start by marking a few points of distance 4 cm from P using the compass. How can this be done? Reprint 2025-26 4 cm P Q 4 cm R You will have to open up the compass against a ruler (see Fig. 8.2) such that the distance between the tip of the compass and the pencil is 4 cm. Now, try to get the full curve. Hint: Keep the point of the compass fixed moving only the pencil. What is the shape of the curve? It is a circle! Take a point on the circle. What will be its distance from P—equal to 4 cm, less than 4 cm or greater than 4 cm? Similarly, what will be the distance between P and another point on the circle? As shown in the figure, the point P is called the centre of the circle and the distance between the centre and any point on the circle is called the radius of the circle. Playing with Constructions Chapter 8_Playing with Constructions.indd 189 13-08-2024 16:38:44 Having explored the use of a compass, go ahead and recreate the images in Fig. 8.1. Can you make the figures look as good as the figures shown there? Try again if you want to! Also, has the use of instruments made the construction easier? Now try constructing the following figures. radius 10 2 3 4 5 6 7 Fig. 8.2 Reprint 2025-26 P centre 189 Ganita Prakash | Grade 6 1. A Person Construct How will you draw this? This figure has two components. You might have figured out a way of drawing the first part. For drawing the second part, see this. Chapter 8_Playing with Constructions.indd 190 13-08-2024 16:38:44 190 The challenge here is to find out where to place the tip of the compass and the radius to be taken for drawing this curve. You can fix a radius in the compass and try placing the tip of the Reprint 2025-26 2. Wavy Wave compass in different locations to see which point works for getting the curve. Use your estimate where to keep the tip. Construct this. As the length of the central line is not specified, we can take it to be of any length. Let us take AB to be the central line such that the length of AB is 8 cm. We write this as AB = 8 cm. Here, the first wave is drawn as a half circle. ? A X ? 8 cm Playing with Constructions B Chapter 8_Playing with Constructions.indd 191 13-08-2024 16:38:44 Figure it Out 1. What radius should be taken in the compass to get this half circle? 2. Take a central line of a different length and try to draw the wave 3. Try to recreate the figure where the waves are smaller than a half circle (as appearing in the neck of the figure, ‘A Person’). The challenge here is to get both the waves to be identical. This may be tricky! What should be the length of AX? on it. Reprint 2025-26 Try This 191 Ganita Prakash | Grade 6 3. Eyes 8.2 Squares and Rectangles Now, let us look at some basic figures having straight lines in their boundary. Make other artwork of your choice with a ruler and a compass. How do you draw these eyes with a compass? For a hint, go to the end of the chapter. Fig. 8.3 Chapter 8_Playing with Constructions.indd 192 13-08-2024 16:38:45 192 What shapes are these? Yes, these are our familiar squares and rectangles. But what makes them squares and rectangles? Consider this rectangle ABCD. The points A, B, C and D are the corners of the rectangle. Lines AB, BC, CD and DA are its sides. Its angles are ∠A, ∠B, ∠C and ∠D. The blue sides AB and CD are called opposite sides, as they lie opposite to each other. Likewise, AD and BC is the other pair of opposite sides. Reprint 2025-26 A D C Fig. 8.4 B Recall that, in a rectangle: R1) The opposite sides are equal in length, and R2) All the angles are 90°. As in the case of rectangles, the corners and sides are defined for a square in the same manner. A square satisfies the following two properties: S1) All the sides are equal, and S2) All the angles are 90o . See the rectangle in Fig. 8.4 and the name given to it: ABCD. This rectangle can also be named in other ways—BCDA, CDAB, DABC, ADCB, DCBA, CBAD and BADC. So, can a rectangle be named using any combination of the labels around its corners? No! For example, it cannot be named ABDC or ACBD. Can you see what names are allowed and what names are not? In a valid name, the corners occur in an order of travel around the rectangle, starting from any corner. Which of the following is not a name for this square? 1. PQSR 2. SPQR 3. RSPQ 4. QRSP S R Q P Playing with Constructions Chapter 8_Playing with Constructions.indd 193 13-08-2024 16:39:01 Rotated Squares and Rectangles Here is a square piece of paper having all its sides equal in length and all angles equal to 90°. It is rotated as shown in the figure. Is it still a square? Let us check if the rotated paper still satisfies the properties of a square. • Are all the sides still equal? Yes. • Are all the angles still 90°? Yes. Reprint 2025-26 193 Ganita Prakash | Grade 6 Rotating a square does not change its lengths and angles. Therefore, this rotated figure satisfies both the properties of a square and so, it is a square. By the same reasoning, a rotated rectangle is still a rectangle. Figure it Out 1. Draw the rectangle and four squares configuration (shown in 2. Identify if there are any squares in this collection. Fig. 8.3) on a dot paper. What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates. Use measurements if needed. A B C D Chapter 8_Playing with Constructions.indd 194 13-08-2024 16:39:01 194  Think: Is it possible to reason out if the sides are equal or not, and if the angles are right or not without using any measuring instruments in the above figure? Can we do this by only looking at the position of corners in the dot grid? 3. Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that their corners are on the dots. Verify if the squares and rectangles that you have drawn satisfy their respective properties. Reprint 2025-26 8.3 Constructing Squares and Rectangles Now, let us start constructing squares and rectangles. How would you construct a square with a side of 6 cm? For help, you can see the following figures. A square PQRS of side length 6 cm is constructed. 6 cm S P 6 cm Q P 6 cm Q Step 3 Method 1 Step 1 Step 2 Mark a point to draw a perpendicular to PQ through P. Playing with Constructions Chapter 8_Playing with Constructions.indd 195 13-08-2024 16:39:01 Method 2 This can also be done using a compass. P 6 cm Q 90° Reprint 2025-26 Mark S on the perpendicular such that PS = 6 cm using a ruler. 195 Ganita Prakash | Grade 6 R Step 4 Draw a perpendicular to line segment PQ through Q. Step 5 If we had used the compass, then the next point can easily be marked using it! 6 cm S R Can you see why PS should be 6 cm long? P 6 cm Q S Chapter 8_Playing with Constructions.indd 196 13-08-2024 16:39:02 196 6 cm 6 cm P Q 6 cm Step 6 R S P 6 cm Q 90º 90º Reprint 2025-26 How long is the side RS and what are the measures of ∠R and ∠S? P Q 5 mm 8.4 An Exploration in Rectangles Construct a rectangle ABCD with AB = 7 cm and BC = 4 cm. Imagine X to be a point that can be moved anywhere along the side AD. Similarly, imagine Y to be a point that can be moved anywhere along the side BC. Note that X can also be placed on the end point A or D. Similarly, Y can also be placed on the end point B or C. 4 cm X A Construct 1. Draw a rectangle with sides of length 4 cm and 6 cm. After 2. Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if 3. Is it possible to construct a 4-sided figure in which— • all the angles are equal to 90º but • opposite sides are not equal? drawing, check if it satisfies both the rectangle properties. it satisfies both the rectangle properties. B C Y 2 cm 1 cm A X Playing with Constructions B Try This Chapter 8_Playing with Constructions.indd 197 13-08-2024 16:39:02 D = X A D B = Y C Y 1 cm C Reprint 2025-26 A = X D D B = Y C 197 Ganita Prakash | Grade 6 At which positions will the points X and Y be at their closest? When do you think they will be the farthest? What does your intuition say? Discuss with your classmates. Now, verify your guesses by placing the points X and Y on the sides and measure how near or far they are. The distance between X and Y can be obtained by measuring the length of the line XY. How does the minimum distance between the points X and Y compare to the length of AB? Change the positions of X and Y to check if there are other positions where they are at their nearest or farthest. You could construct multiple copies of the rectangle and try out various positions of X and Y. How will you keep track of the lengths XY for different positions of X and Y? Here is one way of doing it. Suppose here are some of the positions of X and Y that you have considered: • When X is 2 cm away from A and Y is 4 cm away from B, XY = ___ cm ___ mm and so on. Is there a shorthand way of writing it down? In all the sentences, only the position of X, Y and the length XY changes. So we could write this as: • When X is 5 mm away from A and Y is 3 cm away from B, XY = ___ cm __ mm • When X is 1 cm away from A and Y is 1 cm away from B, XY = ___ cm ___ mm Math Talk Chapter 8_Playing with Constructions.indd 198 13-08-2024 16:39:02 198 Distance of X from A Distance of Y from B Length of XY Reprint 2025-26 Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively? For example, as in the cases like these:  In each of these cases, observe 1. how the length XY compares to that of AB and 2. the shape of the 4-sided figure ABYX. How does the farthest distance between X and Y compare with the length of AC? BD? and so on. Construct Breaking Rectangles Construct a rectangle that can be divided into 3 identical squares as shown in the figure. Distance of X from A Distance of Y from B Length of XY 1 cm 5 mm 1 cm 5 mm 5 mm 5 mm 1 cm 1 cm Playing with Constructions Chapter 8_Playing with Constructions.indd 199 14-08-2024 15:08:49 If this seem difficult, let us simplify the problem. What about constructing a rectangle that can be divided into two identical squares? Can you try it? Solution Explore It is wise to first plan and then construct. But how do we plan? Can you think of a way? Reprint 2025-26 199 Ganita Prakash | Grade 6 One way is to visualise the final figure by drawing a rough diagram of it. What can we infer from this figure? Can you identify the equal sides? Since, the two squares are identical, AB = BC and FE = ED Since ABEF and BCDE are squares, all the sides in each of the squares are equal. This is written as— AF = AB = BE = FE BE = BC = CD = ED So, all the shorter lines are equal! A convention is followed to represent equal sides. It is done by putting a ‘|’ on the line. Refer to the rough figure. Using this analysis, can you try constructing it? Remember, all that was asked for is a rectangle that can be divided into two identical squares and with no measurements imposed. To draw the rectangle ACDF, one could assign any length to AF. For example, if we assign AF = 4 cm, then what must the length of AC be? A B C F E D Chapter 8_Playing with Constructions.indd 200 13-08-2024 16:39:02 200 In fact, one could proceed by drawing AF without even measuring its length using a ruler. We could then construct a line perpendicular to AF that is long enough to contain the other side. As, AB = AF, we need to somehow transfer the length of AF Explore: Can the rectangle now be completed? Reprint 2025-26 With this idea, try constructing a rectangle that can be divided into three identical squares. Give the lengths of the sides of a rectangle that cannot be divided into— • two identical squares; • three identical squares. to get the point B. How do we do it without a ruler? Can it be done using a compass? Observe, how the length of AF is measured using a compass. Use it to mark out the points B and C, and complete the rectangle. Construct A F F A Playing with Constructions B Chapter 8_Playing with Constructions.indd 201 13-08-2024 16:39:02 1. A Square within a Rectangle Construct a rectangle of sides 8 cm and 4 cm. How will you construct a square inside, as shown in the figure, such that the centre of the square is the same as the centre of the rectangle? 4 cm Reprint 2025-26 8 cm 201 Ganita Prakash | Grade 6 Hint: Draw a rough figure. What will be the sidelength of the square? What will be the distance between the corners of the square and the outer rectangle? 2. Falling Squares Now, try this. 4 cm 4 cm Each is a Square of side 4 cm 4 cm Square of side 5 cm Square of side 3 cm Make sure that the squares are aligned the way they are shown. Chapter 8_Playing with Constructions.indd 202 13-08-2024 16:39:02 202 3. Shadings Construct this. Choose measurements of your choice. Note that the larger 4-sided figure is a square and so are the smaller ones. Reprint 2025-26 Square of side 7 cm Hint: Think where the centre of the circle should be. 4. Square with a Hole 5. Square with more Holes 6. Square with Curves This is a square with 8 cm sidelengths. Hint: Think where the tip of the compass can be placed to get all the 4 arcs to bulge uniformly from each of the sides. Try it out! Observe that the circular hole is the same as the centre of the square. Playing with Constructions Try This Chapter 8_Playing with Constructions.indd 203 13-08-2024 16:39:02 8.5 Exploring Diagonals of Rectangles and Squares Consider a rectangle PQRS. Join PR and QS. These two lines are called the diagonals of the rectangle. Compare the lengths of the diagonals. First predict the answer. Then construct a Reprint 2025-26 S P R a b c d e f g h Q 203 Ganita Prakash | Grade 6 rectangle marking the points as shown and measure the diagonals. In rectangle PQRS, the right angles at P and R are referred to as opposite angles. The other pair of opposite angles are the right angles at Q and S. Observe that a diagonal divides each of the pair of opposite angles into two smaller angles. In the figure, the diagonal PR divides angle R into two smaller angles which we simply call g and h. The diagonal also divides angle P into c and d. Are g and h equal? Are c and d equal? First predict the answers, and then measure the angles. What do you observe? Identify pairs of angles that are equal. How should the rectangle be constructed so that the diagonal divides the opposite angles into equal parts? How will you record your observations? First, identify the parameters that need to be tracked. They are the sides of the rectangle and the 8 angles formed by the two diagonals. Are there any other measurements that you would want to keep track of? Sides A B C D E F G H Explore Chapter 8_Playing with Constructions.indd 204 13-08-2024 16:39:02 204 In your experimentation, did you consider the case when all four sides of the rectangle are equal? That is, did you consider the case of a square? See what happens in this special case!   What general laws did you observe with respect to the angles and sides? Try to frame and discuss them with your classmates. How can one be sure if the laws that you have observed will always be true? Reprint 2025-26 Math Talk Construct Solution Let us start with a rough diagram. In what order should its parts be drawn? We will briefly sketch a possible order of construction. Step 1 1. Construct a rectangle in which one of the diagonals divides the opposite angles into 60° and 30°. D A B C Playing with Constructions Chapter 8_Playing with Constructions.indd 205 13-08-2024 16:39:02 AB is drawn with an arbitrary length. What is the next point that can be located? A B A B Reprint 2025-26 205 Ganita Prakash | Grade 6 Step 2 Step 3 We know the line on which D lies. Draw a line through A perpendicular to AB. A 60o B C B C Chapter 8_Playing with Constructions.indd 206 13-08-2024 16:39:03 206 Now ∠A is divided into two angles. One measures 60°. Check what the other angle is. There are at least two ways of finding the point D— • One uses the fact that all the angles of a rectangle are right angles. • The other uses the fact that opposite sides are equal. A Reprint 2025-26 60o Step 4 Method 1 Method 2 Draw a line perpendicular to BC at C to get the point D. C C D A 60o 30o B C D Playing with Constructions Chapter 8_Playing with Constructions.indd 207 13-08-2024 16:39:03 We have seen how to construct rectangles when their sides are given. But what do we do if a side and a diagonal is given? Using a compass, mark the point D such that AD = BC. Join CD to get the required rectangle. A A 60o 60o B B Reprint 2025-26 207 Ganita Prakash | Grade 6 Solution Step 1 Step 2 Draw a perpendicular to line DC at the point C. Let us call this line l. 2. Construct a rectangle where one of its sides is 5 cm and the length of a diagonal is 7 cm. Let us draw a rough diagram. Let us decide the steps of construction. Which line can be drawn first? The base CD measuring length 5 cm can be easily constructed. Next? D 5 cm C l A B D 5 cm C 7 cm Chapter 8_Playing with Constructions.indd 208 13-08-2024 16:39:03 208 This is easy as we know that this line is perpendicular to the base. The point B should be somewhere on this line l. We know that it is at a distance of 7 cm from the point D. How do we spot it? What else do we know about the position of B? D C Reprint 2025-26 5 cm One of the ways of marking B is by taking a ruler and trying to move it around to get a point on line l that is 7 cm from point D. However, this requires trial and error. There is another efficient method which doesn’t involve trial and error. For this, instead of trying to get that one required point of distance 7 cm from D, let us explore a way of getting all the points of distance 7 cm from D. Step 3 Method 1 We know what this shape is! D C l Playing with Constructions Chapter 8_Playing with Constructions.indd 209 13-08-2024 16:39:03 Construct a circle of radius 7 cm with point D as the centre. Can you spot the point B here? Remember that it is 7 cm away from point D and on the line l. Consider the point at which the circle and the line intersect. What is its distance from point D? If needed, check your figure. What do you observe? The point where the circle intersects the line l is the required point B. Reprint 2025-26 209 Ganita Prakash | Grade 6 Method 2 To locate the point B, was it necessary to draw the entire circle? We can see that only the arc near the line l is needed. So, the third step can also be done as shown in the figure below. Having marked the three points of the rectangle, we only need to complete it. Recall that we were in a similar situation in the previous problem also. We saw two methods of completing the rectangle from here. We could follow any one of those methods. Step 4 D C l A C B Chapter 8_Playing with Constructions.indd 210 13-08-2024 16:39:03 210 Construct perpendiculars to DC and BC passing through D and B, respectively. The point where these lines intersect is the fourth point A. Check if ABCD is indeed a rectangle satisfying properties R1 and R2. D Reprint 2025-26 90° 90° 8.6 Points Equidistant from Two Given Points  Construct  Construct House Recreate this figure. Note that all the lines forming the border of the house are of length 5 cm. Solution The first task is to identify in what sequence the lines and curve will have to be drawn. 1. Construct a rectangle in which one of the diagonals divides the 2. Construct a rectangle in which one of the diagonals divides the 3. Construct a rectangle one of whose sides is 4 cm and the diagonal 4. Construct a rectangle one of whose sides is 3 cm and the diagonal opposite angles into 50° and 40°. opposite angles into 45° and 45°. What do you observe about the sides? is of length 8 cm. is of length 7 cm. 5 cm Playing with Constructions B 5 cm 5 cm 1 cm A C 5 cm Chapter 8_Playing with Constructions.indd 211 13-08-2024 16:39:03 Step 1 5 cm 5 cm B D 2 cm Reprint 2025-26 1 cm 5 cm C E D 2 cm 5 cm E 211 Ganita Prakash | Grade 6 Can you complete the figure? Try! We need to locate the point A that is of distance 5 cm from the points B and C. You might have realised that this can be done using a ruler. However, this leads to a lot of trial and error. This construction can be further simplified. How? If you have guessed that this can be done by the use of compass, you are right! Go ahead and explore how the point A can be located without trial and error. There is a similarity between the problem of finding point A in this problem and point B at step 3 of the second solved example of the previous section (see page 209). Step 2 5 cm 5 cm B E 1 cm C Chapter 8_Playing with Constructions.indd 212 13-08-2024 16:39:03 212 Draw a curve that has all its points of 5 cm from the point B; the circle centred at B should be with 5 cm radius. Does this help in spotting the point A? Construct and explore in the figure. The point A can be located by finding the correct point on the circle that is of distance 5 cm from the point C. Again, this can be done using a ruler. But can we use a compass for this? Reprint 2025-26 D 2 c m 5 cm Step 3 Method 1 Take a radius of 5 cm in the compass and with C as the centre, draw a circle. Are you able to spot the point A? Check the figure in your notebook. What do you observe? See the point at which both the circles intersect. How far is it from the point B? How far is it from C? 5 cm 5 cm B D 2 c m 5 cm E 1 cm C Playing with Constructions Chapter 8_Playing with Constructions.indd 213 13-08-2024 16:39:03 Thus, this is the point A! Think Was it necessary to draw two full circles to get the point A? We only needed part of both the circles. Method 2 So the point A could have been obtained just by drawing arcs of radius 5 cm from points B and C. Reprint 2025-26 213 Ganita Prakash | Grade 6 Join A to B and A to C by straight lines. Having obtained point A, what remains is the construction of the remaining arc. How do we do it? Step 4 Can we use the fact that A is of distance 5 cm from both B and C? Take 5 cm radius in the compass and from A, draw the arc touching B and C as shown in the figure. 5 cm 5 cm B D 2 cm E 1 cm 5 cm A C Chapter 8_Playing with Constructions.indd 214 12-12-2024 11:09:41 214 The house is ready! 5 cm B D 5 cm 5 cm Reprint 2025-26 2 cm 5 cm 1 cm A C 5 cm E but is not a square? If such a figure exists, can you construct it? Hints A) Eyes (from 8.1 Artwork and Construct above (page no. 215). Part of the construction is shown earlier. Observe it carefully. You will see two horizontal lines drawn lightly. In geometric constructions, one often constructs supporting curves or figures that are not part of the given figure but help in constructing it. Construct 1. Construct a bigger house in which all the sides are of length 7 cm. 2. Try to recreate ‘A Person’, ‘Wavy Wave’, and ‘Eyes’ from the section 3. Is there a 4-sided figure in which all the sides are equal in length ‘Artworkʼ, using ideas involved in the ‘House’ construction. A Playing with Constructions Chapter 8_Playing with Constructions.indd 215 13-08-2024 16:39:03 The technique to draw the upper and the lower curves of the eye is the same as that used in the figure, ‘A Personʼ. Points A and B are the locations where the tip of the compass is placed when drawing the curves of the eye. Note that the upper curve and the lower curve should together form a symmetrical figure. For this to happen, where should these points A and B be placed? Make a good estimate. B Reprint 2025-26 215 Ganita Prakash | Grade 6 Try to get the eyes as symmetrical and identical as possible. This might need many trials. B) (From Construct above (page no. 211). For the purpose of construction, let us take the side lengths to be of 5 cm. Consider this figure. We need to identify only one more point to make this a 4-sided figure. That point, let us call it D, should be 5 cm from both B and C. How can such a point be found? Can any of the ideas used in the ‘House’ problem be used here? All the points of a circle are at the same distance from its centre. This distance is called the radius of the circle. A Summary 5 cm 5 cm C B Chapter 8_Playing with Constructions.indd 216 14-08-2024 15:11:23 216 A compass can be used to construct circles and their parts. A rough diagram can be useful in planning how to construct a given figure. A rectangle can be constructed given the lengths of its sides or that of one of its sides and a diagonal. Reprint 2025-26 Page no. 188 Section 8.1 Think: Imagine marking all the points of 4 cm distance from the point P. How would they look? Ans. On joining, they are forming a circle. Page no. 191 Section 8.1 Figure it out Q.1. What radius should be taken in the compass to get this half circle? What should be the length of AX? Ans. 2 cm. AX = 4 cm. Q.2. Take a central line of a different length and try to draw the wave on it. Ans. Q.3. Try to recreate the figure where the waves are smaller than a half circle (as appearing in the neck of the figure ‘A Person’). The challenge here is to get both the waves to be identical. This may be tricky! Ans. CHAPTER 8 — SOLUTIONS Playing with Constructions 4 cm Page no. 193 Section 8.2 Q. Which of the following is not a name for this square? 1. PQSR 2. SPQR [1] 3. RSPQ 4. QRSP Ans. PQSR is not a name for the given square. Page no. 194 Section 8.2 Figure it out Draw a rectangle & then leave one dot distance diagonally to place the four smaller squares. Q.2. Identify if there are any squares in this collection. Use measurements if needed. Ans. A is a square. Ans. Yes, In the above figure without measuring one can reason out for equal sides & right angles. Q.3. Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that their corners are on the dots. Verify if the squares and rectangles that you have drawn satisfy their respective properties. Ans. Think: Is it possible to reason out if the sides are equal or not, and if the angles are right or not without using any measuring instruments in the above figure? Can we do this by only looking at the position of corners in the dot grid? Yes, the position of the points on the grid paper makes it possible [2] Page no. 197 Section 8.3 Construct Q.1. Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it satisfies both the rectangle properties. Ans. A = B = C = D = 90 Q.2. Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both the rectangle properties. Ans. Q.3. Is it possible to construct a 4-sided figure in which— • all the angles are equal to 90º but • opposite sides are not equal? Ans. No. Page no. 198 Section 8.4 AB = CD = 4 cm & AD = BC = 6 cm. P = Q = R = S = 90 PQ = SR = 10 cm & PS = QR = 2 cm Ans. Distance of X from A Distance of Y from B Length of XY Distance of X from A Distance of Y from B Length of XY 5 mm 1 cm 2 cm Q. Is there a shorthand way of writing it down? In all the sentences, only the position of X, Y and the length XY changes. So we could write this as: 3 cm 1 cm 4 cm [3] 7.4 cm 7 cm 7.3 cm Page No. 199 5 cm 1 cm 4 cm And so on. Ans. Page no. 199 Q. In each of these cases, observe i. how the length XY compares to that of AB and ii. the shape of the 4-sided figure ABYX. Ans. i. XY = AB ii. ABYX is a rectangle Q. How does the farthest distance between X and Y compare with the length of AC? BD? Ans. The farthest distance between X and Y is equal to AC or BD. Construct Q. Construct a rectangle that can be divided into 3 identical squares as shown in the figure. Ans. Hint: Take the length of the rectangle three times its breadth. Distance of X from A Distance of Y from B Length of XY 5 mm 1 cm 1 cm 5 mm Distance of X from A Distance of Y from B Length of XY 5 mm 1 cm 1 cm 5 mm Q. Have you checked what happens to the length XY when X and Y are placed at the same distance away from A and B, respectively? For example, as in the cases like these: 5 mm 1 cm 1 cm 5 mm 7 cm 7 cm 7 cm Page no. 201 Give the lengths of the sides of a rectangle that cannot be divided into — • two identical squares; • three identical squares Ans. • Lengths of sides of a rectangle that cannot be divided into two identical squares: Length = 4 cm and Breadth = 2.5cm (Try for more!) • Length of sides of a rectangle that cannot be divided into three identical squares: Length = 7cm and Breadth = 2cm (Try for other possibilities.) [4] Page no. 201 Q.4. Square with a Hole Ans. The centre of the circle should be at the meeting point of two line segments connecting opposite vertices. Page no. 204 Section 8.5 Explore Q. How should the rectangle be constructed so that the diagonal divides the opposite angles into equal parts? Ans. The rectangle should be constructed in such a manner that the two adjacent sides are equal i.e. the rectangle becomes a square. Page no. 211 Construct Q.1. Construct a rectangle in which one of the diagonals divides the opposite angles into 50°and 40°. Ans. Q.2. Construct a rectangle in which one of the diagonals divides the opposite angles into 45° and 45°. What do you observe about the sides? Ans. [5] Q.3. Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm. Ans. Q.4. Construct a rectangle one of whose sides is 3 cm and the diagonal is of length 7 cm. Ans. [6] Page no. 215 Construct Q.1. Construct a bigger house in which all the sides are of length 7 cm. Ans. [7] Q.2. Try to recreate ‘A Person’, ‘Wavy Wave’ and ‘Eyes’ from the section Artwork, using ideas involved in the ‘House’ construction. Ans. Q.3. Is there a 4-sided figure in which all the sides are equal in length but is not a square? If such a figure exists, can you construct it? Ans. [8]" class_6,9,Symmetry,ncert_books/class_6/Ganita_Prakash/fegp109.pdf,"9 Look around you—you may find many objects that catch your attention. Some such things are shown below: SYMMETRY Flower Butterfly Chapter 9_Symmetry.indd 217 13-08-2024 17:05:22 There is something beautiful about the pictures above. The flower looks the same from many different angles. What about the butterfly? No doubt, the colours are very attractive. But what else about the butterfly appeals to you? In these pictures, it appears that some parts of the figure are repeated and these repetitions seem to occur in a definite pattern. Can you see what repeats in the beautiful rangoli figure? In the Rangoli Pinwheel Reprint 2025-26 Ganita Prakash | Grade 6 rangoli, the red petals come back onto themselves when the flower is rotated by 90˚ around the centre and so do the other parts of the rangoli. What about the pinwheel? Can you spot which pattern is repeating? Hint: Look at the hexagon first. Now, can you say what figure repeats along each side of the hexagon? What is the shape of the figure that is stuck to each side? Do you recognise it? How do these shapes move as you move along the boundary of the hexagon? What about the other pictures—what is it about those structures that appeals to you and what are the patterns in those structures that repeat? On the other hand, look at this picture of clouds. There is no such repetitive pattern. We can say that the first four figures are symmetrical and the last one is not symmetrical. A symmetry refers to a part or parts of a figure that are repeated in some definite pattern. Clouds Chapter 9_Symmetry.indd 218 13-08-2024 17:05:23 218 What are the symmetries that you see in these beautiful structures? Taj Mahal Gopuram Reprint 2025-26 9.1 Line of Symmetry Figure (a) shows the picture of a blue triangle with a dotted line. What if you fold the triangle along the dotted line? Yes, one half of the triangle covers the other half completely. These are called mirror halves! What about Figure (b) with the four puzzle pieces and a dotted line passing through the middle? Are they mirror halves? No, when we fold along the line, the left half does not exactly fit over the right half. A line that cuts a figure into two parts that exactly overlap when folded along that line is called a line of symmetry of the figure.  Figure it Out (a) (b) Symmetry Chapter 9_Symmetry.indd 219 13-08-2024 17:05:23 1. Do you see any line of symmetry in the figures at the start of the chapter? What about in the picture of the cloud? 2. For each of the following figures, identify the line(s) of symmetry if it exists. Reprint 2025-26 219 Ganita Prakash | Grade 6 Figures with more than one line of symmetry Does a square have only one line of symmetry? Take a square piece of paper. By folding, find all its lines of symmetry. Here are the different folds giving different lines of symmetry. • Fold the paper into half vertically. • Fold it again into half horizontally (i.e., you have folded it twice). Now open out the folds. Fold 3 Fold 4 Fold 1 Fold 2 Chapter 9_Symmetry.indd 220 13-08-2024 17:05:25 220 Again fold the square into half (for a third time now), but this time along a diagonal, as shown in the figure. Again, open it. Vertical Fold Reprint 2025-26 Horizontal Fold Fold it into half (for the fourth time), but this time along the other diagonal, as shown in the figure. Open out the fold. Is there any other way to fold the square so that the two halves overlap? How many lines of symmetry does the square shape have? Thus, figures can have multiple lines of symmetry. The figures below also have multiple lines of symmetry. Can you find them all? We saw that the diagonal of a square is also a line of symmetry. Let us take a rectangle that is not a square. Is its diagonal a line of symmetry? First, see the rectangle and answer this question. Then, take a rectangular piece of paper and check if the two parts overlap by folding it along its diagonal. What do you observe? Symmetry Chapter 9_Symmetry.indd 221 13-08-2024 17:05:25 Reflection So far we have been saying that when we fold a figure along a line of symmetry, the two parts overlap completely. We could also say that the part of the figure on one side of the line of symmetry is reflected by the line to the other side; similarly, the part of the figure on the other side of the line of symmetry is reflected to the first side! Let us understand this by labeling some points on the figure. The figure shows a square with its corners labeled A, B, C and D. Let us first consider the vertical line of symmetry. When we reflect Reprint 2025-26 221 Ganita Prakash | Grade 6 the square along this line, the points B, C on the right get reflected to the left side and occupy the positions occupied earlier by A, D. What happens to the points A, D? A occupies the position occupied by B and D that of C! What if we reflect along the diagonal from A to C? Where do points A, B, C and D go? What if we reflect along the horizontal line of symmetry? A figure that has a line or lines of symmetry is thus, also said to have reflection symmetry. Generating shapes having lines of symmetry So far we have seen symmetrical figures and asymmetrical figures. How does one generate such symmetrical figures? Let us explore this. A D B C Chapter 9_Symmetry.indd 222 13-08-2024 17:05:25 222 Ink Blot Devils You enjoyed doing this earlier in Class 5. Take a piece of paper. Fold it in half. Open the paper and spill a few drops of ink (or paint) on one half. Now press the halves together and then open the paper again. • What do you see? • Is the resulting figure symmetric? • If yes, where is the line of symmetry? • Is there any other line along which it can be folded to produce two identical parts? • Try making more such patterns. Reprint 2025-26 Paper Folding and Cutting Here is another way of making symmetric shapes! In these two figures, a sheet of paper is folded and a cut is made along the dotted line shown. Draw a sketch of how the paper will look when unfolded. Do you see a line of symmetry in this figure? What is it? Make different symmetric shapes by folding and cutting. There are more ways of folding and cutting pieces of paper to get symmetric shapes! Use thin rectangular coloured paper. Fold it several times and create some intricate patterns by cutting the paper, like the one shown here. Identify the lines of symmetry in the repeating design. Use such decorative paper cut-outs for festive occasions. Figure it Out Punching Game The fold is a line of symmetry. Punch holes at different locations of a folded square sheet of paper using a punching machine and create different symmetric patterns. Symmetry Chapter 9_Symmetry.indd 223 13-08-2024 17:05:26 Reprint 2025-26 223 Ganita Prakash | Grade 6 a. b. c. d. 1. In each of the following figures, a hole was punched in a folded square sheet of paper and then the paper was unfolded. Identify the line along which the paper was folded. Figure (d) was created by punching a single hole. How was the paper folded? 2. Given the line(s) of symmetry, find the other hole(s): 3. Here are some questions on paper cutting. Consider a vertical fold. We represent it this way: a. b. c. d. e. Chapter 9_Symmetry.indd 224 13-08-2024 17:05:27 224 Reprint 2025-26 Vertical Fold 4. After each of the following cuts, predict the shape of the hole when the paper is opened. After you have made your prediction, make the cutouts and verify your answer. a. b. c. Similarly, a horizontal fold is represented as follows: Horizontal Fold Symmetry Chapter 9_Symmetry.indd 225 13-08-2024 17:05:27 d. Reprint 2025-26 225 Ganita Prakash | Grade 6 5 Suppose you have to get each of these shapes with some folds and a single straight cut. How will you do it? 6. How many lines of symmetry do these shapes have? a. a. The hole in the centre is a square. b. The hole in the centre is a square. Note: For the above two questions, check if the 4-sided figures in the centre satisfy both the properties of a square. l Chapter 9_Symmetry.indd 226 13-08-2024 17:05:27 226 b. A triangle with equal sides and equal angles. Reprint 2025-26 7. Trace each figure and draw the lines of symmetry, if any: c. A hexagon with equal sides and equal angles. Symmetry Chapter 9_Symmetry.indd 227 13-08-2024 17:05:28 Reprint 2025-26 227 Ganita Prakash | Grade 6 10. Draw the following. In each case, the figure should contain at least one curved boundary. a. A figure with exactly one line of symmetry. b. A figure with exactly two lines of symmetry. c. A figure with exactly four lines of symmetry. 8. Find the lines of symmetry for the kolam below. 9. Draw the following. a. A triangle with exactly one line of symmetry. b. A triangle with exactly three lines of symmetry. c. A triangle with no line of symmetry. Is it possible to draw a triangle with exactly two lines of symmetry? Chapter 9_Symmetry.indd 228 14-08-2024 15:12:38 228 11. Copy the following on squared paper. Complete them so that the blue line is a line of symmetry. Problem (a) has been done for you. (d) (e) (f ) (a) (b) (c) Reprint 2025-26 12. Copy the following drawing on squared paper. Complete each one of them so that the resulting figure has the two blue lines as lines of symmetry. (d) (e) (f ) Hint: For (c) and (f), see if rotating the book helps! (a) (b) (c) (a) (b) (c) A B A B C C Symmetry Chapter 9_Symmetry.indd 229 13-08-2024 17:05:28 (d) (e) (f) D D Reprint 2025-26 E F E F 229 Ganita Prakash | Grade 6 9.2 Rotational Symmetry The paper windmill in the picture looks symmetrical but there is no line of symmetry! However, if you fold it, the two halves will not exactly overlap. On the other hand, if you rotate it by 90° about the red point at the centre, the windmill looks exactly the same. We say that the windmill has rotational symmetry. When talking of rotational symmetry, there is always a fixed point about which the object is rotated. This fixed point is called the centre of rotation. Will the windmill above look exactly the same when rotated through an angle of less than 90°? No! An angle through which a figure can be rotated to look exactly the same is called an angle of rotational symmetry, or just an angle of symmetry, for short. 13. Copy the following on a dot grid. For each figure draw two more lines to make a shape that has a line of symmetry. Chapter 9_Symmetry.indd 230 13-08-2024 17:05:28 230 Reprint 2025-26 180˚ 180˚ For the windmill, the angles of symmetry are 90° (quarter turn), 180° (half turn), 270° (three-quarter turn) and 360° (full turn). Observe that when any figure is rotated by 360°, it comes back to its original position, so 360° is always an angle of symmetry. Thus, we see that the windmill has 4 angles of symmetry. Do you know of any other shape that has exactly four angles of symmetry? How many angles of symmetry does a square have? How much rotation does it require to get the initial square? We get back a square overlapping with itself after 90° of rotation. This takes point A to the position of point B, point B to the position of point C, point C to the position of point D, and point D back to the position of point A. Do you know where to mark the centre of rotation? A B D C Imaginary reference line A D B C Square after rotation Line after rotation Initial position Symmetry Chapter 9_Symmetry.indd 231 13-08-2024 17:05:28 What are the other angles of symmetry? Line after rotation Initial position B A C D 180o C B D Reprint 2025-26 Line after rotation 270o 360o A B A Initial position C D 231 Ganita Prakash | Grade 6 Example: Find the angles of symmetry of the following strip. Solution: Let us rotate the strip in a clockwise direction about its centre. A rotation of 180° results in the figure above. Does this overlap with the original figure. No. Why? Another rotation through 180° from this position gives the original shape. This figure comes back to its original shape only after one complete rotation through 360°. So, we say that this figure does not have rotational symmetry. Rotational Symmetry of Figures with Radial Arms Consider this figure, a picture with 4 radial arms. How many angles of symmetry does it have? What are they? Note that the angle between adjacent central dotted lines is 90°. Can you change the angles between the radial arms so that the figure still has 4 angles of symmetry? Try drawing it. To check if the figure drawn indeed has 4 angles of symmetry, you could draw the figure on two different pieces of paper. Cut out the radial arms from one of the papers. Keep the figure on the paper fixed and rotate the cutout to check for rotational symmetry. How will you modify the figure above so that it has only two angles of symmetry? 90O Chapter 9_Symmetry.indd 232 13-08-2024 17:05:28 232 Reprint 2025-26 We have seen figures having 4 and 2 angles of symmetry. Can we get a figure having exactly 3 angles of symmetry? Can you use radial arms for this? Let us try with 3 radial arms as in the figure below. How many angles of symmetry does it have and what are they? Here is a figure with three radial arms. Here is one way: Symmetry Chapter 9_Symmetry.indd 233 13-08-2024 17:05:29 Trace and cut out a copy of this figure. By rotating the cutout over this figure determine its angles of rotation. We see that only a full turn or a rotation of 360° will bring the figure back into itself. So this figure does not have rotational symmetry as 360 degrees is its only angle of symmetry. However, can anything in the figure be changed to make it have 3 angles of symmetry? Reprint 2025-26 233 Ganita Prakash | Grade 6 Can it be done by changing the angles between the dotted lines? If a figure with three radial arms should have rotational symmetry, then a rotated version of it should overlap with the original. Here are rough diagrams of both of them. If these two figures must overlap, what can you tell about the angles? Observe that ∠A must overlap ∠B, ∠B must overlap ∠C and ∠C must overlap ∠A. So, ∠A = ∠B = ∠C. What must this angle be? We know that a full turn has 360 degrees. This is equally distributed amongst these three angles. So each angle must be 360° 3 = 120°. So, the radial arms figure with 3 arms shows rotational symmetry when the angle between the adjacent dotted lines is 120°. Use paper cutouts to verify this observation. Now how many angles of rotation does the figure have and what are they? C B A B A C Chapter 9_Symmetry.indd 234 13-08-2024 17:05:29 234 Initial position After 120° rotation After 240° (120° + 120°) rotation Note: The colours have been added to show the rotations. Reprint 2025-26 After 360° (120° + 120°+ 120°) rotation Can you draw a figure with radial arms that has a) exactly 5 angles of symmetry, b) 6 angles of symmetry? Also find the angles of symmetry in each case. Hint: Use 5 radial arms for the first case. What should the angle between two adjacent radial arms be? Consider a figure with radial arms having exactly 7 angles of symmetry. What will be its smallest angle of symmetry? Is the number of degrees a whole number in this case? If not, express it as a mixed fraction. Let us find the angles of symmetry for other kinds of figures. Let us explore more figures. 1. Find the angles of symmetry for the given figures about the point marked •. Figure it Out (a) (b) (c) Symmetry Chapter 9_Symmetry.indd 235 13-08-2024 17:05:29 2. Which of the following figures have more than one angle of symmetry? Reprint 2025-26 235 Ganita Prakash | Grade 6 Let us list down the angles of symmetry for all the cases above. • Angles of symmetry when there are exactly 2 of them: 180°, 360°. • Angles of symmetry when there are exactly 3 of them: 120°, 240°, 360°. • Angles of symmetry when there are exactly 4 of them: 90°, 180°, 270°, 360°. 3. Give the order of rotational symmetry for each figure: Chapter 9_Symmetry.indd 236 13-08-2024 17:05:30 236 Do you observe something common about the angles of symmetries in these cases? The first set of numbers are all multiples of 180. The second are all multiples of 120. The third are all multiples of 90. In each case, the angles are the multiples of the smallest angle. You may wonder and ask if this will always happen. What do you think? True or False • Every figure will have 360 degrees as an angle of symmetry. Reprint 2025-26 Is there a smallest angle of symmetry for all figures? It turns out that this is the case for most figures, except for the most symmetric shapes like the circle, whose symmetries we now discuss. Symmetries of a circle The circle is a fascinating figure. What happens when you rotate a circle clockwise about its centre? It coincides with itself. It does not matter what angle you rotate it by! So, for a circle, every angle is an angle of symmetry. Now take a point on the rim of the circle and join it to the centre. Extend the segment to a diameter of the circle. Is that diameter a line of reflection symmetry? It is. Every diameter is a line of symmetry! • If the smallest angle of symmetry of a figure is a natural number in degrees, then it is a factor of 360. Symmetry Chapter 9_Symmetry.indd 237 13-08-2024 17:05:32 Like wheels, we can find other objects around us having rotational symmetry. Find them. Some of them are shown below: Fan Flower Wheel Reprint 2025-26 237 Ganita Prakash | Grade 6 Figure it Out 1. Colour the sectors of the circle below so that the figure has i) 3 angles of symmetry, ii) 4 angles of symmetry, iii) what are the possible numbers of angles of symmetry you can obtain by colouring the sectors in different ways? 2. Draw two figures other than a circle and a square that have both reflection symmetry and rotational symmetry. 3. Draw, wherever possible, a rough sketch of: a. A triangle with at least two lines of symmetry and at least two angles of symmetry. b. A triangle with only one line of symmetry but not having rotational symmetry. c. A quadrilateral with rotational symmetry but no reflection symmetry. d. A quadrilateral with reflection symmetry but not having rotational symmetry. 4. In a figure, 60° is the smallest angle of symmetry. What are the other angles of symmetry of this figure? 5. In a figure, 60° is an angle of symmetry. The figure has two angles of symmetry less than 60°. What is its smallest angle of symmetry? 6. Can we have a figure with rotational symmetry whose smallest angle of symmetry is: a. 45°? b. 17°? Try This Chapter 9_Symmetry.indd 238 13-08-2024 17:05:32 238 Reprint 2025-26 7. This is a picture of the new Parliament Building in Delhi. a. Does the outer boundary of the picture have reflection symmetry? If so, draw the lines of symmetries. How many are they? b. Does it have rotational symmetry around its centre? If so, find the angles of rotational symmetry. 8. How many lines of symmetry do the shapes in the first shape sequence in Chapter 1, Table 3, the Regular Polygons, have? What number sequence do you get? 9. How many angles of symmetry do the shapes in the first shape sequence in Chapter 1, Table 3, the Regular Polygons, have? What number sequence do you get? 10. How many lines of symmetry do the shapes in the last shape sequence in Chapter 1, Table 3, the Koch Snowflake sequence, have? How many angles of symmetry? 11. How many lines of symmetry and angles of symmetry does Ashoka Chakra have? Symmetry Chapter 9_Symmetry.indd 239 13-08-2024 17:05:37 Playing with Tiles a. Use the colour tiles given at the end of the book to complete the following figure so that it has exactly 2 lines of symmetry. b. Use 16 such tiles to make figures that have exactly: 1 line of symmetry 2 lines of symmetry c. Use these tiles in making creative symmetric designs. Reprint 2025-26 239 Ganita Prakash | Grade 6 Chapter 9_Symmetry.indd 240 13-08-2024 17:05:37 240 Reprint 2025-26 Draw a 6 by 6 grid. Two players take turns covering two adjacent squares by drawing a line. The line can be placed either way: horizontally or vertically. The lines cannot overlap. The game goes on till a player is not able to place any more lines. The player who is not able to place a line loses. With what strategy can one play to win this game? When a figure is made up of parts that repeat in a definite pattern, we say that the figure has symmetry. We say that such a figure is symmetrical. A line that cuts a plane figure into two parts that exactly overlap when folded along that line is called a line of symmetry or axis of symmetry of the figure. A figure may have multiple lines of symmetry. Sometimes a figure looks exactly the same when it is rotated by an angle about a fixed point. Such an angle is called an angle of symmetry of the figure. A figure that has an angle of symmetry strictly between 0 and 360 degrees is said to have rotational symmetry. The point of the figure about which the rotation occurs is called the centre of rotation. A figure may have multiple angles of symmetry. Some figures may have a line of symmetry but no angle of symmetry, while others may have angles of symmetry but no lines of symmetry. Some figures may have both lines of symmetry as well as angles of symmetry. Game Summary Not allowed Symmetry Chapter 9_Symmetry.indd 241 13-08-2024 17:05:38 Reprint 2025-26 241 Section 9.1 Page No. 219 Q1. Do you see any line of symmetry in the figures at the start of the chapter? What about in the picture of the cloud? Ans. Yes, there are 6, 4 and 1 lines of symmetry in the figures of flower, rangoli and butterfly respectively. There is no line of symmetry in the figures of pinwheel and cloud. Q2. For each of the following figures, identify the line(s) of symmetry if it exists. Ans. Section 9.1 Page no. 221 Q. Is there any other way to fold the square so that the two halves overlap? How many lines of symmetry does the square shape have? Ans. No, there is no other way to fold the square. Figure it Out CHAPTER 9 — SOLUTIONS Symmetry The square shape has 4 lines of symmetry. Q. We saw that the diagonal of a square is also a line of symmetry. Let us take a rectangle that is not a square. Is its diagonal a line of symmetry? Ans. No, Rectangle’s diagonal is not a line of symmetry. Page no. 222 Q. What if we reflect along the diagonal from A to C? Where do points A, B, C and D go? What if we reflect along the horizontal line of symmetry? Ans. If we reflect along the diagonal from A to C, D occupies the position occupied by B earlier. A and C remain at the same place. If we reflect along the horizontal line of symmetry, D and C occupies the position earlier occupied by A and B respectively. [1] Section 9.1 Page no. 223 Q1 In each of the following figures, a hole was punched in a folded square sheet of paper and then the paper was unfolded. Identify the line along which the paper was folded. Figure (d) was created by punching a single hole. How was the paper folded? Ans. Q2. Given the line(s) of symmetry, find the other hole(s): Ans. For figure (d), paper was folded vertically and then horizontally or vice versa. Q4. After each of the following cuts, predict the shape of the hole when the paper is opened. After you have made your prediction, make the cutouts and verify your answer. [2] Q5. Suppose you have to get each of these shapes with some folds and a single straight cut. How will you do it? Note : For the above two questions, check if the 4-sided figures in the centre satisfy both the properties of a square. Ans. a. At first fold the paper horizontally then vertically. a. The hole in the centre is square. b. The hole in the centre is a square. Now cut a small square at the center (all sides closed corner). b. At first fold the paper horizontally then vertically. Now at closed corner, cut along with a slanting line. [3] Q6. How many lines of symmetry do these shapes have? Ans. a. b. A triangle with equal sides and equal angles. c. A hexagon with equal sides and equal angles. Q7. Trace each figure and draw the lines of symmetry, if any: [4] Ans. Q8. Find the lines of symmetry for the kolam below. Ans. Q9. Draw the following. Is it possible to draw a triangle with exactly two lines of symmetry? a. A triangle with exactly one line of symmetry b. A triangle with exactly three lines of symmetry c. A triangle with no line of symmetry [5] Ans. a. Q10. Draw the following. In each case, the figure should contain at least one curved boundary. Ans. a. A figure with exactly one line of Symmetry b. A figure with exactly two lines of symmetry c. A figure with exactly four lines of symmetry b. c. No, it is not possible to draw a triangle with exactly two lines of symmetry. [6] Q11. Copy the following on squared paper. Complete them so that the blue line is a line of symmetry. Problem (a) has been done for you. Hint: For (c) and (f), see if rotating the book helps! Ans. [7] Q12. Copy the following drawing on squared paper. Complete each one of them so that the resulting figure has the two blue lines as lines of symmetry. Ans. Q13. Copy the following on a dot grid. For each figure draw two more lines to make a shape that has a line of symmetry. Ans. [8] Section 9.2 Page no. 235 Hint: Use 5 radial arms for the first case. What should the angle between two adjacent radial arms be? Ans. Can you draw a figure with radial arms that has a) exactly 5 angles of symmetry, b) 6 angles of symmetry? Also find the angles of symmetry in each case. Angles of symmetry = 72°, 144°, 216°, 288°, 360°. Angles of symmetry = 60°, 120°, 180°, 240°, 300°, 360°. [9] Let us find the angles of symmetry for other kinds of figures. Ans. 360° ÷ 7 = 513 7 No, its smallest angle of symmetry is not a whole number. Page no. 235 Figure it out Q1. Find the angles of symmetry for the given figures about the point marked. Ans. a. Angles of symmetry = 90°, 180°, 270°, 360° Q2. Which of the following figures have more than one angle of symmetry? Ans. The following figures have more than one angle of symmetry– Consider a figure with radial arms having exactly 7 angles of symmetry. What will be its smallest angle of symmetry? Is the number of degrees a whole number in this case? If not, express it as a mixed faction. b. Angle of symmetry = 360° c. Angles of symmetry = 180°, 360° [10] Q3. Give the order of rotational symmetry for each figure: Ans. Orders of rotational symmetry Page no. 236 Ans. Yes, the angles of symmetry are always the multiples of the smallest angle. For example, the second angle is twice the amount of rotation of the first rotation. In each case, the angles are the multiples of the smallest angle. You may wonder and ask if this will always happen. What do you think? • Every figure will have 360 degrees as an angle of symmetry. • If the smallest angle of symmetry of a figure is a natural number in degrees, then it is a factor of 360. Ans. True True or False True [11] Section 9.2 Page no. 238 Figure it out Q1. Color the sectors of the circle below so that the figure has i) 3 angles of symmetry, ii) 4 angles of symmetry, iii) what are the possible numbers of angles of symmetry you can obtain by coloring the sectors in different ways? Ans. Q2. Draw two figures other than a circle and a square that have both reflection symmetry and rotational symmetry. Ans. (i) 3 Angles of symmetry (ii) 4 angles of symmetry (iii) 12 angles of symmetry are possible to obtain. Numbers of line symmetry = 4 Order of rotational symmetry = 4 Numbers of line symmetry = 4 Order of rotational symmetry = 4 [12] Q3. Draw, wherever possible, a rough sketch of Ans. a. a. A triangle with at least two lines of symmetry and at least two angles of symmetry. b. A triangle with only one line of symmetry but not having rotational symmetry. c. A quadrilateral with rotational symmetry but no reflection symmetry. d. A quadrilateral with reflection symmetry but not having rotational symmetry. b. 3 lines of symmetry 3 angles of symmetry 1 line of symmetry No rotational symmetry c. No line of symmetry 2 angles (180°, 360°) [13] d. A quadrilateral with reflection symmetry but not having rotational symmetry Q4. In a figure, 60° is the smallest angle of symmetry. What are the other angles of symmetry of the figure? Ans. Other angles of symmetry = 120°, 180°, 240°, 300°, 360°. Q5. In the figure, 60° in an angle of symmetry. The figure has two angles of symmetry less then 60°. What is its smallest angle of symmetry? Ans. The smallest angle of symmetry = 20° Q6. Can we have a figure with rotational symmetry whose smallest angle of symmetry is Ans. a. yes, as 360° is a multiple of 45°. a. 45°? b. 17°? Q7. This is a picture of the new Parliament Building in Delhi. Ans. a. Yes, the outer boundary of the picture has 3 lines of symmetry. a. Does the outer boundary of the picture have reflection symmetry? If so, draw the lines of symmetries. How many are they? b. Does it have rotational symmetry around its centre? If so, find the angles of rotational symmetry. b. No, as 360° is not a multiple of 17°. b. Yes, the outer boundary has rotational symmetry. The angles of rotational symmetry are 120°, 240°, 360°. [14] Q8. How many lines of symmetry do the shapes in the first shape sequence in Chapter 1, Table 3, the Regular Polygons, have? What number sequence do you get? Ans. Q10. How many lines of symmetry do the shapes in the last shape sequence in Chapter 1, Table 3, the Koch Snowflake sequence, have? How many angles of symmetry? Ans. Number of lines of symmetry : 3, 6, 6,6,6 Angles of symmetry : 3, 6, 6,6,6 Q11. How many lines of symmetry and angles of symmetry does Ashoka Chakra have? Ans. Number of lines of symmetry = 24 It is a counting number sequence. Number of angles of symmetry = 24 Regular Polygon No. of line symmetry Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon 3 4 5 6 7 8 9 10 [15]" class_6,10,The Other Side of Zero,ncert_books/class_6/Ganita_Prakash/fegp110.pdf,"10 More and more numbers! Recall that the very first numbers we learned about in the study of mathematics were the counting numbers 1, 2, 3, 4, … Then we learned that there are even more numbers! For example, there is the number 0 (zero), representing nothing, which comes before 1. The number 0 has a very important history in India and now in the world. For example, around the world we learn to write numbers in the Indian number system using the digits 0 to 9, allowing us to write numbers however large or however small using just these 10 digits. We then learned about more numbers that exist between the numbers 0, 1, 2, 3, 4, … , such as 1 2 , 3 2 , and 13 6 . These are called fractions. But are there still more numbers? Well, 0 is an additional number that we didn’t know about earlier, and it comes before 1 and is less than 1. Are there perhaps more numbers that come before 0 and are less than 0? Phrased another way, we have seen the number line: THE OTHER SIDE OF ZERO Integers Chapter 10_The Other Side of Zero.indd 242 13-08-2024 17:32:05 However, this is actually only a number ‘ray’, in the language we learned earlier in geometry; this ray starts at 0 and goes forever to the right. Do there exist numbers to the left of 0, so that this number ray can be completed to a true number line? That is what we will investigate in this chapter! 0 1 2 3 4 5 6 7 8 9 10 Reprint 2025-26 Can there be a number less than 0? Can you think of any ways to have less than 0 of something? 10.1 Bela’s Building of Fun Children flock to Bela’s ice cream factory to see and taste her tasty ice cream. To make it even more fun for them, Bela purchased a multistoried building and filled it with attractions. She named it Bela’s Building of Fun. Observe that some of the floors in the ‘Building of Fun’ are below the ground. What are the shops that you find on these floors? What is there on the ground floor? A lift is used to go up and down between the floors. It has two buttons: ‘+’ to go up and ‘–’ to go down. Can you spot the lift? To go to the Art Centre from the ‘Welcome Hall’, you must press the ‘+’ button twice. To go down two floors, you must press the ‘–’ button twice, which we write as – – or –2. So if you press +1 (i.e., if you press the ‘+’ button once), then you will go up one floor and if you press –1 (i.e., if you press the ‘–’ button once), then you will go down 1 floor. But this was no ordinary building! We say that the button press is + + or +2. The Other Side of Zero Chapter 10_The Other Side of Zero.indd 243 13-08-2024 17:32:06 What do you press to go four floors up? What do you press to go three floors down? Lift button presses and numbers: +++ is written as + 3 – – – – is written as – 4 Reprint 2025-26 243 Ganita Prakash | Grade 6 Numbering the floors in the building of fun Entry to the ‘Building of Fun’ is at the ground floor level and is called the ‘Welcome Hall’. Starting from the ground floor, you can reach the Food Court by pressing + 1 and can reach the Art Centre by pressing + 2. So, we can say that the Food Court is on Floor + 1 and that the Art Centre is on Floor + 2. Starting from the ground floor, you must press –1 to reach the Toy Store. So, the Toy Store is on Floor –1 similarly starting from the ground floor, you must press – 2 to reach the Video Games shop. So, the Video Games shop is on Floor – 2. Did you notice that + 3 is the floor number of the Book Store, but it is also the number of floors you move when you press + 3? Similarly, – 3 is the floor number but it is also the number of floors you go down when you press – 3, i.e., when you press – – – . A number with a ‘+’ sign in front is called a positive number. A number with a ‘–’ sign in front is called a negative number. The ground floor is called Floor 0. Can you see why? Number all the floors in the Building of Fun. Chapter 10_The Other Side of Zero.indd 244 13-08-2024 17:32:07 244 In the ‘Building of Fun’, the floors are numbered using the ground floor, Floor 0, as a reference or starting point. The floors above the ground floor are numbered with positive numbers. To get to them from the ground floor, one must press the ‘+’ button some number of times. The floors below the ground are numbered with negative numbers. To get to them from the ground floor, one must press the ‘–’ button some number of times. Zero is neither a positive nor a negative number. We do not put a ‘+’ or ‘–’ sign in front of it. Reprint 2025-26 Addition to keep track of movement Start from the Food Court and press + 2 in the lift. Where will you reach? ____________ The starting floor is + 1 (Food Court) and the number of button presses is + 2. Therefore, you reach the target floor (+ 1) + (+ 2) = + 3 (Book Store).  Figure it Out We can describe this using an expression: Starting Floor + Movement = Target Floor. 1. You start from Floor +2 and press –3 in the lift. Where will you reach? Write an expression for this movement. 2. Evaluate these expressions (you may think of them as Starting 3. Starting from different floors, find the movements required to Floor + Movement by referring to the Building of Fun). a. (+1)+(+4) = _______ b. (+4)+(+1) = _______ c. (+4)+(–  3) = _______ d. (–1)+(+2) = _______ e. (–1)+(+1) = _______ f. 0+(+2) = _________ g. 0+(–2) = _________ reach Floor –5. For example, if I start at Floor +2, I must press –7 to reach Floor –5. The expression is (+2) + (–7) = –5. Find more such starting positions and the movements needed to The Other Side of Zero Chapter 10_The Other Side of Zero.indd 245 13-08-2024 17:32:07 Combining button presses is also addition Gurmit was in the Toy Store and wanted to go down two floors. But by mistake he pressed the ‘+’ button two times. He realised his mistake and quickly pressed the ‘–’ button three times. How many floors below or above the Toy Store will Gurmit reach? Gurmit will go one floor down. We can show the movement resulting from combining button presses as an expression: (+2)+(–3) = –1. reach Floor –5 and write the expressions. Reprint 2025-26 245 Ganita Prakash | Grade 6 Evaluate these expressions by thinking of them as the resulting movement of combining button presses: Back to zero! On the ground floor, Basant is in a great hurry and by mistake he presses +3. What can he do to cancel it and stay on the ground floor? He can cancel it by pressing – 3. That is, (+3) + (– 3) = 0. We call – 3 the inverse of +3. Similarly, the inverse of – 3 is +3. If Basant now presses +4 and then presses – 4 in the lift, where will he reach? Here is another way to think of the concept of inverse. If you are at Floor +4 and you press its inverse – 4, then you are back to zero, the ground floor! If you are at Floor – 2 and press its inverse +2, then you go to (– 2) + (+2) = 0, again the ground floor!   Write the inverses of these numbers: a. (+1)+(+4) = _____________ b. (+4)+(+1) = _____________ c. (+4)+(–  3)+(–2) = _______ d. (–1)+(+2)+(–3) = _______ Figure it out +4, –4, –3, 0, +2, –1. Chapter 10_The Other Side of Zero.indd 246 13-08-2024 17:32:08 246 Comparing numbers using floors   Connect the inverses by drawing lines. Who is on the lowest floor? 1. Jay is in the Art Centre. So, he is on Floor +2. 2. Asin is in the Sports Centre. So, she is on Floor ___. 3. Binnu is in the Cinema Centre. So, she is on Floor ____. 4. Aman is in the Toys Store. So, he is on Floor ____. +5 –9 Reprint 2025-26 –7 +8 –8 –5 +9 +7 Floor +3 is lower than Floor +4. So, we write +3 < +4. We also write +4 > +3. Should we write –3 < – 4 or – 4 < – 3? Floor – 4 is lower than Floor – 3. So, – 4 < – 3. It is also correct to write – 3 > –4 Figure it Out 1. Compare the following numbers using the Building of Fun and fill in the boxes with < or >. 2. Imagine the Building of Fun with more floors. Compare the numbers and fill in the boxes with < or >: 3. If Floor A = –12, Floor D = –1 and Floor E = +1 in the building shown on the right as a line, find the numbers of Floors B, C, F, G, and H. a. –2 +5 b. –5 +4 c. –5 –3 d. +6 –6 e. 0 –4 f. 0 +4 a. –10 –12 b. +17 –10 c. 0 –20 d. +9 –9 e. –25 –7 f. +15 –17 Notice that all negative number floors are below Floor 0. So, all negative numbers are less than 0. All the positive number floors are above Floor 0. So, all positive numbers are greater than 0. The Other Side of Zero C 0 +1 E F G H D –1 Chapter 10_The Other Side of Zero.indd 247 13-08-2024 17:32:08 Subtraction to find which button to press In earlier classes, we understood the meaning of subtraction as ‘take away’. For example, there are 10 books on the shelf. I take away 4 books. How many are left on the shelf? We can express the answer using subtraction: 10 – 4 = 6 or ‘Ten take away four is six.’ 4. Mark the following floors of the building shown on the right. a. –7 b. –  4 c. +3 d. –  10 Reprint 2025-26 B A –12 247 Ganita Prakash | Grade 6 You may also be familiar with another meaning of subtraction which is related to comparison or making quantities equal. For example, consider this situation: I have `10 with me and my sister has `6. Now, I can ask the question: ʻHow much more money should my sister get in order to have the same amount as me?ʼ Here, we see the connection between ‘finding the missing number to be added’ and subtraction. For subtraction of positive and negative numbers, we will use this meaning of subtraction as ‘making equal’ or ‘finding the missing number to be added’. In general, when there are two unequal quantities, subtraction can indicate the change needed to make the quantities equal. Subtraction shows how much the starting quantity should change in order to become the target quantity. In the context of different floor levels, what is the change required to reach the Target Floor from the Starting Floor? Notice that the change needed may be positive (for an increase) or negative (for a decrease). Teacher’s Note   Evaluate 15 – 5, 100 – 10 and 74 – 34 from this perspective. We can write this in two ways: 6 + ? = 10 Or 10 – 6 = ? Chapter 10_The Other Side of Zero.indd 248 13-08-2024 17:32:08 248 Your starting floor is the Art Centre and your target floor is the Sports Centre. What should be your button press? You need to go three floors up, so you should press + 3. We can write this as an expression using subtraction: Target Floor – Starting Floor = Movement needed. In the above example, the starting floor is + 2 (Art Centre) and the target floor is + 5. The button press to get to + 5 from + 2 is + 3. Therefore, (+ 5) – (+ 2) = + 3 Reprint 2025-26 Explanation Recall the connection between addition and subtraction. For 3+ ?=5, we can find the missing number using subtraction: 5–3=2. That is, subtraction is the same as finding the missing number to be added. We know that— If the movement needed is to be found, then, So, More examples: a. If the Target Floor is – 1 and Starting Floor is – 2, what button should you press? You need to go one floor up, so, you should press + 1. Expression: (– 1) – (– 2) = (+1) b. If the Target Floor is – 1 and Starting floor is +3, what button should you press? You need to go four floors down, so, you should press – 4. Expression: (– 1) – (+ 3) = (– 4) c. If the Target Floor is +2 and Starting Floor is – 2, what button should you press? You need to go four floors up, so, you should press +4. Expression: (+ 2) – (– 2) = (+ 4) Target Floor – Starting Floor = ? = Movement needed Starting Floor + Movement needed = Target Floor Starting Floor + ? = Target Floor The Other Side of Zero Chapter 10_The Other Side of Zero.indd 249 13-08-2024 17:32:08 Figure it Out Complete these expressions. You may think of them as finding the movement needed to reach the Target Floor from the Starting Floor. a. (+1) –(+4) = _______ b. (0) –(+2) = _________ c. (+4) –(+1) = _______ d. (0)– (–2) = _________ e. (+4) – (–3) = _______ f. (–4)–(–3) = ________ g. (–1) –(+2) = _______ h. (–2) – (–2) = ________ i. (–1) – (+1) = _______ j. (+3)– (–3) = ________ Reprint 2025-26 249 Ganita Prakash | Grade 6 Adding and subtracting larger numbers The picture shows a mine, a place where minerals are extracted by digging into the rock. The truck is at the ground level, but the minerals are present both above and below the ground level. There is a fast moving lift which moves up and down in a mineshaft carrying people and ore. Some of the levels are marked in the picture. The ground level is marked 0. Levels above the ground are marked by positive numbers and levels below the ground are marked by negative numbers. The number indicates how many metres above or below the ground level it is. In the mine, just like in the Building of Fun: Starting Level + Movement = Target Level For example, (+ 40) + (+ 60) = + 100 (– 90) + (– 55) = – 145 Target Level – Starting Level = Movement needed For example, (+ 40) – (– 50) = + 90 (– 90) – (+ 40) = – 130 +1000 +600 Chapter 10_The Other Side of Zero.indd 250 13-08-2024 17:32:08 250 Bela’s Building of Fun had only six floors above and five floors below. That is numbers – 5 to + 6. In the mine above, we have numbers from – 200 to + 180. But we can imagine larger buildings or mineshafts. Just as positive numbers + 1, + 2, + 3, ... keep going up without an end, similarly, negative numbers – 1, – 2, – 3, ... keep going down. Positive and negative numbers, with zero, are called integers. They go both ways from 0: … – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … How many negative numbers are there? Reprint 2025-26 – 1000 – 1000 0 +1000 +900 +800 +700 +600 +500 +400 +300 +200 +100 – 100 – 200 – 300 – 400 Complete these expressions. Check your answers by thinking about the movement in the mineshaft. Adding, subtracting, and comparing any numbers To add and subtract even larger integers, we can imagine even larger lifts! In fact, we can imagine a lift that can extend forever upwards and forever downwards, starting from Level 0. There does not even have to be any building or mine around — just an ‘infinite lift’! For example, suppose we want to carry out the subtraction + 2000 – (–200). We can imagine a lift with 2000 levels above the ground and 200 below the ground. Recall that, To go from the Starting Floor –200 to the Target Floor + 2000, we must press + 2200 (+ 200 to get to zero, and then + 2000 more after that to get to + 2200). Therefore, (+ 2000) – (– 200) = + 2200. Notice that (+ 2000) + (+ 200) is also + 2200. We can use this imagination to add and subtract any integers we like. Figure it Out a. (+40)+ ______ = +200 b. (+40)+_______=–200 c. (–50)+ ______ = +200 d. (–50)+_______=–200 e. (–200) – (–40) = _______ f. (+200)– (+40) = _______ g. (–200) –(+40) = _______ Target Level – Starting Level = Movement needed The Other Side of Zero Chapter 10_The Other Side of Zero.indd 251 13-08-2024 17:32:08 – 500 – 600 – 700 – 800 – 900 Try evaluating the following expressions by similarly drawing or imagining a suitable lift: a. –125+(–30) b. +105– (–55) c. +105+(+55) d. +80–(–150) e. +80+(+150) f. –99– (–200) g. –99+(+200) h. +1500– (–1500) Reprint 2025-26 251 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 Ganita Prakash | Grade 6 In the above example, we saw that + 2000 – (– 200) = + 2000 + (+ 200) = + 2200. In other words, subtracting a negative number is the same as adding the corresponding positive number. That is, we can replace subtraction of a negative number by addition of a positive number! In the other exercises that you did above, did you notice that subtracting a negative number was the same as adding the corresponding positive number? Take a look at the ‘infinite lift’ above. Does it remind you of a number line? In what ways? Back to the number line The ‘infinite lift’ we saw above looked very much like a number line, didn’t it? In fact, if we rotate it by 90°, it basically becomes a number line. It also tells us how to complete the number ray to a number line, answering the question that we had asked at the beginning of the chapter. To the left of 0 are the negative numbers –1, –2, –3, … Usually we drop the ‘+’ sign on positive numbers and simply write them as 1, 2, 3, … Math Talk Chapter 10_The Other Side of Zero.indd 252 13-08-2024 17:32:09 252 0 1 2 3 4 5 6 7 8 9 10 Instead of traveling along the number line using a lift, we can simply imagine walking on it. To the right is the positive (forward) direction, and to the left is the negative (backward) direction. Smaller numbers are now to the left of bigger numbers and bigger numbers are to the right of smaller numbers. So, 2 < 5; –3 < 2; and –5 < –3. If, from 5 you wish to go over to 9, how far must you travel along the number line? Reprint 2025-26 0 1 2 3 4 5 6 7 8 9 10 Now, from 9, if you wish to go to 3, how much must you travel along the number line? You must move 6 steps backward, i.e., you must move –6. Hence, we write 9 + (–6) = 3. You must travel 4 steps. That is why 5 + 4 = 9. (Remember: Starting Number + Movement = Target Number) The corresponding subtraction statement is 9 –5 = 4. (Remember: Target Number – Starting Number = Movement needed) (Remember again : Starting Number + Movement = Target Number) The corresponding subtraction statement is 3–9=–6. (Remember again: Target Number – Starting Number = Movement needed) Now, from 3, if you wish to go to –2, how far must you travel? The Other Side of Zero Chapter 10_The Other Side of Zero.indd 253 13-08-2024 17:32:09 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 You must travel –5 steps, i.e., 5 steps backward. Thus, 3+(–5)=–2. The corresponding subtraction statement is: –2 –3=–5. Figure it Out 1. Mark 3 positive numbers and 3 negative numbers on the number line above. 2. Write down the above 3 marked negative numbers in the following boxes: Reprint 2025-26 253 Ganita Prakash | Grade 6 Using the unmarked number line to add and subtract Just as you can do additions, subtractions and comparisons with small numbers using the number line above, you can also do them with large numbers by imagining an ‘infinite number line’ or drawing an ‘unmarked number line’ as follows: This line shows only the position of zero. Other numbers are not marked. It can be convenient to use this unmarked number line to add and subtract integers. You can show, or simply imagine, the scale of the number line and the positions of numbers on it. For example, this unmarked number line (UNL) shows the addition problem: 85 + (– 60) = ? +25 +85 We then can visualise that 85 + (– 60) = 25 3. Is 2 > – 3? Why? Is –2 < 3? Why? 4. What are a. –5+0 b. 7+(–7) c. –10+20 d. 10 –20 e. 7–(–7) f. –8–(–10)? 0 0 – 60 Chapter 10_The Other Side of Zero.indd 254 13-08-2024 17:32:09 254 The following UNL shows a subtraction problem which can also be written as a missing addend problem: (–100)–(+250) = ? or 250 + ? = –100. In this way, you can carry out addition and subtraction problems, with positive and negative numbers, on paper or in your head using an unmarked number line. We can then visualise that ? = –350 in this problem. – 100 0 + 250 Reprint 2025-26 ? Converting subtraction to addition and addition to subtraction Recall that Target Floor – Starting Floor = Movement needed First method: Looking at the number line, we see we need to move –5 (i.e., 5 in the backward direction). Therefore, – 3 – 2 = – 5. The movement needed is –5. Second method: Break the journey from 2 to –3 into two parts. a. From 2 to 0, the movement is 0–2=–2. b. From 0 to –3, the movement is –3–0=–3. The total movement is the sum of the two movements: – 3 + (– 2) = – 5. Look at the two coloured expressions. There is no subtraction in the second one! If we start at 2 and wish to go to –3, what is the movement needed? Use unmarked number lines to evaluate these expressions: a. –125 + (–30) = _______ b. +105 – (–55) = _______ c. +80 – (–150) = _______ d. –99 – (–200) = _______ Target Floor = Starting Floor + Movement needed or – 3 0 2 The Other Side of Zero – 5 Chapter 10_The Other Side of Zero.indd 255 13-08-2024 17:32:09 In this way, we can always convert subtraction to addition. The number that is being subtracted can be replaced by its inverse and then added instead. Similarly, a number that is being added can be replaced by its inverse and then subtracted. In this way, we can also always convert addition to subtraction. Examples: a. (+7) – (+5)=(+7)+(–5) b. (–3) –(+8)=(–3)+(–8) c. (+8) –(–2)=(+8)+(+2) d. (+6) – (–9)=(+6)+(+9) Reprint 2025-26 255 Ganita Prakash | Grade 6 10.2 The Token Model Using tokens for addition In Bela’s Building of Fun, the lift attendant is bored. To entertain himself, he keeps a box containing lots of positive (red) and negative (green) tokens. Each time he presses the ‘+’ button, he takes a positive token from the box and puts it in his pocket. Similarly, each time he presses the ‘–’ button, he takes a negative token and puts it in his pocket. He starts on the ground floor (Floor 0) with an empty pocket. After one hour, he checks his pocket and finds 5 positive and 3 negative tokens. On which floor is he now? He must have pressed ‘+’ five times and ‘–’ 3 times and (+5)+(–3)=+2. So, he is at Floor +2 now. Here is another way to do the calculation. A positive token and a negative token cancel each other, because the value of this pair of tokens together is zero. These two tokens in his pocket meant that he pressed ‘+’ once and ‘–’ once, respectively, and these cancel each other. We say that a positive and a negative token make a ʻzero pairʼ. When you remove all the zero pairs, you are left with two positive tokens, so (+5) + (–3) = +2. We can perform any such addition using tokens! Chapter 10_The Other Side of Zero.indd 256 13-12-2024 16:07:51 256 Example: Add+ 5 and – 8. Reprint 2025-26 From the picture, we see that we can remove five zero pairs, and we are then left with – 3. Therefore (+ 5) + (– 8) = – 3. Using tokens for subtraction We have seen how to perform addition of integers with positive tokens and negative tokens. We can also perform subtraction using tokens! Example: Let us subtract: (+5) – (+4). Figure it Out a. b. 1. Complete the additions using tokens. a. (+6)+(+4) b. (–3) + (–2) c. (+5)+(–7) d. (–2) + (+6) 2. Cancel the zero pairs in the following two sets of tokens. On what floor is the lift attendant in each case? What is the corresponding addition statement in each case? The Other Side of Zero Chapter 10_The Other Side of Zero.indd 257 13-08-2024 17:32:24 This is easy to do. From 5 positives take away 4 positives to see the result. Example: Let us subtract: (–7) – (–5). Is (– 7) – (– 5) the same as (– 7)+(+5)? Example: Let us subtract: (+5)–(+6). Put down 5 positives. But there are not enough tokens to take out 6 positives! Reprint 2025-26 (–7) – (–5) = –2 (+ 5) – (+ 4) = + 1 257 Ganita Prakash | Grade 6 To get around this issue, we can put out an extra zero pair (a positive and a negative), knowing that this does not change the value of the set of tokens. Now, we can take out 6 positives! See what is left: We conclude that (+5) – (+6) =–1. Example: + 4 – (–6). Start with 4 positives. We have to take out 6 negatives from these. But there are not enough negatives. This is not a problem. We add some zero pairs as this does not change the value of the set of tokens. But how many zero pairs? We have to take away 6 negatives so we put down 6 zero pairs:  Figure it Out 1. Evaluate the following differences using tokens. Check that you get the same result as with other methods you now know: a. (+10) –(+7) b. (–8)– (–4) c. (–9) –(–4) d. (+9) –(+12) e. (–5)– (–7) f. (–2) –(–6) 2. Complete the subtractions: a. (–5) –(–7) b. (+10) –(+13) c. (–7) –(–9) d. (+3) – (+8) e. (–2) –(–7) f. (+3)–(+15) Chapter 10_The Other Side of Zero.indd 258 13-08-2024 17:32:36 258 Now we can take away 6 negatives: Therefore, +4– (–6)=+10. Reprint 2025-26 10.3 Integers in Other Places Credits and debits Suppose you open a bank account at your local bank with the `100 that you had been saving over the last month. Your bank balance therefore, starts at `100. Then you make `60 at your job the next day and you deposit it in your account. This is shown in your bank passbook as a ‘credit’.  Figure it Out Your new bank balance is _______. Your bank balance is now ______. The next day you pay your electric bill of `30 using your bank account. This is shown in your bank passbook as a ‘debit’. 1. Try to subtract: –3– (+5). How many zero pairs will you have to put in? What is the result? 2. Evaluate the following using tokens. a. (–3)–(+10) b. (+8)– (–7) c. (–5)–(+9) d. (–9) – (+10) e. (+6) –(–4) f. (–2)–(+7) The Other Side of Zero Chapter 10_The Other Side of Zero.indd 259 13-08-2024 17:32:36 Yes, some banks do allow your account balance to become negative, temporarily! Some banks also charge you an additional amount if your balance becomes negative, in the form of ‘interest’ or a ‘fee’. Your strategic large purchase the previous day allows you to make, `200 at your business the next day. What is your bank balance now? ______ The next day you make a major purchase for your business of `150. Again this is shown as a debit. Is this possible? Reprint 2025-26 259 Ganita Prakash | Grade 6 You can think of ‘credits’ as positive numbers and ‘debits’ as negative numbers. The total of all your credits (positive numbers) and debits (negative numbers) is your total bank account balance. This can be positive or negative! In general, it is better to try to keep a positive balance in your bank account! As you can see, positive and negative numbers along with zero are extremely useful in the world of banking and accounting. Geographical cross sections  Figure it Out What is your balance now? ______ 1. Suppose you start with `0 in your bank account, and then you have credits of `30, `40, and `50, and debits of `40, `50, and `60. What is your bank account balance now? 2. Suppose you start with `0 in your bank account, and then you have debits of `1, 2, 4, 8, 16, 32, 64, and 128, and then a single credit of `256. What is your bank account balance now? 3. Why is it generally better to try and maintain a positive balance in your bank account? What are circumstances under which it may be worthwhile to temporarily have a negative balance? Chapter 10_The Other Side of Zero.indd 260 13-08-2024 17:32:36 260 We measure the height of geographical features like mountains, plateaus, and deserts from ‘sea level’. The height at sea level is 0m. Heights above sea level are represented using positive numbers and heights below sea level are represented using negative numbers. Reprint 2025-26 Ask what a geographical cross section is by showing the figure in this page. It is like imagining a vertical slice taken out at some location on the Earth. This is what would be seen from a side view. Discuss the notion of ‘sea level’ for measuring heights and depths in geography.  Figure it Out 1. Looking at the geographical cross section, fill in the respective heights: a. b. c. d. e. f. g. G Sea level -1500 -1000 -500 0 500 1000 1500 Height (m) A Teacher’s Note B C D E The Other Side of Zero F Chapter 10_The Other Side of Zero.indd 261 13-08-2024 17:32:37 2. Which is the highest point in this geographical cross section? Which is the lowest point? 3. Can you write the points A, B, …, G in a sequence of decreasing order of heights? Can you write the points in a sequence of increasing order of heights? 4. What is the highest point above sea level on Earth? What is its height? 5. What is the lowest point with respect to sea level on land or on the ocean floor? What is its height? (This height should be negative). Reprint 2025-26 261 Ganita Prakash | Grade 6 Temperature During summer time you would have heard in the news that there is a ‘heat wave’. What do you think will be the temperature during the summer when you feel very hot? In the winter we have cooler or colder temperatures. What has been the maximum temperature during the summer and the minimum temperature during the winter last year in your area? Find out. When we measure temperature, we use Celsius as the unit of measure (°C). The thermometers below are showing 40°C and 15°C temperatures.  Figure it Out 1. Do you know that there are some places in India where temperatures can go below 0°C? Find out the places in India where temperatures sometimes go below 0°C. What is common among these places? Why does it become colder there and not in other places? 2. Leh in Ladakh gets very cold during the winter. The following is a table of temperature readings taken during different times of the day and night in Leh on a day in November. Match the temperature with the appropriate time of the day and night. ˚C 100 –20 20 30 40 50 60 70 80 90 0 10 ˚C 100 0 10 20 30 40 50 60 70 80 90 Math Talk Chapter 10_The Other Side of Zero.indd 262 13-08-2024 17:32:37 262 Temperature Time 14°C 02:00 a.m. –2°C 02:00 p.m. –4°C 11:00 a.m. 8°C 11:00 p.m. Reprint 2025-26 –10 –10 –20 10.4 Explorations with Integers A hollow integer grid There is something special about the numbers in these two grids. Let us explore what that is. Top row: Bottom row: Left column: Right column: Talk about thermometers and how they are used to measure temperature. Bring a laboratory thermometer to the class and measure the temperature of hot water and cold water. Point out to children that there are markings in the thermometer that are below 0°C. Have a discussion on what 0°C indicates, namely, the freezing point of water. 4 + (–1) + (–3)  = 0 (– 1) + (– 1) + 2    = 0 4 + (– 3) + (– 1)  = 0 (– 3) + 1 + 2   = 0 – 3 1 – 1 – 1 2 4 – 1 – 3 Teacher’s Note 5 + (– 3) + (– 5)  = ____ (– 8) + (– 2) + 7   = ____ 5 + 0   + (– 8)  = ____ (– 5) + (– 5)  + 7   = ____ – 8 – 2 7 5 – 3 – 5 0 – 5 The Other Side of Zero Chapter 10_The Other Side of Zero.indd 263 13-08-2024 17:32:37 In each grid, the numbers in each of the two rows (the top row and the bottom row) and the numbers in each of the two columns (the leftmost column and the rightmost column) add up to give the same number. We shall call this sum as the ‘border sum’. The border sum of the first grid is ‘0’.  Figure it Out 1. Do the calculations for the second grid above and find the border sum. Reprint 2025-26 263 Ganita Prakash | Grade 6 An amazing grid of numbers! Below is a grid having some numbers. Follow the steps as shown until no number is left. 3. For the last grid above, find more than one way of filling the numbers to get border sum –4. 4. Which other grids can be filled in multiple ways? What could be the reason? 5. Make a border integer square puzzle and challenge your classmates. 2. Complete the grids to make the required border sum: –10 9 Border sum is +4 Border sum is – 2 Border sum is – 4 – 2 – 1 – 5 4 – 7 – 6 – 10 – 1 3 4 0 9 1 2 – 2 7 –5 6 8 Strike out the row and column of the Circle any unstruck number –2 Circle any number chosen number –5 7 –5 Chapter 10_The Other Side of Zero.indd 264 13-08-2024 17:32:37 264 When there are no more unstruck numbers, STOP. Add the circled numbers. In the example below, the circled numbers are – 1, 9, –7, –2. If you add them, you get –1. – 2 – 1 – 5 4 – 7 – 6 – 10 – 1 3 4 0 9 1 2 – 2 7 – 2 – 1 – 5 4 – 7 – 6 – 10 – 1 3 4 0 9 1 2 – 2 7 Reprint 2025-26 – 2 – 1 – 5 4 – 7 – 6 – 10 – 1 3 4 0 9 1 2 – 2 7 – 2 – 1 – 5 4 – 7 – 6 – 10 – 1 3 4 0 9 1 2 –  2 7 Figure it Out Figure it Out 1. Write all the integers between the given pairs, in increasing order. a. 0 and –7 b. –4 and 4 c. –8 and –15 d. –30 and –23 2. Give three numbers such that their sum is –8. 3. There are two dice whose faces have these numbers: –1, 2, –3, 4, –5, 6. The smallest possible sum upon rolling these dice is –10 = (–5) + (–5) and the largest possible sum is 12 = (6)+(6). Some numbers between (–10) and (+12) are not possible to get by adding numbers on these two dice. Find those numbers. 4. Solve these: 1. Try afresh, choose different numbers this time. What sum did you get? Was it different from the first time? Try a few more times! 2. Play the same game with the grids below. What answer did you get? 3. What could be so special about these grids? Is the magic in the numbers or the way they are arranged or both? Can you make more such grids? – 11 – 8 – 5 – 2 – 20 – 7 – 14 – 11 – 2 1 4 7 7 10 13 16 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 The Other Side of Zero Try This Chapter 10_The Other Side of Zero.indd 265 13-08-2024 17:32:37 5. Find the years below. a. From the present year, which year was it 150 years ago? ________ b. From the present year, which year was it 2200 years ago? _______ Hint: Recall that there was no year 0. 8 + (– 13) (– 8) – (– 13) (13) – 8 13 – (– 8) 8 – 13 (– 8) – (13) (– 13) – (– 8) (– 13) + (– 8) Reprint 2025-26 265 Ganita Prakash | Grade 6 10.5 A Pinch of History c. What will be the year 320 years after 680 BCE? ________ 6. Complete the following sequences: a. (–40), (–34), (–28), (–22), _____, ______, ______ b. 3, 4, 2, 5, 1, 6, 0, 7, _____, _____, _____ c. _____, ______, 12, 6, 1, (–3), (–6), _____, ______, ______ 7. Here are six integer cards: (+1), (+7), (+18), (–5), (–2), (–9). You can pick any of these and make an expression using addition(s) and subtraction(s). Here is an expression: (+18)+(+1)–(+7) – (–2) which gives a value (+14). Now, pick cards and make an expression such that its value is closer to (– 30). 8. The sum of two positive integers is always positive but a (positive integer) – (positive integer) can be positive or negative. What about a. (positive) – (negative) b. (positive) + (negative) c. (negative) + (negative) d. (negative) – (negative) e. (negative) – (positive) f. (negative) + (positive) 9. This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string? Chapter 10_The Other Side of Zero.indd 266 13-12-2024 16:08:29 266 Like general fractions, general integers (including zero and the negative numbers) were first conceived of and used in Asia, thousands of years ago, before they eventually spread across the world in more modern times. The first known instances of the use of negative numbers occurred in the context of accounting. In one of China’s most important mathematical works, The Nine Chapters on Mathematical Art (Jiuzhang Suanshu)—which was completed by the first or second century CE—positive and negative numbers were represented using red and black rods, much like the way we represented them using green and red tokens! Reprint 2025-26 There was a strong culture of accountancy also in India in ancient times. The concept of credit and debit was written about extensively by Kautilya in his Arthaśhāstra (c. 300 BCE), including the recognition that an account balance could be negative. The explicit use of negative numbers in the context of accounting is seen in a number of ancient Indian works, including in the Bakśhālī manuscript from around the year 300 CE, where a negative number was written using a special symbol placed after the number (rather than before the number as we do today). The first general treatment of positive numbers, negative numbers, and zero—all on an equal footing as equally-valid numbers on which one can perform the basic operations of addition, subtraction, multiplication, and even division—was given by Brahmagupta in his Brāhma-sphuṭa-siddhānta in the year 628 CE. Brahmagupta gave clear and explicit rules for operations on all numbers—positive, negative, and zero—that essentially formed the modern way of understanding these numbers that we still use today! Some of Brahmagupta’s key rules for addition and subtraction of positive numbers, negative numbers and zero are given below: Brahmagupta’s Rules for Addition (Brāhma-sphuṭa-siddhānta 18.30, 628 CE) 1. The sum of two positives is positive (e.g., 2 + 3 = 5). 2. The sum of two negatives is negative. To add two negatives, add the numbers (without the signs), and then place a minus sign to obtain the result (e.g., (– 2) + (– 3) = – 5). 3. To add a positive number and a negative number, subtract the smaller number (without the sign) from the greater number (without the sign), and place the sign of the greater number to obtain the result (e.g., – 5 + 3 = – 2, 2 + (– 3) = – 1 and – 3 + 5 = 2). 4. The sum of a number and its inverse is zero (e.g., 2 + (– 2) = 0). 5. The sum of any number and zero is the same number (e.g., – 2 + 0 = – 2 and 0 + 0 = 0). The Other Side of Zero Chapter 10_The Other Side of Zero.indd 267 13-08-2024 17:32:41 Reprint 2025-26 267 Ganita Prakash | Grade 6 Brahmagupta’s Rules for Subtraction (Brāhma-sphuṭa-siddhānta 18.31-18.32) 1. If a smaller positive is subtracted from a larger positive, the result is positive (e.g., 3 – 2 = 1). 2. If a larger positive is subtracted from a smaller positive, the result is negative (e.g., 2– 3 = – 1). 3. Subtracting a negative number is the same as adding the corresponding positive number (e.g., 2 – (– 3) = 2 + 3). 4. Subtracting a number from itself gives zero (e.g., 2 – 2 = 0 and – 2 – (– 2) = 0). 5. Subtracting zero from a number gives the same number (e.g., – 2 – 0 = – 2 and 0 – 0 = 0). Subtracting a number from zero gives the number’s inverse (e.g., 0 – (– 2) = 2). Once you understand Brahmagupta’s rules, you can do addition and subtraction with any numbers whatsoever — positive, negative, and zero! Figure it Out 1. Can you explain each of Brahmagupta’s rules in terms of Bela’s Building of Fun, or in terms of a number line? 2. Give your own examples of each rule. Chapter 10_The Other Side of Zero.indd 268 13-08-2024 17:32:41 268 Brahmagupta was the first to describe zero as a number on an equal footing with positive numbers as well as with negative numbers, and the first to give explicit rules for performing arithmetic operations on all such numbers, positive, negative, and zero—forming what is now called a ring. It would change the way the world does mathematics. However, it took many centuries for the rest of the world to adopt zero and negative numbers as numbers. These numbers were transmitted to, accepted by, and further studied by the Arab world by the 9th century, before making their way to Europe by the 13th century. Reprint 2025-26 Surprisingly, negative numbers were still not accepted by many European mathematicians even in the 18th century. Lazare Carnot, a French mathematician in the 18th century, called negative numbers ‘absurd’. But over time, zero as well as negative numbers proved to be indispensable in global mathematics and science, and are now considered to be critical numbers on an equal footing with and as important as positive numbers—just as Brahmagupta had recommended and explicitly described way back in the year 628 CE! This abstraction of arithmetic rules on all numbers paved the way for the modern development of algebra, which we will learn about in future classes. There are numbers that are less than zero. They are written with a ‘–’ The numbers ..., – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, ... are called integers. The numbers 1, 2, 3, 4, ... are called positive integers and the numbers …, – 4, – 3, – 2, – 1 are called negative integers. Zero (0) is neither positive nor negative. Every given number has another number associated to it which when added to the given number gives zero. This is called the additive inverse of the number. For example, the additive inverse of 7 is – 7 and the additive inverse of – 543 is 543. Addition can be interpreted as Starting Position + Movement = Target Position. Addition can also be interpreted as the combination of movements or increases/decreases: Movement 1 + Movement 2 = Total Movement. Subtraction can be interpreted as Target Position – Starting Position = Movement. sign in front of them (e.g., – 2), and are called negative numbers. They lie to the left of zero on the number line. Summary The Other Side of Zero Chapter 10_The Other Side of Zero.indd 269 13-08-2024 17:32:43 Reprint 2025-26 269 Ganita Prakash | Grade 6 In general, we can add two numbers by following Brahmagupta’s Rules for Addition: a. If both numbers are positive, add the numbers and the result is a positive number (e.g., 2 + 3 = 5). b. If both numbers are negative, add the numbers (without the signs), and then place a minus sign to obtain the result (– 2 + (– 3) = – 5). c. If one number is positive and the other is negative, subtract the smaller number (without the sign) from the greater number (without the sign), and place the sign of the greater number to obtain the result (e.g., – 5 + 3 = – 2). d. A number plus its additive inverse is zero (e.g., 2 + (– 2) = 0). e. A number plus zero gives back the same number (e.g., – 2 + 0 = – 2). We can subtract two integers by converting the problem into an addition problem and then following the rules of addition. Subtraction of an integer is the same as the addition of its additive inverse. Integers can be compared: … – 3 < – 2 < – 1 < 0 < +1 < +2 < +3 < ... Smaller numbers are to the left of larger numbers on the number line. We can give meaning to positive and negative numbers by interpreting them as credits and debits. We can also interpret positive numbers as distances above a reference point like the ground level. Similarly, negative numbers can be interpreted as distances below the ground level. When measuring temperatures in degrees Celsius, positive temperatures are those above the freezing point of water, and negative temperatures are those below the freezing point of water. Chapter 10_The Other Side of Zero.indd 270 13-08-2024 17:32:43 270 Reprint 2025-26 Rules • This is a two player game. Each player has 1 pawn. Both players start at 0. Players can reach either – 50 or + 50 to win but need not decide or fix this before or during play. • Each player rolls two dice at a time. One dice has numbers from + 1 to + 6 and the other dice has numbers from – 1 to – 6. • After each roll of the two dice, the player can add or subtract them in any order and then move the steps that indicate the result. A positive result means moving towards + 50 and a negative result means moving towards – 50. Integers: Snakes and Ladders The Other Side of Zero Chapter 10_The Other Side of Zero.indd 271 13-08-2024 17:32:44 Reprint 2025-26 271 Le a r n i n g Mat e r i al Sh e e ts Chapter 10_The Other Side of Zero.indd 272 13-08-2024 17:32:44 Reprint 2025-26 Learning Material Sheets.indd 273 06-08-2024 11:15:09 Reprint 2025-26 Learning Material Sheets.indd 274 06-08-2024 11:15:10 Reprint 2025-26 Learning Material Sheets.indd 275 06-08-2024 11:15:10 Reprint 2025-26 Learning Material Sheets.indd 276 06-08-2024 11:15:10 Reprint 2025-26 Note: Cut each shape along the white border. A Tangram C B D E Learning Material Sheets.indd 277 06-08-2024 11:15:10 F G Reprint 2025-26 Learning Material Sheets.indd 278 06-08-2024 11:15:10 Reprint 2025-26 Note: Cut each shape along the white border. 1 U N IT 2 2 1 2 2 3 3 3 Fraction Wall 2 4 3 4 4 4 3 5 4 5 5 5 3 6 4 6 5 6 6 6 3 7 4 7 5 7 6 7 7 7 4 8 5 8 6 8 7 8 8 8 4 9 5 9 6 9 7 9 8 9 9 9 6 10 8 10 9 10 10 10 5 10 7 10 Learning Material Sheets.indd 279 06-08-2024 11:15:10 1 3 1 4 Reprint 2025-26 1 5 2 5 1 6 2 6 1 7 2 7 1 8 2 8 3 8 1 9 2 9 3 9 1 10 2 10 3 10 4 10 Learning Material Sheets.indd 280 06-08-2024 11:15:10 Reprint 2025-26 Note: Cut the tiles along the white border. Learning Material Sheets.indd 281 06-08-2024 11:15:10 Reprint 2025-26 Learning Material Sheets.indd 282 06-08-2024 11:15:10 Reprint 2025-26 Notes Learning Material Sheets.indd 283 06-08-2024 11:15:10 Reprint 2025-26 Notes Learning Material Sheets.indd 284 06-08-2024 11:15:10 Reprint 2025-26 Page no. 243 Q. What do you press to go four floors up? What do you press to go three floors down? Ans. • + + + + or + 4 • - - - or – 3 Page No. 244 Q. Number all the floors in the Building of Fun. Ans. +3: Book Store Page No. 245 Q.1. You start from Floor + 2 and press – 3 in the lift. Where will you reach? Write an expression for this movement. Figure it out +2: Art Centre +1: Food Court 0: Welcome Hall -1: Toy Store -2: Video Game. CHAPTER 10 — SOLUTIONS The Other Side of Zero Ans. (+2) + (-3) = -1; Toy store. Q.2. Evaluate these expressions (you may think of them as Starting Floor + Movement by referring to the Building of Fun). Ans. a. (+5) b. (+5) c. (+1) d. (+1) a. (+ 1) + (+ 4) = _______ b. (+ 4) + (+ 1) = _______ c. (+ 4) + (– 3) = _______ d. (– 1) + (+ 2) = _______ e. (– 1) + (+ 1) = _______ f. 0 + (+ 2) = _________ g. 0 + (– 2) = _________ e. 0 f. (+2) g. (-2) [1] Q.3. Starting from different floors, find the movements required to reach Floor – 5. For example, if I start at Floor + 2, I must press – 7 to reach Floor – 5. The expression is (+ 2) + (– 7) = – 5. Find more such starting positions and the movements needed to reach Floor – 5 and write the expressions. Ans. Starting floor + Movement = Target Floor Page No. 246 Figure it out Q. Evaluate these expressions by thinking of them as the resulting movement of combining button presses: Ans. a. (+5) b. (+5) c. (-1) d. (-2) Q. Write the inverses of these numbers: Ans. Inverse of +4 = -4 a. (+ 1) + (+ 4) = _____________ b. (+ 4) + (+ 1) = _____________ c. (+ 4) + (– 3) + (– 2) = _______ d. (– 1) + (+ 2) + (– 3) = _______ +1 + (-6) = (-5) -2 + (-3) = (-5) 0 + (-5) = (-5) (Try more possibilities) +4, –4, –3, 0, +2, –1. Q. Connect the inverses by drawing lines. Inverse of -4 = +4 Inverse of -3 = +3 Inverse of 0 = 0 Inverse of +2 = (-2) Inverse of -1 = +1 (+5) (-7) (–8) (+9) (–9) (+8) (-5) (+7) [2] Ans. (+5) (-7) (–8) (+9) Ans. Binnu is on the lowest floor. Page No. 247 Figure it out Q.1. Compare the following numbers using the Building of Fun and fill in the boxes with < or >. Ans. a. (–2) +5 b. (–5) + 4 c. (–5) – 3 Q. Who is on the lowest floor? 1. Jay is in the Art Centre. So, he is on Floor +2. 2. Asin is in the Sports Centre. So, she is on Floor ___. 3. Binnu is in the Cinema Centre. So, she is on Floor ____. 4. Aman is in the Toys Shop. So, he is on Floor ____. (–9) (+8) (-5) (+7) 1. +2 2. +5 3. -3 4. -1 a. Ans. a. < b. < c. < d. > e. > f. < Q.2. Imagine the Building of Fun with more floors. Compare the numbers and fill in the boxes with < or >: a. (–10) (–12) b. + 17 (–10) c. 0 (–20) d. + 9 (–9) e. (–25) (–7) f. + 15 (–17) Ans. a. > b. > c. > d. > e. < f. > d. + 6 – 6 e. 0 – 4 f. 0 + 4 < < < < < < < [3] Q.3. If Floor A = – 12, Floor D = – 1 and Floor E = + 1 in the building shown on the right as a line, find the numbers of Floors B, C, F, G and H. Ans. B = -9 C = -6 F = +2 G = +6 H = +11 Q.4. Mark the following floors of the building shown on the right. Ans. a. P = -7 b. Q = - 4 c. R = +3 d. S = -10 Page 248 Q. Evaluate 15 – 5, 100 – 10 and 74 – 34 from this perspective. Page No. 249 Figure it out Q. Complete these expressions. You may think of them as finding the movement needed to reach the Target Floor from the Starting Floor. • 15 -5 can be expressed as 5 + ? = 15 . So, ? = 10 • 100 – 10 can be expressed as 10 + ? = 100. So, ? = 90 • 74 – 34 can be expressed as 34 + ? = 74. So, ? = 40 a. (+ 1) – (+ 4) = _______ b. (0) – (+ 2) = _________ c. (+ 4) – (+ 1) = _______ d. (0) – (– 2) = _________ e. (+ 4) – (– 3) = _______ f. (– 4) – (– 3) = ________ g. (– 1) – (+ 2) = _______ h. (– 2) – (– 2) = ________ i. (– 1) – (+1) = _______ j. (+ 3) – (– 3) = ________ a. -7 b. -4 c. +3 d. -10 +5 G Q P C S R B D E A -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 +2 +3 +4 +8 -2 -1 0 +1 +6 +7 Ans. a. (-3) b. (-2) c. (+3) d. (+2) e. (+7) f. (-1) g. (-3) h. (0) i. (-2) j. (+6) [4] Page No. 251 Q. Complete these expressions. Ans. a. (+160) b. (-240) c. (+250) d. (-150) Page No. 252 Q. In the other exercises that you did above, did you notice that subtracting a negative number was the same as adding the corresponding positive number? Ans. Yes. The infinite lift reminds us of a number line. Q. If, from 5 you wish to go over to 9, how far must you travel along the number line? Ans. 4 steps. Q. Now, from 9, if you wish to go to 3, how much must you travel along the number line? Ans. One must travel 6 steps backward or move (-6) from 9 to go to 3. Q. Now, from 3, if you wish to go to – 2, how far must you travel? Ans. One must travel 5 steps backward or move (-5) from 3 to go to -2. Figure it out a. (+ 40) + ______ = + 200 b. (+ 40) + _______ = – 200 c. (– 50) + ______ = + 200 d. (– 50) + _______ = – 200 e. (– 200) – (– 40) = _______ f. (+ 200) – (+ 40) = _______ g. (– 200) – (+ 40) = _______ e. (-160) f. (+160) g. (-240) Page No. 253 Figure it out Q1. Mark 3 positive numbers and 3 negative numbers on the number line above. Ans. Mark A = 2, B = 5, C = 8 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 R Q P A C P = -1, Q = -3, R = -7 (Try other possibilities) [5] B Q2. Write down the above 3 marked negative numbers in the following boxes: Ans. -7, -3, -1 Q3. Is 2 > – 3? Why? Is – 2 < 3? Why? Ans. Yes, 2 > -3. On the number line 2 is on the right side of (-3). Q4. What are (i) – 5 + 0 (ii) 7 + (– 7) (iii) – 10 + 20 (iv) 10 – 20 (v) 7 – (– 7) Ans. (i) – 5 + 0 = -5(ii) 7 + (– 7) = 0 (iii) – 10 + 20 =10 Page No. 255 Q. Use unmarked number lines to evaluate these expressions: Ans. (vi) – 8 – (– 10)? a. – 125 + (– 30) = _______ b. + 105 – (– 55) = _______ c. + 80 – (– 150) = _______ d. – 99 – (– 200) = _______ a. – 125 + (– 30) = -155 b. + 105 – (– 55) = 160 Yes, -2 < 3, as 3 is on the right of -2. (iv) 10 – 20 = -10 (v) 7 – (– 7) = 14 (vi) – 8 – (– 10) = 2 < < < Page No. 257 Figure it out Q.1. Complete the additions using tokens. Ans. a. +10 c. + 80 – (– 150) = 230 d. – 99 – (– 200) = 101 a. (+ 6) + (+ 4) b. (– 3) + (– 2) c. (+ 5) + (– 7) d. (– 2) + (+ 6) b. -5 [6] Q.2. Cancel the zero pairs in the following two sets of tokens. On what floor is the lift attendant in each case? What is the corresponding addition statement in each case? a. b. Ans. a. (+3) + (-5) = (-2) b. (+6) + (-3) = (+3) Page No. 258 Figure it out Q.1. Evaluate the following differences using tokens. Check that you get the same result as with other methods you now know: Ans. a. a. (+ 10) – (+ 7) b. (– 8) – (– 4) c. (– 9) – (– 4) d. (+ 9) – (+ 12) e. (– 5) – (– 7) f. (– 2) – (– 6) c. (-2) d. (+4) Attendant is at (-2) Floor Attendant is at (+3) Floor (+10) – (+7) = (+3) b. (-8) – (-4) = (-8) + (+4) = (-4) c. (-9) – (-4) = (-9) + (+4) = (-5) d. (+9) – (+12) = (-3) e. [7] Try for other methods. Q.2. Complete the subtractions: Ans. a. +2 Section 10.2 Page No. 259 a. (– 5) – (– 7) b. (+ 10) – (+ 13) c. (– 7) – (– 9) d. (+ 3) – (+ 8) e. (– 2) – (– 7) f. (+ 3) – (+ 15) (-5) - (-7) = (+2) f. (-2) – (-6) = +4 b. -3 c. +2 d. (-5) e. +5 f. (-12) Q.1. Try to subtract: – 3 – (+ 5). Ans. -8 Q.2. Evaluate the following using tokens. Ans. a. -13 b. +15 c. -14 d. -19 e. +10 f. -9 Figure it out How many zero pairs will you have to put in? What is the result? a. (– 3) – (+ 10) b. (+ 8) – (– 7) c. (– 5) – (+ 9) d. (– 9) – (+ 10) e. (+ 6) – (– 4) f. (– 2) – (+ 7) We have to put five zero pairs. [8] Section 10.3 Page No. 259 Q. Your new bank balance is _______. Ans. ₹ 100 + ₹ 60 = ₹ 160 Q. Your bank balance is now ______. Ans. ₹ 160 - ₹ 30 = ₹ 130 Q. What is your bank balance now? ______ Is this possible? Ans. ₹ 130 - ₹ 150 = - ₹ 20 Page No. 260 Q. What is your balance now? ______ Ans. ₹ 180 Section 10.3 Page 260 Q1. Suppose you start with 0 rupees in your bank account, and then you have credits of ₹30, ₹40, and ₹50, and debits of ₹40, ₹50, and ₹60. What is your bank account balance now?] Ans. (+30) + (+40) + (+50) – (40) – (50) – (60) Balance amount in Bank Account = ₹ (-30) Figure it Out Q2. Suppose you start with 0 rupees in your bank account, and then you have debits of ₹1, 2, 4, 8, 16, 32, 64, and 128, and then a single credit of ₹256. What is your bank account balance now? Ans. (1) – (2) – (4) – (8) – (16) – (32) – (64) – (128) + (256) Balance amount in the Bank Account = ₹ 1 [9] Page 261 Q.1. Looking at the geographical cross section fill in the respective heights: A B C D E. F. G. Ans. Approximate heights are: A. (+1500) m Q.2. Which is the highest point in this geographical cross-section? Which is the lowest point? Ans. Highest point is A & the lowest point is D. Q.3. Can you write the points A, B, …., G in a sequence of decreasing order of heights? Can you write the point in a sequence of increasing order of heights? Ans. A, E, C, G, F, B, D. (Decreasing order) Figure it Out B. (-500) m C. (+300) m D. (-1200) m E. (+1200) m F. (-200) m G. (+100) m D, B, F, G, C, E, A. (Increasing order) Q.4. What is the highest point above sea level on Earth? What is its height? Ans. Mount-Everest. Its height above sea level is 8848m. Q.5. What is the lowest point with respect to sea level on land or on the ocean floor? What is its height? (This height should be negative). Ans. The lowest know point on the Earth is the challenger deep, located in the Mariana trench in the Pacific Ocean. Its depth is approximately -10994 m. [10] Page 262 Q.2 Leh in Ladakh gets very cold during the winter. The following is a table of temperature readings taken during different times of the day and night in Leh on a day in November. Match the temperature with the appropriate time of the day and night. Ans. Section 10.4 Page 263 Q.1 Do the calculations for the second grid above and find the border sum. Ans. -3, -3, -3, -3 Q.2 Complete the grids to make the required border sum: Ans. One of the ways: Figure it Out Figure it Out Temperature Time 14°C 02:00 p.m. 8°C 11:00 a.m –2°C 11:00 p.m. –4°C 02:00 a.m. -10 10 4 5 -5 9 -10 5 Border sum +4 6 8 -16 11 -5 -19 -2 19 Border sum -2 7 -2 -9 -3 -5 -8 -6 10 Border sum -4 Think of other ways. [11] Page 265 Q.2 Q.1 Write all the integers between the given pairs, in increasing order. Ans. a. -6,-5,-4,-3,-2,-1 b. -3,-2,-1,0,1,2,3 c. -14,-13,-12,-11,-10,-9 d. -29,-28,-27,-26,-25,-24 Q.2 Give three numbers such that their sum is –8. Ans. One of the combinations is -5,7,-10. Q.3 There are two dice whose faces have these numbers: –1, 2, –3, 4, –5, 6. The smallest possible sum upon rolling these dice is –10 = (–5) + (–5) and the largest possible sum is 12 = (6)+(6). Some numbers between (–10) and (+12) are not possible to get by adding numbers on these two dice. Find those numbers. Ans. -9,-7,-5,0,2,7,9,11 Figure it Out a. 0 and –7 b. –4 and 4 c. –8 and –15 d. –30 and –23 Think of other combinations. Figure it Out -8 -14 Q.4 Solve these: Ans. -5 -21 -5 -21 -5 5 5 21 [12] Q.5 Find the years below. Hint: Recall that there was no year 0. Ans. c. -360 BCE Q.6 Complete the following sequences: Ans. a. -16, -10, -4 Q.7 Here are six integer cards: (+1), (+7), (+18), (–5), (–2), (–9). You can pick any of these and make an expression using addition(s) and subtraction(s). Here is an expression: (+18)+(+1)–(+7) – (–2) which gives a value (+14). Now, pick cards and make an expression such that its value is closer to (– 30). Ans. One of the ways is (-2) + (-9) – (+18) – (+1) = -30 Q.8 The sum of two positive integers is always positive but a (positive integer) – (positive integer) can be positive or negative. What about a. (positive) – (negative) b. (positive) + (negative) c. (negative) + (negative) d. (negative) – (negative) e. (negative) – (positive) f. (negative) + (positive) a. From the present year, which year was it 150 years ago? ________ b. From the present year, which year was it 2200 years ago? _______ c. What will be the year 320 years after 680 BCE? ________ a. (–40), (–34), (–28), (–22), _____, ______, ______ b. 3, 4, 2, 5, 1, 6, 0, 7, _____, _____, _____ c. _____, ______, 12, 6, 1, (–3), (–6), _____, ______, ______ b. -1,8, -2 c. 27, 19, … -8,-9,-9 Ans. a. positive b. can be positive or negative c. negative d. can be positive or negative e. negative f. can be positive or negative Q.9 This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string? Ans. 20 [13]" class_7,1,Large Numbers Around Us,ncert_books/class_7/gegp1dd/gegp101.pdf,"1 1.1 A Lakh Varieties! Eshwarappa is a farmer in Chintamani, a town in Karnataka. He visits the market regularly to buy seeds for his rice field. During one such visit he overheard a conversation between Ramanna and Lakshmamma. Ramanna said, “Earlier our country had about a lakh varieties of rice. Farmers used to preserve different varieties of seeds and use them to grow rice. Now, we only have a handful of varieties. Also, farmers have to come to the market to buy seeds”. Lakshmamma said, “There is a seed bank near my house. So far, they have collected about a hundred indigenous varieties of rice seeds from different places. You can also buy seeds from there.” You may have heard the word ‘lakh’ before. Do you know how big one lakh is? Let us find out. Eshwarappa shared this incident with his daughter Roxie and son Estu . Estu was surprised to know that there were about one lakh varieties of rice in this country. He wondered “One lakh! So far I have only tasted 3 varieties. If we tried a new variety each day, would we even come close to tasting all the varieties in a lifetime of 100 years?” What do you think? Guess. LARGE NUMBERS AROUND US Chapter-1.indd 1 4/12/2025 12:13:49 PM Ganita Prakash | Grade 7 But how much is one lakh? Observe the pattern and fill in the boxes given below. The smallest 4-digit number is The smallest 5-digit number is The smallest 6-digit number is 1,00,000 The largest 3-digit number is 999 The largest 4-digit number is The largest 5-digit number is 99,995 1,00,000 is read as “One Lakh” 99,996 99,998 +1 +1 +1 Chapter-1.indd 2 4/12/2025 12:13:49 PM What if a person ate 3 varieties of rice every day? Will they be able to taste all the lakh varieties in a 100 year lifetime? Find out. Estu said, “We know how many days there are in a year — 365, if we ignore leap years. If we live for y years, the number of days in our lifetime will be 365 × y.” 2 Roxie and Estu found that if they ate one variety of rice a day, they would come nowhere close to a lakh in a lifetime! Roxie suggests, “What if we ate 2 varieties of rice every day? Would we then be able to eat 1 lakh varieties of rice in 100 years?” Choose a number for y. How close to one lakh is the number of days in y years, for the y of your choice? Figure it Out 1. According to the 2011 Census, the population of the town of Chintamani was about 75,000. How much less than one lakh is 75,000? 2. The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000? 3. By how much did the population of Chintamani increase from 2011 to 2024? Getting a Feel of Large Numbers You may have come across interesting facts like these: It is not always easy to get a sense of how big these measurements are. But, we can get a better sense of their size when we compare them with something familiar. Let us see an example. Look at the picture on the right. Somu is 1 metre tall. If each floor is about four times his height, what is the approximate height of the building? Which is taller — The Statue of Unity or this building? How much taller? ____________m. • The world’s tallest statue is the ‘Statue of Unity’ in Gujarat depicting Sardar Vallabhbhai Patel. Its height is about 180 metres. • Kunchikal waterfall in Karnataka is said to drop from a height of about 450 metres. Large Numbers Around Us Chapter-1.indd 3 4/12/2025 12:13:50 PM We can see that the height of the Statue of Unity is close to 4 times the height of Somu’s building. How much taller is the Kunchikal waterfall than Somu's building? ____________m. How many floors should Somu’s building have to be as high as the waterfall? ____________ . 500 400 300 200 100 0 me te rs 3 Ganita Prakash | Grade 7 Is One Lakh a Very Large Number? Eshwarappa asked Roxie and Estu, “Is a lakh big or small?” Roxie feels that 1 lakh is a large number: Estu, however, thinks it is not that big: How do you view a lakh — is a lakh big or small? Reading and Writing Numbers • “We had one lakh varieties of rice — that is a lot.” • “Living 1 lakh days would mean living for 274 years — that is a really long time!” • “If 1 lakh people stood shoulder to shoulder in a line they could stretch as far as 38 kilometres.” • “Do you know that the cricket stadium in Ahmedabad has a seating capacity of more than 1 lakh? One lakh people in such a small area!” • “Most humans have 80,000 to 1,20,000 hairs on their heads. Imagine, 1 lakh hairs fit in such a tiny space!” • “I heard that there are some species of fish where a female fish can lay almost one lakh eggs at once very comfortably. Some even lay tens of lakhs of eggs at a time.” Chapter-1.indd 4 4/12/2025 12:13:50 PM We have already been using commas for 5-digit numbers like 45,830 in the Indian place value system. As numbers grow bigger, using commas helps in reading the numbers easily. We use a comma in between the digits representing the “ten thousands” place and the “one lakh” place as you have seen just before (1,00,000). The number name of 12,78,830 is twelve lakh seventy eight thousand eight hundred thirty. Similarly, the number 15,75,000 in words is fifteen lakh seventy five thousand. Write each of the numbers given below in words: 4 (a) 3,00,600 (b) 5,04,085 (c) 27,30,000 (d) 70,53,138 Write the corresponding number in the Indian place value system for each of the following: 1.2 Land of Tens In the Land of Tens, there are special calculators with special buttons. 1. The Thoughtful Thousands only has a +1000 button. How many times should it be pressed to show: 2. The Tedious Tens only has a +10 button. How many times should it be pressed to show: Note to the Teacher: Encourage students to make connections between these facts. For example, can the whole population of Chintamani fit in the stadium? How can we imagine the line of 38 km, having a lakh people, sitting next to each other in the stadium? (a) One lakh twenty three thousand four hundred and fifty six (b) Four lakh seven thousand seven hundred and four (c) Fifty lakhs five thousand and fifty (d) Ten lakhs two hundred and thirty five (a) Three thousand? 3 times (b) 10,000? ____________ (c) Fifty three thousand? ___________ (d) 90,000? ______________ (e) One Lakh? ________________ (f) ____________? 153 times (g) How many thousands are required to make one lakh? Large Numbers Around Us Chapter-1.indd 5 4/12/2025 12:13:51 PM 3. The Handy Hundreds only has a +100 button. How many times should it be pressed to show: (a) Five hundred? _____________ (b) 780? _________ (c) 1000? _________ (d) 3700? ________ (e) 10,000? ___________ (f) One lakh? _____________ (g) ____________? 435 times (a) Four hundred? ___________times (b) 3,700? __________ 5 Ganita Prakash | Grade 7 4. Creative Chitti is a different kind of calculator. It has the following buttons: +1, +10, +100, +1000, +10000, +100000 and +1000000. It always has multiple ways of doing things. “How so?”, you might ask. To get the number 321, it presses +10 thirty two times and +1 once. Will it get 321? Alternatively, it can press +100 two times and +10 twelve times and +1 once. 5. Two of the many different ways to get 5072 are shown below: These two ways can be expressed as: (a) (50 × 100) + (7 × 10) + (2 × 1) = 5072 (c) 10,000? __________ (d) Fifty three thousand? __________ (e) 90,000? __________ (f) 97,600? __________ (g) 1,00,000? __________ (h) _________? 582 times (i) How many hundreds are required to make ten thousand? (j) How many hundreds are required to make one lakh? (k) Handy Hundreds says, “There are some numbers which Tedious Tens and Thoughtful Thousands can’t show but I can.” Is this statement true? Think and explore. +10,00,000 Buttons 5072 Chapter-1.indd 6 4/12/2025 12:13:51 PM Find a different way to get 5072 and write an expression for the same. Figure it Out For each number given below, write expressions for at least two different ways to obtain the number through button clicks. Think like Chitti and be creative. (a) 8300 6 (b) 40629 (c) 56354 (b) (3×1000) + (20×100) + (72 ×1) = 5072 +1,00,000 +10,000 +1,000 3 +100 50 20 +10 7 +1 2 72 Creative Chitti has some questions for you — Create questions like these and challenge your classmates. Systematic Sippy is a different kind of calculator. It has the following buttons: +1, +10, +100, +1000, +10000, +100000. It wants to be used as minimally as possible. How can we get the numbers (a) 5072, (b) 8300 using as few button clicks as possible? Find out which buttons should be clicked and how many times to get the desired numbers given in the table. The aim is to click as few buttons as possible. Here is one way to get the number 5072. This method uses 23 button clicks in total. Is there another way to get 5072 using less than 23 button clicks? Write the expression for the same. (d) 66666 (e) 367813 (a) You have to make exactly 30 button presses. What is the largest 3-digit number you can make? What is the smallest 3-digit number you can make? (b) 997 can be made using 25 clicks. Can you make 997 with a different number of clicks? +10,00,000 +1,00,000 Buttons 5072 +10,000 +1,000 5 Large Numbers Around Us Chapter-1.indd 7 4/12/2025 12:13:51 PM Figure it Out 1. For the numbers in the previous exercise, find out how to get each number by making the smallest number of button clicks and write the expression. 2. Do you see any connection between each number and the corresponding smallest number of button clicks? 3. If you notice, the expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so. +100 0 +10 6 +1 12 7 Math Talk Ganita Prakash | Grade 7 What if we press the +10,00,000 button ten times? What number will come up? How many zeroes will it have? What should we call it? The number will be 100 lakhs, which is also called a crore. 1 crore is written as 1,00,00,000 — it is 1 followed by seven zeroes. 1.3 Of Crores and Crores! The table below shows some numbers according to both the Indian system and the American system (also called the International system) of naming numerals and placing commas. Observe the placement of commas in both systems. Notice that in the Indian system, commas are placed to group the digits in a 3-2-2-2... pattern from right to left (thousands, lakhs, crores, etc.). In the American system, the digits are grouped uniformly in a 3-3-3-3… pattern from right to left (thousands, millions, billions, etc.). The Indian system of naming numbers is also followed in Bhutan, Nepal, Sri Lanka, Pakistan, Bangladesh, Maldives, Afghanistan, and Myanmar. The words lakh and crore originate from the Sanskrit words lakṣha (लक्ष) and koṭi (कोटि). The American system is also used in many countries. Observe the number of zeroes in 1 lakh and 1 crore. 1 lakh, written in numbers would be 1 followed by 5 zeroes. 1 crore, written in numbers would be 1 followed by 7 zeroes. A lakh is a hundred times a thousand, a crore is a hundred times a lakh and an arab is a hundred times a crore (i.e., a hundred thousand is 1 lakh, 100 lakhs is 1 crore, and 100 crores is 1 arab). 1,000 One thousand 1,000 One thousand 10,000 Ten thousand 10,000 Ten thousand 1,00,000 One lakh 100,000 Hundred thousand 10,00,000 Ten lakhs 1,000,000 One million 1,00,00,000 One crore 10,000,000 Ten million 10,00,00,000 Ten crores 100,000,000 Hundred million 1,00,00,00,000 One arab or One hundred crores 1,000,000,000 One billion Indian System American System Chapter-1.indd 8 4/12/2025 12:13:52 PM How many zeros does a thousand lakh have? _____ 8 How many zeros does a hundred thousand have? ____ The number 9876501234 can be easily read by placing commas first: Figure it Out 1. Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems: 2. Write the following numbers in Indian place value notation: 3. Compare and write ‘<’, ‘>’ or ‘=’: We shall come across even bigger numbers in later grades. (a) 9,87,65,01,234 → 9 arab 87 crore 65 lakhs 1 thousand and 234 or 987 crore 65 lakh 1 thousand 234 (in the Indian system). (b) 9,876,501,234 → 9 billion 876 million 501 thousand 234 (in the American system). (a) 4050678 (b) 48121620 (c) 20022002 (d) 246813579 (e) 345000543 (f) 1020304050 (a) One crore one lakh one thousand ten (b) One billion one million one thousand one (c) Ten crore twenty lakh thirty thousand forty (d) Nine billion eighty million seven hundred thousand six hundred (a) 30 thousand ____ 3 lakhs (b) 500 lakhs ______ 5 million (c) 800 thousand ____ 8 million (d) 640 crore ______ 60 billion Large Numbers Around Us Chapter-1.indd 9 4/12/2025 12:13:52 PM 1.4 Exact and Approximate Values 1 lakh people visited the book fair. Ohh, so many people! If I had not gone, they would have written “99,999 people visited the book fair last week”. 9 Ganita Prakash | Grade 7 What do you think of this conversation? Have you read or heard such headlines or statements? Very often, exact numbers are not required and just an approximation is sufficient. For example, according to the 2011 census, the population of Chintamani town is 76,068. Instead, saying that the population is about 75,000 is enough to give an idea of how big the quantity is. Bappi, call and check if Toofan Express is on time. What number did you call? I called the rounded off number, of course! My teacher told us that efficient people deal with rounded off numbers. Chapter-1.indd 10 4/12/2025 12:13:54 PM There are situations where it makes sense to round up a number (rounding up is when the approximated number is more than the actual number). For example, if a school has 732 people including students, teachers and staff: the principal might order 750 sweets instead of 700 sweets. There are situations where it is better to round down (rounding down is when the approximated number is less than the actual number). For example, if the cost of an item is ₹470, the shopkeeper may say that the cost is around ₹450 instead of saying it is around ₹500. Think and share situations where it is appropriate to (a) round up, (b) round down, (c) either rounding up or rounding down is okay and (d) when exact numbers are needed. 10 Nearest Neighbours With large numbers it is useful to know the nearest thousand, lakh or crore. For example, the nearest neighbours of the number 6,72,85,183 are shown in the table below. Similarly, write the five nearest neighbours for these numbers: I have a number for which all five nearest neighbours are 5,00,00,000. What could the number be? How many such numbers are there? Roxie and Estu are estimating the values of simple expressions. 1. 4,63,128 + 4,19,682, Roxie: “The sum is near 8,00,000 and is more than 8,00,000.” Estu: “The sum is near 9,00,000 and is less than 9,00,000.” (a) 3,87,69,957 (b) 29,05,32,481 (a) Are these estimates correct? Whose estimate is closer to the sum? Nearest thousand 6,72,85,000 Nearest ten thousand 6,72,90,000 Nearest lakh 6,73,00,000 Nearest ten lakh 6,70,00,000 Nearest crore 7,00,00,000 Large Numbers Around Us Math Talk Chapter-1.indd 11 4/12/2025 12:13:54 PM 2. 14,63,128 – 4,90,020 Roxie: “The difference is near 10,00,000 and is less than 10,00,000.” Estu: “The difference is near 9,00,000 and is more than 9,00,000.” (b) Will the sum be greater than 8,50,000 or less than 8,50,000? Why do you think so? (c) Will the sum be greater than 8,83,128 or less than 8,83,128? Why do you think so? (d) Exact value of 4,63,128 + 4,19,682 = ___________ (a) Are these estimates correct? Whose estimate is closer to the difference? (b) Will the difference be greater than 9,50,000 or less than 9,50,000? Why do you think so? 11 Ganita Prakash | Grade 7 Populations of Cities Observe the populations of some Indian cities in the table below. Rank City Population (2011) Population (2001) Note to the Teacher: Ask students questions like — “what could the numbers be if the sum had to be less than 8,50,000."" 1 Mumbai 1,24,42,373 1,19,78,450 2 New Delhi 1,10,07,835 98,79,172 3 Bengaluru 84,25,970 43,01,326 4 Hyderabad 68,09,970 36,37,483 5 Ahmedabad 55,70,585 35,20,085 6 Chennai 46,81,087 43,43,645 7 Kolkata 44,86,679 45,72,876 (c) Will the difference be greater than 9,63,128 or less than 9,63,128? Why do you think so? (d) Exact value of 14,63,128 – 4,90,020 = __________ Chapter-1.indd 12 4/12/2025 12:13:54 PM 12 10 Pune 31,15,431 25,38,473 11 Jaipur 30,46,163 23,22,575 12 Lucknow 28,15,601 21,85,927 13 Kanpur 27,67,031 25,51,337 14 Nagpur 24,05,665 20,52,066 8 Surat 44,67,797 24,33,835 9 Vadodara 35,52,371 16,90,000 From the information given in the table, answer the following questions by approximation: 1. What is your general observation about this data? Share it with the class. 2. What is an appropriate title for the above table? 3. How much is the population of Pune in 2011? Approximately, by how much has it increased compared to 2001? 4. Which city’s population increased the most between 2001 and 2011? 5. Are there cities whose population has almost doubled? Which are they? 6. By what number should we multiply Patna’s population to get a number/population close to that of Mumbai? 15 Indore 19,60,631 14,74,968 16 Thane 18,18,872 12,62,551 17 Bhopal 17,98,218 14,37,354 18 Visakhapatnam 17,28,128 13,45,938 19 Pimpri-Chinchwad 17,27,692 10,12,472 20 Patna 16,84,222 13,66,444 Large Numbers Around Us Chapter-1.indd 13 4/12/2025 12:13:54 PM 1.5 Patterns in Products Roxie and Estu are playing with multiplication. They encounter an interesting technique for multiplying a number by 10, 100, 1000, and so on. A Multiplication Shortcut Roxie evaluated 116 × 5 as follows: 116 × 5 = 116 × 10 2 = 58 × 10 = 580. 58 13 Ganita Prakash | Grade 7 Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10? Figure it Out 1. Find quick ways to calculate these products: 2. Calculate these products quickly. (a) 25 × 12 = _____________ How Long is the Product? In each of the following boxes, the multiplications produce interesting patterns. Evaluate them to find the pattern. Extend the multiplications based on the observed pattern. Estu evaluated 824 × 25 as follows: (a) 2 × 1768 × 50 (b) 72 × 125 [Hint: 125 = 1000 8 ] (c) 125 × 40 × 8 × 25 (b) 25 × 240 = _____________ (c) 250 × 120 = _____________ (d) 2500 × 12 =_____________ (e) ______×______= 120000000 824 × 25 = 824 × 100 4 = 20600. 206 Math Talk Chapter-1.indd 14 4/12/2025 12:13:54 PM 14 11 × 11 = 111 × 111 = 1111 × 1111 = 3 × 5 = 33 × 35 = 333 × 335 = 66 × 61 = 666 × 661 = 6666 × 6661 = 101 × 101 = 102 × 102 = 103 × 103 = Observe the number of digits in the two numbers being multiplied and their product in each case. Is there any connection between the numbers being multiplied and the number of digits in their product? Roxie says that the product of two 2-digit numbers can only be a 3- or a 4-digit number. Is she correct? Should we try all possible multiplications with 2-digit numbers to tell whether Roxie’s claim is true? Or is there a better way to find out? Can multiplying a 3-digit number with another 3-digit number give a 4-digit number? Can multiplying a 4-digit number with a 2-digit number give a 5-digit number? Observe the multiplication statements below. Do you notice any patterns? See if this pattern extends for other numbers as well. She explains her reasoning: “We want to know about the number of digits in the product of two 2-digit numbers. To know the smallest such product I took 10 × 10, so all other products will be greater than 100. To know the greatest such product I multiplied the smallest 3-digit numbers (100 × 100) to get 10,000; so the product of all the 2-digit multiplications will be less than 10,000.” Large Numbers Around Us Chapter-1.indd 15 4/12/2025 12:13:54 PM 12-digit × 13-digit = or 1-digit × 1-digit = 1-digit or 2-digit 2-digit × 1-digit = 2-digit or 3-digit 2-digit × 2-digit = 3-digit or 4-digit 3-digit × 3-digit = 5-digit or 6-digit 5-digit × 5-digit = or 8-digit × 3-digit = or 15 Ganita Prakash | Grade 7 Fascinating Facts about Large Numbers Some interesting facts about large numbers are hidden below. Calculate the product or quotient to uncover the facts. Once you find the product or quotient, read the number in both Indian and American naming systems. Share your thoughts and questions about the fact with the class after you discover each number. 1250 × 380 ______________ is the number of kīrtanas composed by Purandaradāsa according to legends. Purandaradāsa was a composer and singer in the 15th century. His kīrtanas spanned social reform, bhakti and spirituality. He systematised methods for teaching Carnatic music which is followed to the present day. How many years did he live to compose so many songs? At what age did he start composing songs? If he composed 4,75,000 songs, how many songs per year did he have to compose? Chapter-1.indd 16 4/15/2025 5:18:03 PM 16 2100 × 70,000 _______________ is the approximate distance in kilometers, between the Earth and the Sun. This distance keeps varying throughout the year. The farthest distance is about 152 million kilometers. As you did before, divide the given numbers to uncover interesting facts about division. Share your thoughts and questions with the class after you uncover each number. How did they measure the distance between the Earth and the Sun? 6400 × 62,500 _________________ is the average number of litres of water the Amazon river discharges into the Atlantic Ocean every second. The river's flow into the Atlantic is so much that drinkable freshwater is found even 160 kilometers into the open sea. Large Numbers Around Us Chapter-1.indd 17 4/12/2025 12:13:55 PM 13,95,000 ÷ 150 _________________ is the distance (in kms) of the longest single-train journey in the world. The train runs in Russia between Moscow and Vladivostok. The duration of this journey is about 7 days. The longest train route in India is from Dibrugarh in Assam to Kanyakumari in Tamil Nadu; it covers 4219 kms in about 76 hours. 17 Ganita Prakash | Grade 7 Adult blue whales can weigh more than 10,50,00,000 ÷ 700 _________________ kilograms. A newborn blue whale weighs around 2,700 kg, which is similar to the weight of an adult hippopotamus. The heart of a blue whale was recorded to be nearly 700 kg. The tongue of a blue whale weighs as much as an elephant. Blue whales can eat up to 3500 kg of krill every day. The largest known land animal, Argentinosauras, is estimated to weigh 90,000 kgs. 52,00,00,00,000 ÷ 130 _________________ was the weight, in tonnes, of global plastic waste generated in the year 2021. Large Number Fact Chapter-1.indd 18 4/12/2025 12:13:55 PM 1.6 Did You Ever Wonder…? Estu is amused by all these interesting facts about large numbers. While thinking about these, he came up with an unusual question, “Could the entire population of Mumbai fit into 1 lakh buses?” What do you think? How can we find out? Let us assume a bus can accommodate 50 people. Then 1 lakh buses can accommodate 1 lakh × 50 = 50 lakh people. 18 In a single gram of healthy soil there can be 100 million to 1 billion bacteria and 1 lakh to 1 million fungi, which can support plants’ growth and health. Share such large-number facts you know / come across with your class. The population of Mumbai is more than 1 crore 24 lakhs (we saw this in an earlier table). So, the entire population of Mumbai cannot fit in 1 lakh buses. The RMS Titanic ship carried about 2500 passengers. Can the population of Mumbai fit into 5000 such ships? Inspired by this strange question, Roxie wondered, “If I could travel 100 kilometers every day, could I reach the Moon in 10 years?” (The distance between the Earth and the Moon is 3,84,400 km.) How far would she have travelled in a year? How far would she have travelled in 10 years? Is it not easier to perform these calculations in stages? You can use this method for all large calculations. Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometers every day. (You had written down the distance between the Earth and the Sun in a previous exercise.) Make necessary reasonable assumptions and answer the questions below: Large Numbers Around Us Chapter-1.indd 19 4/12/2025 12:13:55 PM Think and create more such fun questions and share them with your class. Figure it Out 1. Using all digits from 0 – 9 exactly once (the first digit cannot be 0) to create a 10-digit number, write the — (a) If a single sheet of paper weighs 5 grams, could you lift one lakh sheets of paper together at the same time? (b) If 250 babies are born every minute across the world, will a million babies be born in a day? (c) Can you count 1 million coins in a day? Assume you can count 1 coin every second. 19 Ganita Prakash | Grade 7 2. The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eighty five, which has 43 letters. Give a 7-digit number name which has the maximum number of letters. 3. Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist? 4. Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible. 5. The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also share a letter ‘t’. How far do you have to count to find two consecutive numbers which do not share an English letter in common? 6. Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10. (a) What would the 1000th digit be? At which number would it occur? 7. A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers: (a) Largest multiple of 5 (b) Smallest even number (b) What number would contain the millionth digit? (c) When would you have written the digit ‘5’ for the 5000th time? (a) 20,800 (b) 92,100 Try This Try This Chapter-1.indd 20 4/12/2025 12:13:55 PM 8. How many lakhs make a billion? 9. You are given two sets of number cards numbered from 1 – 9. Place a number card in each box below to get the (a) largest possible sum (b) smallest possible difference of the two resulting numbers. 20 (c) 1,20,500 (d) 65,30,000 (e) 70,25,700 10. You are given some number cards; 4000, 13000, 300, 70000, 150000, 20, 5. Using the cards get as close as you can to the numbers below using any operation you want. Each card can be used only once for making a particular number. (e) 20,90,800: 11. Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick. 12. Grey-headed albatrosses have a roughly 7-feet wide wingspan. They are known to migrate across several oceans. Albatrosses can cover about 900 – 1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take to cross the Pacific Ocean approximately? 13. A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour. 14. Bald eagles are known to fly as high as 4500 – 6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,000 – 12,800 m. How many times bigger are these heights compared to Somu’s building? (a) 1,10,000: Closest I could make is 4000 × (20 + 5) + 13000 = 1,13,000 (b) 2,00,000: (c) 5,80,000: (d) 12,45,000: Chapter-1.indd 21 4/12/2025 12:13:58 PM Ganita Prakash | Grade 7 • We came across large numbers — lakhs, crores and arabs; millions and billions. We learnt how to read and write these numbers in the Indian and American/International naming systems. (a) 1 lakh is 1 followed by 5 zeroes: 1,00,000 (b) 1 crore is 1 followed by 7 zeroes: 1,00,00,000 (c) 1 million is 1 followed by 6 zeroes: 1,000,000 (which is also ten lakhs) (d) 1 arab is 1 followed by 9 zeroes: 1,000,000,000 (which is also 100 crore or 1 billion) • We generally round up or round down large numbers. Many times it is enough just to know roughly how big or small something is. • To get a sense of large numbers or quantities, we can check how many times bigger they are compared to numbers or quantities that are more familiar. • We saw how to factorise numbers and regroup them to make multiplications simpler. • We carried out interesting thought experiments such as  — “Would one be able to watch 1000 movies in a year?” SUMMARY Chapter-1.indd 22 4/12/2025 12:13:58 PM 22 You can either use toothpicks or matchsticks, or just write the digits in this way, using lines to represent sticks. To make the digit 7, three sticks are needed. Write or make the number 5108. How many sticks are required? We can write digits as shown in the image below: 0123456789 1. Make or write the number 42,019. It would require exactly 23 sticks. 2. Starting with 42,019, add or write two more sticks, and make a bigger number. One example is 42,078. What other numbers bigger than 42,019 can you make in this way? 3. Preetham wants to insert the digit ‘1’ somewhere among the digits ‘4’, ‘2’, ‘0’, ‘1’ and ‘9’. Where should he place the digit ‘1’ to get the biggest possible number? 4. What other numbers can he make by placing the digit ‘1’? 1. Make or write the number 63,890. 2. Starting with 63,890, rearrange exactly four sticks and make a bigger number. One example is 88,078. What other numbers bigger than 63,890 can you make in this way? Toothpick Digits Chapter-1.indd 23 4/12/2025 12:14:00 PM Make your own questions and challenge each other. 1. Make any number using exactly 24 sticks or lines. 2. What is the biggest number that can be made using 24 sticks or lines? 3. What is the smallest number that can be made using 24 sticks or lines? Page No. 3 Figure it out 1. According to the 2011 Census, the population of the town of Chintamani was about 75,000. How much less than one lakh is 75,000? Ans: 75,000 is 25,000 less than 1 lakh. 2. The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000? Ans: 1,06,000 is 6,000 more than 1 lakh. 3. By how much did the population of Chintamani increase from Ans: Increase in population from 2011 to 2024 is 31,000. Ans: 40 m Which is taller — The Statue of Unity or this building? How much taller? ____________m. Ans: The Statue of Unity is 140 m taller than the building. 2011 to 2024? Look at the picture on the right. Somu is 1 metre tall. If each floor is about four times his height, what is the approximate height of the building? (Refer textbook for the picture) Large Numbers Around Us Chapter – 1 Page No. 4 Write each of the numbers given below in words: (a) 3,00,600 (b) 5,04,085 How tall is the Kunchikal waterfall compared to the building? 410 m. How many floors should Somu’s building have to be as high as the waterfall? approx 113 floors . Ans: Three lakh six hundred Ans: Five lakh four thousand eighty five 1 Page No. 5 Write the corresponding number for each of the following: Thoughtful thousand (c) 27,30,000 (d) 70,53,138 (a) One lakh twenty-three thousand four hundred and fifty-six. (b) Four lakh seven thousand seven hundred and four. (c) Fifty lakhs five thousand and fifty. (d) Ten lakhs two hundred and thirty-five. (a) Three thousand? 3 times (b) 10,000? 10 times (c) Fifty three thousand? 53 times (d) 90,000? 90 times Ans: Twenty seven lakh thirty thousand Ans: Seventy lakh fifty three thousand one hundred thirty eight Ans: 1,23,456 Ans: 4,07,704 Ans: 50,05,050 Ans: 10,00,235 Tedious tens (e) One Lakh? 100 times (f) 1,53,000 ? 153 times (g) How many thousands are required to make one lakh? 100 (a) Five hundred? 50 times (b) 780? 78 times (c) 1000? 100 times (d) 3700? 370 times (e) 10,000? 1000 times (f) One lakh? 10,000 times 2 Handy Hundreds Page No. 6 Find a different way to get 5072 and write an expression for the same. (g) 4350 ? 435 times (a) Four hundred? 4 times (b) 3,700? 37 times (c) 10,000? 100 times (d) Fifty three thousand? 530 times (e) 90,000? 900 times (f) 97,600? 976 times (g) 1,00,000? 1000 times (h) 58200 ? 582 times (i) How many hundreds are required to make ten thousand ? 100 (j) How many hundreds are required to make one lakh? 1000 (k) Handy Hundreds says, “There are some numbers which Tedious Tens and Thoughtful Thousands can’t show but I can”. Is this Ans: No statement true? Think and explore. Buttons 5072 +10,00,000 +1,00,000 +10,000 +1,000 5 4 1 +100 10 40 +10 7 7 6 +1 2 2 12 3 Figure it out 1. For each number given below, write expressions for at least two Ans: Some ways are given below. Try more ways. (a) 8300 (b) 40629 Ans: (40 × 1000) + (6 × 100) + (2 × 10) + (9 × 1) (c) 56354 Ans: (56 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) (d)66666 Ans: (66 × 1000) + (6 × 100) + (66 × 1) (e)367813 different ways to obtain the number through button clicks. Think like Chitti and be creative. Ans: (8 × 1000) + (3 × 100) Ans: (3 × 100000) + (6 × 10000) + (7 × 1000) ) + (8 × 100) + (1 × 10) + (3 × 1) or, (5 × 1000) + (33 × 100) or, (30 × 1000) + (106 × 100) +(29 × 1) or, (46 × 1000) + (103 × 100) + (54 × 1) or, (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) Page No. 7 Creative Chitti has some questions for you — (a)You have to make exactly 30 button presses. What is the largest 3- digit number you can make? What is the smallest 3-digit number you can make? Ans: 993 is the largest 3-digit number that can be made using 30 button presses. or, (36 × 10000) + (78 × 1000) + (13 × 1) 102 is the smallest 3-digit number that can be made using 30 button presses. 4 (b) 997 can be made using 25 clicks. Can you make 997 with a different number of clicks? Figure it out 1. For the numbers in the previous exercise, find out how to get each Ans: How can we get the numbers (a) 5072, (b) 8300, using as few button clicks as possible? Ans: 997 can also be made as – (a) 5072 = (5 × 1000) + (7 × 10) + (2 × 1) (b) 8300 = (8 × 1000) + (3 × 100) number by making the smallest number of button clicks and write the expression. Ans: (a) 8300 = (8 × 1000) + (3 × 100) (8 × 100) + (19 × 10) + (7 × 1) Number of clicks = 8 + 19 + 7 Number of clicks = 14 Number of clicks = 11 Number of clicks = 11 = 34 (b) 40629 = (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) (c) 56354 = (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) (d) 66666 = (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) (f) 367813 = (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) Number of clicks = 4 + 6 + 2 + 9 = 21 Number of clicks = 5 + 6 + 3 + 5 + 4 = 23 Number of clicks = 6 + 6 + 6 + 6 + 6 = 30 Number of clicks = 3 + 6 + 7 + 8 + 1 + 3 = 28 5 Page No. 8 Page No. 9 (a) 4050678 40,50,678 Four Millions Fifty Thousand Six Hundred Seventy-Eight (b) 48121620 4,81,21,620 Forty Eight Millions One Hundred Twenty One Thousand Six Hundred Twenty (c) 20022002 2,00,22,002 2. Do you see any connection between each number and the corresponding smallest number of button clicks? Figure it out 1. Read the numbers and write their number names in both the Indian and American systems: Ans: The smallest number of buttons click for each number is the sum of Sl. No. Indian System American System How many zeros does a thousand lakh have? 8 zeros How many zeros does a hundred thousand have? 5 zeros its digit. Forty Lakh Fifty Thousand Six Hundred Seventy-Eight Four Crore Eighty-One Lakh Twenty-One Thousand Six Hundred Twenty 4,050,678 48,121,620 Twenty millions twenty two thousand and two (d) 246813579 24,68,13,579 Two Hundred Forty Six Millions Eight Hundred Thirteen Thousand Five Hundred Seventy Nine (e) 345000543 34,50,00,543 345,000,543 Two Crore Twenty Two Thousand Two Twenty Four Crore Sixty Eight Lakh Thirteen Thousand Five Hundred Seventy Nine 6 20,022,002 246,813,579 Page No. 10 Thirty four crore fifty lakh five hundred forty three Three hundred forty five millions five hundred forty three (f) 1020304050 1,02,03,04,050 2. Write the following numbers in Indian place value notation: (a)One crore one lakh one thousand Ten Ans: 1,01,01,010 3. Compare and write ‘<’, ‘>’ or ‘=’: (a) 30 thousand __<__ 3 lakhs (b) 500 lakhs ___>___ 5 million (c) 800 thousand __<__ 8 million (d) 640 crore ___<___ 60 billion (b)One billion one million one thousand one Ans: 1,00,10,01,001 (c) Ten crore twenty lakh thirty thousand forty Ans: 10,20,30,040 (d)Nine billion eighty million seven hundred thousand six hundred Ans: 9,08,07,00,600 One Arab two crore three lakh four thousand fifty 1,020,304,050 One billion twenty millions three hundred four thousand fifty Ans: (a) Buying food/fruits for a group. (b) Estimating time remaining to leave for the school (catching the school bus). (c) Distance estimation between places, especially far off places. (d) Handling money in banks. Find more such situations. Think and share situations where it is appropriate to (a) round up, (b) round down, (c) either rounding up or rounding down is okay and (d) when exact numbers are needed. 7 Page No. 11 Page No. 13 1. What is your general observation about this data? Share with the class. Ans: One of the observations is – Most cities shown in the table have seen a significant rise in population from 2001 to 2011. Think of more observations. Similarly, write the five nearest neighbours for these numbers: From the information given in the table answer the following questions by approximation: (Refer table on P. 12–13 (a) 3,87,69,957 (b) 29,05,32,481 Nearest thousand 3,87,70,000 Nearest ten thousand 3,87,70,000 Nearest lakh 3,88,00,000 Nearest ten lakh 3,90,00,000 Nearest crore 4,00,00,000 Nearest thousand 29,05,32,000 Nearest ten thousand 29,05,30,000 Nearest lakh 29,05,00,000 Nearest ten lakh 29,10,00,000 Nearest crore 29,00,00,000 2. What is an appropriate title for the above table? Ans: Population of some Indian cities in 2001 and 2011. 3. How much is the population of Pune in 2011? Approximately, by how much has it increased compared to 2001? Ans: Population of Pune in 2011 is 31,15,431. The population of Pune has approximately increased by 6 lakh. 4. Which city’s population increased the most between 2001 and -2011? Ans: Bengaluru has shown maximum increase in population which is 41,24,644. 8 Page No. 14 5. Are there cities whose population has almost doubled? Which are they? Ans: Yes, they are Bengaluru, Hyderabad, Surat and Vadodara 6. By what number should we multiply Patna’s population to get a number/population close to that of Mumbai? Ans: We need to multiply Patna’s population by 7.39 to get a number close to the population of Mumbai. Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10? Ans: 10 ÷ 2 is the same as 5. Figure it out 1. Find quick ways to calculate these products: (a) 2 × 1768 × 50 = 2 × 50 × 1768 = 100 × 1768 (b) 72 × 125 [Hint: 125 = 1000 8 = 72 × 1000 8 = 72 8 × 1000 = 9 × 1000 (c) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25 = 1000 × 1000 2. Calculate these products quickly. (a) 25 × 12 = _____________ (b) 25 × 240 = _____________ 25 × 12 = 100 4 × 12 25 × 240 = 100 4 × 240 = 100 × 240 4 = 100 × 60 = 100 × 12 4 = 100 × 3 9 Page No. 15 (c) 250 × 120 = _____________ 250 × 120 = 1000 4 × 120 = 1000 × 120 4 = 1000 × 30 (d) 2500 × 12 =_____________ (e) __1200____×_1,00,000_____= 120000000 Observe the number of digits in the two numbers being multiplied and their product in each case. Is there any connection between the numbers being multiplied and the number of digits in their product? Ans: The maximum digits in their product will be the sum of the digits of the two numbers. The minimum digit in their product will be one less than their sum. Roxie says that the product of two 2-digit numbers can only be a 3- or a 4-digit number. Is she correct? Ans: Yes, Roxie is correct. = 2500 × 12 = 10,000 4 × 12 = 10,000 × 12 4 = 10,000 × 3 Should we try all possible multiplications with 2-digit numbers to tell whether Roxie’s claim is true? Or is there a better way to find out? Ans: She can just check the minimum and maximum number of digits possible. Can multiplying a 3-digit number with another 3-digit number give a 4-digit number? Ans: No. Can multiplying a 4-digit number with a 2-digit number give a 5-digit number? Ans: Yes. 10 Observe the multiplication statements below. Do you notice any patterns? See if this pattern extends for other numbers as well. Page No. 16 Page No. 19 12-digit × 13-digit = 24 or 25 5-digit × 5-digit = 9 or 10 8-digit × 3-digit = 10 or 11 Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometers every day. (You had written down the distance between the Earth and the Sun in a previous exercise.) Ans: We cannot reach the sun in a lifetime because it takes approx. 403 years. Make necessary reasonable assumptions and answer the questions below: (a) If a single sheet of paper weighs 5 grams, could you lift one lakh sheets of paper together at the same time? Ans: Weight of one lakh (1,00,000) sheets = 500 kg. Since a person cannot lift 500 kg so, we cannot lift one lakh sheets of paper. (b) If 250 babies are born every minute across the world, will a million babies be born in a day? Ans: Babies born in one day = 3,60,000 babies. 3,60,000 is less than one million (10 lakh) Hence, a million babies cannot born in a day. (c) Can you count 1 million coins in a day? Assume you can count 1 coin every second. Ans: No. Because, Coins counted per day = 86,400 11 Page No. 19 Figure it out 1. Using all digits from 0 – 9 exactly once (the first digit cannot be 0) to create a 10-digit number, write the — (a)Largest multiple of 5 Ans: largest multiple of 5 = 9876543210 2. The number 10,30,285 in words is ten lakhs thirty thousand two hundred eighty-five, which has 42 letters. Give a 7-digit number name which has the maximum number of letters. Ans: One such number is 77,77,777, which has 60 letters. 3. Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist? Ans: One such number is 987654312. Exchanging last two digits, we get 987654321, which is bigger number than the initial number. Try more! 4. Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible. Ans: 5534512345 6. Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, and so on. The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10. (This question can be solved in different ways. One of the way is —) (a)What would the 1000th digit be? At which number would it occur? Ans: Digits from 1 to 9 = 9 digits Digits from 10 to 99 = 99-10+1 = 90 numbers × 2 = 180 digits Total digits from 1 to 99 = 9 + 180 = 189 digits Remaining digits to reach 1000th digit = 1000 – 189 = 811 Number of 3-digit numbers = 811 3 = 270 full numbers + 1 digit left over. The first 3-digit number is 100 270th 3-digit number is 100 + 270 – 1 = 369 (b)Smallest even number Ans: The smallest such even number = 1023456798 12 7. A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers: (a)20,800 Ans: 20,800 = (2 × 10,000) + (8 × 100) Total clicks = 2 + 8 = 10 button clicks. (b)What number would contain the millionth digit? Ans: The millionth digit occurs in the number 185184+1=1,85,185. (c) When would you have written the digit ‘5’ for the 5000th time? Ans: 13995 (b)92,100 Ans: 92,100 = (9 × 10,000) + (21 × 100) Total clicks = 9 + 21 = 30 button clicks. (c) 1,20,500 Ans: 1,20,500 = (12 × 10,000) + (5 × 100) Total clicks = 12 + 5 = 17 button clicks. (d)65,30,000 Ans: 65,30,000 = (653 × 10,000) Total clicks = 653 button clicks. (e) 70,25,700 Ans: 70,25,700 = (702 × 10,000) + (57 × 100) Total clicks = 759 button clicks. The next number is 370. The first digit of 370, which is 3, is the 1000th digit. 8. How many lakhs make a billion? Ans: 10,000 lakh make a billion. 9. You are given two sets of number cards numbered from 1 – 9. Place a number card in each box below to get the (a) largest possible sum (b) smallest possible difference of two resulting numbers. (a)Ans: Sum = 1,00,54,320 9 9 8 8 7 7 6 6 5 5 4 4 13 10.You are given some number cards; 4000, 13000, 300, 70000, 150000, 20, 5. Using the cards get as close as you can to the numbers below using any operation you want. Each card can be used only once for making a particular number. (a) 1,10,000: 4000 × (20 + 5) + 13000 = 1,13,000 (b) 2,00,000: Ans: closest estimate = 1,50,000 + 70,000 – (4000 × 5) = 2,00,000. (c) 5,80,000: Ans: closest estimate = (1,50,000 × 4) – (4000 × 5) = 5,80,000. (d) 12,45,000: Ans: closest estimate = (70,000 × 20) – 1,50,000 – 4,000 – (300 × 5) = 12,44,500. (e) 20,90,800: Ans: closest estimate = (1,50,000 × 14) + 4,000 – 13,000 = 20,91,000. (b)Ans: Difference = 10,22,447 1 1 2 2 3 3 4 9 9 8 8 7 11.Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick. Ans: Height of each coin = 1 mm Height of Statue of Unity = 180 m = 180 × 100 × 10 mm = 1, 80,000 mm. Number of coins to be stacked = 1,80,000 mm ÷ 1 mm = 1,80,000 coins. 12.Grey-headed Albatrosses have a roughly 7-feet wide wingspan. They are known to migrate across several oceans. Albatrosses can cover about 900 – 1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take to cross an ocean approximately? 14 13.A bar-tailed godwit holds the record for the longest recorded nonstop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour. 14.Bald eagles are known to fly as high as 4500 – 6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,000 – 12,800 m. How many times bigger are these heights compared to Somu’s building? Ans: Estimated number of days if it flies 900 km/day = 12,000 ÷ 900 = 13.3 days Estimated number of days if it flies 1000 km/day = 12,000 ÷ 1000 = 12 days Hence it would take approximately 12 to 14 days for a grey headed albatross to complete a 12,000 km trip to cross an ocean. Ans: Distance covered per day = 13,560 km ÷ 11 days = 1232.73 km or 1233 km. Distance covered per hour = 1233 ÷ 24 = 51.36 km or 51 km. Ans: Height of Somu’s building = 40 m (i) Bald eagles’ flight height = 4500 – 6000 m Lower estimate = 4500 ÷ 40 = 112.5 times Upper estimate = 6000 ÷ 40 = 150 times Bald eagles fly about 112 to 150 times higher than Somu’s building. (ii) Mount Everest height = 8850 m Now, 8850 ÷ 40 = 221.25 Mount Everest is approximately 221 times taller than Somu’s building. (iii) Aeroplanes flight height = 10,000 – 12,000 m Lower estimate = 10,000 ÷ 40 = 250 Upper estimate = 12800 ÷ 40 = 320. Airplanes fly about 250 to 320 times as high as Somu’s building. 15" class_7,2,Arithmetic Expressions,ncert_books/class_7/gegp1dd/gegp102.pdf,"2 2.1 Simple Expressions You may have seen mathematical phrases like 13 + 2, 20 – 4, 12 × 5, and 18 ÷ 3. Such phrases are called arithmetic expressions. Every arithmetic expression has a value which is the number it evaluates to. For example, the value of the expression 13 + 2 is 15. This expression can be read as ‘13 plus 2’ or ‘the sum of 13 and 2’. We use the equality sign ‘=’ to denote the relationship between an arithmetic expression and its value. For example: Example 1: Mallika spends ₹25 every day for lunch at school. Write the expression for the total amount she spends on lunch in a week from Monday to Friday. The expression for the total amount is 5 × 25. 5 × 25 is “5 times 25” or “the product of 5 and 25”. Different expressions can have the same value. Here are multiple ways to express the number 12, using two numbers and any of the four operations +, – , × and ÷: ARITHMETIC EXPRESSIONS 13 + 2 = 15. Chapter-2.indd 24 4/12/2025 11:27:34 AM Choose your favourite number and write as many expressions as you can having that value. Comparing Expressions As we compare numbers using ‘=’, ‘<’ and ‘>’ signs, we can also compare expressions. We compare expressions based on their values and write the ‘equal to’, ‘greater than’ or ‘less than’ sign accordingly. For example, 10 + 2, 15 – 3, 3 × 4, 24 ÷ 2. 10 + 2 > 7 + 1 Figure it Out 1. Fill in the blanks to make the expressions equal on both sides of the = sign: 2. Arrange the following expressions in ascending (increasing) order of their values. Example 2: Which is greater? 1023 + 125 or 1022 + 128? Imagining a situation could help us answer this without finding the values. Raja had 1023 marbles and got 125 more today. Now he has 1023 + 125 marbles. Joy had 1022 marbles and got 128 more today. Now he has 1022 + 128 marbles. Who has more? This situation can be represented as shown in the picture on the right. To begin with, Raja had 1 more marble than Joy. But Joy got 3 more marbles than Raja today. We can see that Joy has (two) more marbles than Raja now. That is, because the value of 10 + 2 = 12 is greater than the value of 7 + 1 = 8. Similarly, (a) 13 + 4 = ____ + 6 (b) 22 + ____ = 6 × 5 (c) 8 × ____ = 64 ÷ 2 (d) 34 – ____ = 25 (a) 67 – 19 (b) 67 – 20 (c) 35 + 25 (d) 5 × 11 (e) 120 ÷ 3 13 – 2 < 4 × 3. Arithmetic Expressions 125 125 Raja (1023 + 125) Joy (1022 + 128) 1022 1022 1 1 1 1 Chapter-2.indd 25 4/12/2025 11:27:34 AM Example 3: Which is greater? 113 – 25 or 112 – 24? Imagine a situation, Raja had 113 marbles and lost 25 of them. He has 113 – 25 marbles. Joy had 112 marbles and lost 24 today. He has 112 – 24 marbles. Who has more marbles left with them? Raja had 1 marble more than Joy. But he also lost 1 marble more than Joy did. Therefore, they have an equal number of marbles now. That is, 1023 + 125 < 1022 + 128. 113 – 25 = 112 – 24. Raja (113 – 25) Joy (112 – 24) 112 112 remove 24 25 remove 24 1 Ganita Prakash | Grade 7 Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case. 2.2 Reading and Evaluating Complex Expressions Sometimes, when an expression is not accompanied by a context, there can be more than one way of evaluating its value. In such cases, we need some tools and rules to specify how exactly the expression has to be evaluated. To give an example with language, look at the following sentences: (a) 245 + 289 246 + 285 (b) 273 – 145 272 – 144 (c) 364 + 587 363 + 589 (d) 124 + 245 129 + 245 (e) 213 – 77 214 – 76 (a) Sentence: “Shalini sat next to a friend with toys”. Meaning: The friend has toys and Shalini sat next to her. (b) Sentence: “Shalini sat next to a friend, with toys”. Meaning: Shalini has the toys and she sat with them next to her friend. Chapter-2.indd 26 4/12/2025 11:27:34 AM This sentence without the punctuation could have been interpreted in two different ways. The appropriate use of a comma specifies how the sentence has to be understood. Let us see an expression that can be evaluated in more than one way. Example 4: Mallesh brought 30 marbles to the playground. Arun brought 5 bags of marbles with 4 marbles in each bag. How many marbles did Mallesh and Arun bring to the playground? Mallesh summarized this by writing the mathematical expression — 30 + 5 × 4. 26 Without knowing the context behind this expression, Purna found the value of this expression to be 140. He added 30 and 5 first, to get 35, and then multiplied 35 by 4 to get 140. Mallesh found the value of this expression to be 50. He multiplied 5 and 4 first to get 20 and added 20 to 30 to get 50. In this case, Mallesh is right. But why did Purna get it wrong? Just looking at the expression 30 + 5 × 4, it is not clear whether we should do the addition first or multiplication. Just as punctuation marks are used to resolve confusions in language, brackets and the notion of terms are used in mathematics to resolve confusions in evaluating expressions. Brackets in Expressions In the expression to find the number of marbles — 30 + 5 × 4 — we had to first multiply 5 and 4, and then add this product to 30. This order of operations is clarified by the use of brackets as follows: When evaluating an expression having brackets, we need to first find the values of the expressions inside the brackets before performing other operations. So, in the above expression, we first find the value of 5 × 4, and then do the addition. Thus, this expression describes the number of marbles: Example 5: Irfan bought a pack of biscuits for ₹15 and a packet of toor dal for ₹56. He gave the shopkeeper ₹100. Write an expression that can help us calculate the change Irfan will get back from the shopkeeper. Irfan spent ₹15 on a biscuit packet and ₹56 on toor dal. So, the total cost in rupees is 15 + 56. He gave ₹100 to the shopkeeper. So, he should get back 100 minus the total cost. Can we write that expression as— 30 + (5 × 4 ) = 30 + 20 = 50. 30 + (5 × 4). Arithmetic Expressions Chapter-2.indd 27 4/12/2025 11:27:34 AM Can we first subtract 15 from 100 and then add 56 to the result? We will get 141. It is absurd that he gets more money than he paid the shopkeeper! We can use brackets in this case: Evaluating the expression within the brackets first, we get 100 minus 71, which is 29. So, Irfan will get back ₹29. 100 – (15 + 56). 100 – 15 + 56 ? 27 Ganita Prakash | Grade 7 Terms in Expressions Suppose we have the expression 30 + 5 × 4 without any brackets. Does it have no meaning? When there are expressions having multiple operations, and the order of operations is not specified by the brackets, we use the notion of terms to determine the order. Terms are the parts of an expression separated by a ‘+’ sign. For example, in 12+7, the terms are 12 and 7, as marked below. We will keep marking each term of an expression as above. Note that this way of marking the terms is not a usual practice. This will be done until you become familiar with this concept. Now, what are the terms in 83 – 14? We know that subtracting a number is the same as adding the inverse of the number. Recall that the inverse of a given number has the sign opposite to it. For example, the inverse of 14 is –14, and the inverse of –14 is 14. Thus, subtracting 14 from 83 is the same as adding –14 to 83. That is, Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples. Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Note that 6 × 5, 4 × 6 are single terms as they do not have any ‘+’ sign. In the following table, some expressions are given. Complete the table. Try This Thus, the terms of the expression 83 – 14 are 83 and –14. 83 – 14 = 83 + – 14 12 + 7 = 12 + 7 Chapter-2.indd 28 4/12/2025 11:27:34 AM All subtractions in an expression are converted to additions in this manner to identify the terms. Here are some more examples of expressions and their terms: –18 – 3 = – 18 + – 3 6 × 5 + 3 = 6 × 5 + 3 2 – 10 + 4 × 6 = 2 + – 10 + 4 × 6 28 Now we will see how terms are used to determine the order of operations to find the value of an expression. We will start with expressions having only additions (with all the subtractions suitably converted into additions). Does changing the order in which the terms are added give different values? Swapping and Grouping Let us consider a simple expression having only two terms. Example 6: Madhu is flying a drone from a terrace. The drone goes 6 m up and then 4 m down. Write an expression to show how high the final position of the drone is from the terrace. The drone is 6 – 4 = 2 m above the terrace. Writing it as sum of terms: 23 – 2 × 4 + 16 + + Expression Expression as the sum of its terms Terms 28 + 19 – 8 + + 13 – 2 + 6 13 + – 2 + 6 13, – 2, 6 4 + 15 – 9 + + 5 + 6 × 3 5 + 6 × 3 Arithmetic Expressions Chapter-2.indd 29 4/12/2025 11:27:34 AM It doesn’t in this case. We already know that swapping the terms does not change the sum when both the terms are positive numbers. Will this also hold when there are terms having negative numbers as well? Take some more expressions and check. Will the sum change if we swap the terms? – 4 + 6 = 2 6 + – 4 = 2 29 Ganita Prakash | Grade 7 Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Thus, in an expression having two terms, swapping them does not change the value. Now consider an expression having three terms: (–7) + 10 + (–11). Let us add these terms in the following two different orders: (adding the last two terms and then adding their sum to the first term) What do you see? The sums are the same in both cases. Again, we know that while adding positive numbers, grouping them in any of the above two ways gives the same sum. Will this also hold when there are terms having negative numbers as well? Take some more expressions and check. Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics? (adding the first two terms and then adding their sum to the third term) Term 1 + Term 2 = Term 2 + Term 1 – 7 + 10 + – 11 – 7 + 10 + – 11 Try This Try This Chapter-2.indd 30 4/12/2025 11:27:34 AM Thus, grouping the terms of an expression in either of the following ways gives the same value. Let us consider the expression (–7) + 10 + (–11) again. What happens when we change the order and add –7 and –11 first, and then add this sum to 10? Will we get the same sum as before? We see that adding the terms of the expression (–7) + 10 + (–11) in any order gives the same sum of –8. 30 Term 1 + Term 2 + Term 3 = Term 1 + Term 2 + Term 3 Does adding the terms of an expression in any order give the same value? Take some more expressions and check. Consider expressions with more than 3 terms also. Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Thus, the addition of terms in any order gives the same value. Therefore, in an expression having only additions, it does not matter in what order the terms are added: they all give the same value. Now let us consider expressions having multiplication and division also, without the order of operations specified by the brackets. The values of such expressions are found by first evaluating the terms. Once all the terms are evaluated, they are added. For example, the expression 30 + 5 × 4 is evaluated as follows: Where (3+2) is first evaluated and this sum is multiplied by 5 (= 25). The expression 7 × 8 is evaluated (= 56). This simplifies to 25 + 56 + 3 = 84. Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again? The expression 5 × (3 + 2) + 78 + 3 is evaluated as follows: 30 + 5 × 4 = 30 + 5 × 4 = 30 + 20 = 50 5 × (3 + 2) + 78 + 3 = 5 × (3 + 2) + 7 × 8 + 3 Arithmetic Expressions 774 8611 9055 1022 1342 Try This Chapter-2.indd 31 4/12/2025 11:27:34 AM In mathematics we use the phrase commutative property of addition instead of saying “swapping terms does not change the sum”. Similarly, “grouping does not change the sum” is called the associative property of addition. Swapping the Order of Things in Everyday Life Manasa is going outside to play. Her mother says, “Wear your hat and shoes!” Which one should she wear first? She can wear her hat first and then her shoes. Or she can wear her shoes first and then her hat. Manasa will look exactly the same in both cases. Imagine a different situation: Manasa’s mother says “Wear your socks and shoes!” Now the 31 Ganita Prakash | Grade 7 order matters. She should wear socks and then shoes. If she wears shoes and then socks, Manasa will feel very uncomfortable and look very different. More Expressions and Their Terms Example 7: Amu, Charan, Madhu, and John went to a hotel and ordered four dosas. Each dosa cost ₹23, and they wish to thank the waiter by tipping ₹5. Write an expression describing the total cost. Cost of 4 dosas = 4 × 23 Can the total amount with tip be written as 4 × 23 + 5? Evaluating it, we get If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms. Example 8: Children in a class are playing “Fire in the mountain, run, run, run!”. Whenever the teacher calls out a number, students are supposed to arrange themselves in groups of that number. Whoever is not part of the announced group size, is out. Ruby wanted to rest and sat on one side. The other 33 students were playing the game in the class. The teacher called out ‘5’. Once children settled, Ruby wrote 6 × 5 + 3 (understood as 3 more than 6 × 5) Thus, 4 × 23 + 5 is a correct way of writing the expression. 4 × 23 + 5 = 4 × 23 + 5 = 92 + 5 = 97 Chapter-2.indd 32 4/12/2025 11:27:35 AM Think and discuss why she wrote this. 32 The expression written as a sum of terms is— 6 × 5 + 3 . + + + For each of the cases below, write the expression and identify its terms: If the teacher had called out ‘4’, Ruby would write ____________ If the teacher had called out ‘7’, Ruby would write ____________ Write expressions like the above for your class size. Example 9: Raghu bought 100 kg of rice from the wholesale market and packed them into 2 kg packets. He already had four 2 kg packets. Write an expression for the number of 2 kg packets of rice he has now and identify the terms. which we also write as 100 2 . The number of 2 kg packets he has now is 4 + 100 2 . The terms are— Example 10: Kannan has to pay ₹432 to a shopkeeper using coins of ₹1 and ₹5, and notes of ₹10, ₹20, ₹50 and ₹100. How can he do it? Meaning: 4 notes of ₹100, 1 note of ₹20, 1 note of ₹10 and 2 notes of ₹1 Meaning: 8 notes of ₹50, 1 note of ₹10, 4 notes of ₹5 and 2 notes of ₹1 Identify the terms in the two expressions above. Can you think of some more ways of giving ₹432 to someone? He had 4 packets. The number of new 2 kg packets of rice is 100 ÷ 2, There is more than one possibility. For example, 432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1 4 100 2 + . Arithmetic Expressions Chapter-2.indd 33 4/12/2025 11:27:35 AM Example 11: Here are two pictures. Which of these two arrangements matches with the expression 5 × 2 + 3? 33 Ganita Prakash | Grade 7 This expression 5 × 2 + 3 can be understood as 3 more than 5 × 2, which describes the arrangement on the left. What is the expression for the arrangement in the right making use of the number of yellow and blue squares? Figure it Out 1. Find the values of the following expressions by writing the terms in each case. 2. Write a story/situation for each of the following expressions and find their values. 34 Let us write this expression as a sum of terms. Do you recall the use of brackets? We need to use brackets for this. 2 × (5 + 3) Notice that this arrangement can also be described using— (a) 28 – 7 + 8 (b) 39 – 2 × 6 + 11 (c) 40 – 10 + 10 + 10 (d) 48 – 10 × 2 + 16 ÷ 2 (e) 6 × 3 – 4 × 8 × 5 (a) 89 + 21 – 10 (b) 5 × 12 – 6 (c) 4 × 9 + 2 × 6 5 × 2 + 3 = 10 + 3 = 13 5 + 3 + 5 + 3 5 × 2 + 3 × 2 OR Chapter-2.indd 34 4/12/2025 11:27:35 AM 3. For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression. (a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have. (b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets: (i) for four adults and three children? (ii) for two groups having three adults each? Removing Brackets—I Let us find the value of this expression, 200 – (40 + 3). We first evaluate the expression inside the bracket to 43 and then subtract it from 200. But it is simpler to first subtract 40 from 200: So, Example 12: We also saw this earlier in the case of Irfan purchasing a biscuit packet (₹15) and a toor dal packet (₹56). When he paid ₹100, the change he gets in rupees is: And then subtract 3 from 160: What we did here was 200 – 40 – 3. Notice, that we did not do The change could also have been calculated as follows: (c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture. 200 – (40 + 3) = 200 – 40 – 3. 100 – (15 + 56) = 29. 200 – 40 = 160. 160 – 3 = 157. 200 – 40 + 3. Arithmetic Expressions Border Grill Gap Total Height 2 cm 5 cm 3 cm Chapter-2.indd 35 4/12/2025 11:27:35 AM Notice how upon removing the brackets preceded by a negative sign, the signs of the terms inside the brackets change. Observe (a) First subtract the cost of the biscuit packet (15) from 100: (b) So, to find the change, we need to subtract the cost of toor dal from 85. This is the amount the shopkeeper owes Irfan if he had purchased only the biscuits. As he has purchased toor dal also, its cost is taken from this remaining amount of 85. What we have done here is 100 – 15 – 56. So, 100 – (15 + 56) = 100 – 15 – 56. 100 – 15 = 85. 85 – 56 = 29. 35 Ganita Prakash | Grade 7 the signs of 40 and 3 in the first example, and that of 15 and 56 in the second. Example 13: Consider the expression 500 – (250 – 100). Is it possible to write this expression without the brackets? To evaluate this expression, we need to subtract 250 – 100 = 150 from 500: If we were to directly subtract 250 from 500, then we would have subtracted 100 more than what we needed to. So, we should add back that 100 to 500 – 250 to make the expression take the same value as 500 – (250 – 100). This sequence of operations is 500 – 250+100. Thus, Check that 500 – (250 – 100) is not equal to 500 – 250 – 100. Notice again that when the brackets preceded by a negative sign are removed, the signs of the terms inside the brackets change. In this case, the signs of 250 and – 100 change to – 250 and 100. Example 14: Hira has a rare coin collection. She has 28 coins in one bag and 35 coins in another. She gifts her friend 10 coins from the second bag. Write an expression for the number of coins left with Hira. This can be expressed by 28 + (35 – 10). We know that this is the same as 28 + (35 + (–10)). Since the terms can be added in any order, this expression can simply be written as 28 + 35 + (–10), or 28 + 35 – 10. Thus, When the brackets are NOT preceded by a negative sign, the terms within them do not change their signs upon removing the brackets. Notice the sign of the terms 35 and – 10 in the above expression. 500 – (250 – 100) = 500 – 150 = 350. 500 – (250 – 100) = 500 – 250 + 100. 28 + (35 – 10) = 28 + 35 – 10 = 53. Chapter-2.indd 36 4/12/2025 11:27:35 AM Tinker the Terms I What happens to the value of an expression if we increase or decrease the value of one of its terms? Some expressions are given in following three columns. In each column, one or more terms are changed from the first expression. Go through the example (in the first column) and fill the blanks, doing as little computation as possible. 36 Rather than simply remembering rules for when to change the sign and when not to, you can figure it out for yourself by thinking about the meanings of the expressions. Figure it Out 1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal. 2. Remove the brackets and write the expression having the same value. 54 is one more than 53, so the value will be 1 more than 37. Is –15 one more or one less than –16? 53 + – 16 = 37 53 + – 16 = 37 – 87 + – 16 = 54 + – 16 = 38 53 + – 15 = (a) 24 + (6 – 4) = 24 + 6 _____ (b) 38 + (_____ _____) = 38 + 9 – 4 (c) 24 – (6 +4) = 24 6 – 4 (d) 24 – 6 – 4 = 24 – 6 _____ (e) 27 – (8 + 3) = 27 8 3 (f) 27– (_____ _____) = 27 – 8 + 3 52 is one less than 53, so the value will be 1 less than 37. Is –17 one more or one less than –16? 52 + – 16 = 53 + – 17 = – 88 + – 15 = – 86 + – 18 = – 97 + – 26 = Arithmetic Expressions Chapter-2.indd 37 4/12/2025 11:27:35 AM 3. Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal? (a) (6 + 10) – 2 and 6 + (10 – 2) (b) 16 – (8 – 3) and (16 – 8) – 3 4. In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms. (a) 14 + (12 + 10) (b) 14 – (12 + 10) (c) 14 + (12 – 10) (d) 14 – (12 – 10) (e) –14 + 12 – 10 (f) 14 – (–12 – 10) (c) 27 – (18 + 4) and 27 + (–18 – 4) 37 Ganita Prakash | Grade 7 5. Add brackets at appropriate places in the expressions such that they lead to the values indicated. 6. Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal. 7. Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘–’, and brackets, as necessary, generate expressions to give as many different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0. 8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1. 9. Consider the two expressions: a) 73 – 14 + 1, b) 73 – 14 – 1. For each of these expressions, identify the expressions from the following collection that are equal to it. (a) 319 + 537, 319 – 537, – 537 + 319, 537 – 319 (b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109 (a) 34 – 9 + 12 = 13 (b) 56 – 14 – 8 = 34 (c) –22 – 12 + 10 + 22 = – 22 (a) 423 + ______= 419 + ______ (b) 207 – 68 = 210 – ______ (a) Do you think she always gets the correct answer? Why? (b) Can you think of other similar strategies? Give some examples. Math Talk Chapter-2.indd 38 4/12/2025 11:27:35 AM Removing Brackets—II Example 15: Lhamo and Norbu went to a hotel. Each of them ordered a vegetable cutlet and a rasgulla. A vegetable cutlet costs ₹43 and a rasgulla costs ₹24. Write an expression for the amount they will have to pay. As each of them had one vegetable cutlet and one rasagulla, each of their shares can be represented by 43 + 24. What about the total amount they have to pay? Can it be described by the expression: 2 × 43 + 24? 38 (a) 73 – (14 + 1) b) 73 – (14 – 1) (c) 73 + (– 14 + 1) d) 73 + (– 14 – 1) This expression means 24 more than 2 × 43. But, we want an expression which means twice or double of 43 + 24. We can make use of brackets to write such an expression: 2 × (43 + 24). So, we can say that together they have to pay 2 × (43 + 24). This is also the same as paying for two vegetable cutlets and two rasgullas: 2 × 43 + 2 × 24. Therefore, If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid? Example 16: In the Republic Day parade, there are boy scouts and girl guides marching together. The scouts march in 4 rows with 5 scouts in each row. The guides march in 3 rows with 5 guides in each row (see the figure below). How many scouts and guides are marching in this parade? The number of boy scouts marching is 4 × 5. The number of girl guides marching is 3 × 5. The total number of scouts and guides will be 4 × 5 + 3 × 5. This can also be found by first finding the total number of rows, i.e., 4 + 3, and then multiplying their sum by the number of children in each row. Thus, the number of boys and girls can be found by (4 + 3) × 5. Therefore, 4 × 5 + 3 × 5 = (4 + 3) × 5. Computing these expressions, we get Writing it as sum of terms gives: 2 × (43 + 24) = 2 × 43 + 2 × 24. 2 × 43 + 24 2 × (43 + 24) Arithmetic Expressions ₹ 43 ₹ 43 2 × 43 + 2 × 24 ₹ 24 ₹ 24 Chapter-2.indd 39 4/12/2025 11:27:36 AM 4 × 5 + 3 × 5 = 4 × 5 + 3 × 5 = 20 + 15 = 35 3 × 5 4 × 5 (4+3) × 5 39 Ganita Prakash | Grade 7 (4 + 3) × 5 = 7 × 5 = 35 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why? Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5? The observations that we have made in the previous two examples can be seen in a general way as follows. Consider 10 × 98 + 3 × 98. This means taking the sum of 10 times 98 and 3 times 98. Swapping the numbers in the products above, this property can be seen in the following form: Similarly, let us consider the expression 14 × 10 – 6 × 10. This means subtracting 6 times 10 from 14 times 10. Clearly, this is the same as 10 + 3 = 13 times 98. Thus, Writing this equality the other way, we get 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 + 98 98 × 10 + 9 × 83 = 98 (10 + 3), and 10 × 98 +3 × 98 = (10 + 3) × 98. (10 + 3) 98 = 10 × 98 + 3 × 98. 98 (10 + 3) = 98 × 10 + 98 × 3. 10 times 3 times 14 times Chapter-2.indd 40 4/16/2025 4:24:23 PM This property can be nicely summed up as follows: The multiple of a sum (difference) is the same as the sum (difference) of the multiples. Tinker the Terms II Let us understand what happens when we change the numbers occurring in a product. 40 Clearly, this is 14 – 6 = 8 times 10. Thus, 14 × 10 – 6 × 10 = (14 – 6) × 10, (14 – 6) × 10 = 14 × 10 – 6 × 10 or Example 17: Given 53 × 18 = 954. Find out 63 × 18. As 63 × 18 means 63 times 18, 63 × 18 = (53 + 10) × 18 = 53 ×18 + 10×18 = 954 + 180 = 1134. Example 18: Find an effective way of evaluating 97 × 25. 97 × 25 means 97 times 25. We can write it as (100 – 3) × 25 We know that this is the same as the difference of 100 times 25 and 3 times 25: Use this method to find the following products: Which other products might be quicker to find like the ones above? Figure it Out 1. Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal. Find this value. Is this quicker than the multiplication procedure you use generally? (a) 95 × 8 (b) 104 × 15 (c) 49 × 50 (a) 3 × (6 + 7) = 3 × 6 + 3 × 7 97 × 25 = 100 × 25 – 3 × 25 Arithmetic Expressions Math Talk Chapter-2.indd 41 4/12/2025 11:27:36 AM (b) (8 + 3) × 4 = 8 × 4 + 3 × 4 (c) 3 × (5 + 8) = 3 × 5 3 × ____ (d) (9 + 2) × 4 = 9 × 4 2 ×____ (e) 3 × (____ + 4) = 3 ____+____ (f) (____+ 6) × 4 = 13 × 4 + ____ (g) 3 × (____+____) = 3 × 5 + 3 × 2 (h) (____+____)×____= 2 × 4 + 3 × 4 (i) 5 × (9 – 2) = 5 × 9 – 5 × ____ (j) (5 – 2) × 7 = 5 × 7 – 2 × ____ (k) 5 × (8 – 3) = 5 × 8 5 × ____ (l) (8 – 3) × 7 = 8 × 7 3 × 7 41 Ganita Prakash | Grade 7 (p) (____–____) × ____= 17 × 7 – 9 × 7 2. In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions. 3. Here is one way to make 14: _2_ × ( _1_ + _6_ ) = 14. Are there other ways of getting 14? Fill them out below: 4. Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions. (m) 5 × (12 –____) =____ 5 ×____ (n) (15 –____) × 7 =____ 6 × 7 (o) 5 × (____–____) = 5 × 9 – 5 × 4 (a) (8 – 3) × 29 (3 – 8) × 29 (b) 15 + 9 × 18 (15 + 9) × 18 (c) 23 × (17 – 9) 23 × 17 + 23 × 9 (d) (34 – 28) × 42 34 × 42 – 28 × 42 (a) _____× (_____+_____) = 14 (b) _____× (_____+_____) = 14 (c) _____× (_____+_____) = 14 (d) _____× (_____+_____) = 14 Chapter-2.indd 42 4/12/2025 11:27:36 AM Figure it Out 1. Read the situations given below. Write appropriate expressions for each of them and find their values. 42 (a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market. 2. Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario? 3. Find different ways of evaluating the following expressions: 4. Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning. (b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year? (c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat? (a) 5 × 2 × 8 (b) (7 – 2) × 8 (c) 8 × 7 (d) 7 × 2 × 8 (e) 7 × 5 – 2 (f) (7 + 2) × 8 (g) 7 × 8 – 2 × 8 (h) (7 – 5) × 8 (a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 (b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 (a) 49 – 7 + 8 49 – 7 + 8 Arithmetic Expressions Chapter-2.indd 43 4/12/2025 11:27:36 AM (b) 83 × 42 – 18 83 × 40 – 18 (c) 145 – 17 × 8 145 – 17 × 6 (d) 23 × 48 – 35 23 × (48 – 35) (e) (16 – 11) × 12 –11 × 12 + 16 × 12 (f) (76 – 53) × 88 88 × (53 – 76) (g) 25 × (42 + 16) 25 × (43 + 15) (h) 36 × (28 – 16) 35 × (27 – 15) 43 Ganita Prakash | Grade 7 5. Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression. 5. Choose a number and create ten different expressions having that value. • We have been reading and evaluating simple expressions for quite some time now. Here we started by revising the meaning of some simple expressions and their values. • We learnt how to compare certain expressions through reasoning instead of bluntly evaluating them. • To help read and evaluate complex expressions without confusion, we use terms and brackets. • When an expression is written as a sum of terms, changing the order of the terms or grouping the terms does not change the value of the expression. This is because the “commutative property of addition” and the “associative property of addition”, respectively. • To evaluate expressions within brackets, we saw that when we remove brackets preceded by a negative sign, the terms within the bracket change their sign. • We also learnt about the “distributive property” — multiplying a number with an expression inside brackets is equal to the multiplying the number with each term in the bracket. (a) 83 – 37 – 12 (i) 84 – 38 – 12 (b) 93 + 37 × 44 + 76 (ii) 84 – (37 + 12) (iii) 83 – 38 – 13 (iv) – 37 + 83 –12 (i) 37 + 93 × 44 + 76 (ii) 93 + 37 × 76 + 44 (iii) (93 + 37) × (44 + 76) (iv) 37 × 44 + 93 + 76 SUMMARY Chapter-2.indd 44 4/12/2025 11:27:36 AM 44 Using three 3’s along with the four operations (addition, subtraction, multiplication, and division) and brackets as needed we can create several expressions. For example, (3 + 3)/3 = 2, 3 + 3 – 3 = 3, 3 × 3 + 3 = 12, and so on. Using four 4’s, create expressions to get all values from 1 to 20. Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between  – 10 and +10. Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100. What other similar interesting questions can you ask? Expression Engineer! Arithmetic Expressions Chapter-2.indd 45 4/16/2025 4:25:29 PM 45 Page No. 24 Ans: Let us choose the number 24. The arithmetic expressions for the number can be as follows: 12 + 12 = 24; 4 × 6 = 24; 48 ÷ 2 = 24, 34 – 10 = 24, 20 + 4 = 24 etc. Now you choose your number and form arithmetic expressions for that number. Page No. 25 Figure it Out 1. Fill in the blanks to make the expressions equal on both sides of the = sign: 2. Arrange the following expressions in ascending (increasing) order of their values. Choose your favourite number and write as many expressions as you can having that value. (a) 13 + 4 = _____ + 6 (b) 22 + ______ = 6 × 5 (c) 8 × ______ = 64 ÷ 2 (d) 34 – ______ = 25 Ans: 1. (a) 11 (b) 8 (c) 4 (d) 9 Arithmetic Expressions Chapter – 2 Page No. 26 Ans: 2. 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 + 25 Use ‘>’ or ‘<’ ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case. Ans: (a) > (b) = (c) < (d) < (e) < (a) 67 – 19 (b) 67 – 20 (c) 35 + 25 (d) 5 × 11 (e) 120 ÷ 3 1 Page No. 28 – 29 Page No. 29 Page No. 30 Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples. Ans: Let us take the numbers. 18 and 10 18 – 10 = 8 and 18 + (– 10) = 8 Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Ans: Refer chapter 10 of class VI Mathematics Textbook. Does changing the order in which the terms are added give different values? Ans: No, For example, 14 + 10 +( – 5) = 19 or (– 5) + 10 + 14 = 19 Will this also hold when there are terms having negative numbers as well? Take some more expressions and check. Ans: Yes, For example: (– 4) + (– 2) = – 6 or (– 2) + (– 4) = – 6 (– 6) + (– 8) = – 14 or (– 8) + (– 6) = – 14 Write some more examples. Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Ans: 2 Page No. 31 Will this also hold when there are terms having negative numbers as well? Take some more expressions and check. Ans: Yes For example, Other way Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics? Ans: Yes. Consider the expression: − 6 + (− 7) + (– 13) Does adding the terms of an expression in any order give the same value? Take some more expressions and check. Consider expressions with more than 3 terms also. Ans: Yes, adding the terms of an expression in any order gives the same value Consider an expression with 3 terms 9 + 4 + 6 Order 1: 9 + 4 + 6 = (9 + 4) + 6 = 13 + 6 = 19 Order 2: 9 + 4 + 6 = 9 + (4 + 6) = 9 + 10 = 19 An expression with 4 terms (including a negative number) Consider the expression: 15 + (– 8) + 5 + 12 Order 1: (15 + (– 8)) + 5 + 12 = (7 + 5) + 12 = 12 + 12 = 24 Order 2: 15 + 5 + 12 + (– 8) = (15 + 5 + 12) + (– 8) 3 Page No. 32 If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms. Ans: The expression for the total cost = 7 × 23 + 5 = 161 + 5 = ₹166. The terms in the expression 7 × 23 + 5 are 7 × 23, 5. Page No. 33 Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again? Ans: No, she can add fourth number, 9055 to the sum she got (11749) to get the correct sum of the list of given numbers. That is 11749 + 9055 = 20804 (Using Associative Property) = ((15 + 5) + 12) + (– 8) = (20 + 12) + (– 8) = 32 + (– 8) = 24 For each of the cases below, write the expression and identify its terms: (a) If the teacher had called out ‘4’, Ruby would write _________ (b) If the teacher had called out ‘7’, Ruby would write _________ Ans: (a) 8 × 4 + 1 Terms: 8 × 4, 1 (b) 4 × 7 + 5 Terms: 4 × 7, 5 Write an expression like the above for your class size. 4 Page No. 34 – 35 Ans: (a) 28 – 7 + 8 Terms: 28, – 7, and 8 Value: 29 (b) 39 – 2 × 6 + 11 Terms: 39, – 2 × 6 and 11 Value: 38 (c) 40 – 10 + 10 + 10 = 40 + (– 10) + 10 + 10 Terms: 40, – 10, 10 and 10 Value: 50 (d) 48 – 10 × 2 + 16 ÷ 2 = 48 + (– 10 × 2) + (16 ÷ 2) Terms: 48, – 10 × 2, 16 ÷ 2 Value: 36 (e) 6 × 3 – 4 × 8 × 5 = (6 × 3) + (– 4 × 8 × 5) Terms: 6 × 3, 4 × 8 × 5 Value: – 142 Identify the terms in the two expressions above. Ans: Way 1: 432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 Terms: 4 × 100, 1 × 20, 1 × 10, and 2 × 1 Can you think of some more ways of giving ₹ 432 to someone? Ans: 40 × 10 + 3 × 10 + 2 × 1 Terms: 40 × 10, 3 × 10 and 2×1 Try for some more expressions. Way 2: 432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1 Terms: 8 × 50, 1 × 10, 4 × 5, and 2 × 1 Figure it Out 1. Find the values of the following expressions by writing the terms in each case. (a)28 – 7 + 8 (b)39 – 2 × 6 + 11 (c) 40 – 10 + 10 + 10 (d) 48 – 10 × 2 + 16 ÷ 2 (e) 6 × 3 – 4 × 8 × 5 5 Ans: 2. Write a story/situation for each of the following expressions and find their values. (а) 89 + 21 – 10 (b) 5 × 12 – 6 (c) 4 × 9 + 2 × 6 3. For each of the following situations, write the expression describing the situation, identify its terms, and find the value of the expression. (а) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have. (a) 89 + 21 – 10; (b) 5 × 12 – 6 (c) 4 × 9 + 2 × 6 One of the stories could be; A library had 89 books on its shelves. The librarian bought 21 new books and added them to the collection. Later, 10 books were borrowed by students. How many books remain on the shelves? Value of the expression: 100 Make a story of your own for the given expression. Value of the expression: 54 Make a story of your own for the given expression. Value of the expression: 48 6 Ans: (a) Expression describing the situation = 2 × 100 + 100 2 Terms: 2 × 100, 100 2 Value: 250 gold coins. (b) (i) Expression = 4 × 40 + 3 × 20 Terms: 4 × 40, 3 × 20 Value: ₹220 (ii) Expression = 2 × (3 × 40). Terms: 2 × (3 × 40) Value: ₹240 (c) Expression = 7 × 5 + 6 × 2 + 2 × 3 Terms: 7 × 5, 6 × 2, 2 × 3 Total height: 53 cm (b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of the tickets? (c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture. (i) for four adults and three children? (ii) for two groups having three adults each? Page No. 37 Some expressions are given in the following three columns. In each column, one or more terms are changed from the first expression. Go through the example (in the first column) and fill in the blanks, doing as little computation as possible. 7 Ans: 8 Page No. 37 – 38 Figure it Out 1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal. Ans: 2. Remove the brackets and write the expression having the same value. (a) 14 + (12 + 10) (b) 14 – (12 + 10) (c) 14 + (12 – 10) (d) 14 – (12 – 10) (e) – 14 + 12 – 10 (f) 14 – ( – 12 – 10) Ans: (a) and (f) (a) 24 + (6 – 4) = 24 + 6 _________ (b) 38 + (_________ ________) = 38 + 9 – 4 (c) 24 – (6 + 4) = 24 6 – 4 (d) 24 – 6 – 4 = 24 – 6 _________ (e) 27 – (8 + 3) = 27 _________ 8 _________ 3 (f) 27 – (_________ ________) = 27 – 8 + 3 (a) 24 + (6 – 4) = 24 + 6 4 (b) 38 + (9 4) = 38 + 9 – 4 (c) 24 – (6 + 4) = 24 6 – 4 (d) 24 – 6 – 4 = 24 – 6 4 (e) 27 – (8 + 3) = 27 – 8 – 3 (f) 27 – (8 3) = 27 – 8 + 3 3. Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal? (a) (6 + 10) – 2 and 6 + (10 – 2) (b) 16 – (8 – 3) and (16 – 8) – 3 (c) 27 – (18 + 4) and 27 + (– 18 – 4) 9 4. In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms. (a)319 + 537, 319 – 537, – 537 + 319, 537 – 319 (b)87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109 Ans: (a) 319 – 537 and – 537 + 319 have the same terms So, 319 – 537 and – 537 + 319 have the same value. (b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109 and also, 87 – 46 + 109 and (87 – 46) + 109 have the same terms and so have equal values. 5. Add brackets at appropriate places in the expressions such that they lead to the values indicated. Ans: (a) (6 + 10) – 2 = 6 + 10 – 2 = 14 6 + (10 – 2) = 6 + 10 – 2 = 14 So, (6 + 10) – 2 = 6 + (10 – 2) (b) 16 – (8 – 3) and (16 – 8) – 3 16 – (8 – 3) = 16 – 8 + 3 = 11 (16– 8) – 3 = 16 – 8 – 3 = 5 So, 16 – (8 – 3) ≠ (16 – 8) – 3 (c) 27 – (18 + 4) and 27 + ( – 18 – 4) 27 – (18 + 4) = 27 – 18 – 4 = 5 27 + ( – 18 – 4) = 27 – 18 – 4 = 5 So, 27 – (18 + 4) = 27 + ( – 18 – 4) (a) 34 – 9 + 12 = 13 (b) 56 – 14 – 8 = 34 (c) – 22 – 12 – 10 + 22 = – 22 Ans: (a) 34 – (9 + 12) = 34 – 21 = 13 (b) (56 – 14) – 8 = 42 – 8 = 34 (c) – 22 – (12 + 10) + 22 = – 22 – 22 + 22 = – 22 10 6. Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal. 7. Using the numbers 2, 3, and 5, and the operators ‘+’ and ‘ – ‘, and brackets, as necessary, generate expressions to give as many different values as possible. 8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1 Ans: (a) 423 + 419 = 419 + 423 (Commutative Property) (b) 207 – 68 = 207 + 3 – 3 – 68 = 207 + 3 – (3 + 68) = 210 – 71 (Associative Property) Ans: Some of the expressions are 1. 3 – 2 – 5 = – 4 2. 5 + 2 + 3 = 10 3. 5 + 2 – 3 = 4 4. 2 – (3 + 5) = – 6 (a) 423 + ________ = 419 + ________ (b) 207 – 68 = 210 – ________ For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0 Form more expressions. (a) Do you think she always gets the correct answer? Why? (b) Can you think of other similar strategies? Give some examples. Ans: (a) Yes 36 – 9 = 36 – (10 – 1) = 36 – 10 + 1 = 26 + 1 (b) Yes To subtract 8: Subtract 10 and add 2 (since – 8 = – 10 + 2). Ex: 55 – 8 = 55 – (10 – 2) = 55 – 10 + 2 More such strategies can be thought of. 11 Page No. 39 Page No. 40 Ans: Expression: 3 × (43 + 24). Ans: Expressions (b) and (c) are equal to the expression 73 – 14 + 1, and expressions (a) and (d) are equal to the expression 73 – 14 – 1. 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why? Ans: 5 × 4 + 3 = (5 × 4) + 3 = 23. 5 × (4 + 3) = 5 × 7 = 35 So, 5 × 4 + 3 ≠ 5 × (4 + 3) Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5? Ans: Yes, all three expressions are equal. 9. Consider the two expressions: (a) 73 – 14 + 1 (b) 73 – 14 – 1 For each of these expressions, identify the expressions from the following collection that are equal to it. (a) 73 – (14 + 1) (b) 73 – (14 – 1) (c) 73 + (– 14 + 1) (d) 73 + (– 14 – 1) If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid? Page No. 41 Use this method to find the following products: (a) 95 × 8 (b) 104 × 15 (c) 49 × 50 Is this quicker than the multiplication procedure you use generally? 12 Ans: (a) 95 × 8 = (100 – 5) × 8 = 100 × 8 – 5 × 8 = 800 – 40 = 760 Which other products might be quicker to find, like the ones above? Ans: This method is useful when one of the numbers is close to a multiple of 10, 50, 100, 1000, etc. For example: (b) 104 × 15 = (100 + 4) × 15 = 100 × 15 + 4 × 15 Try further. (c) 49 × 50 = (50 – 1) × 50 = 50 × 50 – 50 × 1 Try further. Yes, this procedure is quicker than the multiplication procedure we generally use. 98 × 7 = (100 – 2) × 7 ; 103 × 9 = (100 + 3) × 9 49 × 5 = (50 – 1) × 5 Find some more! Page No. 41 – 42 Figure it Out 1. Fill in the blanks with numbers and boxes by signs, so that the expressions on both sides are equal. (а) 3 × (6 + 7) = 3 × 6 + 3 × 7 (b) (8 + 3) × 4 = 8 × 4 + 3 × 4 (c) 3 × (5 + 8) = 3 × 5 3 × ________ (d) (9 + 2) × 4 = 9 × 4 2 × ________ (e) 3 × ( ________ + 4) = 3 ________ + ________ (f) (________ + 6) × 4 = 13 × 4 + ________ (g) 3 × (________ + ________) = 3 × 5 + 3 × 2 (h) (________ + ________) × ________ = 2 × 4 + 3 × 4 13 Ans: (c) 3 × (5 + 8) = 3 × 5 3 × 8 (i) 5 × (9 – 2) = 5 × 9 – 5 × ________ (j) (5 – 2) × 7 = 5 × 7 – 2 × ________ (k) 5 × (8 – 3) = 5 × 8 5 × ________ (l) (8 – 3) × 7 = 8 × 7 3 × 7 (m) 5 × (12 – ________) = ________ 5 × ________ (n) (15 – ________) × 7 = ________ 6 × 7 (o) 5 × (________ – ________) = 5 × 9 – 5 × 4 (p) (________ – ________) × ________ = 17 × 7 – 9 × 7 (d) (9 + 2) × 4 = 9 × 4 2 × 4 (e) 3 × (10 + 4) = 3×10 + 3 × 4 (f) (13 + 6) × 4 = 13 × 4 + 6 × 4 (g) 3 × (5 + 2) = 3 × 5 + 3 × 2 (h) (2 + 3) × 4 = 2 × 4 + 3 × 4 (i) 5 × (9 – 2) = 5 × 9 – 5 × 2 (j) (5 – 2) × 7 = 5 × 7 – 2 × 7 (k) 5 × (8 – 3) = 5 × 8 5 × 3 (l) (8 – 3) × 7 = 8 × 7 3 × 7 (m) 5 × (12 – 3) = 5 × 12 5 × 3 – + + – – (n) (15 – 6) × 7 = 15 × 7 6 × 7 (o) 5 × (9 – 4) = 5 × 9 – 5 × 4 (p) (17 – 9) × 7 = 17 × 7 – 9 × 7 [In (e), (m) you may put any number other than 10 and 3 respectively.] – 14 Ans: Ans: 2. In the boxes below, fill in ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out, and not by evaluating the expressions. 3. Here is one way to make 14: 2 × (1 + 6) = 14. Are there other ways of getting 14? Fill them out below: (a) > : since 8 – 3 > 3 – 8 (b) < (c) < (d) = [Try to reason in other parts] (a) (8 – 3) × 29 (3 – 8) × 29 (b) 15 + 9 × 18 (15 + 9) × 18 (c) 23 × (17 – 9) 23 × 17 + 23 × 9 (d) (34 – 28) × 42 34 × 42 – 28 × 42 (a) ________ × (________ + ________) = 14 (b) ________ × (________ + ________) = 14 (c) ________ × (________ + ________) = 14 (d) ________ × (________ + ________) = 14 (a) 2 × (5 + 2) = 14 (b) 2 × (3 + 4) = 14 (c) 7 × (1 + 1) = 14 (d) 2 × (6 + 1) = 14 4. Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions. (I) (II) 15 Page No. 42 – 44 Ans: Figure it Out 1. Read the situations given below. Write appropriate expressions for each of them and find their values. (a) The district market in Begur operates on all seven days of the week. Rahim supplies 9 kg of mangoes each day from his orchard, and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the number of mangoes supplied by them in a week to the local district market. (b) Binu earns ₹ 20,000 per month. She spends ₹ 5,000 on rent, ₹ 5,000 on food, and ₹ 2,000 on other expenses every month. What is the amount Binu will save by the end of the year? (c) During the daytime, a snail climbs 3 cm up a post, and during the night, while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on top. In how many days will the snail get the treat? For I: Way1: (5 × 4) + (4 × 8) = 52 Way 2: 2 × (4 + 8 + 4) + (8 + 4 + 8) = 52 For II: Way 1: (8 × 5) + (8 × 6) = 88 Way 2: 8 × (5 + 6) = 88 Find more ways. Ans: (a) Expression: 7 × (9 + 11) kg Value: 140 kg (b) Expression: 12 × 20,000 – 12 × (5000 + 5000 + 2000) Amount Binu will save by the end of the year = ₹96,000 (c) In the daytime the snail climbs 3 cm up the post and slips down by 2 cm at night. So, the distance climbed by the snail on the post in full day cycle = 3 – 2 = 1 cm ∴ The distance climbed by the snail in 7 days = 7 cm The height of the post is 10 cm. On the 8th day the snail will climb i.e. 7 × (3 – 2) + 3 × 1 = 10 cm 16 Ans: Expressions describing the scenario (total stories) = (b) and (g). 2. Melvin reads a two – page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario? (a) 5 × 2 × 8 (b) (7 – 2) × 8 (c) 8 × 7 (d) 7 × 2 × 8 (e) 7 × 5 – 2 (f) (7 + 2) × 8 (g) 7 × 8 – 2 × 8 (h) (7 – 5) × 8 3. Find different ways of evaluating the following expressions: (a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 (b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 Ans: (a) Way 1 : 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 + 3 + 5 + 7 + 9) + (– 2 – 4 – 6 – 8 – 10) (b) Way 1: 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) Way 2: 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) +(9 – 10) Try more ways. 4. Compare the following pairs of expressions using ‘<’, ‘>’, or ‘=,’ or by reasoning. Way 2: 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 = (1 + 1 + 1 + 1 + 1) + (– 1 – 1 – 1 – 1 – 1) Try more ways. (a) 49 – 7 + 8 49 – 7 + 8 (b) 83 × 42 – 18 83 × 40 – 18 (c) 145 – 17 × 8 145 – 17 × 6 (d) 23 × 48 – 35 23 × (48 – 35) (e) (16 – 11) × 12 – 11 × 12 + 16 × 12 (f) (76 – 53) × 88 88 × (53 – 76) 17 Ans: 5. Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression that is equal to the given expression. (a) 83 – 37 – 12 (i) 84 – 38 – 12 (ii) 84 – (37 + 12) (iii) 83 – 38 – 13 (iv) – 37 + 83 – 12 6. Choose a number and create ten different expressions having that value. (a) = (b) > (c) < (d) > (e) = (f) > (g) = (h) > (b) 93 + 37 × 44 + 76 (i) 37 + 93 × 44 + 76 (ii) 93 + 37 × 76 + 44 (iii) (93 + 37) × (44 + 76) (iv) 37 × 44 + 93 + 76 Ans: (a) (i) and (iv) (b) (iv) (g) 25 × (42 + 16) 25 × (43 + 15) (h) 36 × (28 – 16) 35 × (27 – 15) Ans: Let’s choose the number 26. Here are ten different expressions with Try for some other numbers. the value 26: 1. 10 + 16 2. 30 – 4 3. 13 × 2 4. 52 2 5. 5 + (3 × 7) 6. (5 × 5) + 1 18" class_7,3,A Peek Beyond the Point,ncert_books/class_7/gegp1dd/gegp103.pdf,"3 3.1 The Need for Smaller Units Sonu’s mother was fixing a toy. She was trying to join two pieces with the help of a screw. Sonu was watching his mother with great curiosity. His mother was unable to join the pieces. Sonu asked why. His mother said that the screw was not of the right size. She brought another screw from the box and was able to fix the toy. The two screws looked the same to Sonu. But when he observed them closely, he saw they were of slightly different lengths. Sonu was fascinated by how such a small difference in lengths could matter so much. He was curious to know the difference in lengths. He was also curious to know how little the difference was because the screws looked nearly the same. In the following figure, screws are placed above a scale. Measure them and write their length in the space provided. A PEEK BEYOND THE POINT Chapter-3.indd 46 4/12/2025 11:39:13 AM Which scale helped you measure the length of the screws accurately? Why? What is the meaning of 2 7 10 cm (the length of the first screw)? numbers is divided into 10 equal parts. To get the length 2 7 10 cm, we go from 0 to 2 and then take seven parts of 1 10. The length of the screw is 2 cm and 7 10 cm. Similarly, we can make sense of the length 3 2 10 cm. We read 2 7 10 cm as two and seven-tenth centimeters, and 3 2 10 cm as three and two-tenth centimeters. As seen on the ruler, the unit length between two consecutive Between 2 cm and 3 cm _______________ More than 2 1 2 cm but less than 3 cm 2 7 10 cm A Peek Beyond the Point _______________ _______________ Chapter-3.indd 47 4/12/2025 11:39:14 AM Can you explain why the unit was divided into smaller parts to measure the screws? Measure the following objects using a scale and write their measurements in centimeters (as shown earlier for the lengths of the screws): pen, sharpener, and any other object of your choice. Write the measurements of the objects shown in the picture: 47 Ganita Prakash | Grade 7 As seen here, when exact measures are required we can make use of smaller units of measurement. 3.2 A Tenth Part The length of the pencil shown in the figure below is 3 4 10 units, which can also be read as 3 units and four one-tenths, i.e., (3 × 1) + (4 × 1 10) units. Chapter-3.indd 48 4/12/2025 11:39:14 AM = 10 times 1 10 = 10 × 1 10 = 1 unit This length is the same as 34 one-tenths units because 10 one-tenths units make one unit. 48 1 10 + 1 10 + 1 10 + 1 10 + 1 10 + 1 10 + 1 10 + 1 10 + 1 10 + 1 10 = 1 + 1 + 1 + 4 10 (3 and 4 one-tenths) A few numbers with fractional units are shown below along with how to read them. For the objects shown below, write their lengths in two ways and read them aloud. An example is given for the USB cable. (Note that the unit length used in each diagram is not the same). 4 1 10 ‘four and one-tenth’ 4 10 ‘four one-tenths’ or ‘four-tenths’ 41 10 ‘forty-one one-tenths’ or ‘forty-one tenths’ 41 1 10 ‘forty-one and one-tenth’ The length of the USB cable is 4 and 8 10 units or 48 10 units. 34 × 1 10 = 34 10 = 10 10 + 10 10 + 10 10 + 4 10 (34 one-tenths) A Peek Beyond the Point Chapter-3.indd 49 4/16/2025 4:28:12 PM Arrange these lengths in increasing order: (a) 9 10 (b) 1 7 10 (c) 130 10 (d) 13 1 10 (e) 10 5 10 (f) 7 6 10 (g) 6 7 10 (h) 4 10 49 Ganita Prakash | Grade 7 Arrange the following lengths in increasing order: 4 1 10, 4 10, 41 10, 41 1 10. Sonu is measuring some of his body parts. The length of Sonu’s lower arm is 2 7 10 units, and that of his upper arm is 3 6 10 units. What is the total length of his arm? To get the total length, let us see the lower and upper arm length as 2 units and 7 one-tenths, and 3 units and 6 one-tenths, respectively. So, there are (2 + 3) units and (7 + 6) one-tenths. Together, they make 5 units and 13 one-tenths. But 13 one-tenths is 1 unit and 3 one-tenths. So, the total length is 6 units and 3 one-tenths. (a) (2 + 3) + ( 7 10 + 6 10) (b) 2 7 10 = 6 3 10 + 3 6 10 = (2 + 3) + ( 13 10) = 5 + 13 10 = 5 + 10 10 + 3 10 = 5 + 1 + 3 10 = 6 + 3 10 Chapter-3.indd 50 4/12/2025 11:39:14 AM Or, both the lengths can be converted to tenths and then added: 50 27 10 + 35 10 = 62 10 62 10 is the same as 60 one-tenths ( 60 10) and 2 one-tenths ( 2 10), which is (c) 27 one-tenths and 35 one-tenths is 62 one-tenths = 5 13 10 = 6 3 10 equal to 6 units and 2 one-tenths, i.e., 6 2 10. The lengths of the body parts of a honeybee are given. Find its total length. Head: 2 3 10 units Thorax: 5 4 10 units Abdomen: 7 5 10 units The length of Shylaja’s hand is 12 4 10 units, and her palm is 6 7 10 units, as shown in the picture. What is the length of the longest (middle) finger? evaluating (12 + 4 10) – (6 + 7 10). This can be done in different ways. For example, The length of the finger can be found by (a) 12 + 4 10 – 6 – 7 10 = (12 – 6) + ( 4 10 – 7 10) = 6 – 3 10 = 5 + 1 – 3 10 Discuss what is being done here and why. 12 4 10 units 6 7 10 units Head Thorax abdomen A Peek Beyond the Point Chapter-3.indd 51 4/12/2025 11:39:14 AM As in the case of counting numbers, it is convenient to start subtraction from the tenths. We cannot remove 7 one-tenths from 4 (b) 12 4 10 11 14 10 = 5 + 10 10 – 3 10 = 5 + 7 10 = 5 7 10 – 6 7 10 – 6 7 10 = 5 7 10 51 Ganita Prakash | Grade 7 one-tenths. So we split a unit from 12 and convert it to 10 one-tenths. Now, the number has 11 units and 14 one-tenths. We subtract 7 onetenths from 14 one-tenths and then subtract 6 units from 11 units. Try computing the difference by converting both lengths to tenths. A Celestial Pearl Danio’s length is 2 4 10 cm, and the length of a Philippine Goby is 9 10 cm. What is the difference in their lengths? How big are these fish compared to your finger? Observe the given sequences of numbers. Identify the change after each term and extend the pattern: The length of a sheet of paper was 8 9 10 units, which can also be said as 8 units and 9 one-tenths. It is folded in half along its length. What is its length now? Celestial Pearl Danio Philippine Goby (a) 4, 4 3 10 , 4 6 10 , _________, _________, _________, _________ (b) 8 2 10 , 8 7 10 , 9 2 10 , _________, _________, _________, _________ (c) 7 6 10 , 8 7 10 , _________, _________, _________, _________ (d) 5 7 10 , 5 3 10 , _________, _________, _________, _________ Chapter-3.indd 52 4/12/2025 11:39:15 AM 3.3 A Hundredth Part 52 (e) 13 5 10 , 13, 12 5 10 , _________, _________, _________, _________ (f) 11 5 10 , 10 4 10 , 9 3 10 , _________, _________, _________, _________ length is between 4 4 10 units and 4 5 10 units. But we cannot state its exact measurement, since there are no markings. Earlier, we split a unit into 10 one-tenths to measure smaller lengths. We can do something similar and split each one-tenth into 10 parts. What is the length of this smaller part? How many such smaller parts make a unit length? As shown in the figure below, each one-tenth has 10 smaller parts, and there are 10 one-tenths in a unit; therefore, there will be 100 smaller parts in a unit. Therefore, one part’s length will be 1 100 of a unit. We can say that its A Peek Beyond the Point Chapter-3.indd 53 4/12/2025 11:39:17 AM We can see that it ends at 4 4 10 5 100, read as 4 units and 4 one-tenths and 5 one-hundredths. How many one-hundredths make one-tenth? Can we also say that the length is 4 units and 45 one-hundredths? Math Talk Returning to our question, what is the length of the folded paper? 53 Ganita Prakash | Grade 7 Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 0. Fill the lengths in the empty boxes. The length of the wire in the first picture is given in three different ways. Can you see how they denote the same length? For the lengths shown below write the measurements and read out the measures in words. 1 3 10 1 10 2 10 20 100 1 100 99 100 130 100 0 1 2 1 1 10 4 100 One and one-tenth and four-hundredths 1 14 100 One and fourteen-hundredths 114 100 One Hundred and Fourteen-hundredths Chapter-3.indd 54 4/12/2025 11:39:18 AM 54 In each group, identify the longest and the shortest lengths. Mark each length on the scale. (a) 3 10 , 3 100 , 33 100 (b) 3 1 10 , 30 10 , 1 3 10 A Peek Beyond the Point Chapter-3.indd 55 4/12/2025 11:39:18 AM (c) 45 100 , 54 100 , 5 10 , 4 10 (d) 3 6 10 , 3 6 100 , 3 6 10 6 100 55 Ganita Prakash | Grade 7 What will be the sum of 15 3 10 4 100 and 2 6 10 8 100 ? This can be solved in different ways. Some are shown below. (e) 8 10 2 100 , 9 100 , 1 8 100 (f) 7 3 10 5 100 , 7 5 10 , 7 41 100 (g) 65 10 15 100 , 5 87 100 , 5 7 100 (a) Method 1 (15 + 2) + ( 3 10 + 6 10) + ( 4 100 + 8 100) = 17 + 9 10 + 12 100 = 17 + 9 10 + 1 10 + 2 100 10 hundredths is the same as 1 tenth. Chapter-3.indd 56 4/12/2025 11:39:20 AM 56 (b) Method 2 = 17 + 10 10 + 2 100 = 18 2 100 . + 2 6 10 8 100 = 17 9 10 12 100 = 17 10 10 2 100 = 18 2 100 15 3 10 4 100 Are both these methods different? Observe the addition done below for 483 + 268. Do you see any similarities between the methods shown above? One can also find the sum 15 3 10 4 100 + 2 6 10 8 100 by converting to hundredths, as follows. (c) (15 + 2) + ( 34 100 + 68 100) (d) ( 1534 100 ) + ( 268 100) = 17 + 1 + 2 100 = 18 2 100 = 1802 100 = 1802 100 + 2 100 (400 + 80 + 3) + (200 + 60 + 8) = (400 + 200) + (80 + 60) + (3 + 8) = 600 + 140 + 11 = 600 + 150 + 1 = 700 + 50 + 1 = 751 = 17 + 102 100 15 is the same as 1500 hundredths and 2 is the same as 200 hundredths. 100 hundredths is same as 1 unit. A Peek Beyond the Point Math Talk Chapter-3.indd 57 4/12/2025 11:39:20 AM What is the difference: 25 9 10 – 6 4 10 7 100 ? One way to solve this is as follows: = 18 2 100 – 6 4 10 25 9 10 – 6 4 10 7 100 25 8 10 10 100 = 19 4 10 3 100 – 6 4 10 7 100 25 8 10 10 100 57 Ganita Prakash | Grade 7 Solve this by converting to hundredths. Observe the subtraction done below for 653 – 268. Do you see any similarities with the methods shown above? Figure it Out Find the sums and differences: What is the difference 15 3 10 4 100 –2 6 10 8 100 ? One way to solve this is as follows: 15 3 10 4 100 15 2 10 14 100 14 12 10 14 100 – 2 6 10 8 100 – 2 6 10 8 100 – 2 6 10 8 100 (600 + 50 + 3) – (200 + 60 + 8) = (600 – 200) + (50 – 60) + (3 – 8) = (600 – 200) + (40 – 60) + (13 – 8) = (600 – 200) + (40 – 60) + 5 = (500 – 200) + (140 – 60) + 5 = 300 + 80 + 5 = 385 = 2 6 10 6 100 Math Talk Math Talk Chapter-3.indd 58 4/12/2025 11:39:20 AM 3.4 Decimal Place Value You may have noticed that whenever we need to measure something more accurately, we split a part into 10 (smaller) equal parts ― we split a unit into 10 one-tenths and then split each one-tenth into 10, onehundredths and then we use these smaller parts to measure. 58 (a) 3 10 + 3 4 100 (b) 9 5 10 7 100 + 2 1 10 3 100 (c) 15 6 10 4 100 + 14 3 10 6 100 (d) 7 7 100 – 4 4 100 (e) 8 6 100 – 5 3 100 (f) 12 6 100 2 100 – 9 10 9 100 Can we not split a unit into 4 equal parts, 5 equal parts, 8 equal parts, or any other number of equal parts instead? Yes, we can. The example below compares how the same length is represented when the unit is split into 10 equal parts and when the unit is split into 4 equal parts. If an even more precise measure is needed, each quarter can further be split into four equal parts. Each part then measures 1 16 of a unit, i.e., 16 such parts make 1 unit. Then why split a unit into 10 parts every time? The reason is the special role that 10 plays in the Indian place value system. For a whole number written in the Indian place value system — for example, 281 — the place value of 2 is hundreds (100), that of 8 is tens (10), and that of 4 is one (1). Each place value is 10 times bigger than the one immediately to its right. Equivalently, each place value is 10 times smaller than the one immediately to its left: 10 ones make 1 ten, 10 tens make 1 hundred, 10 hundreds make 1 thousand, and so on. A Peek Beyond the Point Chapter-3.indd 59 4/12/2025 11:39:20 AM In order to extend this system of writing numbers to quantities smaller than one, we divide one into 10 equal parts. What does this give? It gives one-tenth. Further dividing it into 10 parts gives one-hundredth, and so on. 10,000 1000 100 10 1 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 59 Ganita Prakash | Grade 7 Can we extend this further? What will the fraction be when 1 100 is split into 10 equal parts? It will be 1 1000 , i.e., a thousand such parts make up a unit. 10,000 1000 100 10 1 1 10 1 100 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 × 10 ÷ 10 Chapter-3.indd 60 4/12/2025 11:39:20 AM values at each step, we can also extend to the right of 1 1000 , getting smaller place values at each step. 60 Just as when we extend to the left of 10,000, we get bigger place 1,00,000 10,000 1000 100 10 1 1 10 1 100 1 1000 1 10,000 This way of writing numbers is called the “decimal system” since it is based on the number 10; “decem” means ten in Latin, which in turn is cognate to the Sanskrit daśha meaning 10, with similar words for 10 occurring across many Indian languages including Odia, Konkani, Marathi, Gujarati, Hindi, Kashmiri, Bodo, and Assamese. We shall learn about other ways of writing numbers in later grades. How Big? We already know that a hundred 10s make 1000, and a hundred 100s make 10000. We can ask similar questions about fractional parts: Make a few more questions of this kind and answer them. × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 ÷ 10 (a) How many thousandths make one unit? (b) How many thousandths make one tenth? (c) How many thousandths make one hundredth? (d) How many tenths make one ten? (e) How many hundredths make one ten? A Peek Beyond the Point Chapter-3.indd 61 4/16/2025 4:51:22 PM Notation, Writing and Reading of Numbers We have been writing numbers in a particular way, say 456, instead of writing them as 4 × 100 (4 hundreds) + 5 × 10 (5 tens) + 6 × 1 (6 ones). Similarly, can we skip writing tenths and hundredths? Can the quantity 4 2 10 be written as 42 (skipping the 1 10 in 2 × 1 10) ? If yes, how would we know if 42 means 4 tens and 2 units or it means 4 units and 2 tenths? Similarly, 705 could mean: (a) 7 hundreds, and 0 tens and 5 ones (700 + 0 + 5) (b) 7 tens and 0 units and 5 tenths (70 + 0 + 5 10) (c) 7 units and 0 tenths and 5 hundredths (7 + 0 10 + 5 100) 61 Math Talk Ganita Prakash | Grade 7 Since these are different quantities, we need to have distinct ways of writing them. To identify the place value where integers end and the fractional parts start, we use a point or period (‘.’) as a separator, called a decimal point. The above quantities in decimal notation are then: 7 tens and 5 tenths (70 + 0 + 5 10) 70.5 These numbers, when shown through place value, are as follows: Decimal number Hundreds Tens Units Tenths Hundredths 7 units and 5 hundredths 70.5 7 × 10 0 × 1 5 × 1 10 7.05 7 × 1 0 × 1 10 5 × 1 100 705 7 × 100 0 × 10 5 × 1 7 hundreds and 5 ones (700 + 0 + 5) 705 (7 + 0 + 5 100) Quantity Decimal Notation 7.05 Chapter-3.indd 62 4/12/2025 11:39:20 AM 62 number of tens number of units number of one-tents × 10 × 1 × 1 10 × 1 100 number of one-hundredths value system to numbers also having fractional parts. Just as 705 means 7 × 100 + 5 × 1, the number 70.5 means 7 × 10 + 5 × 1 10 , and 7.05 means 7 × 1 +5 × 1 100. We have seen how to write numbers using the decimal point (‘.’). But how do we read/say these numbers? We know that 705 is read as seven hundred and five. 70.5 is read as seventy point five, short for seventy and five-tenths. 7.05 is read as seven point zero five, short for seven and five hundredths. 0.274 is read as zero point two seven four. We don’t read it as zero point two hundred and seventy four as 0.274 means 2 one-tenths and 7 one-hundredths and 4 one-thousandths. Make a place value table similar to the one above. Write each quantity in decimal form and in terms of place value, and read the number: Thus decimal notation is a natural extension of the Indian place (a) 2 ones, 3 tenths and 5 hundredths (b) 1 ten and 5 tenths (c) 4 ones and 6 hundredths (d) 1 hundred, 1 one and 1 hundredth (e) 8 100 and 9 10 (f) 5 100 A Peek Beyond the Point Chapter-3.indd 63 4/12/2025 11:39:20 AM (h) 2 1 100 , 4 1 10 and 7 7 1000 In the chapter on large numbers, we learned how to write 23 hundreds. 23 hundreds = 23 × 100 = 2000 + 300 = 2300. Similarly, 23 tens would be: 23 tens = 23 × 10 = 200 + 30 = 230. Thousands Hundreds Tens Units (g) 1 10 2 3 0 0 23 63 Ganita Prakash | Grade 7 How can we write 234 tenths in decimal form? Write these quantities in decimal form: (a) 234 hundredths, (b) 105 tenths. 3.5 Units of Measurement Length Conversion We have been using a scale to measure length for a few years. We already know that 1 cm = 10 mm (millimeters). Hundreds Tens Units Tenths Hundredths 234 tenths = 234 10 = 200 10 + 30 10 + 4 10 = 20 + 3 + 4 10 = 23.4. Thousands Hundreds Tens Units 2 3 4 2 3 0 23 234 Chapter-3.indd 64 4/12/2025 11:39:20 AM How many cm is 1 mm? How many cm is (a) 5 mm? (b) 12 mm? 64 1 mm = 1 10 cm = 0.1 cm (i.e., one-tenth of a cm). 5 mm = 5 10 cm = 0.5 cm 12 mm = 10 mm + 2 mm = 1.2 cm. = 1 cm + 2 10 cm How many mm is 5.6 cm? Since each cm has 10 mm, 5.6 cm (5 cm + 0.6 cm) is 56 mm. Fill in the blanks below (mm <–> cm) The illustration below shows how small some things are! Try taking an approximate measurement of each. ________ = 0.9 cm 134 mm =__________ ________ = 203.6 cm 12 mm = 1.2 cm 56 mm = 5.6 cm 70 mm = _______ 0.5 mm 1 mm A Peek Beyond the Point 1.5 mm Chapter-3.indd 65 4/12/2025 11:39:22 AM We also know that 1 m = 100 cm. Based on this, we can say that • The three blue stripes represent the typical relative sizes of pen strokes: fine stroke, medium stroke, and bold stroke. • A human hair is about 0.1 mm in thickness. • The thickness of a newspaper can range from 0.05 to 0.08 mm. • Mustard seeds have a thickness of 1 – 2 mm. • The smallest ant species discovered so far, Carabera Bruni, has a total length of 0.8 – 1 mm. They are found in Sri Lanka and China. • The smallest land snail species discovered so far, Acmella Nana, has a shell diameter of 0.7 mm. They are found in Malaysia. 1 cm = 1 100 m = 0.01 m. 65 Ganita Prakash | Grade 7 How many m is (a) 10 cm? (b) 15 cm? Fill in the blanks below (cm <–> m): How many mm does 1 meter have? Can we write 1 mm = 1 1000 m? Here, we have some more interesting facts about small things in nature! 10 cm = 1 10 m = 0.1 m Since each cm is one-hundredth of a meter, 15 cm can be written as 15 cm = 15 100 m = 10 100 m + 5 100 m = 1 10 m + 5 100 m = 0.15 m. 36 cm = _______ 50 cm = _______ _______ = 0.89 m 4 cm = _______ 325 cm = _______ ________ = 2.07 m Chapter-3.indd 66 4/12/2025 11:39:23 AM 66 1 cm 1 cm 0 Weight Conversion Let us look at kilograms (kg). We know that 1 kg = 1000 gram (g). We can say that How many kilograms is 5 g? How many kilograms is 10 g? • The egg of a hummingbird typically is 1.3 cm long and 0.9 cm wide. • The Philippine Goby is about 0.9 cm long. It can be found in the Philippines and other Southeast Asian countries. • The smallest known jellyfish, Irukandji, has a bell size of 0.5 – 2.5 cm. Its tentacles can be as long as 1 m. They are found in Australia. Its venom can be fatal to humans. • The Wolfi octopus, also known as the Star-sucker Pygmy Octopus, is the smallest known octopus in the world. Their typical size is around 1 – 2.5 cm and they weigh less than 1 gm. They are found in the Pacific Ocean. 10 g = 10 1000 kg = 1 100 kg = 0.010 kg. As each gram is one-thousandth of a kg, 254 g can be written as 1 g = 1 1000 kg = 0.001 kg. 5 g = 5 1000 kg = 0.005 kg. A Peek Beyond the Point Chapter-3.indd 67 4/16/2025 4:29:32 PM Fill in the blanks below (g <–> kg) = ( 200 1000 + 50 1000 + 4 1000) kg = ( 2 10 + 5 100 + 4 1000) kg 254 g = 254 1000 kg = 0.254 kg. 67 Ganita Prakash | Grade 7 Look at the picture below showing different quantities of rice. Starting from the 1g heap, subsequent heaps can be found that are 10 times heavier than the previous heap/packets. The combined weight of rice in this picture is 11.111 kg. Rupee ─ Paise conversion Also, 465 g = _______ 68 g = _________ 1560 g = ________ 704 g = _______ ________ = 0.56 kg _______ = 2.5 kg 1 gram = 1000 milligrams (mg). So, 1 mg = 1 1000 g = 0.001 g. Chapter-3.indd 68 4/12/2025 11:39:25 AM You may have heard of ‘paisa’. 100 paise is equal to 1 rupee. As we have coins and notes for rupees, coins for paise were also used commonly until recently. There were coins for 1 paisa, 2 paise, 3 paise, 5 paise, 10 paise, 20 paise, 25 paise, and 50 paise. All denominations of 25 paise and less were removed from use in the year 2011. But we still see paise in bills, account statements, etc. 1 rupee = 100 paise 1 paisa = 1 100 rupee = 0.01 rupee As each paisa is one-hundredth of a rupee, 68 75 paise = 75 100 rupee Fill in the blanks below (rupee <–> paise) Discuss with adults at home/school the prices of different products and services during their childhood. Try to find old coins and stamps. Let us consider the decimal number 1.4. It is equal to 1 unit and 4 tenths. This means that the unit between 1 and 2 is divided into 10 Try This During the 1970s, a masala dosa cost just 50 paise, one could buy a banana for 20-25 paise, a handful of peppermints were available for 2 paise or 3 paise, and a kg of rice cost ₹2.45. 10 p = ___________ _______p = ₹ 0.05 ________p = ₹ 0.36 _________ = ₹ 0.50 99 p = _________ 250 p = _________ = ( 7 100 + 5 100) rupee = ( 70 100 + 5 100) rupee = 0.75 rupee. A Peek Beyond the Point Chapter-3.indd 69 4/12/2025 11:39:25 AM 3.6 Locating and Comparing Decimals 69 Ganita Prakash | Grade 7 equal parts, and 4 such parts are taken. Hence, 1.4 lies between 1 and 2. Draw the number line and divide the unit between 1 and 2 into 10 equal parts. Take the fourth part, and we have 1.4 on the number line. Name all the divisions between 1 and 1.1 on the number line. Identify and write the decimal numbers against the letters. There is Zero Dilemma! Sonu says that 0.2 can also be written as 0.20, 0.200; Zara thinks that putting zeros on the right side may alter the value of the decimal number. What do you think? We can figure this out by looking at the quantities these numbers represent using place value. 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 5 5.1 5.3 5.4 1 A 1.04 B C D 1.1 1.2 1.3 Chapter-3.indd 70 4/12/2025 11:39:26 AM We can see that 0.2, 0.20, and 0.200 are all equal as they represent the same quantity, i.e., 2 tenths. But 0.2, 0.02, and 0.002 are different. 70 Decimal number Units Tenths Hundredths Thousandths 0.200 0 2 0 0 0.002 0 0 0 2 0.20 0 2 0 0.02 0 0 2 0.2 0 2 Can you tell which of these is the smallest and which is the largest? Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50? Observe the number lines in Figure (a) below. At each level, a particular segment of the number line is magnified to locate the number 4.185. Identify the decimal number in the last number line in Figure (b) denoted by ‘?’. Make such number lines for the decimal numbers: (a) 9.876 (b) 0.407. 4.18 4.185 4.19 4.1 0 4 5 (a) (b) 4.2 10 0 10 A Peek Beyond the Point ? Chapter-3.indd 71 4/12/2025 11:39:26 AM In the number line shown below, what decimal numbers do the boxes labelled ‘a’, ‘b’, and ‘c’ denote? The box with ‘b’ corresponds to the decimal number 7.5; are you able to see how? There are 5 units between 5 and 10, divided into 10 equal parts. Hence, every 2 divisions make a unit, and so every division is 1 2 unit. What numbers do ‘a’ and ‘c’ denote? 5 a b c 10 71 Ganita Prakash | Grade 7 Using similar reasoning find out the decimal numbers in the boxes below. Which is larger: 6.456 or 6.465? To answer this, we can use the number line to locate both decimal numbers and show which is larger. This can also be done by comparing the corresponding digits at each place value, as we do with whole numbers. This comparison is visualised step by step below. Note that the visualisation below is not to scale. 6.456 6.465 6 4.3 8 f g h d Both numbers have 6 units. Both numbers have 6 units and 4 tenths. e 4.8 8.1 Chapter-3.indd 72 4/12/2025 11:39:26 AM 72 6.456 6.465 6 6.4 6.456 6.465 6 6.4 6.45 6.46 Both numbers have 6 units and 4 tenths, but the first number has only 5 hundredths, whereas the second number has 6 hundredths. We start by comparing the most significant digits (digits with the highest place value) of the two numbers. If the digits are the same, we compare the next smaller place value. We keep going till we find a position where the digits are not equal. The number with the larger digit at this position is the greater of the two. Why can we stop comparing at this point? Can we be sure that whatever digits are there after this will not affect our conclusion? Which decimal number is greater? Closest Decimals Consider the decimal numbers 0.9, 1.1, 1.01, and 1.11. Identify the decimal number that is closest to 1. Let us compare the decimal numbers. Arranging these in ascending order, we get 0.9 < 1 < 1.01 < 1.1 < 1.11. Among the neighbours of 1, 1.01 is 1/100 away from 1 whereas 0.9 is 10/100 away from 1. Therefore, 1.01 is closest to 1. Which of the above is closest to 1.09? Which among these is closest to 4: 3.56, 3.65, 3.099? (a) 1.23 or 1.32 (b) 3.81 or 13.800 (c) 1.009 or 1.090 A Peek Beyond the Point Math Talk Chapter-3.indd 73 4/12/2025 11:39:26 AM Which among these is closest to 1: 0.8, 0.69, 1.08? In each case below use the digits 4, 1, 8, 2, and 5 exactly once and try to make a decimal number as close as possible to 25. 73 Math Talk Ganita Prakash | Grade 7 3.7 Addition and Subtraction of Decimals Priya requires 2.7 m of cloth for her skirt, and Shylaja requires 3.5m for her kurti. What is the total quantity of cloth needed? Earlier, we saw how to add 2 7 10 + 3 5 10 (also shown below). Can you carry out the same addition using decimal notation? It is shown below. Share your observations. The total quantity of cloth needed is 6.2 m. How much longer is Shylaja’s cloth compared to Priya’s? the differences 3 5 10 – 2 7 10 and 3.5m – 2.7m are computed. We have to find the sum of 2.7m + 3.5m. We have to find the difference of 3.5m – 2.7m. Again, observe how 3 5 10 2 7 10 + 3 5 10 = 5 12 10 = 6 2 10 2 15 10 2.7 + 3.5 = 6.2 3.5 1 3.5 2 1 Chapter-3.indd 74 4/12/2025 11:39:26 AM As you can see, the standard procedure for adding and subtracting whole numbers can be used to add and subtract decimals. A detailed view of the underlying place value calculation is shown below for the sum 75.345 + 86.691. Its compact form is shown next to it. 74 – 2 7 10 – 2 7 10 = 0 8 10 = 0.8 – 2.7 – 2.7 = + 1 × 100 1 × 10 1 × 1 1 × 100 Write the detailed place value computation for 84.691 – 77.345, and its compact form. Figure it Out 1. Find the sums 2. Find the differences (c) 2.15 + 5.26 (d) 9.01 + 9.10 (e) 29.19 + 9.91 (f) 0.934 + 0.6 (g) 0.75 + 0.03 (h) 6.236 + 0.487 (a) 5.6 – 2.3 (b) 18 – 8.8 (a) 5.3 + 2.6 (b) 18 + 8.8 7 × 10 5 × 1 3 × 1 10 4 × 1 100 5 × 1 1000 8 × 10 6 × 1 6 × 1 10 9 × 1 100 1 × 1 1000 16 × 10 12 × 1 10 × 1 10 13 × 1 100 6 × 1 1000 1 × 1 10 A Peek Beyond the Point 11 1 75.345 +86.691 =162.036 Try This Chapter-3.indd 75 4/12/2025 11:39:26 AM Decimal Sequences Observe this sequence of decimal numbers and identify the change after each term. 4.4, 4.8. 5.2, 5.6, 6.0, … We can see that 0.4 is being added to a term to get the next term. Continue this sequence and write the next 3 terms. (c) 10.4 – 4.5 (d) 17 – 16.198 (e) 17 – 0.05 (f) 34.505 – 18.1 (g) 9.9 – 9.09 (h) 6.236 – 0.487 75 Ganita Prakash | Grade 7 Similarly, identify the change and write the next 3 terms for each sequence given below. Try to do this computation mentally. Make your own sequences and challenge your classmates to extend the pattern. Estimating Sums and Differences Sonu has observed sums and differences of decimal numbers and says, “If we add two decimal numbers, then the sum will always be greater than the sum of their whole number parts. Also, the sum will always be less than 2 more than the sum of their whole number parts.” Let us use an example to understand what his claim means: If the two numbers to be added are 25.936 and 8.202, the claim is that their sum will be greater than 25 + 8 (whole number parts) and will be less than 25 + 1 + 8 + 1. What do you think about this claim? Verify if this is true for these numbers. Will it work for any 2 decimal numbers? What about for the sum of 25.93603259 and 8.202? (a) 4.4, 4.45, 4.5, … (b) 25.75, 26.25, 26.75, … (c) 10.56, 10.67, 10.78, … (d) 13.5, 16, 18.5, … (e) 8.5, 9.4, 10.3, … (f) 5, 4.95, 4.90, … (g) 12.45, 11.95, 11.45, … (h) 36.5, 33, 29.5, … Math Talk Chapter-3.indd 76 4/12/2025 11:39:26 AM Similarly, come up with a way to narrow down the range of whole numbers within which the difference of two decimal numbers will lie. 3.8 More on the Decimal System Decimal and Measurement Disasters Decimal point and unit conversion mistakes may seem minor sometimes but they can lead to serious problems. Here are some actual incidents in which such errors caused major issues. 76 Note to the Teacher: Estimating the result before computing may help in identifying if a mistake happens with the calculation. Try This Several incidents have occurred due to incorrect reading of decimal numbers while giving medication. For example, reading 0.05 mg as 0.5 mg can lead to using a medicine 10 times more than the prescribed quantity. It is therefore important to pay attention to units and the location of the decimal point. Deceptive Decimal Notation Sarayu gets a message: “The bus will reach the station 4.5 hours post noon.” When will the bus reach the station: 4:05 p.m., 4:50 p.m., 4:25 p.m.? None of these! Here, 0.5 hours means splitting an hour into 10 equal parts and taking 5 parts out of it. Each part will be 6 minutes (60 minutes/10) long. 5 such parts make 30 minutes. So, the bus will reach the station at 4:30. Here is a short-story of a decimal mishap: A girl measures the width of an opening as 2 ft 5 inches but conveys to the carpenter to make a door 2.5 ft wide. The carpenter makes a door of width 2 ft 6 inches (since 1 ft = 12 inches, 0.5 ft = 6 inches), and it wouldn’t close fully. • In 2013, the finance office of Amsterdam City Council (Netherlands) mistakenly sent out €188 million in housing benefits instead of the intended €1.8 million due to a programming error that processed payments in euro cents instead of euros. (1 euro-cent = 1/100 euro). • In 1983, a decimal error nearly caused a disaster for an Air Canada Boeing 767. The ground staff miscalculated the fuel, loading 22,300 pounds instead of kilograms—about half of what was needed (1 pound ~ 0.453 kg). The plane ran out of fuel mid-air, forcing the pilots to make an emergency landing at an abandoned airfield. Fortunately, everyone survived. A Peek Beyond the Point Chapter-3.indd 77 4/12/2025 11:39:27 AM I said the door’s width should be 2.5 ft. It is 2.5 ft! You can verify. 77 Ganita Prakash | Grade 7 If you watch cricket, you might have noticed decimal-looking numbers like ‘Overs left: 5.5’. Does this mean 5 overs and 5 balls or 5 overs and 3 balls? Here, 5.5 overs means 5 5 6 overs (as 1 over = 6 balls), i.e., 5 overs and 5 balls. Where else can we see such ‘non-decimals’ with a decimal-like notation? A Pinch of History–Decimal Notation Over Time Decimal fractions (i.e., fractions with denominators like 1 10 , 1 100 , 1 1000 , and so on) are used in the works of a number of ancient Indian astronomers and mathematicians, including in the important 8th century works of Śhrīdharāchārya on arithmetic and algebra. Decimal notation, in essentially its modern form, was described in detail in Kitāb al-Fuṣūl fī al-Ḥisāb al Hindī (The Book of Chapters on Indian Arithmetic) by Abūl Ḥassan al-Uqlīdisī, an Arab mathematician, in around 950 CE. He represented the number 0.059375 as 0 ˈ 059375. In the 15th century, to separate whole numbers from fractional parts, a number of different notations were used: • a vertical mark on the last digit of the whole number part (as shown above), In the 16th century, John Napier, a Scottish mathematician, and Christopher Clavius, a German mathematician, used the point/period (‘.’) to separate the whole number and the fractional parts, while François Viète, a French mathematician, used the comma (‘,’) instead. Currently, several countries use the comma to separate the integer part and the fractional part. In these countries, the number 1,000.5 is written as 1 000,5 (space as a thousand separator). But the decimal point has endured as the most popular notation for writing numbers having fractional parts in the Indian place value system. • use of different colours and • a numerical superscript giving the number of fractional decimal places (0.36 would be written as 36 2 ). Math Talk Chapter-3.indd 78 4/15/2025 5:20:44 PM Figure it Out 1. Convert the following fractions into decimals: 78 (a) 5 100 (b) 16 1000 (c) 12 10 (d) 254 1000 2. Convert the following decimals into a sum of tenths, hundredths and thousandths: 3. What decimal number does each letter represent in the number line below? 4. Arrange the following quantities in descending order: 5. Using the digits 1, 4, 0, 8, and 6 make: 6. Will a decimal number with more digits be greater than a decimal number with fewer digits? 7. Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, 0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total weight of the items she bought. (a) 0.34 (b) 1.02 (c) 0.8 (d) 0.362 (a) 11.01, 1.011, 1.101, 11.10, 1.01 (b) 2.567, 2.675, 2.768, 2.499, 2.698 (c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g (d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m (a) the decimal number closest to 30 (b) the smallest possible decimal number between 100 and 1000. 6.4 6.5 6.6 a c b A Peek Beyond the Point Chapter-3.indd 79 4/12/2025 11:39:28 AM 8. Pinto supplies 3.79 L, 4.2 L, and 4.25 L of milk to a milk dairy in the first three days. In 6 days, he supplies 25 litres of milk. Find the total quantity of milk supplied to the dairy in the last three days. 9. Tinku weighed 35.75 kg in January and 34.50 kg in February. Has he gained or lost weight? How much is the change? 10. Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 6.18, 6.17, ____, _____ 11. How many millimeters make 1 kilometer? 12. Indian Railways offers optional travel insurance for passengers who book e-tickets. It costs 45 paise per passenger. If 1 lakh people opt for insurance in a day, what is the total insurance fee paid? 13. Which is greater? (a) 10 1000 or 1 10 ? 79 Ganita Prakash | Grade 7 14. Write the decimal forms of the quantities mentioned (an example is given): 15. Using each digit 0 – 9 not more than once, fill the boxes below so that the sum is closest to 10.5: 16. Write the following fractions in decimal form: (b) One-hundredth or 90 thousandths? (c) One-thousandth or 90 hundredths? (a) 87 ones, 5 tenths and 60 hundredths = 88.10 (b) 12 tens and 12 tenths (c) 10 tens, 10 ones, 10 tenths, and 10 hundredths (d) 25 tens, 25 ones, 25 tenths, and 25 hundredths (a) 1 2 (b) 3 2 (c) 1 4 (d) 3 4 (e) 1 5 (f) 4 5 Try This Chapter-3.indd 80 4/12/2025 11:39:28 AM 80 • We can split a unit into smaller parts to get more exact/accurate measurements. • We extended the Indian place value system and saw that » 1 unit = 10 one-tenths, » 1 tenth = 10 one-hundredths, » 1 hundredth = 10 one-thousandths, » 10 one-hundredths = 1 tenth, » 100 one-hundredths = 1 unit. • A decimal point (‘.’) is used in the Indian place value system to separate the whole number part of a number from its fractional part. • We also learnt how to compare decimal numbers, locate them on the number line, and perform addition and subtraction on them. SUMMARY Page No. 47 Ans: Eraser → 2 4 10 cm, Pencil → 4 5 10 cm → 4 1 2 cm, Chalk → 1 4 10 Write the measurements of the objects shown in the picture. A Peek Beyond the Point Chapter – 3 Page No. 49 Ans: Arrange the lengths in increasing order: (a) (b) 1 (c) (d) 13 (e) 10 (f) 7 (g) 6 (h) 10 , 9 10 , 1 7 10 , 6 7 10 , 7 6 10 , 10 5 10 , 130 10 , 13 1 10 4 1 Page No. 50 Page No. 51 Page No. 52 Ans: 15 2 10 units. Ans: Arrange the following lengths in increasing order: 4 1 10 , 4 10 , 41 10 , 41 1 10 The lengths of the body parts of a honeybee are given. Find its total length. A Celestial Pearl Danio’s length is cm, and the length of a Increasing order: 4 10 , 41 10 = 4 1 10 , 41 1 10 Abdomen = 7 5 10 units Head = 2 3 10 units Thorax = 5 4 10 units Ans: 1 5 10 cm Philippine Goby is cm. What is the difference in their lengths? Observe the given sequences of numbers. Identify the change after each term and extend the pattern: (a) 4, 4 , 4 , _______, _______, _______, _______ (b) 8 , 8 , 9 , _______, _______, _______, _______ (c) 7 , 8 , _______, _______, _______, _______ 2 Page No. 53 Ans: (a) 4, 4 3 10 , 4 6 10 , 4 9 10 , 5 2 10 , 5 5 10 , 5 8 10 → increment of 3 10 each time. (b) 8 2 10 , 8 7 10 , 9 2 10 , 9 7 10 , 10 2 10 , 10 7 10 , 11 2 10 → increment of 5 10 each time. (c) 7 6 10 , 8 7 10 , 9 8 10 , 10 9 10 , 12, 13 1 10 → increment of 1 + 1 10 (d) 5 7 10 , 5 3 10 , 4 9 10 , 4 5 10 , 4 1 10 , 3 7 10 → subtracting: 4 10 each time (e) 13 5 10 , 13 , 12 5 10 , 12, 11 5 10 , 11, 10 5 10 → decreasing by 5 10 (f) 11 5 10 , 10 4 10 , 9 3 10 , 8 2 10 , 7 1 10 , 6 → decreasing by 1 1 10 How many one‑hundredths make one‑tenth? Can we also say that the length is 4 units and 45 one‑hundredths? (d) 5 , 5 , _______, _______, _______, _______ (e) 13 , 13, 12 , _______, _______, _______, _______ (f) 11 , 10 , 9 , _______, _______, _______, _______ Ans: 1 10 = 10 100 So, 10 one‑hundredths make one‑tenth. Yes, we can write it 4 + 45 100 = 4 units and 45 one‑hundredths. 3 Page No. 54 Ans: Ans: Measurement − 5 37 100 In words: Five and thirtyseven-hundredths. Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 0. Fill the lengths in the empty boxes. For the lengths shown below write the measurements and read out the measures in words. Measurement − 15 3 100 In words: Fifteen and three-hundredths. 4 Page No. 55 Ans: Measurement − 7 52 100 In words: Seven and fiftytwo-hundredths. Measurement − 9 80 100 In words: Nine and eighty-hundredths. In each group, identify the longest and the shortest lengths. Mark each length on the scale. 5 (a) Longest = 33 100 Shortest = 3 100 (b) Longest = 3 1 10 Shortest = 1 3 10 (c) Longest = 54 100 Shortest = 4 10 (d) Longest = 3 6 10 6 100 Shortest = 3 6 100 6 Page No. 56 (e) Longest = 1 8 100 Shortest = 9 100 (f) Longest = 7 5 100 Shortest = 7 3 10 5 100 (g) Longest = 65 10 15 100 Shortest = 5 7 100 7 Page No. 58 Figure it Out Find the sums and differences (a) 3 10 + 3 4 100 Ans: 3 34 100 (b) 9 5 10 7 100 + 2 1 10 3 100 Ans: 11 7 10 (c) 15 6 10 4 100 + 14 3 10 6 100 Ans: 30 (d) 7 7 100 – 4 4 100 Ans: 3 3 100 (e) 8 6 100 – 5 3 100 Ans: 3 3 100 (f) 12 6 10 2 100 – 9 10 9 100 Page No. 63 Ans: Ans: 11 63 100 Make a place value table similar to the one above. Write each quantity in decimal form and in terms of place value, and read the number: 8 a b c d Question 2 ones, 3 tenths and 5 hundredths tenths 6 hundredths 1 one and 1 hundredth 1 ten and 5 4 ones and 1 hundred, Hun dre ds – – 2 3 5 – 2 + 3 10 + 5 100 = 2.35 – 1 – 5 – – 10 + 5 10 = 10.5 Ten point five (or ten and five tenths) – – 4 – 6 – 4 + 0 + 6 100 = 4.06 1 – 1 – 1 – 100 + 0 + 1 + 0 + 1 100 Te ns On es Te nt hs Hun dre dth s Th ous an dth s =101.01 De ci mal Fo rm Two point three five (or two and thirty-five hundredths) One hundred hundredths) four and six zero one (or zero six (or Four point one point Re ad as e 8 100 and 9 10 – – – 9 8 – 9 10 + 8 100 = 0.98 9 one hundred hundredths) one and one hundredth) (or ninetyZero point nine eight eight g h f 100 – – – 5 – 5 100 = 0.05 Zero point zero five (or five hundredths) 1 10 – – – 1 – – 1 10 = 0.1 Zero point one (or one tenth) 2 1 10 , 4 1 10 , and 7 7 1000 5 – 2 4 7 0 1 - 1 – - 7 - - 7 + 0 10 + 0 100 + 7 1000 = 7.007 4 + 1 10 = 4.1 13.117 Thirteen 2.01 Seven point Four point one seven. Two point zero zero point one zero one seven and one and Page No. 64 Write these quantities in decimal form: (a) 234 hundredths (b) 105 tenths. Ans: 2.34 Ans: 10.5 10 Page No. 66 Page No. 68 Ans: 1m = 1000 mm Ans: Yes Ans: 12 mm= 1.2 cm 56 mm = 5.6 cm 70 mm = 7.0 cm 9 mm = 0.9 cm 134 mm = 13.4 cm 2036 mm = 203.6 cm Fill in the blanks below (mm < – > cm) Fill in blanks (cm <-> m): How many mm does 1 meter have? Can we write 1 mm = m? 36 cm = 0.36 m 50 cm = 0.5 m 89 cm = 0.89 m 4 cm = 0.04 m 325 cm = 3.25 m 207 cm = 2.07 m Ans: 465 g = 0.465 kg 68 g = 0.068 kg 1560 g = 1.56 kg 704 g = 0.704 kg 560 g = 0.56 kg 2500 g = 2.5 kg Fill in the blanks below (g < – > kg) 11 Page No. 69 Page No. 70 Page No. 71 Ans: The small divisions between 1 and 1.1 on the number line (if the subdivision is of 10 parts) are: 1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08, 1.09 Ans: Fill in the blanks below (rupee < – > paise) Name all the divisions between 1 and 1.1 on the number line. Identify and write the decimal numbers against the letters. 10 p = ₹0.10 5 p = ₹0.05 36 p = ₹0.36 50 p = ₹0.50 99 p = ₹0.99 250 p = ₹2.50 Decimal Number Units Tenths Hundredths Thousandths Ans: Can you tell which of these is the smallest and which is the largest? 0.2 = 0.20 = 0.200 is the largest. 0.200 0 2 0 0 0.002 0 0 0 2 0.20 0 2 0 0 0.02 0 0 2 0 0.2 0 2 0 0 12 Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50? Ans: 4.5, 4.50, and 04.50 are equal. 4.05 and 4.050 are equal. Identify the decimal number in the last number line in Figure (b) denoted by ‘?’ Ans: Ans: (a) (b) 0.002 is the smallest. Make such number lines for the decimal numbers: (a) 9.876 (b) 0.407. 13 Page No. 72 Ans: Using similar reasoning find out the decimal numbers in the boxes below. Ans: In the number line shown below, what decimal numbers do the boxes labelled ‘a’, ‘b’, and ‘c’ denote? 14 Page No. 73 Ans: 3.65 is closest to 4. Ans: 1.08 is closest to 1. Ans: Which decimal number is greater? (a) 1.23 or 1.32 (b) 3.81 or 13.800 (c) 1.009 or 1.090 Ans: (a) 1.32 is larger. (b) 13.800 is larger. (c) 1.090 is larger. Consider the decimal numbers 0.9, 1.1, 1.01 and 1.11 Which of the above is closest to 1.09? Ans: 1.1 is closest to 1.09. Which among these is closest to 4: 3.56, 3.65, 3.099? Which among these is closest to 1: 0.8, 0.69, 1.08? In each case below use the digits 4, 1, 8, 2, and 5 exactly once and try to make a decimal number as close as possible to 25. Page No. 75 Figure it Out 1. Find the sums (a) 5.3 + 2.6 (b) 18 + 8.8 (c) 2.15 + 5.26 (d) 9.01 + 9.10 (e) 29.19 + 9.91 (f) 0.934 + 0.6 (g) 0.75 + 0.03 (h) 6.236 + 0.487 15 Page No. 76 Ans: 4.4, 4.8. 5.2, 5.6, 6.0, 6.4, 6.8, 7.2 Ans: 2. Find the differences Ans: (a) 3.3 (b) 9.2 (c) 5.9 (d) 0.802 (e) 16.95 (f) 16.405 (g) 0.81 (h) 5.749 Continue this sequence 4.4, 4.8. 5.2, 5.6, 6.0, … and write the next 3 terms. Similarly, identify the change and write the next 3 terms for each sequence given below. (a) 7.9 (b) 26.8 (c) 7.41 (d) 18.11 (e) 39.10 (f) 1.534 (g) 0.78 (h) 6.723 (a) 5.6 – 2.3 (b) 18 – 8.8 (c) 10.4 – 4.5 (d) 17 – 16.198 (e) 17 – 0.05 (f) 34.505 – 18.1 (g) 9.9 – 9.09 (h) 6.236 – 0.487 Ans: (a) 4.4, 4.45, 4.5,4.55, 4.6, 4.65 → increment of + 0.05 Try to do this computation mentally. (a) 4.4, 4.45, 4.5, … (b) 25.75, 26.25, 26.75, … (c) 10.56, 10.67, 10.78, … (d) 13.5, 16, 18.5, … (e) 8.5, 9.4, 10.3, … (f) 5, 4.95, 4.90, … (g) 12.45, 11.95, 11.45, … (h) 36.5, 33, 29.5, … (b) 25.75, 26.25, 26.75, 27.25, 27.75, 28.25 → add 0.5 (c) 10.56, 10.67, 10.78, 10.89, 11.0, 11.11→ Add 0.11 (d) 13.5, 16, 18.5, 21.0, 23.5, 26.0 → Add 2.5 16 Page No. 78 Page No. 79 (e) 8.5, 9.4, 10.3, 11.2, 12.1, 13.0 → Add 0.9 (f) 5, 4.95, 4.90, 4.85, 4.80, 4.75 → Subtract 0.05 (g) 12.45, 11.95, 11.45, 10.95, 10.45, 9.95 → Subtract 0.50 (h) 36.5, 33, 29.5, 26.0, 22.5, 19.0 → Subtract 3.5 Figure it out 1. Convert the following fractions into decimals: (a) (b) (c) (d) Ans: 2. Convert the following decimals into a sum of tenths, hundredths Ans: (a) 5 100 = 0.05 (b) 16 1000 = 0.016 (c) 12 10 = 1.2 (d) 254 1000 = 0.254 and thousandths: (a) 0.34 (b) 1.02 (c) 0.8 (d) 0.362 (a) 3 10 + 4 100 (b) 100 100 + 2 100 = 1 + 2 100 3. What decimal number does each letter represent in the number line below? Ans: (c) 8 10 (d) 3 10 + 6 100 + 2 1000 17 4. Arrange the following quantities in descending order: (a) 11.01, 1.011, 1.101, 11.10, 1.01 → Descending order: 11.10, 11.01, 1.101, 1.011, 1.01. 5. Using the digits 1, 4, 0, 8, and 6 make: (a) the decimal number closest to 30 (b) the smallest possible decimal number between 100 and 1000. Ans: Using the digits 1, 4, 0, 8, and 6, we can make: (a) 40.168 (b) 104.68 (b) 2.567, 2.675, 2.768, 2.499, 2.698 → Descending order: 2.768, 2.698, 2.675, 2.567, 2.499. (c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g →Descending order: 4.678, 4.666, 4.656, 4.600, 4.595. (d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m → Descending order: 33.331, 33.313, 33.31, 33.133, 33.13. 6. Will a decimal number with more digits be greater than a decimal number with fewer digits? Ans: No. It is not necessary. For example, 2.05 (2 digits after decimal) vs 2.5 (1 digit after decimal). 2.5 >2.05 7. Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, 0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total weight of the items she bought. Ans: 1.3 kg 8. Pinto supplies 3.79 L, 4.2 L, and 4.25 L of milk to a milk dairy in the first three days. In 6 days, he supplies 25 liters of milk. Find the total quantity of milk supplied to the dairy in the last three days. 18 Ans: 12.76 L 9. Tinku weighed 35.75 kg in January and 34.50 kg in February. Has he gained or lost weight? How much is the change? Ans: He has lost weight. The change in the weight = 1.25 kg 10. Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, ____, _____ Ans: 9.07, 9.06 11. How many millimeters make 1 kilometer? Ans: 1000000 mm 12. Indian Railways offers optional travel insurance for passengers who book e-tickets. It costs 45 paise per passenger. If 1 lakh people opt for insurance in a day, what is the total insurance fee paid? Ans: The total insurance fee paid for 1 lakh passengers is ₹45,000. 13. Which is greater? Ans: (a) or ? (b) One-hundredth or 90 thousandths? (c) One-thousandth or 90 hundredths? (a) 1 10 > 1 100 (b) 90 thousandths > One-hundredth (c) 90 hundredths > One-thousandth Page No. 80 14. Write the decimal forms of the quantities mentioned (an example is given): (a) 87 ones, 5 tenths and 60 hundredths (b) 12 tens and 12 tenths (c) 10 tens, 10 ones, 10 tenths, and 10 hundredths (d) 25 tens, 25 ones, 25 tenths, and 25 hundredths Ans: 19 15. Using each digit 0 – 9 not more than once, fill the boxes below so that the sum is closest to 10.5: Ans: 16. Write the following fractions in decimal form: Ans: (a) 0.5 (b) 1.5 (c) 0.25 (d) 0.75 (e) 0.2 (f) 0.8 (a) 1 2 (b) 3 2 (c) 1 4 (d) 3 4 (e) 1 5 (f) 4 5 (a) 88.10 121.2 (b) 111.1 277.75 20" class_7,4,Expressions using Letter-Numbers,ncert_books/class_7/gegp1dd/gegp104.pdf,"4 4.1 The Notion of Letter-Numbers In this chapter we shall look at a concise way of expressing mathematical relations and patterns. We shall see how this helps us in thinking about these relationships and patterns, and in explaining why they may hold true. Example 1: Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be? _______ Given Aftab’s age, how will you find out Shabnam’s age? Easy: We add 3 to Aftab’s age to get Shabnam’s age. Can we write this as an expression? Shabnam’s age is 3 years more than Aftab’s. In short, this can be written as: Shabnam’s age = Aftab’s age + 3. Such mathematical relations are generally represented in a shorthand form. In the relation above, instead of writing the phrase ‘Aftab’s Age’, the convention is to use a convenient symbol. Usually, letters or short phrases are used for this purpose. Let us say we use the letter a to denote Aftab’s age (we could have used any other letter), and s to denote Shabnam’s age. Then the expression to find Shabnam’s age will be a + 3, which can be written as EXPRESSIONS USING LETTERNUMBERSChapter-4.indd 81 4/11/2025 7:30:05 PM If a is 23 (Aftab’s age in years), then what is Shabnam’s age? s = a + 3. Aftab’s age 4 10 23 ? a Fig. 4.1 Expression for Shabnam’s age 4 + 3 10 + 3 23 + 3 ? + 3 a + 3 Ganita Prakash | Grade 7 Replacing a by 23 in the expression a + 3, we get, s = 23 + 3 = 26 years. Letters such as a and s that are used to represent numbers are called letter-numbers. Mathematical expressions containing letter-numbers, such as the expression a + 3, are called algebraic expressions. Given the age of Shabnam, write an expression to find Aftab’s age. We know that Aftab is 3 years younger than Shabnam. So, Aftab’s age will be 3 less than Shabnam’s. This can be described as If we again use the letter a to denote Aftab’s age and the letter s to denote Shabnam’s age, then the algebraic expression would be: a = s – 3, meaning 3 less than s. Use this expression to find Aftab’s age if Shabnam’s age is 20. Example 2: Parthiv is making matchstick patterns. He repeatedly places Ls next to each other. Each L has two matchsticks as shown in Figure 4.2. How many matchsticks are needed to make 5 Ls? It will be 5 × 2. How many matchsticks are needed to make 7 Ls? It will be 7 × 2. How many matchsticks are needed to make 45 Ls? It will be 45 × 2. Now, what is the relation between the number of Ls and the number of sticks? First, let us describe the relationship or the pattern here. Every L needs 2 matchsticks. So the number of matchsticks needed will be 2 times the number of L’s. This can be written as: Aftab’s age = Shabnam’s age – 3. Fig. 4.2 Chapter-4.indd 82 4/11/2025 7:30:05 PM Now, we can use any letter to denote the number of L’s. Let’s use n. The algebraic expression for the number of matchsticks will be: This expression tells us how many matchsticks are needed to make n L’s. To find the number of matchsticks, we just replace n by the number of L. Example 3: Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60. 82 Number of matchsticks = 2 × Number of L’s 2 × n. How much should she pay if she buys 10 coconuts and 5 kg jaggery? How much should she pay if she buys 8 coconuts and 9 kg jaggery? Write an algebraic expression to find the total amount to be paid for a given number of coconuts and quantity of jaggery. Here, ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery. The total amount to be paid will be: The corresponding algebraic expression can be written as: Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery. Cost of 10 coconuts = 10 × ₹35 Cost of 5 kg jaggery = 5 × ₹60 Total cost = 10 × ₹35 + 5 × ₹60 = ₹350 + ₹300 = ₹650. Let us identify the relationships and then write the expressions. Quantity needed Relationship Expression Cost of coconuts Number of coconuts × 35 c × 35 Cost of jaggery Number of kgs of jaggery × 60 j × 60 Cost of coconuts + Cost of jaggery. c × 35 + j × 60 Expressions Using Letter-Numbers Chapter-4.indd 83 4/11/2025 7:30:05 PM Notice that for different values of ‘c’ and ‘j’, the value of the expression also changes. Writing this expression as a sum of terms we get: Example 4: We are familiar with calculating the perimeters of simple shapes. Write expressions for perimeters. The perimeter of a square is 4 times the length of its side. This can be written as the expression: 4 × q, where q stands for the sidelength. What is the perimeter of a square with sidelength 7 cm? Use the expression to find out. You must have realised how the use of letter-numbers and algebraic expressions allows us to express general mathematical relations in c × 35 + j × 60 83 Ganita Prakash | Grade 7 a concise way. Mathematical relations expressed this way are often called formulas. Figure it Out 1. Write formulas for the perimeter of: 2. Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe. 3. What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table: No. of ₹100 notes No. of ₹20 notes No. of ₹5 notes Expression and total amount (a) triangle with all sides equal. (b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all sidelengths and angle measures are equal) (c) a regular hexagon 3 5 6 8 4 z 6 × 100 + 4 × 20 + 3 × 5 = 695 Chapter-4.indd 84 4/11/2025 7:30:06 PM 84 4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially? 5. Write algebraic expressions using letters of your choice. (a) 10 + 8 + y (b) (10 + 8) × y (c) 10 × 8 × y (d) 10 + 8 × y (e) 10 × y + 8 (a) 5 more than a number (b) 4 less than a number x y z 4.2 Revisiting Arithmetic Expressions We learnt to write expressions as sums of terms and it became easy for us to read arithmetic expressions. Many times they could have been read in multiple ways and it was confusing. We used swapping (adding two numbers in any order) and grouping (adding numbers by grouping them conveniently) to find easy ways of evaluating expressions. Swapping and grouping terms does not change the value of the expression. We also learnt to use brackets in expressions, including brackets with a negative sign outside. We learnt the distributive property (multiple of a sum is the same as sum of multiples). Let us revise these concepts and find the values of the following expressions: 1. 23 – 10 × 2 2. 83 + 28 – 13 + 32 6. Describe situations corresponding to the following algebraic expressions: 7. In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’. (c) 2 less than 13 times a number (d) 13 less than 2 times a number (a) 8 × x + 3 × y (b) 15 × j – 2 × k w – 1 w Expressions Using Letter-Numbers Chapter-4.indd 85 4/11/2025 7:30:06 PM Let us evaluate the first expression, 23 – 10 × 2. First we shall write the terms of the expression. Notice that one of the terms is a number, while the other one has to be converted to a number before adding the two terms. 3. 34 – 14 + 20 4. 42 + 15 – (8 – 7) 5. 68 – (18 + 13) 6. 7 × 4 + 9 × 6 7. 20 + 8 × (16 – 6) 23 – 10 × 2 = 23 + –10 × 2 = 23 + –20 = 3 85 Ganita Prakash | Grade 7 Let us now evaluate the second one. All the terms of this expression are numbers. If we notice the terms, we find that it will be easier to evaluate if we swap and group the terms. Let us now look at the fifth expression. It has brackets with a negative sign outside. This can be evaluated in two ways — by solving the bracket first (like the solution on the left side) or by removing the brackets appropriately (as on the right side). Now, find the values of the other arithmetic expressions. Algebraic expressions also take number values when the letternumbers they contain are replaced by numbers. In Example 1, for finding Shabnam’s age when Aftab is 23 years old, we replaced the letter-number a in the expression a + 3 by 23, and it took the value 26. = 37 = 68 + –(18 + 13) = 68 + –31 83 + 28 + –13 + 32 = 70 + 60 = 130 83 + 28 – 13 + 32 = = 68 + –18 + –13 OR = 68 + –(18 + 13) = 50 + –13 = 37 Chapter-4.indd 86 4/11/2025 7:30:06 PM 4.3 Omission of the Multiplication Symbol in Algebraic Expressions Look at this number sequence: How can we describe this sequence or pattern? Easy: These are the numbers appearing in the multiplication table of 4 (multiples of 4 in an increasing order). What is the third term of this sequence? It is 4 × 3. What is the 29th term of this sequence? It is 4 × 29. Find an algebraic expression to get the nth term of this sequence. Note that here ‘n’ is a letter-number that denotes a position in the sequence. 86 4, 8, 12, 16, 20, 24, 28, ... As it is the sequence of multiples of 4, it can be seen that the nth term will be 4 times n: As a standard practice, we shorten 4 × n to 4n by skipping the multiplication sign. We write the number first, followed by the letter(s). Find the value of the expression 7k when k = 4. The value is 7 × 4 = 28. Find the value that the expression 5m + 3 takes when m = 2. As 5m stands for 5 × m, the value of the expression when m = 2 is 5 × 2 + 3 = 13. Mind the Mistake, Mend the Mistake Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained. 1. Observe each of them and identify if there is a mistake. 2. If you think there is a mistake, try to explain what might have gone wrong. 3. Then, correct it and give the value of the expression. 1 4 If r = 8, then 2r + 1 = 29. If a = – 4, then 10 – a = 6. 2 5 If d = 6, then 3d = 36. If j = 5, then 2j = 10. 4 × n Expressions Using Letter-Numbers 3 6 If m = –6, then 3 (m + 1) = 19. If s = 7, then 3s – 2 = 15. Chapter-4.indd 87 4/11/2025 7:30:06 PM 4.4 Simplification of Algebraic Expressions Earlier we found expressions to find perimeters of different regular figures in terms of their sides. Let us now find an expression to find the perimeter of a rectangle. 7 If f = 3, g = 1 then 2f – 2g = 2. 8 If t = 4, b = 3 then 2t + b = 24. 9 If h = 5, n = 6 then h – (3 – n) = 4. 87 Ganita Prakash | Grade 7 As in the previous cases, we will first describe how to get the perimeter when the length and the breadth of the rectangle are known: Let us use the letter-numbers l and b in place of length and breadth respectively. Let p denote the perimeter of the rectangle. Then we have As we know, these represent numbers, and so the terms of an expression can be added in any order. Hence the above expression can be written as: Notice that the initial expression (l + b + l + b) and the final expression (2l + 2b) that we got for the perimeter look different. However, they are equal since the expression was obtained from the initial one by applying the same rules and operations we do for numbers; they are equal in the sense that they both take the same values when letternumbers are replaced by numbers. For example, if we assign l = 3, b = 4, we get We call the expression 2l + 2b the simplified form of l + b + l + b. Let us see some more examples of simplification. Example 5: Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days. = l + l + b + b As l + l = 2 × l = 2l, and b + b = 2 × b = 2b, we have Find the sum of length + breadth + length + breadth. l + b + l + b = 3 + 4 + 3 + 4 = 14, and 2l + 2b = 2 × 3 + 2 × 4 = 14. p = l + b + l + b p = 2l + 2b. Chapter-4.indd 88 4/11/2025 7:30:06 PM Let us first find the money earned by the sale of pencils. The money earned by selling pencils on Day 1 is 5c. Similarly, the money earned by selling pencils on Day 2 is _____, and Day 3 is ______. The total money earned by the sale of pencils is 5c + 3c + 10c. Can we simplify this expression further and reduce the number of terms? 88 Erasers (Price ‘d’) 4 6 1 Pencils (Price ‘c’) 5 3 10 Day 1 Day 2 Day 3 The expression means 5 times c is added to 3 times c is added to 10 times c. So in total, the letter-number c is added (5 + 3 + 10) times. This is what we have seen as the distributive property of numbers. Thus, If c = ₹50, find the total amount earned by the scale of pencils. Write the expression for the total money earned by selling erasers. Then, simplify the expression. The expression for the total money earned by selling pencils and erasers during these three days is 18c + 11d. Can the expression 18c + 11d be simplified further? There is no way of further simplifying this expression as it contains different letter-numbers. It is in its simplest form. In this problem, we saw the expression 5c + 3c + 10c getting simplified to the expression 18c. Check that both expressions take the same value when c is replaced by different numbers. Example 6: A big rectangle is split into two smaller rectangles as shown. Write an expression describing the area of the bigger rectangle. Another way of seeing the distributive property. (5 + 3 + 10) × c can be simplified to 18 × c = 18c. 5 × c + 3 × c + 10 × c = (5 + 3 + 10) × c Expressions Using Letter-Numbers Chapter-4.indd 89 4/11/2025 7:30:06 PM The areas of the smaller rectangles are 4v sq. units and 3v sq. units. The area of the bigger rectangle can be found in two ways: (i) by directly using its side lengths v and (4 + 3), or (ii) by adding the areas of the smaller rectangles. The first way gives 7v, and the second way gives 4v + 3v. We know that they are equal: 4v + 3v = 7v, and this is the required expression for the area of the bigger rectangle. As earlier, a big rectangle is split into two smaller rectangles as shown below. Write an expression to find the area of the rectangle AEFD. v 4 3 89 Ganita Prakash | Grade 7 Even in this case, the area of rectangle AEFD can be found in two ways: (i) by directly using the side lengths n and (12 – 4), or (ii) subtracting the area of the rectangle EBCF from that of ABCD. The first method gives us 8n, and the second method gives us 12n – 4n, and they are equal, since 12n – 4n = 8n. This is the expression for the area of the rectangle AEFD. Sets of terms such as (5c, c, 10c), (12n, – 4n) that involve the same letter-numbers are called like terms. Sets of terms such as {18c, 11d} are called unlike terms as they have different letter-numbers. As we have seen, like terms can be added together and simplified into a single term. Example 7: A shop rents out chairs and tables for a day’s use. To rent them, one has to first pay the following amount per piece. When the furniture is returned, the shopkeeper pays back some amount as follows. Write an expression for the total number of rupees paid if x chairs and y tables are rented. For x chairs and y tables, let us find the total amount paid at the beginning and the amount one gets back after returning the furniture. n D A 12 Chair ₹40 Table ₹75 Item Amount F E 4 C Amount returned B Chapter-4.indd 90 4/11/2025 7:30:06 PM Describe the procedure to get these amounts. The total amount in rupees paid at the beginning is 40x + 75y, and the total amount returned is 6x + 10y. So, the total amount paid = (40x + 75y) – (6x + 10y). Can we simplify this expression? If yes, how? If not, why not? 90 Chair ₹6 Table ₹10 Math Talk Since the terms can be added in any order, the remaining bracket can be opened and the expression becomes 40x + 75y + – 6x + – 10y We can group the like terms together, This results in The expression (40x + 75y) – (6x + 10y) is simplified to 34x + 65y, which is the total amount paid in rupees. Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)? Example 8: Charu has been through three rounds of a quiz. Her scores in the three rounds are 7p – 3q, 8p – 4q, and 6p – 2q. Here, p represents the score for a correct answer and q represents the penalty for an incorrect answer. What do each of the expressions mean? If the score for a correct answer is 4 (p = 4) and the penalty for a wrong answer is 1 (q = 1), find Charu’s score in the first round. Charu’s score is 7 × 4 – 3 × 1. We can evaluate this expression by writing it as a sum of terms. 7 × 4 – 3 × 1 = 7 × 4 + – 3 × 1 = 28 + – 3 = 25 What are her scores in the second and third rounds? What if there is no penalty? What will be the value of q in that situation? What is her final score after the three rounds? Her final score will be the sum of the three scores: (7p – 3q) + (8p – 4q) + (6p – 2q). Since the terms can be added in any order, we can remove the brackets and write 7p + – 3q + 8p + – 4q + 6p + –2q = 7p + 8p + 6p + – (3q) + – (4q) + – (2q) (by swapping and grouping) = (7 + 8 + 6)p + – (3 + 4 + 2)q = 21p + – 9q = 21p – 9q. Charu’s total score after three rounds is 21p – 9q. Her friend Krishita’s score after three rounds is 23p – 7q. Recalling how we open brackets in an arithmetic expression, we get (40x + 75y) – (6x + 10y) = (40x + 75y) – 6x – 10y 40x + – 6x + 75y + – 10y = (40 – 6)x + (75 – 10)y = 34x + 65y. Expressions Using Letter-Numbers Math Talk Chapter-4.indd 91 4/11/2025 7:30:06 PM 91 Ganita Prakash | Grade 7 Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q. Can we say who scored more? Can you explain why? How much more has Krishita scored than Charu? This can be found by finding the difference between the two scores. 23p – 7q – (21p – 9q) Simplify this expression further. Example 9: Simplify the expression 4 (x + y) – y Using the distributive property, this expression can be simplified to Example 10: Are the expressions 5u and 5 + u equal to each other? The expression 5u means 5 times the number u, and the expression 5 + u means 5 more than the number u. These two being different operations, they should give different values for most values of u. Let us check this. Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take. 4 (x + y) – y = 4x + 4y – y = 4x + 4y + – y = 4x + (4 – 1)y = 4x + 3y. Chapter-4.indd 92 4/11/2025 7:30:06 PM 92 If the expressions 5u and 5 + u are equal, then they should take the u = 8 u = 11 5u u = 5 u = 5 u = 8 u = 11 u = 2 u = 2 10 5 + u 7 same values for any given value of u. But we can see that they do not. So, these two expressions are not equal. Are the expressions 10y – 3 and 10(y – 3) equal? 10y – 3, short for 10 × y – 3, means 3 less than 10 times y, 10(y – 3), short for 10 × (y – 3), means 10 times (3 less than y). Let us compare the values that these expressions take for different values of y. After filling in the two diagrams, do you think the two expressions are equal? Example 11: What is the sum of the numbers in the picture (unknown values are denoted by letter-numbers)? There are many ways to go about it. Here, we show some of them. 17 y = 10 y = 2 10y – 3 y = 7 y = 7 y = 10 y = 2 y = 0 y = 0 –10 Expressions Using Letter-Numbers 10 (y – 3) 3 r s r s 3 3 3 Math Talk Chapter-4.indd 93 4/11/2025 7:30:06 PM The three expressions might seem different. We can simplify each one and see that they all are the same: 2r + 2s + 24. Figure it Out 1. Adding row wise gives: 2. Adding like terms together gives: 3. Adding the upper half and doubling gives: 1. Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see (4 × 3) + (r + s) + (r + s) + (4 × 3) (8 × 3) + (r + r) + (s + s) 2 × (4 × 3 + r + s) 3 3 3 93 3 Math Talk Ganita Prakash | Grade 7 Mind the Mistake, Mend the Mistake Some simplifications of algebraic expressions are done below. The expression on the right-hand side should be in its simplest form. • Observe each of them and see if there is a mistake. • If you think there is a mistake, try to explain what might have gone wrong. 2. Simplify each of the following expressions: • Then, simplify it correctly. Expression Simplest Form Correct Simplest Form that you get the same thing. 5y x x 5y (a) p + p + p + p, p + p + p + q, p + q + p – q, (b) p – q + p – q, p + q – p + q, (c) p + q – (p + q), p – q – p – q (d) 2d – d – d – d, 2d – d – d – c, (e) 2d – d – (d – c), 2d – (d – d) – c, (f) 2d – d – c – c –6 2 2p 3q 3q 2p 2p 3q –2 3 3q 2p –2 3 –5g –5g 5k 5k 5k 5k 5k 5k 5k 5k 5k 5k –5g –5g 5k 5k Chapter-4.indd 94 4/11/2025 7:30:06 PM 94 1. 3a + 2b 5 2.  3b – 2b – b 0 3.  6 (p + 2) 6p + 8 4.  (4x + 3y) – (3x + 4y) x + y 5.  5 – (2 – 6z) 3 – 6z 6.  2 + (x + 3) 2x – 6 7.  2y + (3y – 6) – y + 6 8.  7p – p + 5q – 2q 7p + 3q 9.  5 (2w + 3x + 4w) 10w + 15x + 20w Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have? 4.5 Pick Patterns and Reveal Relationships In the first section we got a glimpse of algebraic expressions and how to use them to describe simple patterns and relationships in a concise and elegant manner. Here, we continue to look for general relationships between quantities in different scenarios, find patterns and, interestingly, even explain why these patterns occur. Remember the importance of describing in simple language, or visualising mathematical relationships, before trying to write them as expressions. Formula Detective Look at the picture given. In each case, the number machine takes in the 2 numbers at the top of the ‘Y’ as inputs, performs some operations and produces the result at the bottom. The machine performs the same operations on its inputs in each case. Find out the formula of this number machine. 5 10.  3j + 6k + 9h + 12 3 (j + 2k + 3h + 4) 11.  4 (2r + 3s + 5) – 20 – 8r – 12s 2 8 1 9 11 10 10 6 4 Expressions Using Letter-Numbers Chapter-4.indd 95 4/11/2025 7:30:06 PM The formula for the number machine above is “two times the first number minus the second number”. When written as an algebraic expression, the formula is 2a – b. The expression for the first set of inputs is 2 × 5 – 2 = 8. Check that the formula holds true for each set of inputs. expression: expression: expression: expression: expression: 8 15 7 10 95 Ganita Prakash | Grade 7 Find the formulas of the number machines below and write the expression for each set of inputs. Now, make a formula on your own. Write a few number machines as examples using that formula. Challenge your classmates to figure it out! 4 5 Note to the Teacher: Not just solving problems but creating new questions is also very much a part of learning and doing mathematics! expression: expression: expression: expression: expression: expression: expression: expression: expression: expression: 5 7 18 18 5 1 7 1 6 0 3 2 10 3 a b 2 8 1 9 11 10 10 a b Chapter-4.indd 96 4/11/2025 7:30:07 PM Algebraic Expressions to Describe Patterns Example 12: Somjit noticed a repeating pattern along the border of a saree. Somjit wonders if there is a way to describe all the positions where the (i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs. 96 A B C D E F Let us start with design C. It appears for the first time at position 3, the second time at position 6. Where would design C appear for the nth time? We can see that this design appears in positions that are multiples of 3. So the nth occurrence of Design C will be at position 3n. Similarly, find the formula that gives the position where the other Designs appear for the nth time. The positions where B occurs are 2, 5, 8, 11, 14, and so on. We can see that the position of the nth appearance of Design B is one less than the position at which Design C appears for the nth time. Thus, the nth occurrence of Design B is at position: Similarly, the expression describing the position at which the design A appears for the nth time is: 3n – 2. Given a position number can we find out the design that appears there? Which Design appears at Position 122? If the position is a multiple of 3, then clearly we have Design C. As seen earlier, if the position is one less than a multiple of 3, it has Design B, and if it is 2 less than a multiple of 3, then it has Design A. Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below. Position no. Quotient on division by 3 Remainder 99 33 0 3n – 1 Expressions Using Letter-Numbers Chapter-4.indd 97 4/11/2025 7:30:07 PM Use this to find what design appears at positions 99, 122, and 148. Patterns in a Calendar Here is the calendar of November 2024. Consider 2 × 2 squares, as marked in the calendar. The numbers in this square show an interesting property. 122 40 2 148 49 1 97 Ganita Prakash | Grade 7 Let us take the marked 2 × 2 square, and consider the numbers lying on the diagonals; 12 and 20; 13 and 19. Find their sums; 12 + 20, 13 + 19. What do you observe? They are equal. Let us extend the numbers in the calendar beyond 30, creating endless rows. 12 13 19 20 Chapter-4.indd 98 4/11/2025 7:30:07 PM Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can we be sure? To be sure of this we cannot check with all 2 × 2 squares as there are an unlimited number of them. Let us consider a 2 × 2 square. Its top left number can be any number. Let us call it ‘a’. Given that we know the top left number, how do we find the other numbers in this 2 × 2 square? As we have been doing, first let us describe the other numbers in words. 98 a ? ? ? So the other numbers in the 2 × 2 square can be represented as shown in the grid. Let us find the diagonal sums; a + (a + 8), and (a + 1) + (a + 7). Let us simplify them. Since the terms can be added in any order, the brackets can be opened. a + (a + 8) = a + a + 8 = 2a + 8 (a + 1) + (a + 7) = a + 1 + a + 7 = a + a + 1 + 7 = 2a + 8 We see that both diagonal sums are equal to 2a + 8 (8 more than 2 times a). Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’. Thus, we have shown that diagonal sums are equal for any value of a, i.e., for any 2 × 2 square! Consider a set of numbers from the calendar (having endless rows) forming under the following shape: • the number to the right of ‘a’ will be 1 more than it. • the number below ‘a’ will be 7 more than it. • the number diagonal to ‘a’ will be 8 more than it. This problem is an example that shows the power of algebraic modelling in verifying whether a pattern will always hold. 14 15 16 8 Expressions Using Letter-Numbers a + 7 a + 8 a a + 1 Chapter-4.indd 99 4/11/2025 7:30:07 PM Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe? We see that the total sum is always 5 times the number in the centre. Will this always happen? How do you show this? [Hint: Consider a general set of numbers that forms this shape. Take the number at the centre to be ‘a’. Express the other numbers in terms of ‘a’.] Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers. 22 99 Math Talk Math Talk Ganita Prakash | Grade 7 Matchstick Patterns Look at the picture below. It is a pattern using matchsticks. Can you identify what the pattern is? We can see that Step 1 has 1 triangle, Step 2 has 2 triangles, Step 3 has 3 triangles, and so on. Can you tell how many matchsticks there will be in the next step, Step 5? It is 11. You can also draw this and see. How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here? What is the general rule to find the number of matchsticks in the next step? We can see that at each step 2 matchsticks are placed to get the next one, i.e., the number of matchsticks increases by 2 every time. Think of a way to use this to find out the number of matchsticks in Step 33 (without continuing to write the numbers). As each time 2 matchsticks are being added, finding out how many 2s will be added in Step 33 will help. Look at the table below and try to find out. 1 2 3 4 Step Number 1 2 3 4 5 6 No. of Matchsticks 3 5 7 9 11 13 Chapter-4.indd 100 4/11/2025 7:30:08 PM The number of matchsticks needed to make 33 triangles (Step 33) is _________. Similarly, find the number of matchsticks needed for Step 84 and Step 108. 100 Step Number 1 2 3 4 5 6 No. of Matchsticks 3 5 7 9 11 13 3 + 2 3 + 2 + 2 3 + 2 + 2 + 2 3 + 2 + 2 + 2 + 2 What could be an expression describing the rule/formula to find out the number of matchsticks at any step? The pattern is such that in Step 10, nine 2s and an added 3 (3 + 2 × 9) gives the number of matchsticks; in Step 11, ten 2s and an added 3 (3 + 2 × 10) gives the number of matchsticks. For step y, what is the expression? This expression gives the number of matchsticks in Step y. Now we can find the number of matchsticks at any step quickly. You might have already noticed that there is a 2 in the first step also, 3 = 1 + 2. Using this, the expression we get is Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same? Both expressions are the same. There is a different way to count, or see the pattern. Let us take a look at the picture again. It is: one less than y (i.e. y – 1) 2s and a 3. Therefore, the expression is We can check by simplifying the expression 3 + 2 × (y – 1). 3 + 2 × (y – 1) = 3 + 2y – 2 3 + 2 × (y – 1). = 2y + 1. 2y + 1. Expressions Using Letter-Numbers Chapter-4.indd 101 4/11/2025 7:30:08 PM Matchsticks are placed in two orientations — (a) horizontal ones at the top and bottom, and (b) the ones placed diagonally in the middle. For example, in step 2 there are 2 matchsticks placed horizontally and 3 matchsticks placed diagonally. What are these numbers in Step 3 and Step 4? How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1? 101 Ganita Prakash | Grade 7 Figure it Out For the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described. If required, assume some values for the unknowns and try to find the relationship. 1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day? 2. Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day? 3. A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights. (a) 30x + 20y (b) (30 + 20) × (x + y) (c) 20x + 30y (d) (30 + 20) × x + y (e) 30x – 20y (a) p + q + r (b) p + q + 2r (c) 2 × (p + q + r) (d) p + q + r + 2 (e) p + q + r + 1 (f) 2 × (p + q) (a) Write an expression describing how far away the snail is from its starting position. (b) What can we say about the snail’s movement if d > u? Chapter-4.indd 102 4/11/2025 7:30:08 PM 102 4. Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks? 5. In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations. ×3 –5 +3 w + 5 ×4 4w + 20 –4 –8 +3 ×4 3w – 6 w - 3 w + 2 Try This 6. A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. 7. Simplify the following expressions: (a) 3a + 9b – 6 + 8a – 4b – 7a + 16 8. Add the expressions given below: 9. Subtract the expressions given below: (a) If t = 4, what is the time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation] (b) 3 (3a – 3b) – 8a – 4b – 16 (c) 2 (2x – 3) + 8x + 12 (d) 8x – (2x – 3) + 12 (e) 8h – (5 + 7h) + 9 (f) 23 + 4(6m – 3n) – 8n – 3m – 18 (a) 4d – 7c + 9 and 8c – 11 + 9d (b) – 6f + 19 – 8s and – 23 + 13f + 12s (c) 8d – 14c + 9 and 16c – (11 + 9d) (d) 6f – 20 + 8s and 23 – 13f – 12s (e) 13m – 12n and 12n – 13m (f) – 26m + 24n and 26m – 24n Expressions Using Letter-Numbers Chapter-4.indd 103 4/11/2025 7:30:08 PM 10. Describe situations corresponding to the following algebraic expressions: 11. Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we (a) 9a – 6b + 14 from 6a + 9b – 18 (b) – 15x + 13 – 9y from 7y – 10 + 3x (c) 17g + 9 – 7h from 11 – 10g + 3h (d) 9a – 6b + 14 from 6a – (9b + 18) (e) 10x + 2 + 10y from –3y +8 – 3x (f) 8g + 4h – 10 from 7h – 8g + 20 (a) 8x + 3y (b) 15x – 2x 103 Ganita Prakash | Grade 7 12. Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares? 13. Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour. get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut? 13 1 1 5 9 10 14 2 2 6 11 15 3 3 7 12 16 4 4 8 Chapter-4.indd 104 4/11/2025 7:30:09 PM 14. Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares? 104 1 2 3 4 5 ................... 12. Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares? 13. Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour. get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut? 15. Numbers are written in a particular sequence in this endless 4-column grid. Do you see any pattern in it? List other patterns that you see. • Algebraic expressions are used in formulas to model patterns and mathematical relationships between quantities, and to make predictions. • Algebraic expressions use not only numbers but also letter-numbers. The rules for manipulating arithmetic expressions also apply to algebraic expressions. These rules can be used to reduce algebraic expressions to their simplest forms. • Algebraic expressions can be described in ordinary language, and vice versa. Patterns or relationships that are easily written using algebra can often be long and complex in ordinary language. This is one of the advantages of algebra. (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). (b) In which row and column will the following numbers appear: (iii) 201 (c) What number appears in row r and column c? (d) Observe the positions of multiples of 3. (i) 124 (ii) 147 SUMMARY Expressions Using Letter-Numbers 13 1 1 5 9 10 14 2 2 6 11 15 3 3 7 12 16 4 4 8 Math Talk 14. Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares? 1 2 3 4 5 ................... Chapter-4.indd 105 4/11/2025 7:30:09 PM 105 Page No. 83 Page No. 84 Figure it Out Ans: Total cost: ₹ 820 Ans: (a) Let length of each side of the triangle = a units Perimeter of Triangle = 3a units (b) Let length of each side of pentagon = a units Perimeter of Regular Pentagon = 5a units (c) Let length of each side of regular hexagon = a units Perimeter of Regular Hexagon = 6a units 1. Write formulas for the perimeter of: (a) triangle with all sides equal. (b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all sidelengths and angle measures are equal) (c) a regular hexagon 2. Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter–number ‘k’ to denote the length in meters of the other pipe. How much should she pay if she buys 8 coconuts and 9 kg jaggery? Expressions Using Letter–Numbers Chapter – 4 Ans: (20 + k) meters 1 No. of ₹100 Notes No. of ₹20 notes No. of ₹5 notes Expression and total amount (in ₹) 3. What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table: Ans: 4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially? 3 5 6 3 × 100 + 5 × 20 + 6 × 5 = 430 6 4 3 6 × 100 + 4 × 20 + 3 × 5 = 695 8 4 z 8 × 100 + 4 × 20 + z × 5 = 880 + 5z x y z x × 100 + y × 20 + z × 5 = 100x + 20y + 5z (a) 10 + 8 + y (b) (10 + 8) × y (c) 10 × 8 × y Ans: (d) 10 + 8 × y 5. Write algebraic expressions using letters of your choice. (a) 5 more than a number (b) 4 less than a number (c) 2 less than 13 times a number (d) 13 less than 2 times a number Ans: Let letter ‘d’ represent the number, then (a) d + 5 (b) d – 4 (c) 13 × d – 2 (d) 2 × d – 13 (d) 10 + 8 × y (e) 10 × y + 8 2 Page No. 85 Page No. 89 6. Describe situations corresponding to the following algebraic expressions: Ans: A shopkeeper sells a pen for ₹x and a notebook for ₹y. Abha buys 8 pens and 3 notebooks. What will be the total cost of her purchase? (b) 15 × j – 2 × k Ans: A factory makes 15 chairs every day. If the factory works for j days, it will make 15 × j chairs. But 2 chairs break daily for k days. How many good chairs remain after j days and k days breakages? Try for more situations 7. In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’. Ans: (a) 8 × x + 3 × y Page No. 92 Ans: 18c = ₹900. If c = ₹50, find the total amount earned by the sale of pencils. Fill the blanks below by replacing the letter–numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take. 3 Page No. 93-94 Figure it Out Ans: Ans: After filling in the two diagrams, do you think the two expressions are equal? The expressions 10y – 3 and 10(y – 3) are not equal. Ans: Some of the ways are Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing. 4 (i) Method 1: Group by Like Terms Method 2: Group by Rows Top row: 5y + (– 6) + x Bottom row: x + 2 + 5y Combine: (5y + 5y) + (x + x) + (– 6 + 2) =10y + 2x – 4 Final Number: 10y + 2x – 4 Try some other ways! (ii) Method 1: Group by Term Type Method 2: Group by Columns Column 1: 2p + 3q Column 2: 3q + 2p Column 3: (– 2) + 3 + 2p + 3q = 1 + 2p + 3q Column 4: 3 + (– 2) + 3q + 2p = 1 + 3q + 2p Final Number: 8p + 12q + 2 Try some other ways! 5y + x + x + 5y + (– 6) + 2 = (5y + 5y) + (x + x) + (– 6 + 2) = 10y + 2x – 4 2p appears 4 times: 8p 3q appears 4 times: 12q Final number: 3 + (– 2) + (– 2) + 3 = 2 Total: 8p + 12q + 2 (iii) Method 1: Group by Term Type Method 2: Group by Columns –5g appears 4 times: –20g 5k appears 12 times: 60k Total: –20g + 60k First column: –5g + 5k + 5k + (–5g) = –10g + 10k Middle columns 2 and 3 all 5k = 8 × 5k = 40k Last column: –5g + 5k + 5k + (–5g) = –10g + 10k Total of all: = –10g + 10k + 40k –10g + 10k = –20g + 60k Try some other ways! 5 (a) p + p + p + p, p + p + p + q, p + q + p – q (b) p − q + p − q, p + q – p + q (c) p + q − (p + q), p – q – p – q (d) 2d – d – d – d, 2d – d – d – c (e) 2d – d − (d − c), 2d − (d − d) – c (f) 2d – d – c – c Ans: (a) 4p, 3p + q, 2p (b) 2p – 2q, 2q (c) 0, –2q (d) – d, – c (e) c, 2d – c (f) d – 2c ● Observe each of them and see if there is a mistake. ● If you think there is a mistake, try to explain what might have gone wrong. ● Then, simplify it correctly. Ans: Sr. No. Expression Simplest Form Correct Simplest form 1 3a + 2b 5 3a + 2b 2 3b – 2b –b 0 0 3 6 (p + 2) 6p + 8 6p + 12 Some simplifications of algebraic expressions are done below. The expression on the right–hand side should be in its simplest form. Simplify each of the following expressions: 4 (4x + 3y) – (3x + 4y) x +5 x – y 5 5 – (2 – 6z) 3 – 6z 3 + 6z 6 2 + (x + 3) 2x – 6 x + 5 7 2y + (3y – 6) –y + 6 5y – 6 8 7p – p + 5q – 2q 7p + 3q 6p + 3q 9 5 (2w + 3x + 4w) 10w + 15x + 20w 30w + 15x 10 3j + 6k + 9h + 12 3 (j + 2k + 3h + 4) 3j + 6k + 9h + 12 11 4 (2r + 3s + 5) – 20 – 8r – 12s 8r + 12s + 20 6 Page No. 95 Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter–numbers these expressions have? Ans: Sr. No. (A) Expression 1 3a + 2b 3a + 2b 2 2 2 3b – 2b –b 0 1 0 3 6 (p + 2) 6p +12 2 1 4 (4x + 3y) – (3x + 4y) x – y 2 2 5 5 – (2 – 6z) 3 + 6z 2 1 6 2 + (x + 3) x + 5 2 1 7 2y + (3y – 6) 5y – 6 2 1 8 7p – p + 5q – 2q 6p + 3q 2 2 (B) Correct Simplest form (C) Number of terms in (B) (D) Number of letters in (B) Page No. 96 Find the formulas of the number machines below and write the expression for each set of inputs. Ans: (i) Formula: Two is subtracted from the sum of the two numbers. Algebraic expression = a + b – 2 9 5 (2w + 3x + 4w) 30w + 15x 2 2 10 3j + 6k + 9h + 12 3j + 6k + 9h + 12 4 3 11 4 (2r + 3s + 5) 8r + 12s + 20 3 2 No. of letters ≤ no of terms 7 (ii) Formula: One is added to the multiplication of two numbers. Algebraic expression= (a × b) + 1 = ab + 1 4th expression: 10 × 3 + 1 = 30 + 1 = 31 5th expression: a × b + 1 Page No. 97 Use this to find what design appears at positions 99, 122, and 148. Ans: 1. Design C appears at position 99. 2. Design B appears at position 122. 3. Design A appears at position 148. Page No. 99 Will this always happen? How do you show this? Ans: Let’s build a similar 3×3 grid with a Centre number a. Here, the sum is a – 7 + a – 1 + a + a +1 + a + 7 = 5a (5times the number in the Centre.) a – 7 a – 1 a a + 1 a + 7 8 Page No. 101. Page No. 102. 1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day? (a) 30x + 20y (b) (30 + 20) × (x + y) (c) 20x + 30y (d) (30 + 20) × x + y (e) 30x – 20y What are these numbers in Step 3 and Step 4? Ans: Step 3: No. of matchsticks placed horizontally-3 No. of matchsticks placed diagonally-4 Step 4: No. of matchsticks placed horizontally-4 No. of matchsticks placed diagonally-5 How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1? Ans: Figure it Out For horizontal placement: 1, 2, 3, 4, … At the yth step, the number of horizontal matchsticks = y. For diagonal placement: 2, 3, 4, 5, … At the yth step, the number of diagonal matchsticks = y + 1. Yes. Ans: (a) 30x + 20y. Ans: (a) p + q + r 2. Pushpita sells two types of flowers on Independence Day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day? (a) p + q + r (b) p + q + 2r (c) 2 × (p + q + r) (d) p + q + r + 2 (e) p + q + r + 1 (f) 2 × (p + q) 9 (b) What can we say about the snail’s movement if d > u? Ans: Ans: 105 + 21z km Ans: 3. A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights. 4. Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks? 5. In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations. (a) Write an expression describing how far away the snail is from its starting position. (a) 10 (u − d) cm (b) If d > u, the snail slips down more than it climbs. This means the snail will never reach the top. Page No. 103 6. A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. (a) If t = 4, what is the 10 Ans: (a) 22 minutes 7. Simplify the following expressions: (a) 3a + 9b – 6 + 8a – 4b – 7a + 16 (b) 3 (3a – 3b) – 8a – 4b –16 (c) 2 (2x – 3) + 8x + 12 (d) 8x – (2x – 3) + 12 (e) 8h – (5 + 7h) + 9 (f) 23 + 4 (6m – 3n) – 8n – 3m –18 Ans: (a) 4a + 5b + 10 (b) a – 13b – 16 (c) 12x + 6 (d) 6x + 15 (e) h + 4 (f) 5 + 21m – 20n 8. Add the expressions given below: (a) 4d – 7c + 9 and 8c – 11 + 9d (b) – 6f + 19 – 8s and – 23 + 13f + 12s (c) 8d – 14c + 9 and 16c – (11 + 9d) (d) 6f – 20 + 8s and 23 – 13f – 12s (e) 13m – 12n and 12n –13m (f) – 26m + 24n and 26m –24n Ans: (a) 13d + c – 2 (b) 7f + 4s – 4 (c) 2c – d – 2 (d) –7f – 4s + 3 (e) 0 (f) 0 (b) The algebraic expression for the time taken to travel from Yahapur to time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation] Vahapur = 4t + 6 minutes. 11 9. Subtract the expressions given below: (a) 9a – 6b + 14 from 6a + 9b – 18 (b) – 15x + 13 – 9y from 7y – 10 + 3x (c) 17g + 9 – 7h from 11 – 10g + 3h (d) 9a – 6b + 14 from 6a – (9b + 18) (e) 10x + 2 + 10y from –3y +8 – 3x (f) 8g + 4h – 10 from 7h – 8g + 20 Ans: (a) – 3a + 15b – 32 (b) 16y + 18x – 23 (c) 10h – 27g + 2 (d) – (3a + 3b + 32) (e) – 13y – 13x + 6 (f) 3h – 16g + 30 10.Describe situations corresponding to the following algebraic expressions: Ans: (a) 8x + 3y One of the examples is–Situation: A fruit seller sells mangoes at ₹8 each and bananas at ₹3 each. If a customer buys x mangoes and y bananas. The total cost would be 8x + 3y. (b) 15x – 2x Situation: A shopkeeper has 15 pencils in a packet. The cost of one pencil is ₹x. Two pencils in the packet are not sold. Find the amount received for this packet. 11.Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut? 12 Ans: If the rope is folded 10 times and cut, we get 10 + 2 = 12 pieces. In the same way, when the rope is folded r times and cut, we get r + 2 pieces. Page No. 104 Ans: To make 10 squares, we need 12.Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares? How many are required to make w squares? 13.Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour. To make w squares the number of matchsticks required is = 4 + 3 (w - 1) 4 + 3 × 9 = 31 matchsticks Ans: For Position 90: The colour is Yellow. Expressions: In general, 4n – 3 positions for red light, 4n – 1 position for green light and 2n position for yellow light. For Position 190: The colour is Yellow For Position 343: The colour is Green 13 Ans: No. of Squares in Step 4 = 17 No. of Squares in Step 10 = 41 No. of Squares in Step 50 = 201 So, general formula = 5 + (n – 1) × 4 = 4n + 1 No. of vertices = 16n + 4. Page No. 105 14.Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares? 15.Numbers are written in a particular sequence in this endless 4 – column grid. Ans: (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). Let r be the row number. Column 1: So, number in the rth row of column 1 = 4 × (r – 1) + 1 14 (b) In which row and column will the following numbers appear: (c) What number appears in row r and column c? (d) Observe the positions of multiples of 3. Column 2: 4 × (r – 1) + 2 Column 3: 4 × (r – 1) + 3 Column 4: 4 × (r – 1) + 4 If c is the column number, then the general formula to generate all numbers is= 4 × (r – 1) + c (i) 124 (ii) 147 (iii) 201 (i) We divide each number by 4 to find its row and column (ii) 147 will appear at row 37 and column 3 (iii) 201 will appear at row 51 and column 1. The number that appears in row r and column c is 4(r – 1) + c. The pattern in the columns in which multiples of 3 appear, follow a repeating sequence of column number: 3, 2, 1, 4, 3, 2, 1, ... as shown. 124 ÷ 4 So, Quotient = 31 and remainder is 0 ∴ 124 = 4 × 31 + 0 or 4 × 30 + 4 Comparing it with 4 × (r – 1) + c, we get r – 1 = 30, c = 4 So, r = 31 and c = 4 15 Do you see any pattern in it? List other patterns that you see. 1. All numbers in Column 4 are multiples of 4. 2. Even numbers always appear in column 2 and column 4. 3. Odd numbers always appear in column 1 and column 3. Find some more! 16" class_7,5,Parallel and Intersecting Lines,ncert_books/class_7/gegp1dd/gegp105.pdf,"5 5.1 Across the Line Take a piece of square paper and fold it in different ways. Now, on the creases formed by the folds, draw lines using a pencil and a scale. You will notice different lines on the paper. Take any pair of lines and observe their relationship with each other. Do they meet? If they do not meet within the paper, do you think they would meet if they were extended beyond the paper? PARALLEL AND INTERSECTING LINES Chapter-5.indd 106 4/11/2025 7:30:37 PM In this chapter, we will explore the relationship between lines on a plane surface. The table top, your piece of paper, the blackboard, and the bulletin board are all examples of plane surfaces. Let us observe a pair of lines that meet each other. You will notice that they meet at a point. When a pair of lines meet each other at a point on a plane surface, we say that the lines intersect each other. Let us observe what happens when two lines intersect. How many angles do they form? In Fig. 5.2, where line l intersects line m, we can see that four angles are formed. Fig. 5.1 Can two straight lines intersect at more than one point? Activity 1 Draw two lines on a plain sheet of paper so that they intersect. Measure the four angles formed with a protractor. Draw four such pairs of intersecting lines and measure the angles formed at the points of intersection. What patterns do you observe among these angles? In Fig. 5.2, if ∠a is 120°, can you figure out the measurements of ∠b, ∠c and ∠d, without drawing and measuring them? We know that ∠a and ∠b together measure 180°, because when they are combined, they form a straight angle which measures 180°. So, if ∠a is 120°, then ∠b must be 60°. Similarly, ∠b and ∠c together measure 180°. So, if ∠b is 60°, then ∠c must be 120°. And ∠c and ∠d together measure 180°. So, if ∠c is 120°, then ∠d must be 60°. Therefore, in Fig. 5.2, ∠a and ∠c measure 120°, and ∠b and ∠d measure 60°. When two lines intersect each other and form four angles, labelled a, b, c and d, as in Fig. 5.2, then ∠a and ∠c are equal, and ∠b and ∠d are equal! c d a b l Fig. 5.2 m Parallel and Intersecting Lines Chapter-5.indd 107 4/11/2025 7:30:37 PM Is this always true for any pair of intersecting lines? Check this for different measures of ∠a. Using these measurements, can you reason whether this property holds true for any measure of ∠a? We can generalise our reasoning for Fig. 5.2, without assuming the values of ∠a. Since straight angles measure 180°, we must have ∠a + ∠b = ∠a + ∠d = 180°. Hence, ∠b and ∠d are always equal. Similarly, ∠b + ∠a = ∠b + ∠c = 180°, so ∠a and ∠c must be equal. Adjacent angles, like ∠a and ∠b, formed by two lines intersecting each other, are called linear pairs. Linear pairs always add up to 180°. 107 Ganita Prakash | Grade 7 Opposite angles, like ∠b and ∠d, formed by two lines intersecting each other, are called vertically opposite angles. Vertically opposite angles are always equal to each other. From the above reasoning, we conclude that whenever two lines intersect, vertically opposite angles are equal. Such a justification is called a proof in mathematics. Figure it Out List all the linear pairs and vertically opposite angles you observe in Fig. 5.3: Measurements and Geometry You might have noticed that when you measure linear pairs, sometimes they may not add up to 180°. Or, when you measure vertically opposite angles they may be unequal sometimes. What are the reasons for this? There could be different reasons: • Variation in the thickness of the lines drawn. The “ideal” line in geometry does not have any thickness! But it is not possible for us to draw lines without any thickness In geometry, we create ideal versions of “lines” and other shapes we see around us, and analyse the relationships between them. For example, we know that the angle formed by a straight line is 180°. So, if another line divides this angle into two parts, both parts should add up to 180°. We arrive at this simply through reasoning and not by measurement. When we measure, it might not be exactly so, for the reasons mentioned above. Still the measurements come out very close to what we predict, because of which geometry finds widespread application in different disciplines such as physics, art, engineering and architecture. Pairs of Vertically Opposite Angles ∠b and ∠d, … • Measurement errors because of improper use of measuring instruments — in this case, a protractor Linear Pairs ∠a and ∠b, … d l m a Fig. 5.3 c b Chapter-5.indd 108 4/11/2025 7:30:37 PM 5.2 Perpendicular Lines Can you draw a pair of intersecting lines such that all four angles are equal? Can you figure out what will be the measure of each angle? 108 If two lines intersect and all four angles are equal, then each angle must be a right angle (90°). Perpendicular lines are a pair of lines which intersect each other at right angles (90°). In Fig. 5.4, we can say that lines l and m are perpendicular to each other. 5.3 Between Lines Observe Fig. 5.5 and describe the way the line segments meet or cross each other in each case, with appropriate mathematical words (a point, an endpoint, the midpoint, meet, intersect) and the degree measure of each angle. For example, line segments FG and FH meet at the endpoint F at an angle 115.3°. S T V Fig. 5.4 l m O Q Parallel and Intersecting Lines Chapter-5.indd 109 4/11/2025 7:30:37 PM Are line segments ST and UV likely to meet if they are extended? Are line segments OP and QR likely to meet if they are extended? Here are some examples of lines we notice around us. U C G 115.3° F A H D X Fig. 5.5 B M J Y I L P R 109 Ganita Prakash | Grade 7 What is common to the lines in the pictures above? They do not seem likely to intersect each other. Such lines are called parallel lines. Name some parallel lines you can spot in your classroom. Parallel lines are often used in artwork and shading. Which pairs of lines appear to be parallel in Fig. 5.6 below? Parallel lines are a pair of lines that lie on the same plane, and do not meet however far we extend them at both ends. a Chapter-5.indd 110 4/11/2025 7:30:38 PM 110 Note to the Teacher: It is important that the lines lie on the same plane. A line drawn on a table and a line drawn on the board may never meet but that does not make them parallel. b c d e i h Fig. 5.6 g f 5.4 Parallel and Perpendicular Lines in Paper Folding Activity 2 Take a plain square sheet of paper (use a newspaper for this activity). • How would you describe the opposite edges of the sheet? They are _________________________ to each other. • How would you describe the adjacent edges of the sheet? The adjacent edges are _________________________ to each other. They meet at a point. They form right angles. • Fold the sheet horizontally in half. A new line is formed (see Fig. 5.7). • How many parallel lines do you see now? How does the new line segment relate to the vertical sides? Parallel and Intersecting Lines Chapter-5.indd 111 4/11/2025 7:30:38 PM • Make one more horizontal fold in the folded sheet. How many parallel lines do you see now? • What will happen if you do it once more? How many parallel lines will you get? Is there a pattern? Check if the pattern extends further, if you make another horizontal fold. • Make a vertical fold in the square sheet. This new vertical line is ___________ to the previous horizontal lines. • Fold the sheet along a diagonal. Can you find a fold that creates a line parallel to the diagonal line? Fig. 5.7 111 Ganita Prakash | Grade 7 Here is another activity for you to try. Notations In mathematics, we use an arrow mark (>) to show that a set of lines is parallel. If there is more than one set of parallel lines (as in Fig. 5. 9), the second set is shown with two arrow marks and so on. Perpendicular lines are marked with a square angle between them. • Take a square sheet of paper, fold it in the middle and unfold it. • Fold the edges towards the centre line and unfold them. • Fold the top right and bottom left corners onto the creased line to create triangles. Refer to Fig. 5.8. • The triangles should not cross the crease lines. • Are a, b and c parallel to p, q and r respectively? Why or why not? Fig. 5.8 Chapter-5.indd 112 4/11/2025 7:30:38 PM 112 Fig. 5.9 90° Figure it Out 1. Draw some lines perpendicular to the lines given on the dot paper in Fig. 5.10. 2. In Fig. 5.11, mark the parallel lines using the notation given above (single arrow, double arrow etc.). Mark the angle between perpendicular lines with a square symbol. (a) How did you spot the perpendicular lines? (b) How did you spot the parallel lines? Fig. 5.10 Parallel and Intersecting Lines Chapter-5.indd 113 4/11/2025 7:30:39 PM 3. In the dot paper following, draw different sets of parallel lines. The line segments can be of different lengths but should have dots as endpoints. Fig. 5.11 113 Ganita Prakash | Grade 7 4. Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper. 5. In Fig. 5.13, which line is parallel to line a —– line b or line c? How do you decide this? (a) Did you find it challenging to draw some of them? (b) Which ones? (c) How did you do it? a b e f Fig. 5.12 c b h c g d Chapter-5.indd 114 4/11/2025 7:30:39 PM From previous exercises we observed that sometimes it is difficult to be sure whether two lines are parallel. To determine this we use the idea of transversals. 114 Note to the Teacher: It is easier to draw vertical and horizontal lines and the ones inclined at 45° (on rectangular dot sheets), but drawing a line parallel to one which has a different orientation is slightly harder. Let students use their intuition for this. a Fig. 5.13 5.5 Transversals We saw what happens when two lines intersect in different ways. Let us explore what happens when one line intersects two different lines. In Fig. 5.14, line t intersects lines l and m. t is called a transversal. Notice that 8 angles are formed when a line crosses a pair of lines. Is it possible for all the eight angles to have different measurements? Why, why not? What about five different angles — 6, 5, 4, 3 and 2? In Fig. 5.14, since ∠1 and ∠3 are vertically opposite angles, they are equal. Are there other pairs of vertically opposite angles? We can see that there are a total of four pairs of vertically opposite angles and in each pair, the angles are equal to each other. Thus, when a transversal intersects two lines, it forms eight angles with a maximum of four distinct angle measures. Fig. 5.14 8 5 t l 1 4 7 6 3 2 m Parallel and Intersecting Lines Chapter-5.indd 115 4/11/2025 7:30:39 PM 5.6 Corresponding Angles In Fig. 5.14, we notice that the transversal t forms two sets of angles — one with line l and another with line m. There are angles in the first set that correspond to angles in the second set based on their position. ∠1 and ∠5 are called corresponding angles. Similarly, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8 are the corresponding angles formed when transversal t intersects lines l and m. Activity 3 Draw a pair of lines and a transversal such that they form two distinct angles. 115 Ganita Prakash | Grade 7 How many distinct angles have formed now? If one angle is 60°, the other angle of the linear pair should be 120°. So, we already have two distinct angles. So, when we draw another line intersecting the transversal t we wish to form only two angles, 60° and 120°. Step 3: Mark a point Y on line t. Step 4: Draw a line m through point Y that forms a 60° angle to line t. This can be done either by copying ∠a with a tracing paper or you can use a protractor to measure the angles. Step 1: Draw a line l and a transversal t intersecting it at point X. Step 2: Measure ∠a formed by lines l and t (let us say it is 60°). l X a = 60° l X a = 60° l X Fig. 5.15 Fig. 5.17 t Fig. 5.16 t Y t Chapter-5.indd 116 4/11/2025 7:30:39 PM What do you observe about lines l and m? Do they appear to be parallel to each other? Yes, they do appear to be parallel to each other. Angles, ∠a and ∠b are corresponding angles formed by the transversal t on lines l and m. These corresponding angles are equal to each other. 116 m b = 60° l Fig. 5.18 t Y X a = 60° Suppose, we have a transversal intersecting two parallel lines. What can be said about the corresponding angles? Activity 4 Fig. 5.19 has a pair of parallel lines l and m (what is the notation used in the figure to indicate they are parallel?) . Line t is the transversal across these two lines. ∠a and ∠b are corresponding angles. Take a tracing paper and trace ∠a on it. Now place this tracing paper over ∠b and see if the angles align exactly. You will observe that the angles match. Check the other corresponding angles in the figure using a protractor. Are all the corresponding angles equal to each other? When the corresponding angles formed by a transversal on a pair of lines are equal to each other, then the pair of lines are parallel to each other. Corresponding angles formed by a transversal intersecting a pair of parallel lines are always equal to each other. From this we can observe: m b l a Fig. 5.19 t Parallel and Intersecting Lines Chapter-5.indd 117 4/11/2025 7:30:39 PM Activity 5 In Fig. 5.20, draw a transversal t to the lines l and m such that one pair of corresponding angles is equal. You can measure the angles with a protractor. l Are you finding it hard to draw a transversal such that the corresponding angles are equal? Fig. 5.20 m 117 Ganita Prakash | Grade 7 5.7 Drawing Parallel Lines Can you draw a pair of parallel lines using a ruler and a set square? Fig. 5.21 shows how you can do it. Draw a line l with a scale. By sliding your set square you can make two lines perpendicular to line l. Are these two lines parallel to each other? How are we sure that they are parallel to each other? What angles are formed between these lines and line l? Since we used a set square, the angles measure 90°. The position of the lines is different, but they make the same angle with l. If line l is seen as a transversal to the two new lines, then the corresponding angles measure 90°. When a pair of lines are not parallel to each other, the corresponding angles formed by a transversal can never be equal to each other. Fig. 5.21 l Chapter-5.indd 118 4/11/2025 7:30:39 PM As we know these are corresponding angles and they are equal, we can be sure that the lines are parallel. Draw two more parallel lines using the long side of the set square as shown in Fig. 5.22. 118 Fig. 5.22 l How do you know these two lines are parallel? Can you check if the corresponding angles are equal? Figure it Out Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method. Making Parallel Lines through Paper Folding Let us try to do the same with paper folding. For a line l (given as a crease), how do we make a line parallel to l such that it passes through point A? We know how to fold a piece of paper to get a line perpendicular to l. Now, try to fold a perpendicular to l such that it passes through point A. Let us call this new crease t. Now, fold a line perpendicular to t passing through A again. Let us call this line m. The lines l and m are parallel to each other. Note to the Teacher: Students should be encouraged to check the equality of corresponding angles both by using the tracing method and using protractors to measure the angles. Pay attention to the language used to make the relationship between corresponding angles and parallel lines. Equality of corresponding angles is both necessary and sufficient for the pair of lines to be parallel to each other. Fig. 5.23 A l Parallel and Intersecting Lines Chapter-5.indd 119 4/11/2025 7:30:39 PM A t l l l Fig. 5.24 l m A t A 119 Ganita Prakash | Grade 7 Why are lines l and m parallel to each other? 5.8 Alternate Angles In Fig. 5.25, ∠d is called the alternate angle of ∠f, and ∠c is the alternate angle of ∠e. You can find the alternate angle of a given angle, say ∠f, by first finding the corresponding angle of ∠f, which is ∠b and then finding the vertically opposite angle of ∠b, which is ∠d. Activity 6 In Fig. 5.25, if ∠f is 120° what is the measure of its alternate angle ∠d? We can find the measure of ∠d if we know ∠b because they are vertically opposite angles. Remember, vertically opposite angles are equal. What is the measure of ∠b? It is 120° because it is the corresponding angle of ∠f. So, ∠d also measures 120°. In fact, ∠f = ∠b irrespective of the measure of ∠f. Why? Because ∠b is the corresponding angle of ∠f. Similarly, ∠b = ∠d irrespective of the measure of ∠b. Why? Because ∠d is the vertically opposite angle of ∠b. So, it must always be the case that ∠f = ∠d Using our understanding of corresponding angles without any measurements, we have justified that alternate angles are always equal. h g m l a b Fig. 5.25 d c e t f Chapter-5.indd 120 4/11/2025 7:30:40 PM Example 1: In Fig. 5.26, parallel lines l and m are intersected by the transversal t. If ∠6 is 135°, what are the measures of the other angles? 120 Alternate angles formed by a transversal intersecting a pair of parallel lines are always equal to each other. Solution: ∠6 is 135°, so ∠2 is also 135°, because it is the corresponding angle of ∠6 and the lines l and m are parallel. ∠8 is 135°, because it is the vertically opposite angle of ∠6. ∠4 is 135° because it is the corresponding angle of ∠8. ∠2 is 135° because is the vertically opposite angle of ∠4. So, ∠2, ∠4, ∠6, and ∠8 are all 135°. ∠5 and ∠6 are a linear pair, together they measure 180°. If ∠6 is 135°, then ∠5 = 180 – 135 = 45° We can similarly find out that ∠1, ∠3, and ∠7 measure 45°. Example 2: In Fig. 5.27, lines l and m are intersected by the transversal t. If ∠a is 120° and ∠f is 70°, are lines l and m parallel to each other? Solution: ∠a is 120°, so ∠b is 60° because ∠a and ∠b form a linear pair. ∠b is a corresponding angle of ∠f. If l and m are parallel, ∠b should be equal to ∠f, however, they are not equal. Therefore, lines l and m are not parallel to each other as the corresponding angles formed by the transversal t are not equal to each other. m 2 3 4 5 6 135° 7 8 1 t h e 120° Fig. 5.26 g Fig. 5.27 d c f 70° a b l m l t Parallel and Intersecting Lines Chapter-5.indd 121 4/11/2025 7:30:40 PM Example 3: In Fig. 5.28, parallel lines l and m are intersected by the transversal t. If ∠3 is 50°, what is the measure of ∠6? m l 50° 1 2 4 3 Fig. 5.28 5 6 78 t 121 Ganita Prakash | Grade 7 Solution: ∠3 is 50°; therefore, ∠2 is 130°, because ∠2 and ∠3 form a linear pair, and linear pairs always add up to 180°. ∠2 and ∠6 are corresponding angles, and they need to be equal since lines l and m are parallel. So, ∠6 is 130°. Angles ∠3 and ∠6 are called interior angles. Example 4: In Fig. 5.29, line segment AB is parallel to CD and AD is parallel to BC. ∠DAC is 65° and ∠ADC is 60°. What are the measures of angles ∠CAB, ∠ABC, and ∠BCD? Solution: Let us observe the parallel lines AB and CD. AD is a transversal of these two lines. Is there a relation between ∠3 and ∠6? You could try to find the relationship by taking different values for ∠3 and see what ∠6 is. Once you find a relation, try to justify it or prove that this relation holds always. You will find that the sum of the interior angles on the same side of the transversal always add up to 180°. D C Fig. 5.29 60° 65° A B Chapter-5.indd 122 4/11/2025 7:30:40 PM We know that the sum of the interior angles formed by a transversal on a pair of parallel lines adds up to 180°. So ∠ADC + ∠DAB = 180° 60° + ∠DAB = 180°. So ∠DAB = 120°. Can we find ∠CAB from this? ∠DAB = ∠DAC + ∠CAB. So 120° = 65° + ∠CAB. So ∠CAB = 55°. Let us observe the parallel line segments AD and BC. They are intersected by a transversal CD. So, ∠ADC + ∠BCD = 180°, because they are interior angles on the same side of the transversal. Since ∠ADC is given as 60°, ∠BCD = 120° Similarly, we find ∠ABC = 60°. Therefore, in Fig. 5.29, ∠CAB = 55°, ∠ABC = 60°, and ∠BCD = 120°. 122 Figure it Out 1. Find the angles marked below. d= g= 58° 122° g° a= 81° 99° d° a° 48° 58° 122° 97° 83° 69° e° 120° b= e= h= b° 52° h° 75° Parallel and Intersecting Lines i= f= 132° f° c= 81° 99° c° 54° 56° i° 70° Chapter-5.indd 123 4/11/2025 7:30:40 PM j= 27° 97° 124° Fig. 5.30 j° 123 Ganita Prakash | Grade 7 2. Find the angle represented by a. 3. In the figures below, what angles do x and y stand for? 4. In Fig. 5.33, ∠ABC = 45° and ∠IKJ = 78°. Find angles ∠GEH, ∠HEF, ∠FED x° 100° 62° 65° 35° a° 110° a° 67° y° 78° 53° x° 42° Fig. 5.32 Fig. 5.31 a° a° Chapter-5.indd 124 4/11/2025 7:30:40 PM 124 I K A 78° 45° C EG D J H F Fig. 5.33 B 5. In Fig. 5.34, AB is parallel to CD and CD is parallel to EF. Also, EA is perpendicular to AB. If ∠BEF = 55°, find the values of x and y. 6. What is the measure of angle ∠NOP in Fig. 5.35? 5.9 Parallel Illusions There do not seem to be any parallel lines here. Or, are there? [Hint: Draw lines parallel to LM and PQ through points N and O.] M 96° L N 40° B 55° A C E x° Fig. 5.34 Fig. 5.35 D y° O Q a° F Parallel and Intersecting Lines 52° P Chapter-5.indd 125 4/11/2025 7:30:40 PM What causes these illusions? 125 Ganita Prakash | Grade 7 • When two lines intersect, they form four angles. The vertically opposite angles are equal and the linear pairs add up to 180°. • When two lines intersect and the angles formed are 90° (i.e., all four angles are equal), the lines are said to be perpendicular to each other. • When two lines never intersect on a plane, they are called parallel lines. • When a line t intersects another pair of lines, it is called a transversal and it forms 2 sets of 4 angles. Each of the 4 angles in the first set has a corresponding angle in the second set. • When a transversal intersects a pair of parallel lines, the corresponding angles are equal. When a transversal intersects a pair of lines and the corresponding angles are equal, then the pair of lines is parallel. • When a transversal intersects a pair of parallel lines, the alternate angles are equal. • The interior angles on the same side formed by a transversal intersecting a pair of parallel lines always add up to 180°. SUMMARY Chapter-5.indd 126 4/11/2025 7:30:40 PM 126 Page No. 107 Page No. 108. Can two straight lines intersect at more than one point? Ans: No What patterns do you observe among these angles? Ans: ∠a = ∠c and ∠b = ∠d. ∠a + ∠b = 180° and ∠b + ∠c = 180°. Find more such relations. Is this always true for any pair of intersecting lines? Ans: Yes Figure it Out List all the linear pairs and vertically opposite angles you observe in Fig. 5.3: Parallel and Intersecting Lines Chapter – 5 Ans: Pairs of Vertically Opposite Angles ∠b and ∠d, ∠a and ∠c Linear Pairs ∠a and ∠b, ∠b and ∠c, ∠c and ∠d, ∠d Fig. 5.3 and ∠a. Page No. 110 Page No. 111 Ans: a, i and h; c and g; d and f; e and b Activity 2 Which pairs of lines appear to be parallel in Fig. 5.6 below? Take a plain square sheet of paper (use a newspaper for this activity). • How would you describe the opposite edges of the sheet? They are _______ to each other. • How would you describe the adjacent edges of the sheet? The adjacent edges are ______ to each other. They meet at a point. They form right angles. • Fold the sheet horizontally in half. A new line is formed (see Fig. 5.7). • How many parallel lines do you see now? How does the new line segment relate to the vertical sides? Fig. 5.6 Ans: Fill in the Blanks ● The opposite edges of the sheet are parallel to each other. ● The adjacent edges of the sheet are perpendicular to each other. • Make one more horizontal fold in the folded sheet. How many parallel lines do you see now? • What will happen if you do it once more? How many parallel lines will you get? Is there a pattern? Check if the pattern extends further, if you make another horizontal fold. • Make a vertical fold in the square sheet. This new vertical line is ___________ to the previous horizontal lines. • Fold the sheet along a diagonal. Can you find a fold that creates a line parallel to the diagonal line? Fig. 5.7 First Horizontal Fold ● We will see three parallel lines. The new line segment is perpendicular to the vertical sides. Subsequent Horizontal Folds ● After one more horizontal fold, we will see a total of five parallel lines. The two original horizontal edges, the first fold line, and the second fold line are all parallel to each other. ● If we fold it horizontally one more time, we will get nine parallel lines. Yes, the pattern is 2 n + 1, where n is the number of folds. Page No. 112 Vertical and Diagonal Folds • This new vertical line is perpendicular to the previous horizontal lines. • Yes, we can find a fold that creates a line parallel to the diagonal line. Here is another activity for you to try. • Take a square sheet of paper, fold it in the middle and unfold it. • Fold the edges towards the centre line and unfold them. • Fold the top right and bottom left corners onto the creased line to create triangles. Refer to Fig. 5.8. • The triangles should not cross the crease lines. • Are a, b and c parallel to p, q and r respectively? Why or why not? Fig. 5.8 Ans: • Line segments a, b and c are parallel to Line segments p, q and r • a and p lie on parallel lines. b, q are both respectively perpendicular • c, r are both parallel because the triangular folds create lines respectively. to these, so are parallel to each other. parallel to the same diagonal. Page No. 113-114 Ans: 1. Draw some lines perpendicular to the lines given on the dot paper in the Figure. Figure it Out Fig. 5.10 2. In the given Fig.5.11, mark the parallel lines using the notation given above (single arrow, double arrow, etc). Mark the angle between perpendicular lines with a square symbol. (a)How did you spot the perpendicular lines? (b)How did you spot the parallel lines? Fig. 5.11 Ans: Ans: Some of the lines are: 3. In the dot paper following, draw different sets of parallel lines. The line segments can be of different lengths but should have dots as endpoints. 4. Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper. (a) The vertical and horizontal lines on the grid paper meet at a 90- degree angle (a right angle) (b) By identifying lines that always remain the same distance apart. Ans: Some of the lines are (a) Yes (b) Line segments e, f, h, and g. (c) Parallel lines are drawn by keeping them equidistant from the given line. 5. In Fig 5.13, which line is parallel to line a — line b or line c? How do you decide this? (a) Did you find it challenging to draw some of them? (b) Which ones? (c) How did you do it? Ans: In the given figure, line c is parallel to line a because these two lines are always at the same distance apart. Fig. 5.13 Page No. 115 Ans: No, because 1 = 3, 2 = 4, 5 = 7, and 6 = 8 (vertically opposite angles are equal). Ans: No, because 2 = 4. Page No. 119 Is it possible for all the eight angles to have different measurements? Why, why not? What about five different angles — 6, 5, 4, 3 and 2? Figure it Out Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method. Ans: Tools required: Ruler, Set-squares (right-angled triangle), Pencil, eraser Steps: 1. Place the set square so that one side is along the line l. 2. Hold the ruler against the other side of the set square so that the ruler won’t move. 3. Slide the set square along the ruler until one side reaches point A. 4. Draw a line along the edge of the set square through point A. Page No. 120. Why are lines l and m parallel to each other? 5. This new line is parallel to line l and passes through point A. Ans: When we fold the paper, first we make line t perpendicular to l passing through A. Then we make another line m perpendicular to t through point A. Since lines l and m are perpendicular to t, so pair of corresponding angles are equal i.e. 900 . Thus, l and m are parallel. Page No. 123-125 1. Find the angles marked below. Figure it Out Ans: a = 48° b = 52° c = 81° d = 99° e = 69° f = 48° g = 122° h = 75° i = 54° j = 97° 2. Find the angle represented by a. Ans: (i) 138° (ii) 118° (iii) 105° (iv) 23° 3. In the figures below, what angles do x and y stand for? Ans: (i) x = 25°, y = 155° (ii) x = 25° 4. In Figure, ABC = 45° and IKJ = 78°. Find angles GEH, HEF, FED. Ans: GEH = 45°; HEF = 57°; FED = 78° 5. In the Figure, AB is parallel to CD, and CD is parallel to EF. Also, EA is 6. What is the measure of angle NOP in the given figure? perpendicular to AB. If BEF = 55°, find the values of x and y. Ans: x = y = 125° [Corresponding Angles] [Hint: Draw lines parallel to LM and PQ through points N and O] Ans: NOP = 108°" class_7,6,Number Play,ncert_books/class_7/gegp1dd/gegp106.pdf,"6 NUMBER PLAY 6.1 Numbers Tell us Things What do the numbers in the figure below tell us? Remember the children from the Grade 6 textbook of mathematics? Now, they call out numbers using a different rule. What do you think these numbers mean? Chapter-6.indd 127 4/12/2025 11:59:04 AM The children rearrange themselves and each one says a number based on the new arrangement. Could you figure out what these numbers convey? Observe and try to find out. Ganita Prakash | Grade 7 The rule is — each child calls out the number of children in front of them who are taller than them. Check if the number each child says matches this rule in both the arrangements. Write down the number each child should say based on this rule for the arrangement shown below. Figure it Out 1. Arrange the stick figure cutouts given at the end of the book or draw a height arrangement such that the sequence reads: (a) 0, 1, 1, 2, 4, 1, 5 (b) 0, 0, 0, 0, 0, 0, 0 (c) 0, 1, 2, 3, 4, 5, 6 (d) 0, 1, 0, 1, 0, 1, 0 (e) 0, 1, 1, 1, 1, 1, 1 Chapter-6.indd 128 4/12/2025 6:16:31 PM 128 2. For each of the statements given below, think and identify if it is Always True, Only Sometimes True, or Never True. Share your reasoning. (f) 0, 0, 0, 3, 3, 3, 3 (a) If a person says ‘0’, then they are the tallest in the group. (b) If a person is the tallest, then their number is ‘0’. (c) The first person’s number is ‘0’. (d) If a person is not first or last in line (i.e., if they are standing somewhere in between), then they cannot say ‘0’. (e) The person who calls out the largest number is the shortest. (f) What is the largest number possible in a group of 8 people? 6.2 Picking Parity Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 number card. The numbers in the boxes should sum to 30. Can you help him find a way to do it? Add a few even numbers together. What kind of number do you get? Does it matter how many numbers are added? Any even number can be arranged in pairs without any leftovers. Some even numbers are shown here, arranged in pairs. Can you figure out which 5 cards add to 30? Is it possible? There are many ways of choosing 5 cards from this collection. Is there a way to find a solution without checking all possibilities? Let us find out. + + + + = 30 Number Play Chapter-6.indd 129 4/12/2025 11:59:05 AM As we see in the figure, adding any number of even numbers will result in a number which can still be arranged in pairs without any leftovers. In other words, the sum will always be an even number. Now, add a few odd numbers together. What kind of number do you get? Does it matter how many odd numbers are added? Odd numbers can not be arranged in pairs. An odd number is one more than a collection of pairs. Some odd numbers are shown below: 129 Ganita Prakash | Grade 7 Can we also think of an odd number as one less than a collection of pairs? This figure shows that the sum of two odd numbers must always be even! This along with the other figures here are more examples of a proof! What about adding 3 odd numbers? Can the resulting sum be arranged in pairs? No. Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6 odd numbers. Let us go back to the puzzle Kishor was trying to solve. There are 5 empty boxes. That means he has an odd number of boxes. All the number cards contain odd numbers. They should add to 30, which is an even number. Since, adding any 5 odd numbers will never result in an even number, Kishor cannot arrange these cards in the boxes to add up to 30. We can see that two odd numbers added together can always be arranged in pairs. Chapter-6.indd 130 4/12/2025 11:59:05 AM Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this possible? Why or why not? As they were born one year apart, their ages will be (two) consecutive numbers. Can their ages be 51 and 52? 51 + 52 = 103. Try some other consecutive numbers and see if their sum is 112. The counting numbers 1, 2, 3, 4, 5, ... alternate between even and odd numbers. In any two consecutive numbers, one will always be even and the other will always be odd! What would be the resulting sum of an even number and an odd number? We can see that their sum can’t be arranged in pairs and thus will be an odd number. 130 Since 112 is an even number, and Martin’s and Maria’s ages are consecutive numbers, they cannot add up to 112. We use the word parity to denote the property of being even or odd. For instance, the parity of the sum of any two consecutive numbers is odd. Similarly, the parity of the sum of any two odd numbers is even. Figure it Out 1. Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums: 2. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins in his piggy bank. He calculated the total and got ₹205. Did he make a mistake? If he did, explain why. If he didn’t, how many coins of each type could he have? 3. We know that: Similarly, find out the parity for the scenarios below: (d) even – even = ___________________ (a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd) (b) Sum of 2 odd numbers and 3 even numbers (c) Sum of 5 even numbers (d) Sum of 8 odd numbers (a) even + even = even (b) odd + odd = even (c) even + odd = odd (e) odd – odd = ___________________ (f) even – odd = ___________________ Number Play Chapter-6.indd 131 4/12/2025 11:59:05 AM Small Squares in Grids In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number. Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product? (g) odd – even = ___________________ 131 Ganita Prakash | Grade 7 Find the parity of the number of small squares in these grids: Parity of Expressions Consider the algebraic expression: 3n + 4. For different values of n, the expression has different parity: Come up with an expression that always has even parity. Come up with expressions that always have odd parity. Come up with other expressions, like 3n + 4, which could have either odd or even parity. The expression 6k + 2 evaluates to 8, 14, 20,... (for k = 1, 2, 3,...) — many even numbers are missing. Some examples are: 100p and 48w – 2. Try to find more. (a) 27 × 13 (b) 42 × 78 (c) 135 × 654 10 34 even n Value of 3n + 4 Parity of the Value 3 13 odd 8 28 even Chapter-6.indd 132 4/12/2025 11:59:05 AM Are there expressions using which we can list all the even numbers? Are there expressions using which we can list all odd numbers? We saw earlier how to express the nth term of the sequence of multiples of 4, where n is the letter-number that denotes a position in the sequence (e.g., first, twenty third, hundred and seventeenth, etc.). What would be the nth term for multiples of 2? Or, what is the nth even number? What is the 100th odd number? 132 Hint: All even numbers have a factor 2. Let us consider odd numbers. To answer this question, consider the following question: What is the 100th even number? This is 2 × 100 = 200. Does this help in finding the 100th odd number? Let us compare the sequence of evens and odds term-by-term. Even Numbers: 2, 4, 6, 8, 10, 12,... Odd Numbers: 1, 3, 5, 7, 9, 11,... We see that at any position, the value at the odd number sequence is one less than that in the even number sequence. Thus, the 100th odd number is 200 – 1 = 199. Write a formula to find the nth odd number. Let us first describe the method that we have learnt to find the odd number at a given position: (b) 2n – 1 Thus, 2n is the formula that gives the nth even number, and 2n – 1 is the formula that gives the nth odd number. 6.3 Some Explorations in Grids Observe this 3 × 3 grid. It is filled following a simple rule — use numbers from 1 – 9 without repeating any of them. There are circled numbers outside the grid. (b) Then subtract 1 from the even number. Writing this in expressions, we get (a) 2n (a) Find the even number at that position. This is 2 times the position number. 4 7 5 6 1 2 Number Play 16 9 Chapter-6.indd 133 4/12/2025 11:59:05 AM Are you able to see what the circled numbers represent? The numbers in the yellow circles are the sums of the corresponding rows and columns. Fill the grids below based on the rule mentioned above: 24 9 12 9 5 13 14 18 12 16 17 4 3 13 17 15 3 9 8 24 15 6 133 20 Ganita Prakash | Grade 7 Make a couple of questions like this on your own and challenge your peers. You might have realised that it is not possible to find a solution for this grid. Why is this the case? The smallest sum possible is 6 = 1 + 2 + 3. The largest sum possible is 24 = 9 + 8 + 7. Clearly, any number in a circle cannot be less than 6 or greater than 24. The grid has sums 5 and 26. Therefore, this is impossible! In the earlier grids which we solved, Kishor noticed that the sum of all the numbers in the circles was always 90. Also, Vidya observed that the sum of the circled numbers for all three rows, or for all three columns, was always 45. Check if this is true in the previous grids you have solved. Why should the row sums and column sums always add to 45? From this grid, we can see that all the row sums added together will be the same as the sum of the numbers 1 – 9. This can be seen for column sums as well. The sum of the numbers 1 – 9 is A square grid of numbers is called a magic square if each row, each column and each diagonal, add up to the same number. This number is called the magic sum. Diagonals are shown in the picture. Trying to create a magic square by randomly filling the grid with numbers may be difficult! This is because there are a large number of ways of filling a 3 × 3 grid using the numbers 1 – 9 without repetition. In fact, it can be found that there are exactly 3,62,880 such ways. Surprisingly, the number of ways to fill in the grid can be found without listing all of them. We will see in later years how to do this. Instead, we should proceed systematically to make a magic square. For this, let us ask ourselves some questions. Try solving the problem below. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. The 3 row sums added together gives 45! So does adding the column sums. 6 1 2 3 9 8 4 7 5 9 11 26 6 3+9+8 6+1+2 4+7+5 21 19 5 Chapter-6.indd 134 4/12/2025 11:59:05 AM 134 1. What can the magic sum be? Can it be any number? 4+6+3 5+2+8 7+1+9 Let us focus, for the moment, only on the row sums. We have seen that in a 3 × 3 grid with numbers 1 – 9, the total of row sums will always be 45. Since in a magic square the row sums are all equal, and they add up to 45, they have to be 15 each. Thus, we have the following observation. Observation 1: In a magic square made using the numbers 1 – 9, the magic sum must be 15. 2. What are the possible numbers that could occur at the centre of a magic square? Let us consider the possibilities one by one. Can the central number be 9? If yes, then 8 must come in one of the other squares. For example, In this, we must have 8 + 9 + other number = 15. But this is not possible! The same issue will occur no matter where we place 8. So, 9 cannot be at the centre. Can the central number be 1? If yes, then 2 should come in one of the other squares. Here, we must have 2 + 1 + other number = 15. But this is not possible because we are only using the numbers 1 – 9. The same issue will occur no matter where we place 1. So, 1 cannot be at the centre, either. Using such reasoning, find out which other numbers 1 – 9 cannot occur at the centre. This exploration will lead us to the following interesting observation. Observation 2: The number occurring at the centre of a magic square, filled using 1 – 9, must be 5. Let us now see where the smallest number 1 and the largest number 9 should come in a magic square. Our second observation tells us that they will have to come in one of the boundary positions. Let us classify these positions into two categories: 8 9 1 2 Number Play Chapter-6.indd 135 4/12/2025 11:59:05 AM 5 135 Ganita Prakash | Grade 7 If yes, then there should exist three ways of adding 1 with two other numbers to give 15. We have 1 + 5 + 9 = 1 + 6 + 8 = 15. Is any other combination possible? Similarly, can 9 can be placed in a corner position? Observation 3: The numbers 1 and 9 cannot occur in any corner, so they should occur in one of the middle positions. Can you find the other possible positions for 1 and 9? Figure it Out We can describe how the numbers within the magic square are related to each other, i.e., the structure of the magic square. 1 1. How many different magic squares can be made using the numbers 1 – 9? 2. Create a magic square using the numbers 2 – 10. What strategy would you use for this? Compare it with the magic squares made using 1 – 9. Can 1 occur in a corner position? For example, can it be placed as follows? Now, we have one full row or column of the magic square! Try completing it! [Hint: First fill the row or columns containing 1 and 9] 1 5 9 1 5 9 15 15 5 15 Math Talk Chapter-6.indd 136 4/12/2025 11:59:06 AM Generalising a 3 × 3 Magic Square 136 3. Take a magic square, and 4. What other operations can be performed on a magic square to yield another magic square? 5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.). (b) double each number In each case, is the resulting grid also a magic square? How do the magic sums change in each case? (a) increase each number by 1 Math Talk Choose any magic square that you have made so far using consecutive numbers. If m is the letter-number of the number in the centre, express how other numbers are related to m, how much more or less than m. [Hint: Remember, how we described a 2 × 2 grid of a calendar month in the Algebraic Expressions chapter]. Once the generalised form is obtained, share your observations with the class. Figure it Out The First-ever 4 × 4 Magic Square The first ever recorded 4 × 4 magic square is found in a 10th century inscription at the Pārśhvanath Jain temple in Khajuraho, India, and is known as the Chautīsā Yantra. 1. Using this generalised form, find a magic square if the centre number is 25. 2. What is the expression obtained by adding the 3 terms of any row, column or diagonal? 3. Write the result obtained by— 4. Create a magic square whose magic sum is 60. 5. Is it possible to get a magic square by filling nine non-consecutive numbers? (a) adding 1 to every term in the generalised form. (b) doubling every term in the generalised form Number Play m Math Talk Try This Chapter-6.indd 137 4/12/2025 6:21:34 PM Chau̐ tīs means 34. Why do you think they called it the Chautīsā Yantra? Every row, column and diagonal in this magic square adds up to 34. Can you find other patterns of four numbers in the square that add up to 34? The first ever recorded 4 × 4 magic square, the Chautīsā Yantra, at Khajuraho, India 16 3 10 5 7 12 1 14 2 13 8 11 9 6 15 4 137 2 7 6 9 5 1 4 3 8 Ganita Prakash | Grade 7 Magic Squares in History and Culture The first magic square ever recorded, the Lo Shu Square, dates back over 2000 years to ancient China. The legend tells of a catastrophic flood on the Lo River, during which the gods sent a turtle to save the people. The turtle carried a 3 × 3 grid on its back, with the numbers 1 to 9 arranged in a magical pattern. Magic squares were studied in different parts of the world at different points of time including India, Japan, Central Asia, and Europe. Indian mathematicians have worked extensively on magic squares, describing general methods of constructing them. The work of Indian mathematicians was not just limited to 3 × 3 and 4 × 4 grids, which we considered above, but also extended to 5 × 5 and other larger square grids. We shall learn more about these in later grades. The occurrence of magic squares is not limited to scholarly mathematical works. They are found in many places in India. The picture to the right is of a 3 × 3 magic square found on a pillar in a temple in Palani, Tamil Nadu. The temple dates back to the 8th century CE. 3 × 3 magic squares can also be found in homes and shops in India. The Navagraha Yantra is one such example shown below. Chapter-6.indd 138 4/12/2025 11:59:06 AM 138 Notice that a different magic sum is associated with each graha. A picture of a Kubera Yantra is shown below: 6.4 Nature’s Favourite Sequence: The Virahāṅka– Fibonacci Numbers! The sequence 1, 2, 3, 5, 8, 13, 21, 34, . . . (Virahāṅka–Fibonacci Numbers) is one of the most celebrated sequences in all of mathematics — it occurs throughout the world of Art, Science, and Mathematics. Even though these numbers are found very frequently in Science, it is remarkable that these numbers were first discovered in the context of Art (specifically, poetry)! The Virahāṅka–Fibonacci Numbers thus provide a beautiful illustration of the close links between Art, Science, and Mathematics. Discovery of the Virahāṅka Numbers The Virahāṅka numbers first came up thousands of years ago in the works of Sanskrit and Prakrit linguists in their study of poetry! In the poetry of many Indian languages, including, Prakrit, Sanskrit, Marathi, Malayalam, Tamil, and Telugu, each syllable is classified as either long or short. A long syllable is pronounced for a longer duration than a short syllable — in fact, for exactly twice as long. When singing such a poem, a short syllable lasts one beat of time, and a long syllable lasts two beats of time. This leads to numerous mathematical questions, which the ancient poets in these languages considered extensively. A number of important mathematical discoveries were made in the process of asking and answering these questions about poetry. One of these particularly important questions was the following. How many rhythms are there with 8 beats consisting of short syllables (1 beat) and long syllables (2 beats)? That is, in how many ways can one 27 20 25 22 24 26 23 28 21 Number Play Chapter-6.indd 139 4/12/2025 11:59:07 AM 139 Ganita Prakash | Grade 7 fill 8 beats with short and long syllables, where a short syllable takes one beat of time and a long syllable takes two beats of time. Here are some possibilities: long long long long short short short short short short short short short long long short long long long short short long . . . Can you find others? Phrased more mathematically: In how many different ways can one write a number, say 8, as a sum of 1’s and 2’s? For example, we have: 8 = 2 + 2 + 2 + 2, 8 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1, 8 = 1 + 2 + 2 + 1 + 2, 8 = 2 + 2 + 1 + 1 + 2, etc. Do you see other ways? Here are all the ways of writing each of the numbers 1, 2, 3, and 4, as the sum of 1’s and 2’s. n = 1 1 1 n = 2 1 + 1 2 2 n = 3 1 + 1 + 1 1 + 2 2 + 1 3 Different Ways Number of Ways Chapter-6.indd 140 4/12/2025 11:59:07 AM Try writing the number 5 as a sum of 1s and 2s in all possible ways in your notebook! How many ways did you find? (You should find 8 different ways!) Can you figure out the answer without listing down all the possibilities? Can you try it for n = 8? Here is a systematic way to write down all rhythms of short and long syllables having 5 beats. Write a ‘1+’ in front of all rhythms having 4 beats, and then a ‘2+’ in front of all rhythms having 3 beats. This gives us all the rhythms having 5 beats: 140 n = 4 1 + 1 + 1 + 1 1 + 1 + 2 1 + 2 + 1 2 + 1 + 1 2 + 2 5 Thus, there are 8 rhythms having 5 beats! The reason this method works is that every 5-beat rhythm must begin with either a ‘1+’ or a ‘2+’. If it begins with a ‘1+’, then the remaining numbers must give a 4-beat rhythm, and we can write all those down. If it begins with a 2+, then the remaining number must give a 3-beat rhythm, and we can write all those down. Therefore, the number of 5-beat rhythms is the number of 4-beat rhythms, plus the number of 3-beat rhythms. How many 6-beat rhythms are there? By the same reasoning, it will be the number of 5-beat rhythms plus the number of 4-beat rhythms, i.e., 8 + 5 = 13. Thus, there are 13 rhythms having 6 beats. Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of 1’s and 2’s in all possible ways. Did you get 13 ways? This beautiful method for counting all the rhythms of short syllables and long syllables having any given number of beats was first given by the great Prakrit scholar Virahāṅka around the year 700 CE. He gave his method in the form of a Prakrit poem! For this reason, the sequence 1, 2, 3, 5, 8, 13, 21, 34, . . . is known as the Virahāṅka sequence, and the numbers in the sequence are known as the Virahāṅka numbers. Virahāṅka was the first known person in history to explicitly consider these important numbers and write down the rule for their formation. Other scholars in India also considered these numbers in the same poetic context. Virahāṅka was inspired by earlier work of the legendary Sanskrit scholar Piṅgala, who lived around 300 BCE. After Virahāṅka, these numbers were also written about by Gopala (c. 1135 CE) and then by Hemachandra (c. 1150 CE). In the West, these numbers have been known as the Fibonacci numbers, after the Italian mathematician who wrote about them in the year 1202 CE — about 500 years after Virahāṅka. As we can see, Fibonacci was not first, nor the second, not even the third person to write about these numbers! Sometimes the term “Virahāṅka–Fibonacci numbers” is used so that everyone understands what is being referred to. So, how many rhythms of short and long syllables are there having 8 beats? We simply take the 8th element of the Virahāṅka sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, ... Thus, there are 34 rhythms having 8 beats. n = 5 1 + 1 + 1 + 1 + 1 1 + 1 + 1 + 2 1 + 1 + 2 + 1 1 + 2 + 1 + 1 1 + 2 + 2 2 + 1 + 1 + 1 2 + 1 + 2 2 + 2 + 1 Number Play Chapter-6.indd 141 4/12/2025 11:59:07 AM 141 Ganita Prakash | Grade 7 Write the next number in the sequence, after 55. We have seen that the next number in the sequence is given by adding the two previous numbers. Check that this holds true for the numbers given above. The next number is 34 + 55 = 89. Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34 , 55, 89, ____, ____, ____, … If you have to write one more number in the sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)? What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities? Today, the Virahāṅka–Fibonacci numbers form the basis of many mathematical and artistic theories, from poetry to drumming, to visual arts and architecture, to science. Perhaps the most stunning occurrences of these numbers are in nature. For example, the number of petals on a daisy is generally a Virahāṅka number. How many petals do you see on each of these flowers? A daisy with 13 petals A daisy with 21 petals A daisy with 34 petals Chapter-6.indd 142 4/12/2025 11:59:08 AM There are many other remarkable mathematical properties of the Virahāṅka– Fibonacci numbers that we will see later, in mathematics as well as in other subjects. These numbers truly exemplify the close connections between Art, Science, and Mathematics. 6.5 Digits in Disguise You have done arithmetic operations with numbers. How about doing the same with letters? 142 In the calculations below, digits are replaced by letters. Each letter stands for a particular digit (0 – 9). You have to figure out which digit each letter stands for. Here, we have a one-digit number that, when added to itself twice, gives a 2-digit sum. The units digit of the sum is the same as the single digit being added. What could U and T be? Can T be 2? Can it be 3? Once you explore, you will see that T = 5 and UT = 15. Let us look at one more example shown on the right. Here K2 means that the number is a 2-digit number having the digit ‘2’ in the units place and ‘K’ in the tens place. K2 is added to itself to give a 3-digit sum HMM. What digit should the letter M correspond to? Both the tens place and the units place of the sum have the same digit. What about H? Can it be 2? Can it be 3? These types of questions can be interesting and fun to solve! Here are some more questions like this for you to try out. Find out what each letter stands for. Share how you thought about each question with your classmates; you may find some new approaches. + T UT T T Number Play + K2 HMM K2 Chapter-6.indd 143 4/12/2025 11:59:08 AM Figure it Out 1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why? These types of questions are called ‘cryptarithms’ or ‘alphametics’. + Z + 3D + KP + C ZOO ED5 PRR 1FF YY B5 KP C1 143 Ganita Prakash | Grade 7 10. Identify the statements that are true. 2. Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not? 3. Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums. 4. Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed. 5. Fill in the following blanks with ‘odd’ or ‘even’: 6. What is the parity of the sum of the numbers from 1 to 100? 7. Two consecutive numbers in the Virahāṅka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence? 8. Angaan wants to climb an 8-step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, 2, 1, 2. In how many different ways can he reach the top? 9. What is the parity of the 20th term of the Virahāṅka sequence? (a) Sum of an odd number of even numbers is ______ (b) Sum of an even number of odd numbers is ______ (c) Sum of an even number of even numbers is ______ (d) Sum of an odd number of odd numbers is ______ e e o o e Chapter-6.indd 144 4/12/2025 6:28:51 PM 11. Solve this cryptarithm: 144 (a) The expression 4m – 1 always gives odd numbers. (b) All even numbers can be expressed as 6j – 4. (c) Both expressions 2p + 1 and 2q – 1 describe all odd numbers. (d) The expression 2f + 3 gives both even and odd numbers. + TA TAT UT In this chapter, we have explored the following: • In the first activity, we saw how to represent information about how a sequence of numbers (e.g., height measures) is arranged without knowing the actual numbers. • We learnt the notion of parity — numbers that can be arranged in pairs (even numbers) and numbers that cannot be arranged in pairs (odd numbers). • We learnt how to determine the parity of sums and products. • While exploring sums in grids, we could determine whether filling a grid is impossible by looking at the row and column sums. We extended this to construct magic squares. • We saw how Virahāṅka numbers were first discovered in history through the Arts. The Virahāṅka sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55, ... • We became math-detectives through cryptarithms, where digits are replaced by letters. I just heard that 14,70,369 people got married last year. SUMMARY Wait, shouldn’t it be an even number? Number Play Chapter-6.indd 145 4/12/2025 11:59:09 AM 145 Page No- 128 Write down the number each child should say based on this rule for the arrangement shown below. The rule is: Each child calls out the number of children standing in front of them who are taller than they are. Let’s assign the numbers from left to right for the 7 children: Child 1: From left → 0 Child 2: From left → 0 Child 3: From left → 1 Child 4: From left → 0 Child 5: From left → 3 Child 6: From left → 0 Child 7: From left → 3 CHAPTER – 6 Number Play 1. Arrange the stick figure cutouts given at the end of the book or draw a height arrangement such that the sequence reads: (a) 0, 1, 1, 2, 4, 1, 5 (b) 0, 0, 0, 0, 0, 0, 0 (c) 0, 1, 2, 3, 4, 5, 6 (d) 0, 1, 0, 1, 0, 1, 0 (e) 0, 1, 1, 1, 1, 1, 1 (f) 0, 0, 0, 3, 3, 3, 3 Figure it out 1 Ans: True, Only Sometimes True, or Never True. Share your reasoning. (a) If a person says ‘0’, then they are the tallest in the group. (b) If a person is the tallest, then their number is ‘0’. (c) The first person’s number is ‘0’. (d) If a person is not first or last in line (i.e., if they are standing somewhere in between), then they cannot say ‘0’. (e) The person who calls out the largest number is the shortest. (f) What is the largest number possible in a group of 8 people? Ans: (a) Only sometimes true (b) Always true (c) Always true (d) Only sometimes true (e) Only sometimes true (f) The largest possible number in a group of 8 people is 7 2. For each of the statements given below, think and identify if it is Always (a) From the cutouts at the back of the book we have → F, C, B, G, A, D, E Try for other subparts May also like to draw this way — Page No- 131 Figure it out 1. Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums: (a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd) (b) Sum of 2 odd numbers and 3 even numbers 2 (d) Sum of 8 odd numbers Ans: (a) Even number (b) Even number (c) Even number (d) Even number 2. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins in his piggy bank. He calculated the total and got ₹205. Did he make a mistake? If he did, explain why. If he didn’t, how many coins of each type could he have? Ans: Lakpa must have made a mistake! The total can never be ₹205 with the given coin counts. 3. We know that: (c) Sum of 5 even numbers (a) even + even = even (b) odd + odd = even (c) even + odd = odd Similarly, find out the parity for the scenarios below: (d) even – even = _____ (e) odd – odd = _____ (f) even – odd = _____ Ans: (g) odd – even = _____ (d) Even (e) Even (f) Odd (g) Odd 3 Ans: Page No- 132 Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product? The parity of the number of small squares in a grid (with m rows and n columns) is: (i) Even, if at least one of the numbers (either m or n) is even. (ii) Odd, if both numbers (m and n) are odd. (iii) Even, if both numbers (m and n) are even. (a) 27 × 13 Find the parity of the number of small squares in these grids: Ans: Come up with an expression that always has odd parity. Ans: Some such expressions are 2m + 1, 8a + 3; 4n + 3, 6n + 5 etc. or even parity. Ans: (b) 42 × 78 (c) 135 × 654 (a) Odd (b) Even (c) Even Come up with other expressions, like 3n + 4, which could have either odd 4 Ans: 2n-1 is the expression that can list all odd numbers. What would be the nth term for multiples of 2? Or, what is the nth even number? Ans: 2n Ans: (i) If n is odd, then 3n + 4 is odd then some expressions of odd parity would be— Some expressions of even parity — 2n + 4, 4n + 6 etc. (ii) If n is even, then 3n + 4 is even then all expression will be even parity — (5n – 2), n + 4 etc. Are there expressions using which we can list all odd numbers? Fill the grids below based on the rule mentioned above: 5n + 2, 9n + 4 etc. Try to find out other possible ways. 5 Page No- 136 1. How many different magic squares can be made using the numbers 1 – 9? Ans: There is exactly one unique magic square with numbers 1 – 9, if we ignore rotations and reflections. Ans: We will add 1 to every number in the magic square in Q.1 2. Create a magic square using numbers 2 – 10. What strategy would you use Figure it out for this? Compare it with the magic squares made using 1 – 9. 8 1 6 3 5 7 4 9 2 9 2 7 4 6 8 5 10 3 Ans: Consider the magic square. 3. Take a magic square, and (a) Increase each number by 1 (b) Double each number In each case, is the resulting grid also a magic square? How do the magic sums change in each case? 8 1 6 3 5 7 4 9 2 6 another magic square? Ans: Operations like Addition, subtraction, multiplication and division can be performed on a magic square to yield another magic square. 4. What other operations can be performed on a magic square to yield This is still a magic square. (a) After increasing each number by 1: (b) After Doubling each number: This is still a magic square In case (a), magic sum is increased by 3 times that constant. (i.e., 1) In case (b), magic sum is multiplied by that constant. (i.e., 2) 16 2 12 6 10 14 8 18 4 9 2 7 4 6 8 5 10 3 numbers (like 2 – 10, 3 – 11, 9 – 17, etc.). Ans: Magic square with numbers 1 – 9: 5. Discuss ways of creating a magic square using any set of 9 consecutive 8 1 6 3 5 7 4 9 2 7 numbers. If m is the letter-number of the number in the centre, express how other numbers are related to m, how much more or less than m. [Hint: Remember, how we described a 2 × 2 grid of a calendar month in the Algebraic Expressions chapter]. Ans: Taking ‘m’ as the letter-number of the number in the centre and expressing other numbers in relation to ‘m’. Magic square with numbers 2 – 10 can be created by adding 1 to each number of the magic square 1 – 9. Magic square with numbers 3 – 11 can be created by adding 2 to each number of the magic square 1 – 9. Magic square with numbers 9 – 17 can be created by adding 8 to each number of the magic square 1 – 9. Choose any magic square that you have made so far using consecutive m + 3 m – 4 m + 1 9 2 7 4 6 8 5 10 3 Page No- 137 Figure it out 1. Using this generalised form, find a magic square if the centre number is 25. Ans: m – 2 m m + 2 m – 1 m + 4 m – 3 28 21 26 23 25 27 24 29 22 8 column, or diagonal? Ans: The expression obtained = 3 × m where m is the letter - number representing the number in the centre. 3. Write the result obtained by (a) Adding 1 to every term in the generalised form. (b) Doubling every term in the generalised form. Ans: (a) Ans: 2. What is the expression obtained by adding the 3 terms of any row, 4. Create a magic square whose magic sum is 60. (b) 2m + 6 2m – 8 2m + 2 2m – 4 2m 2m + 4 2m – 2 2m + 8 2m – 6 m + 4 m – 3 m + 2 m – 1 m + 1 m + 3 m m + 5 m – 2 23 16 21 5. Is it possible to get a magic square by filling nine non-consecutive numbers? Ans: Yes, it is possible. Try some more. 18 20 22 19 24 17 24 3 18 12 27 6 9 15 21 9 Page No- 142 Ans: 55 + 89 = 144, 89 + 144 = 233, 144 + 233 = 3779 The next number will be an even number. Page No- 143-44 1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why? Ans. The bulb will be OFF. Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ____, ____, ____, … If you have to write one more number in the sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)? What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities? The parity patterns — Sequence: 1(O), 2(E), 3(O), 5(O), 8(E), 13(O), 21(O), 34(E), … The pattern is: Odd, Even, Odd, Odd, Even, Odd, Odd, Even…. The group odd, even, odd repeats. 3 rd term = 1st term + 2nd term = O + E = O; 4 th term = 2nd term + 3rd term = E + O = O; 5 th term = 3rd term + 4th term = O + O = E… Figure it out sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not? Ans. Each sheet has two pages, one with an even number and the other with an odd number. Since there are 50 sheets, so there will be fifty pages with an even number and 50 pages with an odd number The sum of the even numbers is even. The sum of the odd numbers taken even number of times is also even. Thus, 6000 being even, it is possible that sum of all the page numbers of the loose sheets could be 6000. 2. Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both 10 Ans: One of the ways is – be zero. Use negative numbers, as needed. Ans: One of the such magic squares is – 3. Here is a 2 × 3 grid. For each row and column, the parity of the sum is 4. Make a 3 × 3 magic square with 0 as the magic sum. All numbers cannot written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums. Try more combinations. 3 – 4 1 (a) Sum of an odd number of even numbers is ______ (b) Sum of an even number of odd numbers is ______ (c) Sum of an even number of even numbers is ______ (d) Sum of an odd number of odd numbers is ______ Ans: (a) Even. (b) Even. 5. Fill in the following blanks with ‘odd’ or ‘even’: – 2 0 2 – 1 4 – 3 11 6. What is the parity of the sum of numbers from 1 to 100? Ans: The parity is even. [Hint– Add numbers from 1 to 100.] 7. Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence? Ans: The next two numbers are: 987 + 1597 = 2584 1597 + 2584 = 4181 The previous two numbers are: 1597 – 987 = 610 987 – 610 = 377 The sequence is …, 377, 610, 987, 1597, 2584, 4181, … 8. Angaan wants to climb an 8 - step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, 2, 1, 2. In how many different ways can he reach the top? Ans: Different Ways Number of ways (c) Even. (d) Odd. 1, 1, 1, 1, 1, 1, 1, 1 1 1, 1, 1, 1, 1, 1, 2 1, 1, 1, 1, 1, 2, 1 . . . 7 So, the total ways in which Angaan reaches the top = 1 + 7 + 15 + 10 + 1 = 34 ways. 1, 1, 1, 1, 2, 2 1, 1, 1, 2, 1, 2 . . . 1, 1, 2, 2, 2 1, 2, 2, 2, 1 . . . 2, 2, 2, 2 1 12 15 10 Ans: Consider the Virahanka sequence given below: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …. The pattern of odd/even in Virahanka numbers is: O, E, O, O, E, O … So the 20th term of the Virahanka sequence is even. Ans: 9. What is the parity of the 20th term of the Virahanka sequence? 10. Identify the true statements. 11. Solve this cryptarithm: (a) The expression 4m – 1 always gives odd numbers. (b) All even numbers can be expressed as 6j – 4. (c) Both expressions 2p + 1 and 2q – 1 describe all odd numbers. (d) The expression 2f + 3 gives both even and odd numbers. (a) True (b) False (c) False (d) False Ans: U = 9, T = 1 and A = 0. 13" class_7,7,A Tale of Three Intersecting Lines,ncert_books/class_7/gegp1dd/gegp107.pdf,"7 A triangle is the most basic closed shape. As we know, it consists of: Observe the symbol used to denote a triangle and how the triangles are named using their vertices. While naming a triangle, the vertices can come in any order. The three sides meeting at the corners give rise to three angles that we call the angles of the triangle. For example, in ∆ABC, these angles are ∠CAB, ∠ABC, ∠BCA, which we simply denote as ∠A, ∠B and ∠C, respectively. B C ∆ ABC • three corner points, that we call the vertex of the triangle, and • three line segments or the sides of the triangle that join the pairs of vertices. Triangles come in various shapes. Some of them are shown below. A A TALE OF THREE INTERSECTING LINES Y Z ∆ YZX X V ∆WUV U W ∆BAU A B U Chapter-7.indd 146 4/11/2025 7:31:55 PM What happens when the three vertices lie on a straight line? 7.1 Equilateral Triangles Among all the triangles, the equilateral triangles are the most symmetric ones. These are triangles in which all the sides are of equal lengths. Let us try constructing them. Construct a triangle in which all the sides are of length 4 cm. How did you construct this triangle and what tools did you use? Can this construction be done only using a marked ruler (and a pencil)? Constructing this triangle using just a ruler is certainly possible. But this might require several trials. Say we draw the base — let us call it AB — of length 4 cm (see the figure below), and mark the third point C using a ruler such that AC = 4 cm. This may not lead to BC also having a length of 4 cm. If this happens, we will have to keep making attempts to mark C till we get BC to be 4 cm long. How do we make this construction more efficient? Recall solving a similar problem in the previous year using a compass (in the Chapter ‘Playing with Constructions’). We had to mark the top point of a ‘house’ which is 5 cm from two other points. The method we used to get that point can also be used here. After constructing AB = 4 cm, we can do the following. Step 1: Using a compass, construct a sufficiently long arc of radius 4 cm from A, as shown in the figure. The point C is somewhere on this arc. How do we mark it? A B 4 cm C 4 cm 4 cm? A Tale of Three Intersecting Lines Chapter-7.indd 147 4/11/2025 7:31:55 PM A B 4 cm 147 Ganita Prakash | Grade 7 Step 2: Construct another arc of radius 4 cm from B. Let C be the point of intersection of the arcs. The construction ensures that both AC and BC are of length 4 cm. Can you see why? Step 3: Join AC and BC to get the required equilateral triangle. A B 4 cm A B 4 cm C Chapter-7.indd 148 4/11/2025 7:31:55 PM 7.2 Constructing a Triangle When its Sides are Given How do we construct triangles that are not equilateral? Construct a triangle of sidelength 4 cm, 5 cm and 6 cm. As in the previous case, this triangle can also be constructed using just a marked ruler. But it will involve several trials. 148 How do we construct this triangle more efficiently? Choose one of the given lengths to be the base of the triangle: say 4 cm. Draw the base. Let A and B be the base vertices, and call the third vertex C. Let AC = 5 cm and BC = 6 cm. Like we did in the case of equilateral triangles, let us first get all the points that are at a 5 cm distance from A. These points lie on the circle whose centre is A and has radius 5 cm. The point C must lie somewhere on this circle. How do we find it? We will make use of the fact that the point C is 6 cm away from B. Construct an arc of radius 6 cm from B. Fig. 7.1 Fig. 7.2 4 cm A B 5 cm A Tale of Three Intersecting Lines Chapter-7.indd 149 4/11/2025 7:31:55 PM A B 5 cm 6 cm Fig. 7.3 149 Ganita Prakash | Grade 7 The required point C is one of the points of intersection of the two circles. The reason why the point of intersection is the third vertex is the same as for equilateral triangles. This point lies on both the circles. Hence its distance from A is the radius of the circle centred at A (5 cm) and its distance from B is the same as the radius of the circle centred at B (6 cm). Let us summarise the steps of construction, noting that constructing full circles is not necessary to get the third vertex (see Fig. 7.2 and 7.3). Step 1: Construct the base AB with one of the side lengths. Let us choose AB = 4 cm (see Fig. 7.1). Step 2: From A, construct a sufficiently long arc of radius 5 cm (see Fig. 7.2). Step 3: From B, construct an arc of radius 6 cm such that it intersects the first arc (see Fig. 7.3). Step 4: The point where both the arcs meet is the required third vertex C. Join AC and BC to get ∆ABC. Construct Construct triangles having the following sidelengths (all the units are in cm): (a) 4, 4, 6 C A B Chapter-7.indd 150 4/11/2025 7:31:55 PM We have seen that triangles having all three equal sides are called equilateral triangles. Those having two equal sides are called isosceles triangles. Figure it Out 1. Use the points on the circle and/or the centre to form isosceles triangles. 150 (b) 3, 4, 5 (c) 1, 5, 5 (d) 4, 6, 8 (e) 3.5, 3.5, 3.5 2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size. Are Triangles Possible for any Lengths? Can one construct triangles having any given sidelengths? Are there lengths for which it is impossible to construct a triangle? Let us explore this. Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle? Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible for these sidelengths. Try to find more sets of lengths for which a triangle construction is impossible. See if you can find any pattern in them. A and B are the centres of circles of the same size A B A, B, and C are the centres of circles of the same size A Tale of Three Intersecting Lines A B C Math Talk Chapter-7.indd 151 4/11/2025 7:31:55 PM We see that a triangle is possible for some sets of lengths and not possible for others. How do we check if a triangle exists for a given set of lengths? One way is to actually try to construct the triangle and check if it is possible. Is there a more efficient way to check this? Triangle Inequality Consider the lengths 10 cm, 15 cm and 30 cm. Does there exist a triangle having these as sidelengths? To tackle this question, let us study a property of triangles. Imagine a small plot of plain land having a tent, a tree, and a pole. Imagine you are at the entrance of the tent and want to go to the tree. Which is the shorter path: (i) the straight-line path to the tree (the red path) or (ii) the straight-line path from the tent to the pole, followed by the straight-line path from the pole to the tree (the yellow path)? 151 Ganita Prakash | Grade 7 Clearly, the direct straight-line path from the tent to the tree is shorter than the roundabout path via the pole. In fact, the direct straight-line path is the shortest possible path to the tree from the tent. Will the direct path between any two points be shorter than the roundabout path via a third point? Clearly, the answer is yes. Can this understanding be used to tell something about the existence of a triangle having sidelengths 10 cm, 15 cm and 30 cm? Let us suppose that there is a triangle for this set of lengths. Remember that at this point we are not sure about the existence of the triangle but we are only supposing that it exists. Let us draw a rough diagram. Chapter-7.indd 152 4/11/2025 7:31:55 PM 152 Fig. 7.4 Does everything look right with this triangle? If this triangle were possible, then the direct path between any two vertices should be shorter than the roundabout path via the third vertex. Is this true for our rough diagram? Let us consider the paths between B and C. What is the length of the roundabout path via the vertex A? It is the sum of the lengths of line segments BA and AC. Is the direct path length shorter than the roundabout path length? Yes. Let us now consider the paths between A and B. Is the direct path length shorter than the roundabout path length? Yes. Finally, consider paths between C and A. Is the direct path length shorter than the roundabout path length? In this case, the direct path is longer, which is absurd. Can such a triangle exist? No. Therefore, a triangle having sidelengths 10 cm, 15 cm and 30 cm cannot exist. We are thus able to see without construction why a triangle for the set of lengths 10 cm, 15 cm and 30 cm cannot exist. We have been able to figure this out through spatial intuition and reasoning. Recall how we used similar intuition and reasoning to discover properties of intersecting and parallel lines. We will continue to do this as we explore geometry. Direct path length = BC = 10 cm Roundabout path length = BA + AC = 15 cm + 30 cm = 45 cm Direct path length = AB = 15 cm Finding the length of the roundabout path via the vertex C, we get Roundabout path length = AC + CB = 30 cm + 10 cm = 40 cm Direct path length = CA = 30 cm Roundabout path length = CB + BA = 10 cm + 15 cm = 25 cm A Tale of Three Intersecting Lines Chapter-7.indd 153 4/11/2025 7:31:55 PM Can we say anything about the existence of a triangle having sidelengths 3 cm, 3 cm and 7 cm? Verify your answer by construction. “In the rough diagram in Fig. 7.4, is it possible to assign lengths in a different order such that the direct paths are always coming out to be shorter than the roundabout paths? If this is possible, then a triangle might exist.” Is such rearrangement of lengths possible in the triangle? Math Talk 153 Ganita Prakash | Grade 7 Figure it Out 1. We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle. 2. Can we say anything about the existence of a triangle for each of the following sets of lengths? You would have realised that using a rough figure and comparing the direct path lengths with their corresponding roundabout path lengths is the same as comparing each length with the sum of the other two lengths. There are three such comparisons to be made. 3. For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens: Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore for different sets of lengths. But this doesn’t happen for the third length: 30 > 10 + 15. (a) 10 km, 10 km and 25 km (b) 5 mm, 10 mm and 20 mm (c) 12 cm, 20 cm and 40 cm 10 < 15 + 30 15 < 10 + 30 Try This Chapter-7.indd 154 4/11/2025 7:31:56 PM Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations? [Hint: Consider the direct lengths in the increasing order.] Given three sidelengths, what do we need to compare to check for the existence of a triangle? When each length is smaller than the sum of the other two, we say that the lengths satisfy the triangle inequality. For example, the set 3, 4, 5 satisfies the triangle inequality whereas, the set 10, 15, 30 does not satisfy the triangle inequality. We have seen that lengths such as 10, 15, 30 that do not satisfy the triangle inequality cannot be the sidelengths of a triangle. 154 Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm? Why do we not need to check the other two sides? This means that all the direct path lengths are less than the roundabout path lengths. Does this confirm the existence of a triangle? If one of the direct path lengths had been longer, we could have concluded that a triangle would surely not exist. But in this case, we can only say that a triangle may or may not exist. For the triangle to exist, the arcs that we construct to get the third vertex must intersect. Is it possible to determine that this will happen without actually carrying out the construction? Visualising the construction of circles Let us imagine that we start the construction by constructing the longest side as the base. Let AB be the base of length 8 cm. The next step is the construction of sufficiently long arcs corresponding to the other two lengths: 4 cm and 5 cm. Instead of just constructing the arcs, let us complete the full circles. Suppose, we construct a circle of radius 4 cm with A as the centre. This satisfies the triangle inequality: 8 < 4 + 5 = 9 A Tale of Three Intersecting Lines Chapter-7.indd 155 4/11/2025 7:31:56 PM Now, suppose that a circle of radius 5 cm is constructed, centred at B. Can you draw a rough diagram of the resulting figure? Note that in the figure below, AX = 4 cm and AB = 8 cm. So, what is BX? Does this length help in visualising the resulting figure? 155 Ganita Prakash | Grade 7 Since BX = 4 cm, and the radius of the circle centred at B is 5 cm, it is clear that the circles will intersect each other at two points. What does this tell us about the existence of a triangle? The points A and B along with either of the points of intersection of the circles will give us the required triangle. Thus, there exists a triangle having sidelengths 4 cm, 5 cm and 8 cm. Figure it Out 1. Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure. Fig. 7.5: Circles intersecting each other at two points Chapter-7.indd 156 4/11/2025 7:31:56 PM We observe from the previous problems that whenever there is a set of lengths satisfying the triangle inequality (each length < sum of the other two lengths), there is a triangle with those three lengths as sidelengths. Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure? 156 (a) 2, 2, 5 (b) 3, 4, 6 (c) 2, 4, 8 (d) 5, 5, 8 (e) 10, 20, 25 (f) 10, 20, 35 (g) 24, 26, 28 We can be sure of the existence of a triangle only if we can show that the circles intersect internally (as in Fig. 7.5) whenever the triangle inequality is satisfied. But are there other possibilities when the two circles are constructed? Let us visualise and study them. The following different cases can be conceived: Which of the above-mentioned cases will lead to the formation of a triangle? Clearly, triangles are formed only when the circles intersect each other internally (Case 3). Note that while constructing the circles, we take Case 1: Circles touch each other (a) the length of the base AB = longest of the given length (b) the radii of the circles to be the smaller two lengths. A B Case 3: Circles intersect each other internally A B Case 2: Circles do not intersect A B A Tale of Three Intersecting Lines Chapter-7.indd 157 4/11/2025 7:31:56 PM Let us study each of these cases by finding the relation between the radii (the smaller two lengths) and AB (longest length). Case 1: Circles touch each other at a point For this case to happen, sum of the two smaller lengths = longest length sum of the two radii = AB or 157 Ganita Prakash | Grade 7 Case 2: Circles do not intersect internally For this case to happen, what should be the relation between the radii and AB? Case 3: Circles intersect each other It can be seen from the figure that, sum of the the two smaller lengths < longest length sum of the two radii < AB A B A B X or Chapter-7.indd 158 4/11/2025 7:31:56 PM 158 A radius 1 radius 2 B Can we use this analysis to tell if a triangle exists when the lengths satisfy the triangle inequality? If the given lengths satisfy the triangle inequality, then the sum of the two smaller lengths is greater than the longest length. This means that this will lead to Case 3 where the circles intersect internally, and so a triangle exists. How will the two circles turn out for a set of lengths that do not satisfy the triangle inequality? Find 3 examples of sets of lengths for which the circles: Frame a complete procedure that can be used to check the existence of a triangle. Conclusion If a given set of three lengths satisfies the triangle inequality, then a triangle exists having those as sidelengths. If the set does not satisfy the triangle inequality, then a triangle with those sidelengths does not exist. AB is composed of one radius and a part of the other. So, (a) touch each other at a point, (b) do not intersect. sum of the two smaller lengths > longest length sum of the two radii > AB, or A Tale of Three Intersecting Lines Chapter-7.indd 159 4/11/2025 7:31:56 PM Figure it Out 1. Check if a triangle exists for each of the following set of lengths: 2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer. 3. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen): (a) 1, 100, 100 (b) 3, 6, 9 (c) 1, 1, 5 (d) 5, 10, 12 (a) 1, 100 (b) 5, 5 (c) 3, 7 159 Ganita Prakash | Grade 7 7.3 Construction of Triangles When Some Sides and Angles are Given We have seen how to construct triangles when their sidelengths are given. Now, we will take up constructions where in place of some sidelengths, angle measures are given. Two Sides and the Included Angle How do we construct a triangle if two sides and the angle included between them are given? Here are some examples of measurements showing the included angle. Construct a triangle ABC with AB = 5 cm, AC = 4 cm and ∠A= 45°. See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those sidelengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible. Let us take one of the given sides, AB, as the base of the triangle. 4 cm 60° 3 cm 6 cm 40° 3 cm 2 cm 5 cm 30° Try This Chapter-7.indd 160 4/11/2025 7:31:56 PM Step 1: Construct a side AB of length 5 cm. Step 2: Construct ∠A = 45° by drawing the other arm of the angle. Step 3: Mark the point C on the other arm such that AC = 4 cm. Step 4: Join BC to get the required triangle. 160 45° A B 5 cm 45° A B 4 cm 5 cm C Figure it Out 1. Construct triangles for the following measurements where the angle is included between the sides: We have seen that triangles do not exist for all sets of sidelengths. Is there a combination of measurements in the case of two sides and the included angle where a triangle is not possible? Justify your answer using what you observe during construction. Two Angles and the Included Side In this case, we are given two angles and the side that is a part of both angles, which we call the included side. Here are some examples of such measurements: Construct a triangle ABC where AB = 5 cm, ∠A = 45° and ∠B = 80°. 40° 50° (a) 3 cm, 75°, 7 cm (b) 6 cm, 25°, 3 cm (c) 3 cm, 120°, 8 cm 5 cm 4 cm 30° 45° A Tale of Three Intersecting Lines 20° 6 cm 50° Math Talk Chapter-7.indd 161 4/11/2025 7:31:56 PM Step 1: Draw the base AB of length 5 cm. Step 2: Draw ∠A and ∠B of measures 45°, and 80° respectively. Step 3: The point of intersection of the two new line segments is the third vertex C. Let us take the given side as the base. 45° 80° A B 5 cm C 45° A B 5 cm 80° 161 C Ganita Prakash | Grade 7 Figure it Out 1. Construct triangles for the following measurements: Do triangles always exist? Do triangles exist for every combination of two angles and their included side? Explore. As in the case when we are given all three sides, it turns out that there is not always a triangle for every combination of two angles and the included side. Find examples of measurements of two angles with the included side where a triangle is not possible. Let us try to visualise such a situation. Once the base is drawn, try to imagine how the other sides should be so that they do not meet. Here are some obvious examples. If the two angles are greater than or equal to a right angle (90°), then it is clear that a triangle is not possible. Now we make one of the base angles an acute angle, say 40°. What are the possible values that the other angle should take so that the lines don’t meet? (a) 75°, 5 cm, 75° (b) 25°, 3 cm, 60° (c) 120°, 6 cm, 30° Math Talk Chapter-7.indd 162 4/11/2025 7:31:56 PM It is clear that if the line from B is “inclined” sufficiently to the right, then it will not meet the line l. 162 (a) Try to find a possible ∠B (marked in the figure) for this to happen. (b) What could be smallest value of ∠B for the lines to not meet? A B 40° l C Visually, it is clear that the line that creates the smallest ∠B has to be the one parallel to l. Let us call this parallel line m. Can you tell the actual value of ∠B be in this case? [Hint: Note that AB is the transversal.] We have seen that when two lines are parallel, the internal angles on the same side of the transversal add up to 180°. So ∠B = 140°. So, for what values of ∠B, does a triangle not exist? Does the length AB play any part here? From the discussion above, it can be seen that the length AB does not play any part in deciding the existence of a triangle. We can say that a triangle does not exist when ∠B is greater than or equal to 140°. Figure it Out 1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category: The blue line is the line with the least rightward bend that doesn’t meet the line ‘l’ (a) 30° (b) 70° (c) 54° A B 40° l C A Tale of Three Intersecting Lines Chapter-7.indd 163 4/11/2025 7:31:56 PM 2. Determine which of the following pairs can be the angles of a triangle and which cannot: Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? Can the sum of the two angles be used for framing this rule? (d) 144° (a) 35°, 150° (b) 70°, 30° (c) 90°, 85° (d) 50°, 150° 163 Ganita Prakash | Grade 7 When the sum of two given angles is less than 180°, a triangle exists with these angles. If the sum is greater than or equal to 180°, there is no triangle with these angles. Let us take two angles, say 60° and 70°, whose sum is less than 180°. Let the included side be 5 cm. What could the measure of the third angle be? Does this measure change if the base length is changed to some other value, say 7 cm? Construct and find out. In general, once the two angles are fixed, does the third angle depend on the included sidelength? Try with different pairs of angles and lengths. The measurements might show that the sidelength has no effect (or a very small effect) on the third angle. With this observation, let us see if we can find the third angle without carrying out the construction and measurement. Try experimenting with different triangles to see if there is a relation between any two angles and the third one. To find this relation, what data will you keep track of and how will you organise the data you collect? Consider a triangle ABC with ∠B= 50° and ∠C= 70°. Let us see how we can find ∠A without construction. A Math Talk Chapter-7.indd 164 4/11/2025 7:31:57 PM We saw that the notion of parallel lines was useful to determine that the sum of any two angles of a triangle is less than 180°. Parallel lines can be used to find the third angle, ∠BAC as well. Let us suppose we construct a line XY parallel to BC through vertex A. We can see new angles being formed here: ∠XAB, and ∠YAC. What are their values? Since the line XY is parallel to BC, ∠XAB = ∠B and ∠YAC = ∠C, because they are alternate angles of the transversals AB and AC. 164 B C 50° 70° B C Fig. 7.6 X Y 50° 70° A Now construct a triangle (taking BC to be of any suitable length) and verify if this is indeed the case. Figure it Out 1. Find the third angle of a triangle (using a parallel line) when two of the angles are: 2. Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out. 3. Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C? Therefore, ∠XAB = 50°, and ∠YAC = 70°. Can we find ∠BAC from this? We know that ∠XAB, ∠YAC and ∠BAC together form 180°. So (a) 36°, 72° (b) 150°, 15° (c) 90°, 30° (d) 75°, 45° ∠XAB + ∠YAC + ∠BAC = 180° 50° + ∠BAC + 70° = 180° 120° + ∠BAC = 180° Thus, ∠BAC = 60° A Tale of Three Intersecting Lines Try This Chapter-7.indd 165 4/11/2025 7:31:57 PM Angle Sum Property What can we say about the sum of the angles of any triangle? Consider a triangle ABC. To find the sum of its angles, we can use the same method of drawing a line parallel to the base: construct a line through A that is parallel to BC. B C 50° A 165 Ganita Prakash | Grade 7 We need to find ∠A + ∠B + ∠C. We know that ∠B = ∠XAB, ∠C = ∠YAC. So, ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC = 180° as together they form a straight angle. Thus we have proved that the sum of the three angles in any triangle is 180°! This rather surprising result is called the angle sum property of triangles. Take some time to reflect upon how we figured out the angle sum property. In the beginning, the relationship between the third angle and the other two angles was not at all clear. However, a simple idea of drawing a line parallel to the base through the top vertex (as in Fig. 7.7) suddenly made the relationship obvious. This ingenious idea can be found in a very influential book in the history of mathematics called ‘The Elements’. This book is attributed to the Greek mathematician Euclid, who lived around 300 BCE. This solution is yet another example of how creative thinking plays a key role in mathematics! There is a convenient way of verifying the angle sum property by folding a triangular cut-out of a paper. Do you see how this shows that the sum of the angles in this triangle is 180°? X A B C Fig. 7.7 Y Chapter-7.indd 166 4/11/2025 7:31:57 PM 166 Exterior Angles The angle formed between the extension of a side of a triangle and the other side is called an exterior angle of the triangle. In this figure, ∠ACD is an exterior angle. Find ∠ACD, if ∠A = 50°, and ∠B = 60°. So, ∠ACB = 70° So, ∠ACD = 180 ° – 70° = 110°, since, ∠ACB and ∠ACD together form a straight angle. Find the exterior angle for different measures of ∠A and ∠B. Do you see any relation between the exterior angle and these two angles? [Hint: From angle sum property, we have ∠A + ∠B + ∠ACB = 180°.] We also have ∠ACD + ∠ACB = 180°, since they form a straight angle. What does this show? From the angle sum property, we know that B C D 50° + 60° + ∠ACB = 180° 110° + ∠ACB = 180° A Exterior Angle A Tale of Three Intersecting Lines Chapter-7.indd 167 4/11/2025 7:31:57 PM 7.4 Constructions Related to Altitudes of Triangles There is another set of useful measurements with respect to a triangle — the height of each of its vertices with respect to the opposite sides. In the world around us, we talk of the heights of various objects: the height of a person, the height of a tree, the height of a building, etc. What do we mean by the word ‘height’? Consider a triangle ABC. What is the height of the vertex A from its opposite side BC, and how can it be measured? Let AD be the line segment from A drawn perpendicular to BC. The length of AD is the height of the vertex A from BC. The line segment AD is said to be one of the ‘altitudes’ of the triangle. The other altitudes B CD Fig. 7.8 A 167 Ganita Prakash | Grade 7 are BE and CF in the figure below: the perpendiculars drawn from the other vertices to their respective opposite sides. Whenever we use the word height of the triangle, we generally refer to the length of the altitude to whatever side we take as base (this altitude is AD in the case of Fig. 7.8). What would the altitude from A to BC be in this triangle? Altitudes Using Paper Folding Cut out a paper triangle. Fix one of the sides as the base. Fold it in such a way that the resulting crease is an altitude from the top vertex to the base. Justify why the crease formed should be perpendicular to the base. A We extend BC and then drop the perpendicular from A to this line. B C B C E F BD C A A Chapter-7.indd 168 4/11/2025 7:31:57 PM Construction of the Altitudes of a Triangle Construct an arbitrary triangle. Label the vertices A, B, C taking BC to be the base. Construct the altitude from A to BC, Constructing the altitude using just a ruler is not accurate. To get a more precise angle of 90°, we use a set square along with a ruler. Can you see how to do this? 168 Step 1: Keep the ruler aligned to the base. Place the set square on the ruler as shown, such that one of the edges of the right angle touches the ruler. Step 2: Slide the set square along the ruler till the vertical edge of the set square touches the vertex A. Step 3: Draw the altitude to BC through A using the vertical edge of the set square. Does there exist a triangle in which a side is also an altitude? Visualise such a triangle and draw a rough diagram. We see that this happens in triangles where one of the angles is a right angle. Triangles having one right angle are called right-angled triangles or simply right triangles. Altitude from A to BC B C B C A Tale of Three Intersecting Lines A A Chapter-7.indd 169 4/11/2025 7:31:58 PM B C A 169 Ganita Prakash | Grade 7 7.5 Types of Triangles In our study of triangles, we have encountered the following types of triangles; equilateral, isosceles, scalene and right-angled triangles. Did you spot any other type of triangle? The classification of triangles as equilateral and isosceles was based on equality of sides. Equilateral triangles have sides of equal length. Isosceles triangles have two sides of equal length. Scalene triangles have sides of three different lengths. Can a similar classification be done based on equality of angles? Is there any relation between these two classifications? We will answer these questions in a later chapter. We used angle measures when classifying a triangle as a rightangled triangle. What are the other types of triangles based on angle measures? Equilateral Triangle Isosceles Triangle Scalene Triangle Equilateral Triangle Isosceles Triangle Scalene Triangle Chapter-7.indd 170 4/11/2025 7:31:58 PM A classification of triangles based on their angle measures is acuteangled, right-angled and obtuse-angled triangles. We have already seen what a right-angled triangle is. It is a triangle with one right angle. Similarly, an obtuse-angled triangle has one obtuse angle. What could an acute-angled triangle be? Can we define it as a triangle with one acute angle? Why not? In an acute-angled triangle, all three angles are acute angles. Figure it Out 1. Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC. 2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY. 170 Math Talk 3. Construct a right-angled triangle ∆ABC with ∠B = 90°, AC = 5 cm. How many different triangles exist with these measurements? 4. Through construction, explore if it is possible to construct an equilateral triangle that is (i) right-angled (ii) obtuse-angled. Also construct an isosceles triangle that is (i) right-angled (ii) obtuse-angled. • Use of a compass simplifies the construction of triangles when the sidelengths are given. • A set of three lengths where the length of each is smaller than the sum of the other two is said to satisfy the triangle inequality. • Sidelengths of a triangle satisfy triangle inequality, and, if a given set of lengths satisfy the triangle inequality, a triangle can be constructed with those sidelengths. • Triangles can be constructed when the following measurements are given: (a) two of the sides and their included angle (b) two angles and the included side • The sum of the angles of a triangle is always 180°. • An altitude of a triangle is a perpendicular line segment from a vertex to its opposite side. • Equilateral triangles have sides of equal length. Isosceles triangles have two sides of equal length. Scalene triangles have sides of three different lengths. • Triangles are classified based on their angle measures as acute-angled, right-angled and obtuse-angled triangles. [Hint: Note that the other measurements can take any values. Take AC as the base. What values can ∠A and ∠C take so that the other angle is 90°?] SUMMARY A Tale of Three Intersecting Lines Try This Chapter-7.indd 171 4/11/2025 7:31:58 PM 171 Ganita Prakash | Grade 7 There is a spider in a corner of a box. It wants to reach the farthest opposite corner (marked in the figure). Since it cannot fly, it can reach the opposite point only by walking on the surfaces of the box. What is the shortest path it can take? Take a cardboard box and mark the path that you think is the shortest from one corner to its opposite corner. Compare the length of this path with that of the paths made by your friends. Hint: Shortest Path in a Box! Chapter-7.indd 172 4/11/2025 7:32:01 PM 172 Page No. 150 Ans: One Such diagram is Try more Page No. 151 1. Use the points on the circle and/or the centre to form isosceles triangles. 2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size. Figure it out A Tale of Three Intersecting Lines Chapter – 7 Ans: Take points C, C1, C2, C3 … on the circle with centre B. The triangles ABC, AC1B, AC2B, … are isosceles triangle. Similarly for points D1, D2, D3 … ∆ADB, ∆AD1B, ∆AD2B, … are isosceles triangles. ∆ABC and ∆ADB are equilateral triangle. Similarly it can be done for the other figure. 1 Page No. 154 1. We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle. Ans: Here sum of the two sides (i.e 3 cm and 4 cm) is less than the third side (i.e 8 cm). Similarly for sides of lengths 2cm, 3cm and 6cm, 2 + 3 < 6. So, in both cases triangles are not possible. Ans: 3. Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore for different sets of lengths. Ans: Some sets are 7, 10, 15; 12, 14, 18 2. Can we say anything about the existence of a triangle for each of the following sets of lengths? Figure it out (a) 10 km, 10 km and 25 km Ans: A triangle does not exist (b) 5 mm, 10 mm and 20 mm Ans: A triangle does not exist (c) 12 cm, 20 cm and 40 cm Ans: A triangle does not exist Page No. 156 1. Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure. Ans: Figure it out (a) 2, 2, 5 Ans: Since 2 + 2 < 5. So, these are not the side lengths of a triangle. 2 Page No. 159 (b) 3, 4, 6 Ans: Since 3 < 4 + 6, 4 < 6 + 3 and 6 < 3 + 4. These are the side lengths of a triangle. (c) 2, 4, 8 Ans: 8 > 2 + 4. (d) 5, 5, 8 Ans: Since 5 < 5 + 8, 8 < 5 + 5. (e) 10, 20, 25 Ans: Since 10 < 20 + 25, 20 < 25 + 10 and 25 < 10 + 20. (f) 10, 20, 35 Ans: 35 > 10 + 20. (g) 24, 26, 28 Ans: Since 24 < 26 + 28, 26 < 28 + 24 and 28 < 24 + 26. These are the side lengths of a triangle. Figure it out So, these are not the side lengths of a triangle. These are the side lengths of a triangle. These are the side lengths of a triangle. So, these are not the side lengths of a triangle. 1. Check if a triangle exists for each of the following set of lengths: (a) 1, 100, 100 Ans: A triangle exists. (b) 3, 6, 9 Ans: A triangle cannot exists. (c) 1, 1, 5 Ans: A triangle cannot be constructed. (d) 5, 10, 12 Ans: A triangle exists. 3 Page No. 161 Ans: There exists an equilateral triangle with sides 50, 50, 50 because each length < sum of the other two lengths. There exists an equilateral 2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? 3. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal We have seen that triangles do not exist for all sets of sidelengths. Is there a combination of measurements in the case of two sides and the (a) 1, 100 Ans: 99.5, 100, 100.7, 100.5, 99.4. (b) 5, 5 Ans: 5, 4, 3, 4.9, 1 (c) 3, 7 Ans: 5, 8, 7, 6.4, 6 Justify your answer. triangle of any side length. values could also be chosen): Try some more lengths. Ans: If the included angle is greater than or equal to 180°, then a triangle is not possible. included angle where a triangle is not possible? Justify your answer using what you observe during construction. (a) 4cm, 210°, 6cm (b) 3cm, 180°, 7cm For example– 4 Page No. 163 144° 30° (b) 70° 54° (a) (c) (d) 1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different 2. Determine which of the following pairs can be the angles of a triangle and which cannot: Figure it out angles for each category: Triangle is possible Triangle is not possible Another angle should be less than 150°. It could be 50° or 90°. Find the third angle in each case. Another angle should be less than 110°. It could be 60° or 80°. Find the third angle in each case. Another angle should be less than 126°. It could be 90° or 64°. Find the third angle in each case. Another angle should be less than 36°. It could be 20° or 35°. Find the third angle in each case. Another angle should be greater than or equal to 150°. It could be 150° or 170°. Another angle should be greater than or equal to 110°. It could be 140° or 120°. Another angle should be greater than or equal to 126°. It could be 134° or 154°. Another angle should be greater than or equal to 36°. It could be 36° or 90°. (a) 35°, 150° Ans: This pair cannot form angles of a triangle. (b) 70°, 30° Ans: This pair of angles can be the angles of a triangle. (c) 90°, 85° Ans: This pair of angles can be the angles of a triangle. (d) 50°, 150° Ans: This pair cannot form angles of a triangle. 5 Page No. 165 Ans: Rules for two angles A and B to form a triangle is 0° < A + B < 180°. 1. Find the third angle of a triangle (using a parallel line) when two of Ans:Consider ∆ABC, Draw line XY parallel to line BC so AB and AC become Figure it out the angles are: Process for (a) is given. Try for rest. (a) 36°, 72° Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? On line XY — transversals. Therefore XAB = B = 36° and YAC = C = 72° (Alternate Angles) (b) 150°, 15° Ans: 15° (c) 90°, 30° Ans: 60° (d) 75°, 45° Ans: 60° XAB + BAC + YAC = 180° So, BAC = 72° 6 2. Can you construct a triangle all of whose angles are equal to 70°? If 3. Here is a triangle in which we know, B = C and A = 50°. Can you find B and C? Ans: Ans: B = C = 65° two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out. • No, such triangle is not possible. • 40° • 60° 7" class_7,8,Working with Fractions,ncert_books/class_7/gegp1dd/gegp108.pdf,"8 8.1 Multiplication of Fractions Aaron walks 3 kilometres in 1 hour. How far can he walk in 5 hours? This is a simple question. We know that to find the distance, we need to find the product of 5 and 3, i.e., we multiply 5 and 3. Distance covered in 1 hour = 3 km. Therefore, Distance covered in 5 hours = 3 + 3 + 3 + 3 + 3 km WORKING WITH FRACTIONS = 5 × 3 km = 15 km. Chapter-8.indd 173 4/12/2025 12:41:31 PM Aaron’s pet tortoise walks at a much slower pace. It can walk only 1 4 kilometre in 1 hour. How far can it walk in 3 hours? Here, the distance covered in an hour is a fraction. This does not matter. The total distance covered is calculated in the same way, as multiplication. Distance covered in 1 hour = 1 4 km. 1 4 km 0 Distance in 1 hour Distance in 3 hour Ganita Prakash | Grade 7 Let us consider a case where the time spent walking is a fraction of an hour. We saw that Aaron can walk 3 kilometres in 1 hour. How far can he walk in 1 5 hours? We continue to calculate the total distance covered through multiplication. Finding the product: Distance in 1 5 hour 0 3 km Therefore, distance covered in 3 hours = 3 × 1 4 km = 1 4 + 1 4 + 1 4 km = 3 4 km. The tortoise can walk 3 4 km in 3 hours. Distance covered in 1 5 hours = 1 5 × 3 km. Distance in 1 hour Chapter-8.indd 174 4/12/2025 12:41:31 PM Distance covered in 1 hour = 3 km. 3 km into 5 equal parts, which is 3 5 km. How far can Aaron walk in 2 5 hours? 174 In 1 5 hours, distance covered is equal to the length we get by dividing This tells us that 1 5 × 3 = 3 5  . Finding the product: 1. We can first find the distance covered in 1 5 hours. 2. Since, the duration 2 5 is twice 1 5 , we multiply this distance by 2 to Here is the calculation. Distance covered in 1 hour = 3 km. 1. Distance covered in 1 5 hour 2. Multiplying this distance by 2, we get Once again, we have — Distance covered = 2 5 × 3 km. = The length we get by dividing 3 km in 5 equal parts = 3 5 km. get the total distance covered. Distance in 1 5 hour Distance in 2 5 hour Distance in 1 hour 2 × 3 5 = 6 5 km. Working with Fractions Chapter-8.indd 175 4/12/2025 12:41:31 PM From this we can see that Discussion We did this multiplication as follows: • First, we divided the multiplicand, 3 , by the denominator of the multiplier, 5, to get 3 5 . 2 5 × 3 = 6 5 . Multiplier Multiplicand 2 5 × 3 175 Ganita Prakash | Grade 7 Thus, whenever we need to multiply a fraction and a whole number, we follow the steps above. Example 1: A farmer had 5 grandchildren. She distributed 2 3 acre of land to each of her grandchildren. How much land in all did she give to her grandchildren? Example 2: 1 hour of internet time costs ₹8. How much will 1 1 4 hours of internet time cost? 1 1 4 hours is 5 4 hours (converting from a mixed fraction). Cost of 5 4 hour of internet time = 5 4 × 8 It costs ₹10 for 1 1 4 hours of internet time. • We then multiplied the result by the numerator of the multiplier, that is 2, to get 6 5 . 5 × 2 3 = 2 3 + 2 3 + 2 3 + 2 3 + 2 3 = 10 3 . = 5 × 8 4 = 5 × 2 = 10. Multiplier Multiplicand 2 5 2 5 2 5 × 3 2 × 3 5 = 6 5 3 ÷ 5 = 3 5 Chapter-8.indd 176 4/12/2025 12:41:31 PM Figure it Out 1. Tenzin drinks 1 2 glass of milk every day. How many glasses of milk 2. A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make ___ km of the water canal. If they work 5 days a week, they can make ___ km of the water canal in a week. 3. Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks? 4. Safia saw the Moon setting on Monday at 10 pm. Her mother, who is 176 does he drink in a week? How many glasses of milk did he drink in the month of January? a scientist, told her that every day the Moon sets 5 6 hour later than 5. Multiply and then convert it into a mixed fraction: (d) 13 11 × 6 So far, we have learnt multiplication of a whole number with a fraction, and a fraction with a whole number. What happens when both numbers in the multiplication are fractions? Multiplying Two Fractions We know, that Aaron’s pet tortoise can walk only 1 4 km in 1 hour. How far can it walk in half an hour? Following our approach of using multiplication to solve such problems, we have, Finding the product: Distance covered in 1 hour = 1 4 km. Distance covered in 1 2 hour = 1 2 × 1 4 km. the previous day. How many hours after 10 pm will the moon set on Thursday? (a) 7 × 3 5 (b) 4 × 1 3 (c) 9 7 × 6 Hour Distance Working with Fractions 1 1 4 1 2 ? Chapter-8.indd 177 4/12/2025 12:41:32 PM dividing 1 4 into 2 equal parts. To find this, it is useful to represent fractions using the unit square to stand for a “whole”. Therefore, the distance covered in 1 2 an hour is the length we get by Unit square as a “whole” 1 4 of the whole 1 4 177 ? 1 4 divided into 2 equal parts Ganita Prakash | Grade 7 If the tortoise walks faster and it can cover 2 5 km in 1 hour, how far will it walk in 3 4 of an hour? Distance covered = 3 4 × 2 5 . Finding the product: (i) First find the distance covered in 1 4 of an hour. (ii) Multiply the result by 3, to get the distance covered in 3 4 of an (i) Distance in km covered in 1 4 of an hour Now we divide this 1 4 into 2 equal parts. What do we get? What fraction of the whole is shaded? hour. = The quantity we get by dividing 2 5 into 4 equal parts. and one of the parts is shaded, we can say that 1 8 of the whole is shaded. So, the distance covered by the tortoise in half an hour is 1 8 km. Since the whole is divided into 8 equal parts This tells us that 1 2 × 1 4 = 1 8 . whole Chapter-8.indd 178 4/12/2025 12:41:32 PM shaded part (in Fig. 8.1) is a region we get when we divide 2 5 into 4 equal parts. 178 Taking the unit square as the whole, the How much of the whole is it? The whole is divided into 5 rows and 4 columns, creating 5 × 4 = 20 equal parts. Number of these parts shaded = 2. So, the distance covered in 1 4 of an hour = 2 20 . Fig. 8.1 2 5 (ii) Now, we need to multiply 2 20 by 3. Distance covered in 3 4 of an hour = 3 × 2 20 Discussion In the case of a fraction multiplied by another fraction, we follow a method similar to the one we used, when we multiplied a fraction by a whole number. We multiplied as follows: Using this understanding, multiply 5 4 × 3 2 . the whole. Since, the fraction 3 2 is one whole and a half, it can be seen as follows: = 6 20 . So, 3 4 × 2 5 = 6 20 = 3 10. First, let us represent 3 2 , taking the unit square as 2 5 ÷ 4 = 2 20 Divide the multiplicand by 4. 3 4 3 × 2 20 = 6 20 =  Divide the multiplicand by 3. 3 4 Multiplier Multiplicand Working with Fractions 3 4 × 2 5 whole Math Talk Chapter-8.indd 179 4/12/2025 12:41:33 PM first divide this fraction 3 2 into 4 equal parts. It can be done as shown in the Fig. 8.2 with the yellow shaded region representing the fraction obtained by dividing 3 2 into 4 equal parts. What is its value? Following the steps of multiplication, we need to We see that the whole is divided into — 2 rows and 4 columns, creating 2 × 4 = 8 equal parts. Number of parts shaded = 3. So the yellow shaded part = 3 8 . Fig. 8.2 179 3 2 Ganita Prakash | Grade 7 product of 5 4 and 3 2 : Connection between the Area of a Rectangle and Fraction Multiplication In the Fig. 8.3, what is the length and breadth of the shaded rectangle? Since we started with a unit square (of side 1 unit), the length and breadth are 1 2 unit and 1 4 unit. What is the area of this rectangle? We see that 8 such rectangles give the square of area 1 square unit. So, the area of each rectangle is 1 8 square units. Do you see any relation between the area and the product of length and breadth? The area of a rectangle of fractional sides equals the product of its sides. In general, if we want to find the product of two fractions, we can find the area of the rectangle formed with the two fractions as its sides. Figure it Out 1. Find the following products. Use a unit square as a whole for representing the fractions: Now, the next step is multiplying this result by 5. This gives the 5 4 × 3 2 = 5 × 3 8 = 15 8 . 1 8 Fig. 8.3 Chapter-8.indd 180 4/12/2025 12:41:33 PM 180 Now, find 1 12 × 1 18. (a) 1 3 × 1 5 (b) 1 4 × 1 3 (c) 1 5 × 1 2 (d) 1 6 × 1 5 Doing this by representing the fractions using a unit square is cumbersome. Let us find the product by observing what we did in the above cases. In each case, the whole is divided into rows and columns. The number of rows is the denominator of the multiplicand, which is 18 in this case. The number of columns is the denominator of the multiplier, which is 12 in this case. Thus, the whole is divided into 18 × 12 equal parts. multiplied, their product is 2. Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations. So, 1 18 × 1 12 = 1 (18×12) = 1 216. Thus, when two fractional units are We express this as: (a) 2 3 × 4 5 (b) 1 4 × 2 3 (c) 3 5 × 1 2 (d) 4 6 × 3 5 1 (product of denominators). 1 b × 1 d = 1 b × d  . Working with Fractions 12 parts 18 parts 1 18 Chapter-8.indd 181 4/12/2025 12:41:33 PM Multiplying Numerators and Denominators Now, find 5 12 × 7 18. Like the previous case, let us find the product by performing the multiplication, step by step. First, the whole is divided into 18 rows and 12 columns creating 12 × 18 equal parts. The value we get by dividing 7 18 into 12 equal parts is 7 (12 × 18) . 12 parts 18 parts 181 7 18 Ganita Prakash | Grade 7 the product. This is (5 × 7) (12 × 18). This formula was first stated in this general form by Brahmagupta in his Brāhmasphuṭasiddhānta in 628 CE. The formula above works even when the multiplier or multiplicand is a whole number. We can simply rewrite the whole number as a fraction with denominator 1. For example, Multiplication of Fractions—Simplifying to Lowest Form Then, we multiply this result by 5 to get So, 5 12 × 7 8 = (5 × 7) (12 × 18) = 35 216. From this we can see that, in general, a b × c d = a × c b × d  . 3 × 3 4 can be written 3 1 × 3 4 3 5 × 4 can be written 3 5 × 4 1 = 3 × 4 5 × 1 = 12 5 . = 3×3 1×4 = 9 4 . And, 5 12 parts 18 parts 7 18 Chapter-8.indd 182 4/12/2025 12:41:33 PM Multiply the following fractions and express the product in its lowest form: Instead of multiplying the numerators (12 and 5) and denominators (7 and 24) first and then simplifying, we could do the following: We see that both the circled numbers have a common factor of 12. We know that a fraction remains the same when the numerator and denominator are divided by the common factor. In this case, we can divide them by 12. 182 12 7 × 5 24 = 12 × 5 7 × 24 12 7 × 5 24 When multiplying fractions, we can first divide the numerator and denominator by their common factors before multiplying the numerators and denominators. This is called cancelling the common factors. A Pinch of History In India, the process of reducing a fraction to its lowest terms — known as apavartana — is so well known that it finds mention even in a non-mathematical work. A Jaina scholar Umasvati (c. 150 CE) used it as a simile in a philosophical work. Figure it Out 1. A water tank is filled from a tap. If the tap is open for 1 hour, 7 10 of the tank gets filled. How much of the tank is filled if the tap is open for Let us use the same technique to do one more multiplication. 14 × 25 15 × 42 = 1 × 5 3 × 3 = 5 9 . 1 12 × 5 7 × 24 = 1 × 5 7 × 2 = 5 14 1 3 3 5 2 14 15 × 25 42 . Working with Fractions Chapter-8.indd 183 4/12/2025 12:41:33 PM 2. The government has taken 1 6 of Somu’s land to build a road. What part of the land remains with Somu now? She gives half of Try This (a) 1 3 hour ____________ (b) 2 3 hour ____________ (c) 3 4 hour ____________ (d) 7 10 hour ____________ (e) For the tank to be full, how long should the tap be running? 183 Ganita Prakash | Grade 7 3. Find the area of a rectangle of sides 3 3 4 ft and 9 3 5 ft. 4. Tsewang plants four saplings in a row in his garden. The distance 5. Which is heavier: 12 15 of 500 grams or 3 20 of 4 kg? Is the Product Always Greater than the Numbers Multiplied? Since, we know that when a number is multiplied by 1, the product remains unchanged, we will look at multiplying pairs of numbers where neither of them is 1. When we multiply two counting numbers greater than 1, say 3 and 5, the product is greater than both the numbers being multiplied. between two saplings is 3 4 m. Find the distance between the first and last sapling. [Hint: Draw a rough diagram with four saplings with distance between two saplings as 3 4 m] the remaining part of the land to her daughter Krishna and 1 3 of it to her son Bora. After giving them their shares, she keeps the remaining land for herself. (a) What part of the original land did Krishna get? (b) What part of the original land did Bora get? (c) What part of the original land did Somu keep for herself? 3 × 5 = 15 Chapter-8.indd 184 4/12/2025 12:41:33 PM than 8. 184 The product, 15, is more than both 3 and 5. But what happens when we multiply 1 4 and 8? 1 4 × 8 = 2 In the above multiplication the product, 2, is greater than 1 4 , but less What happens when we multiply 3 4 and 2 5  ? Let us compare this product 6 20 with the numbers 3 4 and 2 5 . For this, 3 4 × 2 5 = 6 20 let us express 3 4 as 15 20 and 2 5 as 8 20. From this we can see that the product is less than both the numbers. When do you think the product is greater than both the numbers multiplied, when is it in between the two numbers, and when is it smaller than both? [Hint: The relationship between the product and the numbers multiplied depends on whether they are between 0 and 1 or they are greater than 1. Take different pairs of numbers and observe their product. For each multiplication, consider the following questions.] Situation 1 Both numbers are greater than 1 Situation 2 Both numbers are between 0 and 1 Situation 3 One number is between 0 and 1, and one number is greater than 1 Situation Multiplication Relationship (e.g., 4 3 × 4) (e.g., 3 4 × 2 5 ) (e.g., 3 4 × 5) The product (16 3 ) is greater than both the numbers The product (15 4 ) is less than the number greater than 1 and greater than the number between 0 and 1 The product ( 3 10) is less than both the numbers Working with Fractions Chapter-8.indd 185 4/12/2025 12:41:33 PM Create more such examples for each situation and observe the relationship between the product and the numbers being multiplied. What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks: • When one of the numbers being multiplied is between 0 and 1, the product is ____________ (greater/less) than the other number. • When one of the numbers being multiplied is greater than 1, the product is _____________ (greater/less) than the other number. 185 Ganita Prakash | Grade 7 Order of Multiplication We know that 1 2 × 1 4 = 1 8 . In general, note that the area of a rectangle remains the same even if the length and breadth are interchanged. The order of multiplication does not matter. Thus, 1 2 1 8 1 4 Now, what is 1 4 × 1 2  ? That is 1 8 too. a b × c d = c d × a b . 1 4 1 8 1 2 Chapter-8.indd 186 4/12/2025 12:41:34 PM This can also be seen from Brahmagupta’s formula for multiplying fractions. 8.2 Division of Fractions What is 12 ÷ 4? You know this already. But can this problem be restated as a multiplication problem? What should be multiplied by 4 to get 12? That is, 186 4 × ? = 12 Dividend Divisor Quotient 12 ÷ 4 = 3 We can use this technique of converting division into multiplication problems to divide fractions. What is 1 ÷ 2 3  ? Let us rewrite this as a multiplication problem What should be multiplied by 2 3 to get the product 1? If we somehow cancel out the 2 and the 3, we are left with 1. So, Let us try another problem: This is the same as 2 3 × ? = 3. Can you find the answer? We know what to multiply 2 3 by to get 1. We just need to multiply that 2 3 × 3 2 = 1 2 3 × ? = 1 Answer 1 ÷ 2 3 = 3 2 . 3 ÷ 2 3 . Working with Fractions Chapter-8.indd 187 4/12/2025 12:41:34 PM by 3 to get 3. So, So, Rewriting it as a multiplication problem, we have 3 ÷ 2 3 = 3 2 × 3 = 9 2 . What is 1 5 ÷ 1 2 ? 2 3 × 3 2 × 3 = 3 1 2 × ? = 1 5 . Answer 187 Ganita Prakash | Grade 7 Discussion In each of the division problems above, observe how we found the answer. Can we frame a rule that tells us how to divide two fractions? Let us consider the previous problem. In every division problem we have a dividend, divisor and quotient. The technique we have been using to get the quotient is: 1. First, find the number which gives 1 when multiplied by the divisor. We see that the resulting number is a fraction whose numerator is the divisor’s denominator and denominator is the divisor’s numerator. How do we solve this? So, What is 2 3 ÷ 3 5 ? Rewriting this as multiplication, we have How will we solve this? So, 2 3 ÷ 3 5 = 5 3 × 2 3 = 10 9 . 1 5 ÷ 1 2 = 2 × 1 5 = 2 5 . 1 2 × 2 × 1 5 = 1 5 3 5 × 5 3 × 2 3 = 2 3 3 5 × ? = 2 3 . Answer Answer Chapter-8.indd 188 4/12/2025 12:41:34 PM 188 For the divisor 3 5 this fraction is 5 3 . We call 5 3 the reciprocal of 3 5 . When we multiply a fraction by its reciprocal, we get 1. So, the first step in our technique is to find the divisor’s reciprocal. Dividend = 5 3 × 2 3 = 10 9 2 3 ÷ 3 5 Divisor Quotient 2. We then multiply the dividend with this reciprocal to get the quotient. So, a b ÷ c d = d c × a b = d × a c × b. As with methods and formulas for addition, subtraction, and multiplication of fractions that you learnt earlier, this method and formula for division of fractions, in this general form, was first explicitly stated by Brahmagupta in his Brāhmasphuṭasiddhānta (628 CE). So, to evaluate, for example, 2 3 ÷ 3 5 using Brahmagupta’s formula above, we write: Dividend, Divisor and the Quotient When we divide two whole numbers, say 6 ÷ 3, we get the quotient 2. Here the quotient is less than the dividend. This can be rewritten as: 2 3 ÷ 3 5 = 2 3 × 5 3 = 2 × 5 3 × 3 = 10 9 . • Find the reciprocal of the divisor • Multiply this by the dividend to get the quotient. Summarising, to divide two fractions: a b ÷ c d = a b × d c = a × d b × c . 6 ÷ 3 = 2, 2 < 6 Working with Fractions Chapter-8.indd 189 4/12/2025 12:41:34 PM But what happens when we divide 6 by 1 4 ? Here the quotient is greater than the dividend! What happens when we divide 1 8 by 1 4 ? Here too the quotient is greater than the dividend. When do you think the quotient is less than the dividend and when is it greater than the dividend? Is there a similar relationship between the divisor and the quotient? 6 ÷ 1 4 = 24. 1 8 ÷ 1 4 = 1 2 . 189 Ganita Prakash | Grade 7 Use your understanding of such relationships in multiplication to answer the questions above. 8.3 Some Problems Involving Fractions Example 3: Leena made 5 cups of tea. She used 1 4 litre of milk for this. How much milk is there in each cup of tea? volume of milk should be: 1 4 ÷ 5. Writing this as multiplication, we have: We perform the division as follows as per Brahmagupta’s method: The reciprocal of 5 (the divisor) is 1 5 . Leena used 1 4 litres of milk in 5 cups of tea. So, in 1 cup of tea the Milk in 1 cup 5 × (milk per cup) = 1 4 . 1 4 litres of 5 cups Chapter-8.indd 190 4/12/2025 12:41:34 PM Multiplying this reciprocal by the dividend ( 1 4 ), we get Example 4: Some of the oldest examples of working with non-unit fractions occur in humanity’s oldest geometry texts, the Śhulbasūtra. Here is an example from Baudhāyana’s Śhulbasūtra (c. 800 BCE). sides is 1 5 units. 190 So, each cup of tea has 1 20 litre of milk. Cover an area of 7 1 2 square units with square bricks each of whose 1 5 × 1 4 = 1 20. How many such square bricks are needed? Number of bricks = 15 2 ÷ 1 25. The reciprocal of the divisor is 25. Multiplying the reciprocal by the dividend, we get Example 5: This problem was posed by Chaturveda Pṛithūdakasvāmī (c. 860 CE) in his commentary on Brahmagupta’s book Brāhmasphuṭasiddhānta. Four fountains fill a cistern. The first fountain can fill the cistern in a day. The second can fill it in half a day. The third can fill it in a quarter of a day. The fourth can fill the cistern in one fifth of a day. If they all flow together, in how much time will they fill the cistern? Let us solve this problem step by step. In a day, the number of times — • the first fountain will fill the cistern is 1÷ 1 = 1 • the second fountain will fill the cistern is 1 ÷ 1 2 = _____ • the third fountain will fill the cistern is 1 ÷ 1 4 = _____ The total area to be covered is 7 1 2 sq. units = 15 2 sq. units. Each square brick has an area of 1 5 × 1 5 = 1 25 square units. As (Number of bricks) × (Area of a brick) = Total Area, 25 × 15 2 = 25 × 15 2 = 375 2 . Working with Fractions Chapter-8.indd 191 4/12/2025 12:41:34 PM The number of times the four fountains together will fill the cistern in a day is ___ + ____ + ___ + ____ = 12. together is 1 12 days. • the fourth fountain will fill the cistern is 1 ÷ 1 5 = _____ Thus, the total time needed by the four fountains to fill the cistern 191 Ganita Prakash | Grade 7 Fractional Relations Here is a square with some lines drawn inside. What fraction of area of the whole square does the shaded region occupy? area of the whole square. There are different ways to solve this problem. Here is one of them: Let the area of the whole square be 1 square unit. We can see that the top right square (in Fig. 8.5), occupies 1 4 of the Fig. 8.4 Math Talk Chapter-8.indd 192 4/12/2025 12:41:35 PM 192 Area of red square = 1 4 square units. Fig. 8.5 Fig, 8.6 Let us look at this red square. The area of the triangle inside it (coloured yellow) is half the area of the red square. So, the area of the yellow triangle = 1 2 × 1 4 = 1 8 square units. What fraction of this yellow triangle is shaded? yellow triangle. Are you able to see why? The area of shaded part = 3 4 × 1 8 = 3 32 square units. Thus, the shaded region occupies 3 32 of the area of the whole square. In each of the figures given below, find the fraction of the big square that the shaded region occupies. We will solve more interesting problems of this kind in a later chapter. The shaded region occupies 3 4 of the area of the Working with Fractions Chapter-8.indd 193 4/12/2025 12:41:36 PM A Dramma-tic Donation The following problem is translated from Bhāskarāchārya’s (Bhāskara II’s) book, Līlāvatī, written in 1150 CE. a dramma. If you know the mathematics of fractions well, tell me O child, how many cowrie shells were given by the miser to the beggar.” Dramma refers to a silver coin used in those times. The tale says that 1 dramma was equivalent to 1280 cowrie shells. Let’s see what fraction of a dramma the person gave: “O wise one! A miser gave to a beggar 1 5 of 1 16 of 1 4 of 1 2 of 2 3 of 3 4 of (1 2 × 2 3 × 3 4 × 1 5 × 1 16 × 1 4) th part of a dramma. 193 Ganita Prakash | Grade 7 Evaluating it gives 6 7680 . Upon simplifying to its lowest form, we get 6 7680 = 1 1280. So, one cowrie shell was given to the beggar. You can see in the answer Bhāskarāchārya’s humour! The miser had given the beggar only one coin of the least value (cowrie). Around the 12th century, several types of coins were in use in different kingdoms of the Indian subcontinent. Most commonly used were gold coins (called dinars/gadyanas and hunas), silver coins (called drammas/tankas), copper coins (called kasus/panas and mashakas), and cowrie shells. The exact conversion rates between these coins varied depending on the region, time period, economic conditions, weights of coins and their purity. Gold coins had high-value and were used in large transactions and to store wealth. Silver coins were more commonly used in everyday transactions. Copper coins had low-value and were used in smaller transactions. Cowrie shells were the lowest denomination and were used in very small transactions and as change. If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 copper panas, 1 copper pana = 6 mashakas, and 1 pana = 30 cowrie shells, A Pinch of History 1 copper pana = 1 48 gold dinar ( 1 12 × 1 4) 1 cowrie shell =____ copper panas 1 cowrie shell =____ gold dinar. Chapter-8.indd 194 4/12/2025 12:41:36 PM As you have seen, fractions are an important type of number, playing a critical role in a variety of everyday problems that involve sharing and dividing quantities equally. The general notion of non-unit fractions as we use them today — equipped with the arithmetic operations of addition, subtraction, multiplication, and division — developed largely in India. The ancient Indian geometry texts called the Śhulbasūtra — which go back as far as 800 BCE, and were concerned with the construction of fire altars for rituals — used general non-unit fractions extensively, including performing division of such fractions as we saw in Example 3. Fractions even became commonplace in the popular culture of India as far back as 150 BCE, as evidenced by an offhand reference to the reduction of fractions to lowest terms in the philosophical work of the revered Jain scholar Umasvati. 194 General rules for performing arithmetic operations on fractions — in essentially the modern form in which we carry them out today — were first codified by Brahmagupta in his Brāhmasphuṭasiddhānta in 628 CE. We have already seen his methods for adding and subtracting general fractions. For multiplying general fractions, Brahmagupta wrote: “Multiplication of two or more fractions is obtained by taking the product of the numerators divided by the product of the denominators.” (Brāhmasphuṭasiddhānta, Verse 12.1.3) That is, a b × c d = a × c b × d . For division of general fractions, Brahmagupta wrote: “The division of fractions is performed by interchanging the numerator and denominator of the divisor; the numerator of the dividend is then multiplied by the (new) numerator, and the denominator by the (new) denominator.” Bhāskara II in his book Līlāvatī in 1150 CE clarifies Brahmagupta‘s statement further in terms of the notion of reciprocal: “Division of one fraction by another is equivalent to multiplication of the first fraction by the reciprocal of the second.” (Līlāvatī, Verse 2.3.40) Both of these verses are equivalent to the formula: Bhāskara I, in his 629 CE commentary Āryabhaṭīyabhāṣhya on Aryabhata’s 499 CE work, described the geometric interpretation of multiplication of fractions (that we saw earlier) in terms of the division of a square into rectangles via equal divisions along the length and breadth. Many other Indian mathematicians, such as Śhrīdharāchārya (c. 750 CE), Mahāvīrāchārya (c. 850 CE), Caturveda Pṛithūdakasvāmī (c. 860 CE), and Bhāskara II (c. 1150 CE) developed the usage of arithmetic of fractions significantly further. The Indian theory of fractions and arithmetic operations on them was transmitted to, and its usage developed further, by Arab and African mathematicians such as al-Hassâr (c. 1192 CE) of Morocco. The theory was then transmitted to Europe via the Arabs over the next few a b ÷ c d = a b × d c = a × d b × c . Working with Fractions Chapter-8.indd 195 4/12/2025 12:41:36 PM Bhāskara I’s visual explanation that 1 5 x 1 4 = 1 20 1 4 1 20 1 5 195 Ganita Prakash | Grade 7 centuries, and came into general use in Europe in only around the 17th century, after which it spread worldwide. The theory is indeed indispensable today in modern mathematics. Figure it Out 1. Evaluate the following: 2. For each of the questions below, choose the expression that describes the solution. Then simplify it. (a) Maria bought 8 m of lace to decorate the bags she made for school. She used 1 4 m for each bag and finished the lace. How many bags did she decorate? 3 ÷ 7 9 14 4 ÷ 2 2 3 ÷ 2 3 14 6 ÷ 7 3 4 3 ÷ 3 4 7 4 ÷ 1 7 8 2 ÷ 4 15 1 5 ÷ 1 9 1 6 ÷ 11 12 3 2 3 ÷ 1 3 8 (i) 8 × 1 4 (ii) 1 8 × 1 4 (iii) 8 ÷ 1 4 (iv) 1 4 ÷ 8 Chapter-8.indd 196 4/12/2025 12:41:36 PM 196 (b) 1 2 meter of ribbon is used to make 8 badges. What is the (c) A baker needs 1 6 kg of flour to make one loaf of bread. He has length of the ribbon used for each badge? 5 kg of flour. How many loaves of bread can he make? (i) 8 × 1 2 (ii) 1 2 ÷ 1 8 (iii) 8 ÷ 1 2 (iv) 1 2 ÷ 8 (i) 5 × 1 6 (ii) 1 6 ÷ 5 (iii) 5 ÷ 1 6 (iv) 5 × 6 3. If 1 4 kg of flour is used to make 12 rotis, how much flour is used to 4. Pāṭīgaṇita, a book written by Sridharacharya in the 9th century 5. Mira is reading a novel that has 400 pages. She read 1 5 of the pages 6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 3 4 litres of petrol? 7. Amritpal decides on a destination for his vacation. If he takes a 8. Mariam’s grandmother baked a cake. Mariam and her cousins 9. Choose the option(s) describing the product of ( 565 465 × 707 676 ): train, it will take him 5 1 6 hours to get there. If he takes a plane, it make 6 rotis? CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1 ÷ 1 6 , 1 ÷ 1 10 , 1 ÷ 1 13 , 1 ÷ 1 9 , and 1 ÷ 1 2 ” . What should the friend say? yesterday and 3 10 of the pages today. How many more pages does she need to read to finish the novel? will take him 1 2 hour. How many hours does the plane save? finished 4 5 of the cake. The remaining cake was shared equally by Mariam’s three friends. How much of the cake did each friend get? (a) > 565 465 (b) < 565 465 Working with Fractions Chapter-8.indd 197 4/12/2025 12:41:37 PM 10. What fraction of the whole square is shaded? (c) > 707 676 (d) < 707 676 (e) > 1 (f) < 1 197 Ganita Prakash | Grade 7 11. A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source? 12. What is 1 – 1 2 ? • Brahmagupta’s formula for multiplication of fractions: a b × c d = a × c b × d . • When multiplying fractions, if the numerators and denominators have some common factors, we can cancel them first before multiplying the numerators and denominators. • In multiplication — when one of the numbers being multiplied is between 0 and 1, the product is less than the other number. If one of the numbers being multiplied is greater than 1, then the product is greater than the other number. (1 – 1 2) × (1 – 1 3) ? (1 – 1 2) × (1 – 1 3) × (1 – 1 4) × (1 – 1 5) ? (1 – 1 2) × (1 – 1 3) × (1 – 1 4) × (1 – 1 5) × (1 – 1 6) × (1 – 1 7) × (1 – 1 8) × (1 – 1 9) × (1 – 1 10) ? Make a general statement and explain. SUMMARY Fig. 8.7 Chapter-8.indd 198 4/12/2025 12:41:38 PM 198 • The reciprocal of a fraction a b is b a . When we multiply a fraction by its reciprocal, the product is 1. • Brahmagupta’s formula for division of fractions: a b ÷ c d = a b × d c = a × d b × c . • In division — when the divisor is between 0 and 1, the quotient is greater than the dividend. When the divisor is greater than 1, the quotient is less than the dividend. Chess is a popular 2-player strategy game. This game has its origins in India. It is played on an 8 × 8 chequered grid. There are 2 sets of pieces — black and white — one set for each player. Find out how each piece should move and the rules of the game. Here is a famous chess-based puzzle. From its current position, a Queen piece can move along the horizontal, vertical or diagonal. Place 4 Queens such that no 2 queens attack each other. For example, the arrangement below is not valid as the queens are in the line of attack of each other. Now, place 8 queens on this 8 × 8 grid so that no 2 queens attack each other! Chess Puzzles— Non-attacking Queens Working with Fractions Chapter-8.indd 199 4/12/2025 12:41:40 PM 199 Learning Material Sheets Chapter-8.indd 200 4/12/2025 12:41:40 PM Chapter-8.indd 201 4/12/2025 12:41:43 PM Note Chapter-8.indd 202 4/12/2025 12:41:43 PM Page No. 176 Ans: Quantity of milk that Tenzin drinks in a week = 7 2 glasses. 2. A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make ___ km of the water canal. If they work 5 days a week, they can make ___ km of the water canal in a week. Ans: 3. Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks? Ans: 1. Tenzin drinks glass of milk every day. How many glasses of milk Figure it out does he drink in a week? How many glasses of milk did he drink in the month of January? Quantity of milk Tenzin drinks in the month of January which has 31 days = 31 2 glasses. Length of water canal made in 1 day = 1 8 km. Length of water canal made in 5 days = 5 8 km. Oil each family gets in 1 week = 5 3 litres. Working with Fractions CHAPTER – 8 previous day. How many hours after 10 pm will the moon set on Thursday? Ans: Time at which the moon sets on Thursday (i.e. 3 days after Monday) 4. Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a Oil each family gets in 4 weeks = 20 3 litres. scientist, told her that every day the Moon sets hour later than the = 10 + 2 + 1 2 = 12 1 2 am Or 12:30 am Page No. 177 Page No. 180 5. Multiply and then convert it into a mixed fraction: 1. Find the following products. Use a unit square as a whole for representing the fractions: (a) 7 × Ans: 7 × 3 5 = 21 5 = 4 1 5 (b) 4 × Ans: 4 × 1 3 = 4 3 = 11 3 (c) × 6 Ans: 9 7 × 6 = 54 7 = 7 5 7 (d) × 6 Ans: 13 11 × 6 = 78 11 = 7 1 11 Figure it out (a) × Ans: 1 15 (b) × Ans: 1 12 (c) × Ans: 1 10 (d) × Ans: 1 30 Page No. 181 Page No. 183 2. Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations. 1. A water tank is filled from a tap. If the tap is open for 1 hour, of the (a) × Ans: 8 15 (Draw the figure in the same manner as shown in question 1) (b) × Ans: 2 12 = 1 6 (c) × Ans: 3 10 (d) × Ans: 12 30 = 2 5 Figure it out tank gets filled. How much of the tank is filled if the tap is open for Ans: 7 30 part Ans: 14 30 part Ans: 21 40 part (a) hour ____________ (b) hour ____________ (c) hour ____________ Ans: Part of land that remains with somu = 5 6 (a) 5 12 Page No. 184 2. The government has taken of Somu’s land to build a road. What part Ans: 49 100 part Ans: 10 7 hours of the land remains with Somu now? She gives half of the remaining part of the land to her daughter Krishna and of it to her son Bora. After giving them their shares, she keeps the remaining land for herself. (a) What part of the original land did Krishna get? (b) What part of the original land did Bora get? (c) What part of the original land did Somu keep for herself? (d) hour ____________ (e) For the tank to be full, how long should the tap be running? (b) 5 18 (c) 5 36 Ans: 36 sq. ft. Ans: 2 1 4 m Ans: 3 20 of 4 kg is heavier than 12 15 of 500 gm. 3. Find the area of a rectangle of sides 3 ft and 9 ft. 4. Tsewang plants four saplings in a row in his garden. The distance 5. Which is heavier: of 500 grams or of 4 kg? between two saplings is m. Find the distance between the first and last sapling. Page No. 191 Page No. 193 In a day, the number of times — (a) the first fountain will fill the cistern is 1÷ 1 = 1 (b) the second fountain will fill the cistern is 1 ÷ = _____ Ans: 2 times (c) the third fountain will fill the cistern is 1 ÷ = _____ Ans: 4 times (d) the fourth fountain will fill the cistern is 1 ÷ = _____ Ans: 5 times The number of times the four fountains together will fill the cistern in a day is ___ + ____ + ___ + ____ = 12. Ans: The number of times the four fountains together will fill the cistern in a day is = 1 + 2 + 4 + 5 = 12 In each of the figures given below, find the fraction of the big square that the shaded region occupies. Ans: The shaded region of Fig. (a) occupies 3 8 of the area of whole square. The red shaded part of Fig. (b) occupies 1 16 of the area of whole square. (a) (b) Page No. 196 1. Evaluate the following: 2. For each of the questions below, choose the expression that describes Figure it out the solution. Then simplify it. (a) Maria bought 8 m of lace to decorate the bags she made for school. (b) meter of ribbon is used to make 8 badges. What is the length of (c) A baker needs kg of flour to make one loaf of bread. He has 5 kg Ans: the ribbon used for each badge? Ans: 3 ÷ 7 9 = 27 7 14 4 ÷ 2 = 7 4 2 3 ÷ 2 3 = 1 14 6 ÷ 7 3 = 1 4 3 ÷ 3 4 = 16 9 7 4 ÷ 1 7 = 49 4 8 2 ÷ 4 15 = 15 1 5 ÷ 1 9 = 9 5 1 6 ÷ 11 12 = 2 11 3 2 3 ÷ 1 3 8 = 2 2 3 She used m for each bag and finished the lace. How many bags did she decorate? (iii) 8 ÷ 1 4 (iv) 1 2 ÷ 8 Page No. 197 Ans: 1 8 kg 3. If kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis? Ans: of flour. How many loaves of bread can he make? (iii) 5 ÷ 1 6 Ans: 40 Ans: 200 pages. 6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 litres of petrol? Ans: 44 km. Ans: 4 2 3 hour 4. Patiganita, a book written by Sridharacharya in the 9th century C.E., 5. Mira is reading a novel that has 400 pages. She read of the pages 7. Amritpal decides on a destination for his vacation. If he takes a train, 8. Mariam’s grandmother made a cake. Mariam and her cousins finished of the cake. The remaining cake was shared equally by Mariam’s mentions this problem: “Friend, after thinking, what sum will be obtained by adding together yesterday and of the pages today. How many more pages does she need to read to finish the novel? it will take him 5 hours to get there. If he takes a plane, it will take him hour. How many hours does the plane save? three friends. How much of the cake did each friend get? 1 ÷ , 1 ÷ , 1 ÷ , 1 ÷ , and 1 ÷ ”. What should the friend say? Ans: 1 15 of the whole cake. Ans: (a), (c) and (e) are correct. 9. Choose the option(s) describing the product of ( × ): 10.What fraction of the whole square is shaded? Ans: 3 32 of the whole square. 11.A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the figure 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source? 12.What is 1 – ? Ans: 1 2 Ans: 1 3 (1 − 1 2 ) × (1 − 1 3 ) × (1 − 1 4 ) × (1 − 1 5 )? Ans: 1 5 (1 − 1 2 ) × (1 − 1 3 ) × (1 − 1 4 ) × (1 − 1 5 ) × (1 − 1 6 ) × (1 − 1 7 ) × (1 − 1 8 ) × (1 − 1 9 ) × (1 − 1 10 )? Ans: Mango trees = 29 32 and Sugarcane trees = 3 32 (1 − 1 2 ) (1 − 1 2 ) × (1 − 1 3 )? Ans: 1 10 Make a general statement and explain. Ans: (1 − 1 2 ) × (1 − 1 3 ) × (1 − 1 4 ) × (1 − 1 5 ) . . . . . . . . . . . (1 − 1 ) = 1" class_7,9,Geometric Twins,ncert_books/class_7/gegp1dd2/gegp201.pdf,"1 1.1 Geometric Twins The symbol on this signboard needs to be recreated on another board. How do we do it? One way is to trace the outline of this symbol on tracing paper to reconstruct the figure. But this is difficult for big symbols. What else can we do? GEOMETRIC TWINS Chapter-1.indd 1 10/9/2025 11:44:10 AM Can we take some measurements that would allow us to exactly recreate this figure? If yes, what measurements should we take? Are the arm lengths AB and BC sufficient to exactly recreate this figure? Let us name the corner points of this symbol as shown. A C B Ganita Prakash | Grade 7 | Part-II Suppose these lengths are AB = 4 cm, BC = 8 cm. We observe that several such symbols can be constructed with the same lengths. To get the exact replica, would it help to take any other measurement? The measure of ∠ABC, along with the two arm lengths AB and BC, fix the shape and size of this figure. Can you draw the symbol if it is known that AB = 4 cm, BC = 8 cm, and ∠ABC = 80°? These three measurements can help us create an exact replica of the symbol on the signboard. Figures that are exact copies of each other or have the same shape and size are said to be congruent. Congruent figures can be superimposed exactly, one over the other. C A B C A B C A B A C B Chapter-1.indd 2 10/9/2025 11:44:10 AM Two congruent figures are shown below. You could use a tracing paper to trace the first figure and superimpose it on the second one. You will find that they fit exactly, one over the other. Note that while checking for congruence, a figure can be rotated or flipped before superimposing it on the other figure. So, the following pairs of figures are also congruent to each other. 2 Let us get back to the symbol we saw on the signboard. Suppose there are two such symbols that look identical and we need to confirm that they are indeed congruent. Can we use their measurements to verify this? If it is known that both symbols have the same arm lengths, can it be concluded that the two symbols are congruent? We have seen that there can exist several such non-congruent figures with different angles between the given arm lengths. Fixing the angle determines the shape and size of the figure. Thus, if both symbols have the same arm lengths and angle, we can be sure that the figures are congruent. Figure it Out 1. Check if the two figures are congruent. 2. Circle the pairs that appear congruent. Geometric Twins Chapter-1.indd 3 10/9/2025 11:44:10 AM 3. What measurements would you take to create a figure congruent to a given: (a) Circle (b) Rectangle 3 Ganita Prakash | Grade 7 | Part-II 4. How would we check if two figures like the one below are congruent? 1.2 Congruence of Triangles (a) Circles are congruent? (b) Rectangles are congruent? Using this, state how would you check if two — Use this to identify whether each of the following pairs are congruent. Chapter-1.indd 4 10/9/2025 11:44:10 AM Meera and Rabia have been asked to make a cardboard cutout identical to a triangular frame they have in school. They see that the frame is too big to be traced on a paper and replicated. 4 What do you think they can do? Measuring the Sidelengths Can certain measurements of the triangle be used for this? Using a measuring tape, the girls measure the sides of the triangle to be 40 cm, 60 cm, and 80 cm. Then, Rabia takes out her protractor to measure the angles. She is stopped by Meera. Meera: The angles of the triangle are not required! With the side lengths we have measured, we can create a triangle congruent to this one. Do you agree with Meera? Instead of the lengths being 40 cm, 60 cm, and 80 cm, suppose the sidelengths had been 4 cm, 6 cm, 8 cm (this triangle can fit on our page). Is this information sufficient to replicate the triangle with the same size and shape? If yes, can you do so? Rabia: If I were to construct this triangle, I would first draw a line segment having one of the given lengths, say 6 cm, and then draw circles from each of its end points with radii 4 cm and 8 cm. But the circles would intersect at two points, forming two triangles: ∆ABE and ∆ABF Geometric Twins Chapter-1.indd 5 10/17/2025 4:00:25 PM Rabia: Do these two triangles have the same shape and size? If not, then we will not be sure which of these would actually be congruent to the original triangle we are trying to replicate. E F A B AB = 6 cm 5 Ganita Prakash | Grade 7 | Part-II Examine whether ∆ABE and ∆ABF are congruent. For this, you could use one or more of the following methods — tracing and comparing, taking a cutout and superimposing, or observing that AB acts as a line of symmetry due to the ‘sameness’ of the act of construction above and below this line. We see that ∆ABE and ∆ABF are congruent. From this general construction, we can see that all triangles with the same sidelengths are congruent. Hence, Meera was right when she said that the sidelengths are sufficient to construct a congruent triangle. Thus, we have the following result: If two triangles have the same sidelengths, then they are congruent. We call this the SSS (Side Side Side) condition for congruence. Conventions to Express Congruence The two triangles given below are congruent. How can these two triangles be superimposed? Which vertices of ∆XYZ and ∆ABC should we overlap? This has to be done so that the equal sides overlap. Figure out how. Overlapping Vertex A over Vertex X, Vertex B over Vertex Y and Vertex C over Vertex Z will ensure that equal sides overlap, making the triangles fit exactly over each other. C A B Y X Z Chapter-1.indd 6 10/9/2025 11:44:11 AM Are there other ways of overlapping the vertices so that the triangles fit exactly over each other? The fact that these triangles are congruent shows that their respective angles are equal: Thus, when two triangles are congruent, there are corresponding vertices, sides and angles which fit exactly over each other when the triangles are made to overlap. In this case, they are 6 (a) Corresponding Vertices: A and X, B and Y, C and Z ∠A = ∠X, ∠B = ∠Y and ∠C = ∠Z To capture this relation that exists when two triangles are congruent, their congruence is written as follows: By writing this, we mean that: Can you identify a pair of congruent triangles below? Why are they congruent? • the first vertex in the name of ΔABC corresponds to the first vertex in the name of ΔXYZ, • the second vertex in the name of ΔABC corresponds to the second vertex in the name of ΔXYZ, and • similarly with the third vertices in the names of ΔABC and ΔXYZ. By this convention, it is incorrect to write for these two triangles that ΔACB ≅ ΔXYZ. However, another correct way of saying it is ΔACB ≅ ΔXZY. (b) Corresponding Sides: AB and XY, BC and YZ, AC and XZ (c) Corresponding Angles: ∠A and ∠X, ∠B and ∠Y, ∠C and ∠Z A B ∆ABC ≅ ΔXYZ corresponds to corresponds to ∆ABC ≅ ΔXYZ corresponds to Geometric Twins Chapter-1.indd 7 10/9/2025 11:44:11 AM Consider ΔABD and ΔCDB. Since ABCD is a rectangle, we have AB = CD AD = CB If the remaining sides of ΔABD and ΔCDB have the same length then the SSS condition is satisfied, confirming the congruence of the two triangles. Is this the case? D C Fig.1.1 7 Ganita Prakash | Grade 7 | Part-II The remaining side is a common side BD, so the SSS condition holds. Hence, the triangles are congruent. We know the corresponding sides of the two triangles. We have to identify the corresponding vertices. Can they be the following? ΔABD ΔCDB A C B B D D Verify this by superimposing paper cutouts of the triangles obtained from the rectangle ABCD (Fig. 1.1). We see that this correspondence lays the side AB of ΔABD over the side CB of ΔCDB. But these sides need not be equal, and hence, this superimposition will not establish congruence. Identify the correct correspondence of vertices and express the congruence between the two triangles. Figure it Out 1. Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways of expressing this congruence. Chapter-1.indd 8 10/9/2025 11:44:11 AM 2. Determine whether the triangles are congruent. If yes, express the congruence. 8 3.5 cm R E D 5 cm 6 cm J 3.5 cm 5 cm 6 cm A M 3. In the figure below, AB = AD, CB = CD. 4. In the figure below, are ΔDFE and ΔGED congruent to each other? It is given that DF = DG and FE = GE. Can you identify any pair of congruent triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons. A B D C D Geometric Twins Chapter-1.indd 9 10/17/2025 4:01:30 PM Measuring the Angles Instead of measuring the three sidelengths of the triangular frame, if Meera and Rabia measure the three angles, can they recreate the triangle exactly? Suppose the angles are 30°, 70°, and 80°. Can we create an exact copy of the frame with this? As we see, we can draw many triangles with these measurements that are not congruent. F G E 9 Ganita Prakash | Grade 7 | Part-II These triangles are seen to have the same shape, but not the same size. Hence, two triangles that have the same set of angles need not be congruent. Measuring Two Sides and the Included Angle ΔABC and ΔXYZ are two triangles such that Are they congruent? To check this, we need to see if there can exist non-congruent triangles with the given measurements. These measurements correspond to the case of two sides and the included angle. We have seen how to construct a triangle given these measurements. Construct a triangle having the above measurements. Compare it with the triangles constructed by your classmates. Are the triangles all congruent? Explain why all such triangles with these measurements are congruent. Thus, when two sides and the included angle of two triangles are equal, the two triangles are congruent. This is referred to as the SAS (Side Angle Side) condition for congruence. 30° B C AB = XY = 6 cm, AC = XZ = 5 cm, and ∠A = ∠X = 30° 80° 70° A B C 70° 30° 30° 80° A B C 70° 80° A Chapter-1.indd 10 10/9/2025 11:44:11 AM Measuring Two Sides and a Non-included Angle What if two sides and a non-included angle are equal? ΔABC and ΔXYZ are two triangles such that Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out. 10 AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B = ∠Y = 30° Looking at a rough diagram helps in planning the construction. How does one construct a triangle having these measurements? Step 1: Draw the base PQ of length 6 cm. Step 2: Draw a line l from P that makes an angle of 30° with PQ. Step 3: Draw a sufficiently long arc from Q of radius 4 cm cutting the line l. P Q P Q 30° 6 cm 6 cm 30° R 4 cm l l S Geometric Twins Chapter-1.indd 11 10/9/2025 11:44:11 AM How do we find the required triangle from this figure? A point of intersection of the arc and the line l gives the third point of the required triangle. But we see that the arc intersects the line l at two different points R and S. Both ΔPQR and ΔPQS satisfy the given measurements. P Q 6 cm 30° R 11 P Q 30° 6 cm Ganita Prakash | Grade 7 | Part-II Hence, we can draw two non-congruent triangles with the given measurements. This is called the SSA (Side Side Angle) condition. We have seen that SSA condition does not guarantee congruence. We have examined cases using two sides and an angle for determining congruence. Can we use two angles and a side? Let us first take the case of two angles and the included side. Two Angles and the Included Side ΔABC and ΔXYZ are two triangles with, Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out. We have seen how to construct a triangle when we are given two angles and the included side. This construction should make it clear that all the triangles having these measurements must be congruent to each other. Hence, ΔABC ≅ ΔXYZ. This condition is referred to as the ASA (Angle Side Angle) condition for congruence. R BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z = 30°. 4 cm P Q 30° 6 cm s 4 c m Chapter-1.indd 12 10/10/2025 2:42:36 PM In the figure, Point O is the midpoint of AD and BC. What can one say about the lengths AB and CD? 12 D B A C O Are there any other equal sides or angles? We see that the SAS condition (two sides and the included angle) is satisfied, and so we can conclude that the triangles are congruent. What are the Corresponding Vertices? As we need AO and OD to overlap, and BO and CO to overlap for the triangles to exactly fit over each other, the corresponding vertices in ΔAOB and ΔDOC are A and D, O and O (vertex common to both the triangles), and B and C. Thus, AB and DC are corresponding sides as they overlap when the triangles are superimposed. Thus, their lengths are equal. Figure it Out 1. Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence. We have, AO = OD (as O is the midpoint of AD) We also have, ∠AOB = ∠DOC, as they are vertically opposite angles. BO = OC (as O is the midpoint of BC). A ΔAOB ≅ ΔDOC. X Geometric Twins Chapter-1.indd 13 10/10/2025 2:42:36 PM 2. Given that CD and AB are parallel, and AB = CD, what are the other equal parts in this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are the two resulting triangles congruent? If so, express the congruence.) B 5 cm C 47° 7 cm Y 5 cm 47° 7 cm Z 13 Ganita Prakash | Grade 7 | Part-II 3. Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent? 4. Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC. A B C A B C D A D O D Chapter-1.indd 14 10/17/2025 4:02:35 PM Measuring Two Angles and a Non-Included Side The following triangles ΔABC and ΔXYZ are such that ∠A = ∠X = 35°, ∠C = ∠Z = 75°, and BC = YZ = 4 cm. Are the triangles congruent? Give a reason. 14 B C How do we proceed with this problem? Here is a method. What are the measures of ∠B and ∠Y? Does this help in showing that ΔABC and ΔXYZ are congruent? We know that the sum of the angles of a triangle is 180°. So ∠B + 35° + 75° = 180°, or ∠B + 110° = 180° Thus, ∠B = 70°. Similarly, ∠Y is also 70°. Thus, we have ∠B = ∠Y. B C 4 cm 35° 35° A A 75° Fig.1.2 Y Z 4 cm 35° 35° X X 75° Geometric Twins Chapter-1.indd 15 10/9/2025 11:44:11 AM These two triangles now satisfy the ASA condition with ∠B = ∠Y BC = YZ ∠C = ∠Z So, ΔABC ≅ ΔXYZ. B C 4 cm 70° 75° Y Z 4 cm 70° 75° 15 Ganita Prakash | Grade 7 | Part-II In Fig. 1.2, the equalities are between two angles and the non-included side of the two triangles. This condition is referred to as the AAS (Angle Angle Side) condition. As we have seen, the AAS condition guarantees congruence. We have seen that the SSA condition doesn’t always guarantee congruence. However, there are some special cases when SSA does guarantee congruence. Here is one such important case. Measuring Two Sides in a Right Triangle ΔABC and ΔXYZ are right-angled triangles such that BC = YZ = 4 cm, ∠B = ∠Y = 90° and AC = XZ=5cm. Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out. Looking at the rough diagram helps in planning the construction. Step 1: Draw the base QR of length 4 cm. Step 2: Draw a line l perpendicular to QR from Q. Step 3: From R, cut an arc on line l of radius 5 cm. Step 4: Let P be the point at which the arc intersects the line l. Join PR. Q R 4 cm 5 cm P Chapter-1.indd 16 10/9/2025 11:44:11 AM ΔPQR is the required triangle. 16 Q R 4 cm P Consider the downward extension of line l below QR. Would the arc from R meet this line downwards as well (as in the case of triangle construction when the sidelengths are given)? If so, would this lead to a triangle whose size and shape are different from ΔPQR, and yet has the given measurements? It can be seen that the other triangle we get below is also congruent to ΔPQR. Why? Therefore, all triangles having these measurements will be congruent to each other. Thus, we conclude that ΔABC ≅ ΔXYZ. In the case that we have considered, the parts that are equal to their corresponding parts in another triangle are (b) two other sides, one of which is opposite to the right angle. This side is called the hypotenuse. This is called the RHS (Right Hypotenuse Side) condition, and is one more condition for congruence. Conditions that are sufficient to guarantee congruence From the discussions so far, we can see that two triangles are congruent if any of the following conditions are satisfied: 1.3 Angles of Isosceles and Equilateral Triangles Congruence is a very powerful tool for studying properties of geometric figures. Let us use it to discover an important property of isosceles triangles. (a) the right angle (a) SSS condition (b) SAS condition (c) ASA condition (d) AAS condition (e) RHS condition Geometric Twins Chapter-1.indd 17 10/9/2025 11:44:11 AM ΔABC is isosceles with AB = AC, and ∠A = 80. What can we say about ∠B and ∠C? Construct the altitude from A to BC. B C 80° A 17 Ganita Prakash | Grade 7 | Part-II Thus, the triangles satisfy the RHS condition. Hence, ΔADB ≅ ΔADC. This shows that ∠B = ∠C, as they are corresponding parts of congruent triangles. Thus, the angles opposite to equal sides are equal. Can you use this fact to find ∠B and ∠C? Angles in an Equilateral Triangle Equilateral triangles are those in which all the sides have equal lengths. What can we say about their angles? We can use the recently discovered fact that angles opposite to equal sides are equal. A We have, AB = AC (given) ∠ADB = ∠ADC = 90° (from construction) AD is a common side of the two triangles ΔADB and ΔADC. D B C A Chapter-1.indd 18 10/9/2025 11:44:11 AM The sides AB and AC are equal. So ∠B = ∠C. 18 B C B C A Similarly, the sides AB and BC are equal. So ∠A = ∠C. So, all the three angles of an equilateral triangle are equal, just like their sides. What could be their measures? As the three angles should add up to 180°, we have 3 × angle in an equilateral triangle = 180°. So each angle is 60°. Verify this by construction. Thus, just using the notion of congruence, we have deduced that the angles of an equilateral triangle are all 60°. Congruent Triangles in Real Life: Congruent triangles can be seen in various constructions and designs from ancient to modern times. Here are a few examples. World-famous Louvre Museum in Paris World-famous Egyptian Pyramid of Giza Geometric Twins Chapter-1.indd 19 10/9/2025 11:44:13 AM Dome design Rangoli design 19 Ganita Prakash | Grade 7 | Part-II Describe the congruent triangles you see in each picture. Figure it Out 1. ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles. 2. Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent. (a) AB = DE (b) AB = EF BC = EF ∠A = ∠E CA = DF AC = ED (c) AB = DF (d) ∠A = ∠D ∠B = ∠D = 90° ∠B = ∠E AC = FE AC = DF (e) AB = DF ∠B = ∠F AC = DE Rabindra Setu or Howrah Bridge Chapter-1.indd 20 10/9/2025 11:44:13 AM 3. It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a transversal for these two lines. Are there any equal alternate angles?] 20 D B A C O 4. ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent to ΔCDA? A 5. Find ∠B and ∠C, if A is the centre of the circle. 6. Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘|’ are equal to each other and those marked with a double ‘|’ are equal to each other, etc. Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways? A C B C R D 120° A L V B C Geometric Twins B F D Chapter-1.indd 21 10/9/2025 11:44:13 AM U 56° 34° 34° 44° K 34° 68° 46° 44° 90° 30° 56° 98° 21 Ganita Prakash | Grade 7 | Part-II • Figures that have the same shape and size are said to be congruent. These figures can be superimposed so that one fits exactly over the other. • While verifying congruence, a figure can be rotated or flipped to make it fit exactly over the other figure via superimposition. • When two triangles have the same sidelengths, we say that the SSS (Side Side Side) condition is satisfied. The SSS condition guarantees congruence. • When two sides and the included angle of one triangle are equal to the two sides and the included angle of another triangle, we say that the SAS (Side Angle Side) condition is satisfied. The SAS condition also guarantees congruence. • When two angles and the included side of one triangle are equal to the two angles and the included angle of another triangle, we say that the ASA (Angle Side Angle) condition is satisfied. The ASA condition guarantees congruence. Congruence holds even if the side is not included between the angles AAS (Angle Angle Side) condition. • In a right-angled triangle, the side opposite to the right angle is called the hypotenuse. • When a side and a hypotenuse of a right-angled triangle are equal to a side and the hypotenuse of another right-angled triangle, we say that the RHS (Right Hypotenuse Side) condition is satisfied. The RHS condition also guarantees congruence. • Two triangles need not be congruent if two sides and a non-included angle are equal. • In a triangle, angles opposite to equal sides are equal. • The angles in an equilateral triangle are all 60°. SUMMARY Chapter-1.indd 22 10/9/2025 11:44:13 AM 22 Draw lines and split the region consisting of white squares into 6 smaller congruent regions. Expression Engineer! Chapter-1.indd 23 10/9/2025 11:44:15 AM" class_7,10,operations with integers,ncert_books/class_7/gegp1dd2/gegp202.pdf,"OPERATIONS WITH INTEGERS 2 2.1 A Quick Recap of Integers Rakesh’s Puzzle: A Number Game Rakesh gives you a challenge. “I have thought of two numbers”, he says. “Their sum is 25, and their difference is 11.” Can you tell me the two numbers? You don’t need to use any formulas. Just try different pairs of numbers and then check: 1. Do the two numbers add up to 25? 2. Is the difference between them 11? (Remember: the difference means first number – second number.) Write your guesses like this: First Number Second Number Sum Difference Chapter-2.indd 24 10/9/2025 2:27:38 PM Did you find the right pair? Now that you’ve found the correct pair, Rakesh gives you a second challenge: “Think of two numbers whose sum is 25, but their difference is –11.” Use the same method. Try different pairs of numbers and fill in the table again. You will notice that if you swap the numbers from the first 10 15 25 –5 20 5 25 15 19 6 25 13 18 7 25 11 puzzle, you get the answer to Rakesh’s second puzzle. That is, the first number is 7 and the second is 18! Figure it Out Let us try to find a few more pairs of numbers from their sums and differences: Choose a partner and take turns to play this game. In each turn, one of you can think of two integers, and give their sum and difference; the other person must then figure out the integers. After some practice, you can perform this magic trick for your family members and surprise them! Carrom Coin Integers A carrom coin is struck to move it to the right. Each strike moves the coin a certain number of units of distance rightwards based on the force of the strike. To begin with, the coin is at point 0. If the coin is struck twice, with the first strike moving it by 4 units and the second strike moving it by 3 units, what will be the final position of the coin? (a) Sum = 27, Difference = 9 (b) Sum = 4, Difference = 12 (c) Sum = 0, Difference = 10 (d) Sum = 0, Difference = – 10 (e) Sum = – 7, Difference = – 1 (f ) Sum = – 7, Difference = – 13 0 1 2 3 4 5 6 7 8 9 10 11 12 Operations with Integers Chapter-2.indd 25 10/9/2025 2:27:38 PM It is clear that the coin will be 4 + 3 = 7 units from 0. If the coin is struck twice, and if the two movements are known, can you give a formula for the final position of the coin? If the first strike moves the coin ‘a’ units to the right and if the second strike moves the coin ‘b’ units to the right, then the final position is P = a + b, where P is the distance of the coin from the starting point 0. 0 P b a P = a + b 25 Ganita Prakash | Grade 7 | Part-II Now, suppose the coin can be struck to move it in either direction — left or right. The coin is at 0. If it is struck twice (the direction of the two strikes may be the same or different) can you give a formula for the final position of the coin? One way to do this is to consider different cases of the directions of the strikes An efficient way to model this situation is to use positive and negative integers. First, let us model the line on which the coin moves as a number line. Let us consider the rightward movement positive and the leftward movement negative. Suppose the first strike moves the coin rightward by 5 units from 0, and the second strike leftward by 7 units, then we take the First Movement = 5 units. Second Movement = – 7 units. • both are rightward, • both are leftward, • the first one is rightward and the second one is leftward, and • the first one is leftward and the second one is rightward. . . . –5 –4 –3 –2 –1 0 1 2 3 4 5 6 . . . –ve +ve –ve 0 +ve Chapter-2.indd 26 10/9/2025 2:27:38 PM What is the final position of the coin? This can be found by simply adding the two movements: 5 + (– 7) = – 2. So the coin is at –2, or it is 2 units to the left of 0. In general, if the first strike moves the coin ‘a’ units (which is positive if the strike is to the right and negative if the strike is to the left), and the second strike ‘b’ units (which is positive if the strike is to the right and negative if the strike is to the left), then the final position ‘P’, after the two strikes, is again P = a + b. Based on this new model, answer the following questions: 26 1. If the first movement is – 4 and the final position is 5, what is the second movement? By modeling the movements as numbers, both positive and negative, we are able to capture two pieces of information — the distance (magnitude) and the direction (rightwards or leftwards). For example, when we say the movement is – 4, the magnitude is 4 and the direction is leftward. From the figures below, what can you conclude about the magnitudes of a and b compared to each other, and what are their directions? Remember to start from 0. 2. If there are multiple strikes causing movements in the order 1, – 2, 3, – 4, …, – 10, what is the final position of the coin? 1. 2. –ve P 0 +ve b a a b Operations with Integers Try This Chapter-2.indd 27 10/9/2025 2:27:38 PM 3. –ve 0 P +ve P –ve 0 +ve b a 27 Ganita Prakash | Grade 7 | Part-II In addition to the number line, we used the token model to understand integers. We used this model to perform addition and subtraction in Grade 6. Let us do a quick recall. We used green ( ) token to represent positive 1 and red ( ) token to represent negative 1, that is –1. Together they make zero, as they cancel each other out. Find (+7) – (+18). To subtract 18 from 7, i.e., (+7) – (+18), we need to remove 18 positives from 7 positives. How many? We have 7 positives and we need 11 more. So we need to put in 11 zero pairs: We can now remove 18 positives. But there are not enough tokens to remove 18 positives! We put in enough zero pairs so that we can remove 18 positives. Chapter-2.indd 28 17-10-2025 16:22:48 What is left? There are 11 negatives, meaning –11. We had also seen that subtracting a number is the same as adding its additive inverse. Using tokens, argue out the following statements. (a) 7 – 18 = 7 + (– 18) (additive inverse of 18 is – 18) (b) 4 – (– 12) = 4 + 12 (additive inverse of – 12 is 12) 28 Note to the Teacher: Tokens of different shapes may be used for positive and negative numbers for visually challanged students. So, 7 – 18 = –11. Math Talk Additive inverse of an integer a is represented as – a. So the additive inverse of 18 is represented as – (18) = – 18, and the additive inverse of – 18 is represented as – (– 18) = 18. 2.2 Multiplication of Integers We used the token model to represent addition and subtraction of integers. We now explore how to model multiplication of integers using tokens. Suppose we put some positive tokens into an empty bag as shown in the figure. How many positives are in the bag now? There are 8 positives in the bag. We can see this as adding 2 positives to the bag 4 times. Thus, the operation is, 4 × 2 = 8. We have seen this kind of multiplication of positive integers before. Can we use tokens to give meaning to multiplications like 4 × (– 2)? Let us see how. For every new operation, we start with an empty bag. 4 × (–2) can be interpreted as placing 2 negatives into an empty bag 4 times. We use red tokens for negatives, so we place 2 negatives into an empty bag 4 times. There are now 8 red tokens or 8 negatives in the bag, meaning – 8. 4 × (– 2) = (– 8). Two red tokens 4 times = – 8 Two green tokens 4 times = 8 Operations with Integers 4 × (– 2) = – 8 4 × 2 = 8 Chapter-2.indd 29 17-10-2025 16:19:34 Similarly find the values of 4 × (– 6) and 9 × (– 7)? How can we interpret (– 4) × 2? When the multiplier is positive, we place tokens into the bag. When the multiplier is negative, we remove tokens from the bag. So, for (– 4) × 2, we need to remove two positives or two green tokens from the bag 4 times. Multiplier Multiplicand Product 29 Ganita Prakash | Grade 7 | Part-II Why are we trying to remove green tokens and not red tokens? But there are no tokens in the bag, because we start with an empty bag. Just as in the case of subtraction, to remove 2 positives from an empty bag, we need to first place 2 zero pairs inside and then remove the 2 positives. We need to do this 4 times. After removing the positives, 8 negatives are left in the bag. This is – 8. This shows that (– 4) × 2 = – 8. What happens when both the integers in the multiplication are negative? How do we model (–4) × (–2) with tokens? For (– 4) × (– 2), we need to remove 2 negatives from the bag 4 times. Since there are no red tokens in the bag, we need to place 2 zero pairs and remove 2 negatives, and we need to do this 4 times. Remove 2 green tokens from the zero pairs, 4 times (–  4) × 2 = –  8 Math Talk Chapter-2.indd 30 10/9/2025 2:27:50 PM 8 positives are left in the bag. This is + 8. So, – 4 × – 2 = + 8. So far, we have established the following results by using tokens: 4 × 2 = 8, 4 × (– 2) = – 8, (– 4) × 2 = – 8, and (– 4) × (– 2) = 8. 30 Remove 2 red tokens from the zero pairs, 4 times (–  4) × (– 2) = 8 Figure it Out Consider the numbers represented by the following tokens: We can see that all of them represent the number (– 2). Now, take 4 times each of these token sets. That is, place each set into the empty bag 4 times. 1. Using the token interpretation, find the values of: 2. If 123 × 456 = 56088, without calculating, find the value of: 3. Try to frame a simple rule to multiply two integers. (a) 3 × (– 2) (b) (– 5) × (– 2) (c) (– 4) × (– 1) (d) (– 7) × 3 (a) (– 123) × 456 (b) (– 123) × (– 456) (c) (123) × (– 456) (a) (b) (c) Operations with Integers Math Talk Chapter-2.indd 31 17-10-2025 16:24:21 What integer do we get as the final answer in each case? Do we get different answers because the sets look different, or the same answer because they all represent – 2? Check this for 5 × 4, by taking different token sets corresponding to 4. We have seen that – 4 × 2 is the number obtained by removing 2 positive tokens from the empty bag 4 times. We know that removing or subtracting a number is the same as adding its inverse. Using this, can – 4 × 2 be defined through a process of addition of tokens instead of removal of tokens? 31 Math Talk Math Talk Ganita Prakash | Grade 7 | Part-II Since removing 2 positive tokens is the same as adding 2 negative tokens, – 4 × 2 can also be obtained by adding 2 negative tokens to the empty bag 4 times. (–  4) × 2 Patterns in Integer Multiplication We have explored the multiplication of integers in cases where the multiplier is positive, when it is negative, when the multiplicand is positive and when it is negative. Using this understanding, let us construct a sequence of multiplications and observe the patterns. What do you notice in this pattern? Can you describe it? We can see that, when the multiplicand is positive, for every unit decrease in the multiplier the product decreases by the multiplicand. Will this pattern continue when the multiplier goes below zero and becomes a negative number? Yes indeed! The same pattern continues when the multiplier becomes a negative number. What is the pattern when the multiplicand is a negative integer? 4 × 3 = 12 3 × 3 = 9 2 × 3 = 6 1 × 3 = 3 0 × 3 = 0 4 × 3 = 12 3 × 3 = 9 2 × 3 = 6 1 × 3 = 3 0 × 3 = 0 –3 –3 –3 –3 –3 –3 –3 –3 Chapter-2.indd 32 10/9/2025 2:27:53 PM This is the inverse of the previous pattern. When the multiplicand is negative, for every unit decrease of the multiplier, the product increases by the multiplicand. Will this pattern continue when the multiplier goes below zero and becomes a negative integer? Yes! 32 4 × (–3) = –12 3 × (–3) = –9 2 × (–3) = –6 1 × (–3) = –3 0 × (–3) = 0 +3 +3 +3 +3 –1 × 3 = –3 –2 × 3 = –6 –3 × 3 = –9 –3 –3 –3 Even when the multiplier is negative, the same pattern is observed. When the multiplicand is negative, for every unit decrease in the multiplier, the product increases by the multiplicand. We can see from these patterns that, what is true for multiplication when the integers are positive, is also true when the integers are negative. With this understanding of multiplication of integers, let us look at the times 3 tables when the multipliers and multiplicands are positive, and when they are negative. We observe the following: • The magnitude of the product does not change with the change in the signs of the multiplier and the multiplicand. 1 × 3 = 3 2 × 3 = 6 3 × 3 = 9 4 × 3 = 12 5 × 3 = 15 6 × 3 = 18 7 × 3 = 21 8 × 3 = 24 9 × 3 = 27 10 × 3 = 30 –1 × 3 = –3 – 2 × 3 = –6 –3 × 3 = –9 –4 × 3 = –12 –5 × 3 = –15 –6 × 3 = –18 –7 × 3 = –21 –8 × 3 = –24 –9 × 3 = –27 –10 × 3 = –30 1 × –3 = –3 2 × –3 = –6 3 × –3 = –9 4 × –3 = –12 5 × –3 = –15 6 × –3 = –18 7 × –3 = –21 8 × –3 = –24 9 × –3 = –27 10 × –3 = –30 4 × (–3) = –12 3 × (–3) = –9 2 × (–3) = –6 1 × (–3) = –3 0 × (–3) = 0 (–1) × (–3) = 3 (–2) × (–3) = 6 (–3) × (–3) = 9 Operations with Integers –1 × –3 = 3 –2 × –3 = 6 –3 × –3 = 9 –4 × –3 = 12 –5 × –3 = 15 –6 × –3 = 18 –7 × –3 = 21 –8 × –3 = 24 –9 × –3 = 27 –10 × –3 = 30 +3 +3 +3 +3 +3 +3 +3 Chapter-2.indd 33 10/9/2025 2:27:53 PM Figure it Out Find the following products. • When both the multiplier and the multiplicand are positive, the product is positive. • When both the multiplier and the multiplicand are negative, the product is positive. • When one of the multiplier or the multiplicand is positive and the other is negative, their product is negative. (a) 4 × (– 3) (b) (– 6) × (– 3) 33 Ganita Prakash | Grade 7 | Part-II Consider the expression 1 × a. We know that the value of this expression is ‘a’ for all positive integers. Is this true for all negative integers too? Using the token model, we put ‘a’ negatives into the bag just once. After this, the bag contains ‘a’ negatives. For example, if ‘a’ is – 5 (5 negatives), then the bag contains 5 negatives, i.e., –5. So, What is the value of the expression – 1 × a? When ‘a’ is positive, then from our observations on integer multiplication, the product has the same magnitude as ‘a’ but is negative. When ‘a’ is negative, then the product has the same magnitude as ‘a’ but is positive. In each case, we notice that the product is the additive inverse of the multiplicand ‘a’. Thus, – 1 × a = – a (for all integers a). In the case of integers, is the product the same when we swap the multiplier and the multiplicand? Try this for some numbers. Observe the following pairs of multiplications (fill in the blanks where needed): (c) (– 5) × (– 1) (d) (– 8) × 4 (e) (– 9) × 10 (f) 10 × (– 17) 3 × – 4 = –12 – 4 × 3 = –12 – 30 × 12 = _______ 12 × – 30 = _______ 1 × a = a (for all integers a, both positive and negative). Chapter-2.indd 34 10/9/2025 2:27:54 PM What do you notice in these pairs of multiplication statements? The product is the same when we ‘swap’ the multiplier and multiplicand. Earlier, we have seen a similar property with addition. Will this always happen? The magnitude of the product does not change when the multiplier and the multiplicand are swapped. This is because the magnitude of the product depends only on the magnitudes of the multiplier and the multiplicand, and we know that the product of two positive integers does not change when the numbers being multiplied are swapped. 34 –15 × – 8 = 120 – 8 × –15 = 120 14 × – 5 = –70 – 5 × _____ = – 70 Does the sign of the product change if we swap the multiplier and multiplicand? If both are positive or both are negative, the product is positive before and after swapping. So the sign does not change in this case. If one is positive and the other negative, the product is negative before and after swapping. So the sign does not change in this case either. Hence, the product does not change when the multiplier and multiplicand are swapped, whatever their signs may be. Thus, multiplication is commutative for integers. In general, for any two integers, a and b, we can say that a × b = b × a. Brahmagupta’s Rules for Multiplication and Division of Positive and Negative Numbers Just like for addition and subtraction of integers, Brahmagupta in his Brāhmasphuṭasiddhānta (628 CE) also articulated explicit rules for integer multiplication and division. He used the notions of fortune (dhana) for positive values and debt (ṛṇa) for negative values. In his Brāhmasphuṭasiddhānta (18.30-32), Brahmagupta wrote: “The product or quotient of two fortunes is a fortune. The product or quotient of two debts is a fortune. The product or quotient of a debt and a fortune is a debt. The product or quotient of a fortune and a debt is a debt.” This represented the first time that rules for multiplication and division of positive and negative numbers were articulated, and was an important step in the development of arithmetic and algebra! Example 1: An exam has 50 multiple choice questions. 5 marks are given for every correct answer and 2 negative marks for every wrong answer. What are Mala’s total marks if she had 30 correct answers and 20 wrong answers? Solution: We use positive and negative integers. The mark for each correct answer is a positive integer 5 and for each wrong answer is a negative integer – 2. Marks for 30 correct answers = 30 × 5. Marks for 20 wrong answers = 20 × (– 2). Thus the arithmetic expression for 30 correct answers and 20 wrong answers is: 30 × 5 + 20 × (– 2) = 150 + (– 40) = 110. Mala got 110 marks in the exam. Operations with Integers Chapter-2.indd 35 10/17/2025 4:54:57 PM 35 Ganita Prakash | Grade 7 | Part-II What are the maximum possible marks in the exam? What are the minimum possible marks? Example 2: There is an elevator in a mining shaft that moves above and below the ground. The elevator’s positions above the ground are represented as positive integers and positions below the ground are represented as negative integers. (a) The elevator moves 3 metres per minute. If it descends into the shaft from the ground level (0), what will be its position after one hour? (b) If it begins to descend from 15 m above the ground, what will be its position after 45 minutes? Solution: Solution to part (a) of the question: Method 1: We can model this using subtraction. The elevator moves at 3 metres per minute. So in one hour it moves 180 metres (60 × 3). If it started at ground level (0 metres) and descended, we should subtract 180 from 0. Chapter-2.indd 36 10/9/2025 2:27:54 PM So, it will reach the (–180) metre position, which is 180 metres below the ground. Method 2: Let us say that the speed and direction of the elevator are represented by an integer (metres per minute). It is +3 when it is moving up and it is (–3) when moving down. Since the elevator is moving down, the speed is (–3) metres per minute. It moves for 60 minutes. So it goes 36 60 × (–3) = (–180). 0 – 180 = (–180). The position of the elevator after 60 minutes is 180 metres below the ground level. Find the solution to part (b) using Method 1 described above. Solution to part (b) using Method 2: Starting Position = 15. Distance Travelled = The elevator moves down at the speed of 3 metres per minute for 45 minutes, that is, (45 × (–3)). So, Ending Position = 15 + (45 × (–3)) = 15 + (–(45×3)) = 15 + (–135) = (–120). The elevator will be 120 metres below the ground. A Magic Grid of Integers A grid containing some numbers is given below. Follow the steps as shown until no number is left. –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 Strike out the row and the coloumn containing that number Circle any unstruck number Circle any number Operations with Integers Chapter-2.indd 37 10/9/2025 2:27:54 PM When there are no more unstruck numbers, stop. Multiply the circled numbers. An example is shown below. –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 Round 1 –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 Round 2 –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 Round 3 –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 Round 4 37 Ganita Prakash | Grade 7 | Part-II Try again , and choose different numbers this time. What product did you get? Was it different from the first time? Try a few more times with different numbers! Play the same game with the grid below. What answer do you get? Division of Integers We have earlier seen how division can be converted into multiplication. For example, (– 100) ÷ 25 can be reframed as, ‘what should be multiplied to 25 to get (– 100)?’. That is, Similarly, (– 100) ÷ (– 4) can be reframed as, ‘What should be multiplied to (– 4) to get (– 100)?’ (– 4) × ? = (– 100). We know that 25 × (– 4) = (– 100). Therefore, (– 100) ÷ 25 = (– 4). –28 14 –42 21 12 –6 18 –9 20 –10 30 –15 8 –4 12 –6 What is so special about these grids? Is the magic in the numbers or the way they are arranged or both? Can you make more such grids? 25 × ? = (– 100). Try This Chapter-2.indd 38 10/9/2025 2:27:54 PM 38 We know that (– 4) × 25 = (– 100). Therefore, (– 100) ÷ (– 4) = 25. Similarly, we know that (– 25) × (– 2) = 50. Therefore, 50 ÷ (– 25) = (– 2). Can you summarise the rules for integer division looking at the above pattern? In general, for any two positive integers a and b, where b ≠ 0, we can say that a ÷ – b = – (a ÷ b), Figure it Out 1. Find the values of: 2. A freezing process requires that the room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins? 3. A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5 per bag of grey cement sold. [Represent the profit/ loss as integers.] 4. Replace the blank with an integer to make a true statement. (a) 14 × (– 15) (b) – 16 × (– 5) (c) 36 ÷ (– 18) (d) (– 46) ÷ (– 23) (a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss? (b) If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss. – a ÷ b = – (a ÷ b), and – a ÷ – b = a ÷ b. Operations with Integers Chapter-2.indd 39 10/10/2025 2:44:24 PM Expressions Using Integers What is the value of the expression 5 × – 3 × 4? Does it matter whether we multiply 5 × – 3 and then multiply the product with 4, or if we multiply – 3 × 4 first and then multiply the product with 5? (a) (– 3) × _____ = 27 (b) 5 × _____ = (– 35) (c) _____ × (– 8) = (– 56) (d) _____ × (– 12) = 132 (e) _____ ÷ (– 8) = 7 (f) _____ ÷ 12 = – 11 (5 × – 3) × 4 = – 15 × 4 = – 60. 5 × (– 3 × 4) = 5 × – 12 = – 60. 39 Ganita Prakash | Grade 7 | Part-II Take a few more examples of multiplication of 3 integers and check this property. What do you observe? We can see that the product is the same when we ‘group’ the multiplications in these two ways. So, integer multiplication is associative, just like integer addition. In general, for any three integers a, b, and c, a × (b × c) = (a × b) × c. In the expression 5 × – 3 × 4, try to multiply 5 and 4 first and then multiply the product with – 3: (5 × 4) × – 3. 5 × 4 = 20, and 20 × – 3 = – 60. This also gives the same product. Are there orders in which 5 × – 3 × 4 can be evaluated? Will the product be the same in all these cases? Multiply the expression 25 × – 6 × 12 in all the different orders and check if the product is the same in all cases. The product remains the same when 3 or more numbers are multiplied in any order. Look at the following series of multiplications: When –1 is multiplied 2 or 4 times the product is positive. When it is multiplied 3 or 5 times the product is negative. Can you generalise these statements further? – 1 × – 1 = 1 – 1 × – 1 × – 1 = – 1 – 1 × – 1 × – 1 × – 1 = 1 – 1 × – 1 × – 1 × – 1 × – 1 = 1 Chapter-2.indd 40 10/9/2025 2:27:54 PM Using this understanding of multiplication of many integers, can you give a simple rule to find the sign of the product of many integers? Now, consider the expression 5 × (4 + (– 2)). As in the case of positive integers, is this expression equal to 5 × 4 + 5 × (– 2)? We see that it does. Recall that we call this property the distributive property. Check if the distributive property holds for (– 2) × (4 + (– 3)) (that is, if this expression equals (– 2) × 4 + (– 2) × (– 3)), and for a few other such expressions of your choice. What do you observe? We see that the distributive property seems to hold for integers, as well. Will this always happen? 40 Math Talk Pick the Pattern Two pattern machines are given below. Each machine takes 3 numbers, does some operations and gives out the result. Find the operations being done by Machine 1. In the case of positive integers, we used a rectangular arrangement of objects to visually understand why the distributive property holds. We can use the same setup even in the case of integers by using green tokens for positive numbers and red tokens for negative numbers. For example, consider the following rectangular arrangement of tokens — We see that the overall arrangement represents 4 × (2 + (– 3)), and it is clear that this also equals the sum of 4 × 2 and 4 × (–3). 4 × 2 4 × (– 3) Can you visually show the distributive property for an expression like –4 × (2 + (–3))? [Hint: Use the fact that multiplying a number by –4 is adding the inverse of the number 4 times.] Thus, for any integers a, b, and c, we have a × (b + c) = (a × b) + (a × c). Operations with Integers 4 × (2 + (– 3)) Try This Chapter-2.indd 41 10/9/2025 2:28:00 PM 41 Ganita Prakash | Grade 7 | Part-II Written as an expression, this will be a + b – c, where a is the first number, b is the second number, and c is the third number. For example, 5 + 8 – 3 = 10, and (– 4) + (– 1) – (– 6) = 1. So, the result of the last group will be, (– 10) + (– 12) – (– 9) = _______. Find the operations being done by Machine 2 and fill in the blank. Make your own machine and challenge your peers in finding its operations. Figure it Out 1. Find the values of the following expressions: 2. Find the values of the following expressions: 3. Find the integer whose product with (– 1) is: The operation done by Machine 1 is (a) (– 5) × (18 + (– 3)) (b) (– 7) × 4 × (– 1) (c) (– 2) × (– 1) × (– 5) × (– 3) (a) (– 27) ÷ 9 (b) 84 ÷ (– 4) (c) (– 56) ÷ (– 2) (a) 27 (b) – 31 (c) – 1 (d) 1 (e) 0 (first number) + (second number) – (third number). Math Talk Chapter-2.indd 42 10/9/2025 2:28:00 PM 42 5. Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by – 3 and add 1; repeat. An example sequence is shown below. 4. If 47 – 56 + 14 – 8 + 2 – 8 + 5 = – 4, then find the value of – 47 + 56 – 14 + 8 – 2 + 8 – 5 without calculating the full expression. –7 22 11 32 –16 –8 –4 –2 –1 4 2 1 6. In a test, (+ 4) marks are given for every correct answer and (– 2) marks are given for every incorrect answer. 7. Pick the pattern — find the operations done by the machine shown below. Try this with different starting numbers: (– 21), (– 6), and so on. Describe the patterns you observe. (a) Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test? (b) Anil scored (– 10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered? Operations with Integers Math Talk Chapter-2.indd 43 10/10/2025 2:45:48 PM 10. An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins — a+13 pibs coin and a – 9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs + 85 pibs? 8. Imagine you’re in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours. 9. Find 3 consecutive numbers with a product of (a) – 6, (b) 120. 43 Ganita Prakash | Grade 7 | Part-II [Hint: Writing down a few multiples of 13 and 9 can help.] 11. Find the values of: 12. Arrange the expressions given below in increasing order. 13. Given that (– 548) × 972 = – 532656, write the values of: 14. Given that 207 × (– 33 + 7) = – 5382, write the value of – 207 × (33 – 7) = _________. 15. Use the numbers 3, – 2, 5, – 6 exactly once and the operations ‘+’, ‘–’, and ‘×’ exactly once and brackets as necessary to write an expression such that — Yes, we can use 10 coins of +13 pibs and 5 coins of – 9 pibs to make a total of + 85. Using the two denominations, try to get the following totals: (a) + 20 (b) + 40 (c) – 50 (d) + 8 (e) + 10 (f ) – 2 (g) + 1 (h) Is it possible to purchase an item that costs 1568 pibs? (a) (32 × (– 18)) ÷ ((– 36)) (b) (32 ) ÷ ((– 36) × (– 18)) (c) (25 × (– 12)) ÷ ((45) × (– 27)) (d) (280 × (– 7)) ÷ ((– 8) × (– 35)) (a) (– 348) + (– 1064) (b) (– 348) – (– 1064) (c) 348 – (– 1064) (d) (– 348) × (– 1064) (e) 348 × (– 1064) (f) 348 × 964 (a) (– 547) × 972 (b) (– 548) × 971 (c) (– 547) × 971 Try This Chapter-2.indd 44 10/9/2025 2:28:02 PM 16. Fill in the blanks in at least 5 different ways with integers: 44 (a) the result is the maximum possible (b) the result is the minimum possible (a) + × = –36 (b) ( – ) × = 12 (c) ( – ( – )) = –1 • When two integers are multiplied, the product is positive when both the multiplier and multiplicand are positive, or when both are negative. The product is negative if one of them is positive and the other is negative. • When two integers are divided, the quotient is positive when both the dividend and divisor are positive, or both are negative. The quotient is negative when one of them is positive and the other is negative. • Integer multiplication is commutative, i.e., for any two integers a and b, a × b = b × a. • Integer multiplication is associative, i.e., for any three integers, a, b, and c, a × (b × c) = (a × b) × c. • Integer multiplication is distributive over addition, i.e., for any three integers, a, b, and c, a × (b + c) = (a × b) + (a × c). SUMMARY – 9 – 6 – 3 3 3 6 9 – 6 – 4 – 2 2 2 4 6 – 3 – 2 – 1 1 1 2 3 – 3 – 2 – 1 × 1 2 3 3 2 1 1 – 1 – 2 – 3 6 4 2 2 – 2 – 4 – 6 9 6 3 3 – 3 – 6 – 9 Operations with Integers Chapter-2.indd 45 10/17/2025 5:20:17 PM Terhüchü is a game played in Assam and Nagaland. The board has 16 squares and diagonals are marked as shown in the following figure. This is usually roughly scratched on a large piece of stone or just drawn on mud. There are 2 players, and each player has a set of 9 coins placed as shown. The coins in one set look different from those in the other set. Terhüchü 45 Objective The goal is to capture all the opponent’s coins. The first player to do so is the winner. A player may also win by blocking any legal move by their opponent. If a draw seems unavoidable, the player with more coins wins. Gameplay • The starting position of the game is as shown above. • Players take turns. In each turn, they can move a single coin along a line, in any direction, to a neighbouring vacant intersection. Or, if an opponent’s coin is at a neighbouring intersection, and there is a vacant intersection just beyond it, they can jump over the opponent’s coin and land in the vacant intersection. Chapter-2.indd 46 17-10-2025 16:48:58 • If a player jumps over an opponent’s coin, it is considered captured and is removed from the board. Multiple captures in one move are allowed, and the direction can change after each jump. • Inside the triangular corners, which are outside the main square, a coin may skip an intersection and move straight to the next one. That is, it can jump over an empty intersection and go to the one beyond it." class_7,11,finding common ground,ncert_books/class_7/gegp1dd2/gegp203.pdf,"3 3.1 The Greatest of All Sameeksha is building her new house. The main room of the house is 12 ft by 16 ft. She feels that the room would look nice if the floor is covered with square tiles of the same size. She also wants to use as few tiles as possible, and for the length of the tile to be a whole number of feet. What size tiles should she buy? Let us explore how to find the largest square tile that can be used. The breadth of the room is 12 ft and the length is 16 ft. FINDING COMMON GROUND Chapter-3.indd 47 10/9/2025 11:49:05 AM For the tiles to fit the breadth of the room exactly, the side of the tile should be a factor of 12. Similarly, for the tiles to fit the length of the room exactly, the side should be a factor of 16. So the side of the tile should be a factor of both 12 and 16. What are the common factors of 12 and 16? The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 16 are 1, 2, 4, 8, 16. The common factors are 1, 2, and 4. So, the square tiles can have sides 1 ft, 2 ft, and 4 ft. Among these, she should use the largest sized square tile. Can you explain why? Therefore, she needs tiles of size 4 ft. Ganita Prakash | Grade 7 | Part-II How many tiles of this size should she purchase? What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change? The Highest Common Factor (HCF) of two or more numbers, is the highest (or greatest) of their common factors. It is also known as the Greatest Common Divisor (GCD). In the previous problem, 4 is the HCF of 12 and 16. Lekhana purchases rice from two farms and sells it in the market. She bought 84 kg of rice from one farm and 108 kg from the other farm. She wants the rice to be packed in bags, so each bag has rice from only one farm and all bags have the same weight that is a whole number of kg. If she wants to use as few bags as possible, what should the weight of each bag be? To divide 84 kg of rice into bags of equal weight, we need the factors of 84. Similarly, for 108, we need the factors of 108. Factors of 84 — 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. Factors of 108 — 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. Since, Lekhana wants to use bags of the same weight for both farms, the weight of the bag should be a common factor. The common factors of 84 and 108 are 1, 2, 3, 4, 6, and 12. She can use any of these weights to pack rice from both farms in bags of equal weight. But, she wants to minimise the number of bags. Which weight should she choose to minimise the number of bags? We can draw rough diagrams like the one shown on the previous page to visualise the given scenario. It may help in understanding and solving. Try This Chapter-3.indd 48 10/10/2025 2:47:45 PM Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the chapter ‘Prime Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and tries to collect the treasure. In each case below, the two numbers upon which treasures are kept are given. Find the longest jump size (starting from 0) using which Jumpy can land on both the numbers having the treasure. (a) 14 and 30 (b) 7 and 11 (c) 30 and 50 (d) 28 and 42 Is the longest jump size for the numbers the same as their HCF? Explain why it is so. 48 So far, we have been listing all the factors to find the HCF. This can become cumbersome for numbers with many factors, as you would have observed for the numbers 30 and 50, and 28 and 42. Sometimes, we may also miss some factors which can lead to errors. Can this process be simplified? Can it be made more reliable? It turns out that using prime factorisation can simplify the process. We will start by revisiting primes and prime factorisation. Primes Recall that a prime is a number greater than 1 that has only 1 and the number itself as its factors. Last year, we tried to find patterns amongst primes between 1 to 100. We also came across the Sieve of Eratosthenes — a method to list all primes. Prime Factorisation Any number can be written as a product of primes — keep rewriting composite factors until only primes are left. Recall that we call this the prime factorisation of a given number. For example, we can find the prime factorisation of 90 as follows: The number 90 could also have been factorised as 3 × 30 or 2 × 45 or in a few other different ways. Will these all lead to the same prime factors? Remarkably, the resulting prime factors are always the same with perhaps only a change in their order. For example, if we consider factorising 3 × 30 further, we get 90 = 3 × 30 = 3 × 2 × 15 = 3 × 2 × 3 × 5, 90 = 3 × 3 × 2 × 5. Finding Common Ground Chapter-3.indd 49 10/17/2025 4:04:53 PM and we have arrived at the same prime factors, just in a different order. Note that the prime factorisation of a prime number is the prime number itself. Procedure for Prime Factorisation Can you see what is happening below? 3—— 105 5—— 35 7 2—— 30 3—— 15 5 49 Ganita Prakash | Grade 7 | Part-II Each circled number is the product of the numbers that are to its left and below. For example, 30 = 2 × 15, 15 = 3 × 5. Each time, a composite number is factorised so that, at least one factor is prime. We stop when we reach a prime number. Can you write the prime factorisation of 105 and 30 using these two figures? We collect the prime factors along the left and the one at the bottom. We then construct the prime factorisation as shown in the figure. While carrying out factorisation, the circles are usually left out and the steps are carried out in this format: Let us call this method the division method. Try finding the prime factorisation of 1200 using the method above. If we had used the earlier method, our calculation would have been as follows: Which calculation is easier to carry out? Factors of a Number Using Prime Factorisation From the prime factorisation of a number, we can construct all its factors. This can be used to simplify the procedure for finding the HCF of two numbers. Consider the number 840 and its prime factorisation 2 × 2 × 2 × 3 × 5 × 7. 3 105 5 35   7 1200 = 40 × 30 = 5 × 8 × 5 × 6 = … 2 30 3 15   5 105 = 3 × 5 × 7 3—— 105 5—— 35 7 Chapter-3.indd 50 10/10/2025 2:47:45 PM Is 2 × 2 × 7 = 28 a factor of 840? If yes, what should it be multiplied by to get 840? To answer these questions, we can reorder the prime factors of 840 as follows (recall that reordering factors does not change the product): So, 840 = 28 × 30. Thus, (2 × 2 × 7) = 28 is a factor of 840, and it should be multiplied by 30 (2 × 3 × 5) to get 840. 50 840 = (2 × 2 × 7) × (2 × 3 × 5). Similarly, is 2 × 7 = 14 a factor of 840? Why or why not? Is 2 × 2 × 2 a factor of 840? Why or why not? Is 3 × 3 × 3 a factor of 840? Why or why not? using just its prime factors? Find the factors of 225 using prime factorisation. gives us a factor. Let us systematically form these subparts. Prime factors: 3, 5. Combination of two prime factors: 3 × 3 = 9, 5 × 5 = 25, 3 × 5 = 15. Combination of three prime factors: 3 × 3 × 5 = 45, 3 × 5 × 5 = 75. Combination of four prime factors: 3 × 3 × 5 × 5 = 225. Adding 1 to this list of factors, we see that the factors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225. Check that all the factors of 225 occur in this list. Figure it Out List all the factors of the following numbers: (a) 90 (b) 105 Can we use this idea to list down all the possible factors of a number Factorising 225 to its primes, we get 225 = 5 × 5 × 3 × 3. We have seen that any ‘subpart’ of this factorisation Finding Common Ground 5 225 5 45 3 9 3 3 1 Chapter-3.indd 51 10/9/2025 11:49:06 AM (c) 132 (d) 360 (this number has 24 factors) (e) 840 (this number has 32 factors) a number is, the longer its prime factorisation will be”. What do you think of Anshu’s claim? We can see that it is not true. For example, look at the prime factorisations of 96 and 121. After observing a few prime factorisations, Anshu claims “The larger 96 = 2 × 2 × 2 × 2 × 2 × 3 121 = 11 × 11. 51 Ganita Prakash | Grade 7 | Part-II Finding the HCF of Numbers Using Prime Factorisation We now see how to make use of the observations made so far to find common factors and the Highest Common Factor (HCF). Example 1: Find the common factors, and the HCF of 45 and 75. We have seen: Factors of 45 are subparts of factors occurring in 3 × 3 × 5 and factors of 75 are subparts of factors occurring in 3 × 5 × 5. So the common factors should be subparts of both the factorisations. Can you write them down? Calculate the prime factorisation for both numbers: In mathematics, statements or claims made without proof or verification are called ‘conjectures’. Anshu’s claim is a conjecture. We disproved this conjecture by finding a counterexample, i.e., an example where the conjecture is false. While exploring or solving problems you might also get some conjectures. You can try to reason and verify these and also share with the class. 3 × 3 × 5 5 3 × 5 × 5 3 × 3 × 5 3 3 × 5 × 5 75 = 3 × 5 × 5. 45 = 3 × 3 × 5 Chapter-3.indd 52 10/10/2025 3:16:36 PM 3, 5, 3 × 5 = 15 are subparts of both, and hence, they are the common factors along with 1. The highest among them is 15. So, their HCF = 15. Example 2: Find the common factors, and the HCF of 112 and 84. 52 Calculating prime factorisations, we get, 3 × 3 × 5 3 × 5 3 × 5 × 5 112 = 2 × 2 × 2 × 2 × 7 and 84 = 2 × 2 × 3 × 7. The highest among the common factors, HCF, is 2 × 2 × 7 = 28. Example 3: Find the common factors and the HCF of 96 and 275. We have 96 = 2 × 2 × 2 × 2 × 2 × 3 275 = 5 × 5 × 11. There is no subpart that is common amongst these two factorisations. So, 1 is the only common factor. It is also their HCF. Figure it Out Find the common factors and the HCF of the following numbers: Finding the common subparts, we get 2 × 2 × 2 × 2 × 7 7 2 × 2 × 3 × 7 2 × 2 × 2 × 2 × 7 2 × 2 2 × 2 × 3 × 7 2 × 2 × 2 × 2 × 7 2 × 7 2 × 2 × 3 × 7 2 × 2 × 2 × 2 × 7 2 2 × 2 × 3 × 7 2 × 2 × 2 × 2 × 7 2 × 2 × 7 2 × 2 × 3 × 7 Finding Common Ground Chapter-3.indd 53 10/9/2025 11:49:06 AM How do we directly find the HCF without listing all the factors? Example 4: Find the HCF of 30 and 72. We need the largest common subpart to find the HCF. Clearly, it will contain only those primes that occur in both the factorisations: 2 and 3 in this case. How many 2s will it contain? (a) 50, 60 (b) 140, 275 (c) 77, 725 (d) 370, 592 (e) 81, 243 72 = 2 × 2 × 2 × 3 × 3 30 = 2 × 3 × 5 53 Ganita Prakash | Grade 7 | Part-II The prime factorisation of 30 has only one 2. So, the largest common subpart should contain only one 2. How many 3s will it contain? The prime factorisation of 30 has only one 3. So, the largest common subpart should contain only one 3. This has been carried out below. Thus, to find the HCF, we identify the common primes and find the minimum number of times each of them appear in the factorisations of the given numbers. Example 5: Find the HCF of 225 and 750. Common primes: 3 and 5. Let us find the largest common subpart. How many 3s will it contain? 750 contains only one 3, which is the minimum number of 3 across both numbers. So the largest common subpart should contain only one 3. How many 5s will it contain? 225 contains the minimum number of 5s which occurs two times. So, the largest common subpart should contain two 5s, i.e., 5 × 5. HCF = 2 × 3 = 6 30 = 2 × 3 × 5 72 = 2 × 2 × 2 × 3 × 3 225 = 3 × 3 × 5 × 5 750 = 2 × 3 × 5 × 5 × 5 HCF = 3 × 5 × 5 = 75 225 = 3 × 3 × 5 × 5 750 = 2 × 3 × 5 × 5 × 5 Chapter-3.indd 54 10/9/2025 11:49:06 AM To find the HCF of more than 2 numbers, a similar method of finding the largest common subpart of the common primes can be followed. Figure it Out 54 1. Find the HCF of the following numbers: 2. Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not? (a) 24, 180 (b) 42, 75, 24 (c) 240, 378 (d) 400, 2500 (e) 300, 800 3.2 Least, but not Last! Anshu and Guna make torans out of strips of cloth. Multiple strips are placed one next to another to make a toran. Anshu uses strips of length 6 cm and Guna uses strips of 8 cm length. If both have to make torans of the same length, what is the smallest possible length, the torans could be? What is the length of the shortest toran that they can both make? Anshu uses cloth strips of 6 cm each. Any toran he makes will be a multiple of 6. So, the length of the toran could be 6, 12, 18, 24, 30, 36, 42, 48, 54 cm, and so on. Similarly, any toran Guna makes should be a multiple of 8. So, the length of the toran he makes could be 8, 16, 24, 32, 40, 48, 56, 64, 72 cm, and so on. From this, we can see that if both have to make torans of the same length, the length of the toran should be a common multiple of 6 and 8. From the two lists, we can see that 24 and 48 are two of the common multiples of 6 and 8. So, 24 cm and 48 cm are lengths of toran that Anshu and Guna can both stitch. 24 is the smallest among them. So, 24 cm is the length of the shortest toran that both can stitch. 24 is the lowest number among all the common multiples of 6 and 8. What about the largest common multiple? Does such a number exist? A sweet shop gives out free gajak to school children on Mondays. Today is a Monday and Kabamai enjoyed eating the gajak. But she visits the sweet shop once every 10 days. When is the next time she would be able to get free gajak from Sweet shop? (Answer in number of days.) Finding Common Ground Chapter-3.indd 55 10/17/2025 4:06:14 PM Since the shop gives free sweets every Monday, it will give free sweets again after 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, ... days. These are multiples of 7. As we saw before, imagining or visualising the scenario helped us to see that it can be solved using multiples of the strips’ lengths. Gajak is a sweet made from sesame seeds, jaggery and ghee 55 Ganita Prakash | Grade 7 | Part-II These are multiples of 10. When will Kabamai eat free sweets again? It will happen on days common to the sequences of multiples above. It can be seen that this will first happen after 70 days. Notice that, here too, 70 is the lowest among all the common multiples of 7 and 10. For both these problems the solution was the lowest common multiple. The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the first number for which ‘idli-vada’ will be called out: (a) 4 and 6 (b) 7 and 11 (c) 14 and 30 (d) 15 and 55 Is the answer always the LCM of the two numbers? Explain. As in the case of the HCF, the process of finding the LCM by listing down the multiples may get tedious for larger numbers, as you would have seen for questions (c) and (d) above. Kabamai will arrive at the sweet shop again after 10, 20, 30, 40, 50, 60, 70, ... days. Chapter-3.indd 56 10/9/2025 11:49:07 AM Prime factorisation can simplify the process of finding the LCM as well. How do we find the LCM of two numbers using their prime factors? Finding LCM through Prime Factorisation We have seen that every factor of a number is formed by taking a subpart of its prime factorisation. We used this fact to come up with a method to find the HCF of two numbers. In a similar manner, we can come up with a method to find the LCM. We begin by comparing the prime factorisations of a number and a multiple of that number. For example, let us take 36 and its multiple 648 (=36 × 18). 56 What do you observe? We can see that the prime factors of the multiple contain the prime factors of the number along with some more prime factors. Will this happen with every multiple? Can this be used to find the LCM? Example 6: Find the LCM of 14 and 35. Is 2 × 3 × 5 × 7 also a common multiple? What is the lowest among all the common multiples of 14 and 35? It is 2 × 5 × 7 = 70 because 2 × 5 × 7 contains 14 = 2 × 7 as well as 35 = 5 × 7, and removing any number from 2 × 5 × 7 will give a number that is not a common multiple of 14 and 35. Example 7: Find the LCM of 96 and 360. We get, 36 = 2 × 2 × 3 × 3, 648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3). We get, 14 = 2 × 7 35 = 5 × 7. Common multiples should contain each prime factor as a subpart: 2 × 7 as a subpart and 5 × 7 as a subpart. For example, 2 × 7 × 5 × 7 × 3 is a common multiple of 14 and 35. 2 × 2 × 5 × 7 × 7 × 11 is another common multiple. We have, 96 = 2 × 2 × 2 × 2 × 2 × 3 360 = 2 × 2 × 2 × 3 × 3 × 5. Finding Common Ground Chapter-3.indd 57 10/9/2025 11:49:07 AM Every common multiple should contain the prime factors of both 96 and 360. For the LCM, we will need the smallest such number. Let us build the LCM looking at each prime factor. The prime factors appearing in both the numbers are 2, 3 and 5. Now, we shall find out how many occurrences there are for each prime factor. How many 2s should the LCM contain? The factorisation of 96 contains 2 × 2 × 2 × 2 × 2 (five occurrences of 2s) and the factorisation of 360 contains 2 × 2 × 2 (three occurrences of 2s). Choosing five occurrences of 2s as part of LCM will contain both these subparts. 57 Ganita Prakash | Grade 7 | Part-II Choosing more than five occurrences of 2s will give a common multiple; but it will not be the lowest. Are you able to see why? How many 3s should the LCM contain? The factorisation of 96 contains 3 (one occurrence of 3) and the factorisation of 360 contains 3 × 3 (two occurrences of 3s). Choosing two occurrences of 3s as part of LCM will contain both these subparts. How many 5s should the LCM contain? The factorisation of 96 doesn’t have any 5s and the factorisation of 360 has one occurrence of 5. So, we choose one occurrence of 5 to be a part of the LCM. Thus, the LCM of 96 and 360 will be 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440. To build the LCM of two numbers, we can identify all the prime factors and find the maximum number of times each of them occur in either of the factorisations. This process can be extended to find the LCM of two or more numbers. Figure it Out Find the LCM of the following numbers: (a) 30, 72 (b) 36, 54 (c) 105, 195, 65 (d) 222, 370 3.3 Patterns, Properties, and a Pretty Procedure! The HCF of 6 and 18 is 6, which is one of the two given numbers. Chapter-3.indd 58 10/9/2025 11:49:07 AM Find more such number pairs where the HCF is one of the two numbers. How can we describe such pairs of numbers? We can see that it happens when one number is a factor of the other. This also means the other number will be a multiple of the first number! Such a statement describing a pattern or a property that holds in all possible cases is called a general statement. This process is called generalisation. Such a generalisation can also be described using algebra. Let us see how. 58 Be my factor, I’ll be your multiple! If n is a number, then any multiple of n can be written as a positive integer multiplied by n. For example, if we take n and 5n (short for 5 × n), then 5n is a multiple of n, and n is a factor of 5n. The HCF of n and 5n = n. For number pairs satisfying this property (i.e., one of the numbers is the HCF), (a) if m is a number, what could be the other number? (b) if 7k is a number, what could be the other number? Figure it Out (e) Two co-prime numbers Share your observations with the class. 1. Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold. 2. The LCM of 3 and 24 is 24 (it is one of the two given numbers). (a) Two consecutive even numbers (b) Two consecutive odd numbers (c) Two even numbers (d) Two consecutive numbers (a) Find more such number pairs where the LCM is one of the two numbers. (b) Make a general statement about such numbers. Describe such number pairs using algebra. Finding Common Ground Math Talk Chapter-3.indd 59 10/9/2025 11:49:08 AM What happens to the HCF of two numbers if both numbers are doubled? Take some pairs of numbers and explore. Are you able to see why the HCF will also double? 3. Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold. (a) Two multiples of 3 (b) Two consecutive even numbers (c) Two consecutive numbers (d) Two co-prime numbers 59 Ganita Prakash | Grade 7 | Part-II If both numbers are doubled, then both numbers get an extra factor of 2 in their prime factors. This 2 will be included as a factor in the largest common subpart, and so the HCF will double. For example, consider the numbers 270 and 50. Let us double these numbers to get 540 and 100. Consider the following two multiples of 14 — 14 × 6, 14 × 9. What is their HCF? Clearly, 14 is a common factor. Is it also the highest common factor? To see it, let us calculate the prime factorisations. Here are some more numbers where both numbers are multiples of the same number. Find their HCF: HCF = 2 × 5 = 10 270 = 2 × 3 × 3 × 3 × 5 50 = 2 × 5 × 5 HCF = 2 × 2 × 5 =20. 540 = 2 × 2 × 3 × 3 × 3 × 5 100 = 2 × 2 × 5 × 5 14 × 6 = 2 × 7 × 2 × 3 14 14 × 9 = 2 × 7 × 3 × 3 14 HCF = 14 × 3 = 42. Chapter-3.indd 60 10/9/2025 11:49:08 AM (a) 18 × 10, 18 × 15 (b) 10 × 38, 10 × 21 (c) 5 × 13, 5 × 20 (d) 12 × 16, 12 × 20 In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens. Efficient Procedures for HCF and LCM See the procedure on the right. Can you explain how it has been carried out? 60 2 84, 180 2 42, 90 3 21, 45 7, 15 This is similar to the procedure for prime factorisation. At each step, the two numbers are divided by a common prime factor, and the two quotients are written down in the next row. This continues till we get two numbers that do not have any common prime factors. How do we use this to find the HCF of 84 and 180? Explore. [Hint: Observe that 84 = 2 × 2 × 3 × 7, and 180 = 2 × 2 × 3 × 15 similar to prime factorisation] Find the HCF in the following cases. This procedure not only gives the HCF but can also be used to find the LCM! Can you see how? HCF = 2 × 5 × 5 × 3 HCF = 2 × 5 × 7 2 300, 150 5 150, 75 5 30, 15 3 6, 3 2, 1 2 300, 150 5 150, 75 5 30, 15 3 6, 3 2, 1 2 630, 770 5 315, 385 7 63, 77 3 9, 11 3, 11 2 630, 770 5 315, 385 7 63, 77 9, 11 Finding Common Ground Chapter-3.indd 61 10/17/2025 5:10:01 PM Why are these the LCMs? [Hint: Will the product of the factors marked as the LCM of 300 and 150 contain the prime factorisations of both 300 and 150? Is this the smallest such number?] Guna says “I found a better way to factorise to find HCF/LCM. This is faster than what was taught in class!” “For the numbers 300 and 150, I can first directly divide both numbers by 50. The HCF will be 50 × 3. 50 300, 150 3  6,  3 LCM = 2 × 5 × 5 × 3 × 2 × 1 LCM = 2 × 5 × 7 × 3 × 3 × 11  2,  1 61 Ganita Prakash | Grade 7 | Part-II The LCM will be 50 × 3 × 2 × 1”. Anshu also tried to remove the bigger common factors. “For 630 and 770, I will divide both numbers by 10 first. Now, I can divide them by 7. The HCF will be 10 × 7 = 70 The LCM will be 10 × 7 × 9 × 11 = 6930”. Can you see why this works? We need not restrict ourselves to dividing only one prime factor at a time. Both numbers can be divided by whatever common factor we are able to identify. You can try this method for these pairs of numbers. (a) 90 and 150 (b) 84 and 132 Property Involving both the HCF and the LCM Which is greater — the LCM of two numbers or their product? You could analyse the above statement using examples. Then try to reason or prove, why the LCM is never greater than the product of the numbers. [Hint: Is the product also a common multiple of the two numbers?] There is an interesting relation between the product of two numbers and their HCF and LCM. Consider the numbers 105 and 95. Find their LCM. Factorising them into their primes: 105 = 3 × 5 × 7 95 = 5 × 19 10 630, 770 7 63, 77 9, 11 Try This Chapter-3.indd 62 10/17/2025 4:17:17 PM Is the LCM a factor of the product? If yes, what should it be multiplied with to get the product? It can be seen that Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers: (a) 45, 105 (b) 275, 352 (c) 222, 370 62 LCM = 3 × 5 × 7 × 19. Let us consider the product in the factorised form: 105 × 95 = 3 × 5 × 5 × 7 × 19 105 × 95 = LCM × 5. Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF? Why does this happen? Can you give an explanation or proof? [Hint: Consider the prime factorisation of the given numbers. Among their prime factors, some are common to both factorisations, and the rest occur in only one of them. Between the HCF and the LCM, see how the common and non-common prime factors get distributed. In the product, observe how these two kinds of prime factors occur. Compare them.] Explore whether this property holds when 3 numbers are considered. Figure it Out 1. In the two rows below, colours repeat as shown. When will the blue stars meet next? 2. (a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2? (b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2? 3. Find the HCF and LCM of the following (state your answers in the form of prime factorisations): Thus, our observations seem to suggest the following: HCF × LCM = Product of the two numbers. Finding Common Ground Try This Try This Chapter-3.indd 63 10/10/2025 3:23:19 PM 4. Find two numbers whose HCF is 1 and LCM is 66. 5. A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.) (a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11 (b) 45 and 36 63 Ganita Prakash | Grade 7 | Part-II 10. Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be: 6. The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps? 7. Among the numbers below, which is the largest number that perfectly divides both 306 and 36? 8. Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11. 9. Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially? (a) 9 cm (b) 6 cm (c) 4 cm (d) 3 cm (e) 2 cm (a) 36 (b) 612 (c) 18 (d) 3 (e) 2 (f) 360 (a) 72 (b) 90 (c) 45 (d) 3 (e) 36 (f) None of these (a) Less than both numbers (b) In between the two numbers Try This Chapter-3.indd 64 10/9/2025 11:49:08 AM 11. A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit? 12. What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5? 13. Here is a problem posed by the ancient Indian Mathematician 64 (c) Greater than both numbers (d) Less than m × n (e) Greater than m × n Mahaviracharya (850 C.E.). Add together 8 15, 1 20, 7 36, 11 63 and 1 21. What do you get? How can we find this sum efficiently? Math Talk Try This • Last year, we looked at common multiples, and common factors, and were also introduced to the amazing world of primes! • In this chapter, we learnt a method to find the prime factorisation of a number. • Finding all the factors of a number from its prime factorisation is easy but quite tedious — we have to list every possible subpart! • The Highest Common Factor (HCF) is the highest among all the common factors of a group of numbers. Every common factor is contained in the prime factorisation of the number. To find the HCF, we include the minimum number of occurances of each prime across the prime factorisation of all the numbers. • The Lowest Common Multiple (LCM) is the lowest among all the common multiples of a group of numbers. Every common multiple contains the prime factorisation of the numbers. To find the LCM, we include the highest number of occurrences of each prime across the prime factorisations of all the numbers. • We explored more about HCF and LCM; we discovered related properties and patterns when numbers are consecutive, even, co-prime, etc. • We learnt a procedure to get both the HCF and the LCM at the same time! We also saw how to make this even quicker! • We learned some terms that are used when discussing mathematics, such as ‘conjecture’ and ‘generalisation’. SUMMARY Finding Common Ground Chapter-3.indd 65 10/9/2025 11:49:08 AM The largest prime found so far has 4,10,24,320 digits! It was discovered on October 12, 2024. If I start writing this number, how long could it take me? 65 You might have noticed and wondered about these different circle designs around the page numbers on each page! The picture below shows all the designs for the numbers from 1 to 100. Mystery Colours! Chapter-3.indd 66 10/10/2025 3:24:54 PM Try to decode the colour scheme for each number. There are several interesting patterns here. Share your observations with your classmates. Extending this scheme, colour the page numbers from 101 – 110." class_7,12,another peek beyond the point,ncert_books/class_7/gegp1dd2/gegp204.pdf,"4 4.1 A Quick Recap of Decimals Recall that decimals are the natural extension of the Indian place value system to represent decimal fractions ( 1 10, 1 100, 1 1000, and so on) and their sums. For example, 27.53 refers to a quantity that has: • 2 Tens We have already learned how to multiply and divide fractions. In this chapter, we will learn how to perform these operations with decimals. You will see that the procedures for multiplying and dividing decimals are natural extensions of the procedures for multiplying and dividing counting numbers. • 7 Units (Ones) • 5 Tenths • 3 Hundredths ANOTHER PEEK BEYOND THE POINT Chapter-4.indd 67 10/9/2025 2:44:41 PM Jonali and Pallabi play a game. Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi’s answer in the blank spaces. Jonali goes to the market to buy spices. She purchases 50 g of Cinnamon, 100 g of Cumin seeds, 25 g of Cardamom and 250 g of Pepper. Express each of the quantities in kilograms by writing them in terms of fractions as well as decimals. The fractions Jonali gave Pallabi have denominators 10, 100, 1000, and so on. Ganita Prakash | Grade 7 | Part-II Write the following fractions as a sum of fractions and also as decimals: Can you give a simple rule to divide any number by a number of the form 1 followed by zeroes — 10, 100, 1000, etc.? For example, 123 10 , Fraction Expanding the Numerator 847 10000 254 1000 200 1000 + 50 1000 + 4 1000 23 1000 173 100 = 2 10 + 5 100 + 4 1000 Sum of one-tenths, one-hundredths, one-thousands,... Decimals 0.2 + 0.05 + 0.004 0.254 Math Talk Chapter-4.indd 68 10/10/2025 3:26:47 PM 24 100 or 678 1000 ? Look for a pattern in the previous problems. Here is one such rule. Let us consider the example 123 ÷ 10. Step 1: Write the dividend as it is and place a decimal point at the end. Step 2: Count the number of zeroes in the divisor. Step 3: Move the decimal point from Step 1 left by the same number of places as the count from Step 2. Add zeroes in front if needed. 68 123. 10 1 zero 12.3 Examples: 4.2 Decimal Multiplication Example 1: Arshad goes to a stationery shop and purchases 5 pens. If one pen costs ₹9.5 (9 rupees and 50 paisa), how much should he pay the shopkeeper? What operation must we use here? We have to multiply 9.5 by 5, which is the same as adding 9.5, 5 times. That is 9.5 × 5 = 9.5 + 9.5 + 9.5 + 9.5 + 9.5 = 47.5. We can also directly multiply the numbers by converting them into fractions. 9.5 is 95 10 and 5 is 5 1 as a fraction. The cost of 5 pens = 5 1 × 95 10 . Recall that, to find the product of two fractions, we multiply the numerators and multiply the denominators. 5 1 × 95 10 = 5 × 95 1 × 10 24 ÷ 100 = 0.24 678 ÷ 1000 = 0.678 12÷1000 = 0.012 12345÷1000 = 12.345 Another Peek Beyond the Point Chapter-4.indd 69 17-10-2025 16:51:08 The cost of 5 pens is ₹47.5. Example 2: A car travels 12.5 km per litre of petrol. What is the distance covered with 7.5 litres of petrol? We have to multiply 12.5 by 7.5. The distance covered is 93.75 km. The distance covered = 12.5 × 7.5 = 125 10 × 75 10 = 125×75 10×10 = 9375 100 = 93.75 = 475 10 = 47.5. 69 Ganita Prakash | Grade 7 | Part-II Can the product of two decimals be a natural number? Can the product of a decimal and a natural number be a natural number? Example 3: The distance between Ajay's school and his home is 827 m. He walks to school in the morning and then walks back home in the evening, 6 days a week. How much does he walk in a week? Answer in kilometres. Each way between school and home, Ajay walks 827 metres, i.e., 0.827 km. So, in a day he walks, 1.654 × 6 = 1654 1000 × 6 = 1654 × 6 1000 = 9924 1000 = 9.924 Ajay walks 9.924 km a week. Example 4: Find the area of the given rectangle. 5.7 × 13.3 = 57 10 × 133 10 = 7581 100 = 75.81 The area is 75.81 sq cm. Observe the number of digits after the decimal point in the multiplier, the multiplicand and the product. Also note the number of zeroes in the denominator. 0.827 × 2 = 827 1000 × 2 = 827 × 2 1000 = 1654 1000 = 1.654 He goes to school 6 days a week. So, in a week, he walks Area of the rectangle = 5. 7 c m 13.3 cm Math Talk Chapter-4.indd 70 10/9/2025 2:44:42 PM 9.5 × 5 = 95 × 5 10 = 475 10 = 47.5 1 0 1 12.5 × 7.5 = 125 10 × 75 10 = 9375 100 = 93.75 70 Examples No. of digits after the decimal point in: Multiplier Multiplicand Product 1 1 2 Suppose we know that 596 × 248 = 147808, can you immediately write down the product of 5.96 × 24.8? By looking at the above examples, can you frame a rule to multiply two decimals? Multiplication of decimals is the same as the multiplication of their corresponding fractions. When multiplying fractions, we multiply the numerators and denominators, respectively. The product of the numerators = Product of the numbers with the decimal points removed. Since both the denominators are of the form 1, 10, 100, 1000, … , the product of the denominators is also of the form 1 followed by zeroes. The number of zeroes in the product is the sum of the number of zeroes in each denominator. In the product, the decimal point is placed based on the total number of zeroes in the denominator. So, to multiply two decimals, we can multiply the two numbers obtained by removing the decimal point, and then place the decimal point appropriately as shown below. To evaluate 5.96 × 24.8: 1.64 × 6 = 164 100 × 6 = 984 100 = 9.84 2 0 2 5.7 × 13.35 = 57 10 × 1335 100 = 57 × 1335 10 × 100 = 76095 1000 =76.095 1 2 3 Another Peek Beyond the Point Math Talk Chapter-4.indd 71 10/9/2025 2:44:42 PM 2 decimal places 1 decimal place 3 decimal places 2 + 1 = 3 decimal places 5.96 × 24.8 = 147.808 } } } } 596 × 248 = 147808 71 Ganita Prakash | Grade 7 | Part-II Example 5: Let us use the above rule to find the product of 5.8 and 1.24. Let us first multiply 58 and 124. The product is 7192. The sum of the number of digits after the decimal point in the multiplier and multiplicand is 3. So, the product is 7.192. Verify this by converting the multiplier and multiplicand into fractions. Is the Product Always Greater than the Numbers Multiplied? Recall multiplication of two fractions. Unlike counting numbers, when two fractions are multiplied the product is not always greater than or equal to both numbers. Let us examine the case of the product of two decimals. 2.25 × 8 = 18. In the above multiplication, the product (18) is greater than 2.25 and 8. But in, 0.25 × 8 = 2, the product (2) is greater than 0.25 but less than 8. In the case of 0.25 × 0.8 = 0.2, the product (0.2) is less than both 0.25 and 0.8. When is the product of two decimals greater than both the numbers? When is it less than both the numbers? Since decimals are just representations of fractions, the relationship between the numbers multiplied and the product are similar to fractions. 1 digit after decimal point 2 digits after decimal point 5.8 × 1.24 Chapter-4.indd 72 10/9/2025 2:44:42 PM Situation 1 Both numbers are greater than 1 (e.g., 3.4 × 6.5) The product (22.1) is greater than both the numbers. Situation 2 Both numbers are between 0 and 1 (e.g., 0.75 × 0.4) The product (0.3) is less than both the numbers. Situation 3 One number is between 0 and 1 and one number is greater than 1 (e.g., 0.75 × 5) 72 Situation Multiplication Relationship The product (3.75) is less than the number greater than 1 and greater than the number between 0 and 1. Figure it Out 1. Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths and so on: 2. Find the products: 3. Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts? 4. Meenu bought 4 notebooks and 3 erasers. The cost of each book was ₹15.50 and each eraser was ₹2.75. How much did she spend in all? 5. The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by placing 36 rupee coins one over the other? Write the answer in centimeters. 6. The price of 1 kg of oranges is ₹56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why? 7. Dwarakanath purchases notebooks at a wholesale price of ₹23.6 per piece and sells each notebook at ₹30/-. How much profit does he make if he sells 50 books in a week? 8. Given that 18 × 12 = 216, find the products: (a) 6 × 4 tenths = 24 tenths (b) 7 × 0.3 (c) 9 × 5 hundredths (a) 27.34 × 6 (b) 4.23 × 3.7 (c) 0.432 × 0.23 Another Peek Beyond the Point Math Talk Chapter-4.indd 73 10/9/2025 2:44:42 PM 10. Multiplying the following numbers by 10, 100 and 1000 to complete the table. 9. In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications? (a) 18 × 1.2 (b) 18 × 0.12 (c) 1.8 × 1.2 (d) 0.18 × 0.12 (e) 0.018 × 0.012 (f) 1.8 × 12 In which of the cases above is the product less than 1? (a) 7 × 0.6 (b) 0.7 × 0.6 (c) 0.7 × 6 (d) 0.07 × 0.06 73 Ganita Prakash | Grade 7 | Part-II 4.3 Decimal Division Example 6: Anuja has a 3.9 m length of ribbon and she wants to cut it into 10 equal pieces. What is the length of each piece in decimal? Since there are 10 pieces, we can find the length of each piece by dividing 3.9 by 10. So, what is 3.9 ÷ 10? Let’s convert 3.9 into fraction 3.9 = 39 10. Hence, 3.9 ÷ 10 is the same as dividing 39 10 by 10. Recall that, dividing a fraction by a divisor is the same as multiplying the fraction by the reciprocal of the 23.02 0.306 24.67 0.92 5.7 × 10 × 100 × 1000 What is the length of each piece in decimal? Chapter-4.indd 74 10/9/2025 2:44:43 PM divisor. The reciprocal of 10 is 1 10. So, 39 10 ÷ 10 = 39 10 × 1 10 = 39 100 = 0.39. Thus, the length of each piece of ribbon is 0.39 m. What is the length of each piece if the ribbon is cut into 100 equal pieces? 3.9 ÷ 100 = 39 10 × 1 100 = 39 1000 = 0.039. When the ribbon is cut into 100 equal pieces, the length of each piece is 0.039 m. 74 What is 0.039 m in centimetres and millimetres? By looking at these divisions, we can frame a simple rule for dividing decimals by 1, 10, 100, 1000, and so on. When we divide a decimal by 1, 10, 100, 1000, and so on, we can just move the decimal point to the left by as many places as there are zeroes in the divisor! Example 7: Neenu has 29 metres of red ribbon and wants to share it equally with Anu. What is the length of ribbon that each of them will get? Since the ribbon needs to be divided into two equal parts, each girl will get a piece of 29 ÷ 2 metres. If each of them gets 14 m, then 1 m remains. If we divide 1 m among the two, Decimal ÷ 10 ÷ 100 ÷ 1000 ÷ 10000 18.7 1.87 0.187 0.0187 0.00187 21.1 0.13 2.146 Another Peek Beyond the Point 0.0058 Chapter-4.indd 75 10/9/2025 2:44:44 PM each will get another 1 2 m. How do we convert 1 2 into a decimal? It is easy to express a fraction as a decimal if the denominator is 1, 10, 100, 1000, etc. So, can we find a fraction equivalent to 1 2 with such a denominator? 1 2 = 1 × 5 2 × 5 = 5 10 (multiplying the numerator and denominator by 5). 75 Ganita Prakash | Grade 7 | Part-II We know that the fraction 5 10 can be represented as a decimal 0.5. So, each girl will get 14 m and an additional 0.5 m of ribbon. Hence, the length of ribbon each will get is 14 + 0.5 = 14.5 m. Now, what if the ribbon was shared between four friends instead of 2? So, each will get 29 ÷ 4 m, that is 29 4 m. Now, the denominator of the fraction is 4. To convert a fraction to a decimal, it helps if the denominator is of the form 1, 10, 100, 1000, and so on. Can we find a fraction equivalent to 29 4 with such a denominator? Is 4 a factor of 10? No. Is it a factor of 100? Yes. 4 × 25 = 100. So we can get an equivalent fraction of 29 4 by multiplying the numerator and denominator by 25. So each of the 4 friends will get 7.25 m of ribbon. Division Using Place Value We have seen how to divide two counting numbers to get a decimal quotient. We first represented the division as a fraction. Then we found an equivalent fraction, with the denominator being of the form 1, 10, 100, 1000, and so on. It was then easy to represent this equivalent fraction as a decimal. 29 × 25 4 × 25 = 725 100 = 7.25 Chapter-4.indd 76 17-10-2025 16:51:59 Now, let us look at the division using place value procedure to calculate the decimal quotient. Suppose we want to write the quotient 10 3 as a decimal. Can we convert this fraction to an equivalent fraction with a denominator such as 1, 10, 100, 1000, etc.? It is not possible. So, we need a more general method to divide any two counting numbers. Let us see how we can use division using place value for this. Let us start with a quick recap of division using place value. 76 Example 8: Find the value of 1324 ÷ 4. 1324 ÷ 4 Divide 1324 into 4 equal parts. 1 Thousand ÷ 4 Not possible without regrouping. Regroup 1 Thousand into 10 Hundreds. 10 Hundreds + 3 Hundreds = 13 Hundreds. 13 Hundreds ÷ 4 Each part gets 3 Hundreds, and 1 Hundred remains. 13 Hundreds + 2 Tens + 4 Ones 1 Thousand + 3 Hundreds + 2 Tens + 4 Ones 13 Hundreds + 2 Tens + 4 Ones Another Peek Beyond the Point Th H T O Chapter-4.indd 77 10/9/2025 2:44:44 PM Regroup 1 Hundred into 10 Tens. 10 Tens + 2 Tens = 12 Tens. 3 Hundreds 3 Hundreds 3 Hundreds 3 Hundreds 1 Hundred 4) 1324 (0 + 3 + ___ + ___ –12 1 77 Ganita Prakash | Grade 7 | Part-II 12 Tens ÷ 4 Each part gets 3 Tens. 4 Ones ÷ 4 Each part gets 1 Ones. 3 Hundreds 3 Hundreds 3 Tens 3 Hundreds 3 Tens 3 Hundreds 3 Tens 3 Hundreds 3 Tens 3 Tens 12 Tens + 4 Ones 4 Ones 3 Hundreds 3 Tens 3 Hundreds 3 Tens 3 Hundreds 3 Tens Th H T O 4) 1324 (0 + 3 + 3 + ___ –12 12 –12 0 4) 1324 (0 + 3 + 3 + 1 –12 12 –12 04 – 4 0 Th H T O Chapter-4.indd 78 10/9/2025 2:44:44 PM So, 1324 ÷ 4 = 0 Thousands + 3 Hundreds + 3 Tens + 1 Ones = 331. We also call this division using place values as ‘long division’. Division with a Decimal Quotient Now, let us use this understanding of long division to find the value of 1325 ÷ 4. 1325 ÷ 4 Divide 1325 into 4 equal parts. We can follow the same steps as in the previous problem. 78 1 Ones 1 Ones 1 Ones 1 Ones We are left with 1 Ones. It is not clear how to divide 1 Ones into 4 equal parts. But we can regroup this as 10 tenths. 3 Hundreds 1 Ones 3 Tens 5 Ones 3 Hundreds 3 Tens 1 Ones 3 Hundreds 1 Ones 3 Tens 3 Hundreds 3 Tens 1 Ones 10 Tenths 3 Hundreds 1 Ones 1 Ones 3 Tens 3 Hundreds 3 Tens 1 Ones 3 Hundreds Another Peek Beyond the Point 1 Ones 3 Tens 3 Hundreds 4) 1325 (0 + 3 + 3 + 1 –12 12 –12 05 – 4 1 Th H T O 3 Tens 1 Ones Chapter-4.indd 79 17-10-2025 16:53:33 10 Tenths ÷ 4 Each part gets 2 Tenths and 2 Tenths remain. 3 Hundreds 3 Tens 1 Ones 2 Tenths 10 Tenths 3 Hundreds 3 Tens 1 Ones 2 Tenths 3 Hundreds 3 Tens 1 Ones 2 Tenths 2 Tenths 4) 1325 (0 + 3 + 3 + 1 + 2 –12 12 –12 05 – 4 10 – 8 2 Th H T O Te nt hs 3 Hundreds 3 Tens 1 Ones 2 Tenths Tens Ones Tenths 79 Ganita Prakash | Grade 7 | Part-II We are left with 2 tenths. To divide 2 Tenths into 4 equal parts, we have to regroup them as 20 Hundredths. 20 Hundredths ÷ 4 Each part gets 5 Hundredths. 2 Tenths 5 Hundredths 3 Hundreds 3 Tens 1 Ones 20 Hundredths 3 Hundreds 3 Tens 1 Ones 2 Tenths 2 Tenths 5 Hundredths 3 Hundreds 3 Tens 1 Ones 3 Hundreds 3 Tens 1 Ones 2 Tenths 2 Tenths 5 Hundredths 20 Hundredths 3 Hundreds 3 Tens 1 Ones 2 Tenths 4) 1325 (0 + 3 + 3 + 1 + 2 + 5 –12 12 –12 05 – 4 10 – 8 20 – 20 0 Th H T O Ten th s 3 Hundreds 3 Tens 1 Ones 2 Tenths 2 Tenths 5 Hundredths 3 Hundreds 3 Tens 1 Ones 3 Hundreds 3 Tens 1 Ones 2 Tenths Tens Ones Tenths Hundredths H un dr ed th s Chapter-4.indd 80 17-10-2025 16:55:11 So, 1325 ÷ 4 = 0 Thousands + 3 Hundreds + 3 Tens + 1 Ones + 2 Tenths + 5 Hundredths. This we know is 331.25, thus, 1325 ÷ 4 = 331.25. Can we verify this by finding an equivalent fraction for 1325 4 ? To get an equivalent fraction such that the denominator is of the form 1, 10, 100, 1000, and so on, we can multiply the numerator and denominator by 25. 80 1325 × 25 4 × 25 = 33125 100 = 331.25. Thus, the procedure of division using place value can be extended to find quotients with decimal values. Ones can be regrouped as tenths, tenths can be regrouped as hundredths and so on. Example 9 : Find the value of 237 ÷ 8. To divide 2 Hundred into 8 equal parts we need to regrouped them as 20 Tens. 20 Tens + 3 Tens = 23 Tens. 23 Tens ÷ 8 2 Tens, and 7 Tens remain. 8) 237 (0 2 –16 7 7 Tens can be regrouped as 70 Ones. 70 Ones + 7 Ones = 77 Ones. 77 Ones ÷ 8 9 Ones, and 5 Ones remain. To divide 5 Ones into 8 equal parts we need to regroup them as 50 Tenths. When we regroup Ones into Tenths, we place a decimal point in the quotient. 50 Tenths ÷ 8 6 Tenths, and 2 Tenths remain. 8) 237 (0 2 9 . 6 16 77 72 50 48 2 H T O Tenth s 237 ÷ 8 Divide 2 Hundreds + 3 Tens + 7 Ones into 8 equal parts. Tens Remember, when we regroup Ones into Tenths we need to place a decimal point in the quotient. Another Peek Beyond the Point 8) 237 (0 2 9 –16 77 – 72 5 H T O H T O Tens Ones Tens Chapter-4.indd 81 10/9/2025 2:44:44 PM 2 Tenths cannot be divided into 8 equal parts. So we need to regroup them as 20 Hundredths. 20 Hundredths ÷ 8 2 Hundredths, and 4 Hundredths remain. Tenths Ones 8) 237 (0 2 9 . 6 2 16 77 72 50 48 20 16 4 H T O Tenth s H undr edth s Tens Ones Tenths Hundredths 81 Ganita Prakash | Grade 7 | Part-II To divide 4 Hundredths into 8 equal parts we need to regroup them as 40 Thousandths. 40 Thousandths ÷ 8 5 Thousandths. Thus, 237 ÷ 8 = 29.625. Division with a Decimal Dividend Example 10: A shopkeeper has 9.5 kg of sugar and he wants to pack it equally in 4 bags. What is the weight of each bag of sugar? To find the weight of each bag we need to divide 9.5 by 4. 8) 237 (0 2 9 . 6 2 5 16 77 72 50 48 20 16 40 40 0 H T O Tenth s Hun dr ed ths Th ous an dths H und re dth s Tens Ones Tenths Hundredths Thousandths Th ou sand ths Chapter-4.indd 82 10/9/2025 2:44:44 PM Again, we place the decimal point in the quotient before we divide the tenths.Each bag of sugar weighs 2.375 kg. 82 4) 9.5 (2 . 3 7 5 8 15 12 30 28 20 20 0 O Te nt hs Ones Tenths Hundredths Thousandths Example 11: What is the value of 0.06 ÷ 5? 0.06 0 Ones + 0 Tenths + 6 Hundredths 0 Ones ÷ 5 0 Ones. When we move from Ones to Tenths we need to place the decimal point in the quotient. 0 Tenths ÷ 5 0 Tenths. 6 Hundredths ÷ 5 1 Hundredths, and 1 Hundredth remains. We need to regroup 1 Hundredth into 10 Thousandths. 10 Thousandths ÷ 5 2 Thousandths. So, the quotient is 0.012. Figure it Out 1. Find the quotient by converting the denominator into 1, 10, 100 or 1000 and verify the solution by the long division method (division by place value). 2. Choose the correct answer: (a) 18 5 (b) 415 4 (c) 1217 2 (d) 4827 8 (a) 1526 4 = (i) 38.15 (ii) 380.15 (iii) 381.5 (iv) 381.05 Another Peek Beyond the Point 5) 0.06 (0 . 0 1 2 0 00 00 06 5 10 10 0 O Tenth s Hun dre dth s Th ou sandt hs Ones Tenths Hundredths Thousandths Chapter-4.indd 83 10/10/2025 3:27:41 PM 3. What is the quotient? 4. What is the quotient? (b) 3567 8 = (i) 4458.75 (ii) 44.5875 (a) 132 ÷ 4 = (b) 13.2 ÷ 4 = (c) 1.32 ÷ 4 = (d) 0.132 ÷ 4 = (a) 126 ÷ 8 = (b) 12.6 ÷ 8 = (c) 1.26 ÷ 8 = (d) 0.126 ÷ 8 = (e) 0.0126 ÷ 8 = (iii) 445.875 (iv) 4458.75 Remember, when we regroup Ones into Tenths we need to place a decimal point in the quotient. 83 Ganita Prakash | Grade 7 | Part-II Division with a Decimal Divisor Example 12: Ravi went from Pune to Matheran by scooter in 2.5 hours. The distance was 126 km. What was his average speed? We can get the average speed by dividing the distance by the time taken. 126 ÷ 2.5. 126 ÷ 25 10 = 126 × 10 25 = 1260 25 . With the long division procedure we find the quotient is 50.4. So, the average speed at which Ravi travelled was 50.4 km per hour. Example 13: Find 4.68 ÷ 1.3. Again converting the divisor into a fraction we get 4.68 ÷ 13 10 = 4.68 × 10 13 = 46.8 13 . Now, what about 4.68 ÷ 0.13? 4.68 ÷ 0.13 = 4.68 ÷ 13 100 = 4.68 × 100 13 = 468 13 . What do you notice in these cases? When the divisor is a decimal, we can convert it into a counting number by suitably multiplying it by 10, 100, 1000, and so on. We must also multiply the dividend by the same number. Thus, 4.68 0.13 = 4.68 × 100 0.13 × 100 = 468 13 . Once we convert the divisor into a counting number, we can then follow the division using place value procedure to find the quotient. When the divisor is a decimal, we convert the divisor into a fraction. Chapter-4.indd 84 10/9/2025 2:44:45 PM Does This Ever End? Can you calculate 10 ÷ 3? Try dividing using long division. 10 1 Tens + 0 Ones. Step 1: Regroup 1 Tens into 10 Ones. 10 Ones ÷ 3 3 Ones, and 1 Ones remain. Step 2: Regroup 1 Ones as 10 Tenths. 10 Tenths ÷ 3 3 Tenths, and 1 Tenths remain. 84 Step 3: Regroup 1 Tenths as 10 Hundredths. 10 Hundredths ÷ 3 3 Hundredths , and 1 Hundredths remain. Step 4: Regroup 1 Hundredths as 10 Thousandths. 10 Thousandths ÷ 3 3 Thousandths, and 1 Thousandths remain. Regroup 1 Thousandths as 10 TenThousandths. This never seems to end! Each time we divide by 3, there is a remainder of 1. Will this process end? In long division, we see that at each step we get a remainder of 1. So, the process will never end! So, 10 ÷ 3 cannot be expressed using a finite number of digits in the decimal form. There are decimal divisions where the quotient never ends! We will explore such numbers in greater detail in a later class. Can you find the quotients of 10 ÷ 9, and 100 ÷ 11? Now divide 1 by 7 (1 ÷ 7). Will this end? Note all the remainders we get. It starts with 1, then 3, then 2, then 6, and so on. Let us represent this as a chain. 1 3 2 6 4 5 5 4 6 2 3 1 10 ÷ 3 = 3.333 ... Another Peek Beyond the Point 7) 1 ( 0.142857142857... 0 10 7 30 28 20 14 60 56 40 35 50 49 10 7 30 28 20 14 60 56 40 35 50 49 1 3) 10 ( 3.333 ... 9 10 9 10 9 10 9 ... Chapter-4.indd 85 10/9/2025 2:44:45 PM What do you observe? Can you explain why this division never ends? Not only do the remainders repeat in a cycle, the digits of the quotient also repeat in a cycle! 0.142857 142857 14… A Magic Number: 142857 Let us consider the number 142857 that arose when dividing 1 by 7. Multiply 142857 by numbers from 1 to 6. 1 3 ... 85 Ganita Prakash | Grade 7 | Part-II What are the products? What do you notice? You get the same number back, but with the digits cycled around! Multiply 142857 by 7. What do you observe? Are there other such numbers? Yes! To find one such number, you can find 1 ÷ 17 in decimal, and use the repeating block of digits. Are there infinitely many such “cyclic” numbers? That is, can we keep finding more cyclic numbers, or do they eventually stop? In 1927, the Austrian mathematician Emil Artin conjectured (guessed) that there must be infinitely many such numbers. However, even today, nearly a century later, this conjecture remains unsolved—despite a lot of research on the question by many mathematicians! Dividend, Divisor, and Quotient When we divide two counting numbers, the quotient is always less than the dividend. For example, 128 ÷ 4 = 32, and 32 (quotient) < 128 (dividend). But what happens when we divide 128 by 0.4? The quotient is greater than the dividend. Will the quotient be always greater than the dividend when the divisor is a decimal? Try it out with different values of the divisor. Describe the relationship between the dividend, divisor, and the quotient. Create a table for capturing this relationship in different situations, like we did for multiplication. 128 ÷ 0.4 = 320. Math Talk Try This Chapter-4.indd 86 10/10/2025 3:28:11 PM Figure it Out 86 1. Express the following fractions in decimal form: 2. Find the quotients: (a) 2 5 (b) 13 4 (c) 4 50 (d) 5 8 (a) 24.86 ÷ 1.2 (b) 5.728 ÷ 1.52 Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15? 10. 13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive? 3. Evaluate the following using the information 156 × 12 = 1872. 4. Evaluate the following: 5. Find the quotient: 6. A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece? 7. If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side? 8. 3 litres of watermelon juice is shared among 8 friends equally. How much watermelon juice will each get? Express the quantity of juice in millilitres. 9. A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre? (a) 15.6 × 1.2 = __________ (b) 187.2 ÷ 1.2 = __________ (c) 18.72 ÷ 15.6 = __________ (d) 0.156 × 0.12 = __________ (a) 25 ÷ ______ = 0.025 (b) 25 ÷ ______ = 250 (c) 25 ÷ ______ = 2.5 (d) 25 ÷ 10 = 25 × _____ (e) 25 ÷ 0.10 = 25 × ______ (f) 25 ÷ 0.01 = 25 × ______ (a) 2.46 ÷ 1.5 = (b) 2.46 ÷ 0.15 = (c) 2.46 ÷ 0.015 = Another Peek Beyond the Point Chapter-4.indd 87 10/9/2025 2:44:45 PM What pattern do you observe? Why are 2 and 5 related in this way? Math Talk 1 2 × 2 = 0.25 1 2 × 2 × 2 = 0.125 1 2 × 2 × 2 × 2 = 0.0625 1 2 × 2 × 2 × 2 × 2 = ? 1 2 = 0.5 1 5 × 5 = 0.04 1 5 × 5 × 5 = 0.008 1 5 × 5 × 5 × 5 = 0.0016 1 5 × 5 × 5 × 5 × 5 = ? 1 5 = 0.2 87 Ganita Prakash | Grade 7 | Part-II 4.4 Look Before You Leap! Did you know that it takes the Earth 365.2422 days to go around the Sun and not 365 days? For our convenience, we consider 365 days as a year in a calendar. We are talking about Gregorian calendar. This means that, after one calendar year or 365 days, the Earth still needs 0.2422 more days to complete one full revolution around the Sun. This doesn’t seem like much. But what happens after 100 such calendar years? Using our understanding of decimal multiplication, 0.2422 × 100 = 24.22 days. After 100 calendar years, the Earth will need 24.22 more days to complete its 100th revolution around the Sun. In your Science classes, you have learnt that we experience seasons because of the Earth’s tilt in axis and its revolution around the Sun. If our calendar does not accurately indicate the position of the Earth around the Sun, our seasons and our annual calendar will not match! To correct this problem, the idea of a leap year was introduced. Every fourth year, one additional day is added to the calendar year. Making an Adjustment Hey, I have not yet finished! Chapter-4.indd 88 10/9/2025 2:44:46 PM 88 Add an extra day every 4 years. Do you know which month has this extra day? Let us see how this solution works. Looking at the above sequence, the number of days after 4 calendar years is 4 × 365 + 1 = 1461 days. What is the number of days that the Earth needs to make 4 full revolutions around the Sun? With this new scheme of adding one extra day every 4th year, what is the number of days in 100 calendar years? Can you write an expression to calculate that number? Here is one way to form the expression. Each calendar year has 365 days. In 100 calendar years, the number of days is 100 × 365. But years that are divisible by 4 have one extra day. Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 … 365 365 365 366 365 365 365 366 … The year has 366 days The year has 365 days Is the year divisible by 4? 4 × 365.2422 = 1460.9688 days. Yes No Another Peek Beyond the Point Math Talk Chapter-4.indd 89 10/10/2025 3:29:07 PM How many years are divisible by 4 in 100 years? Can you form different expressions for the same question? The actual number of days that the Earth takes to go around the Sun 100 times is, 100 × 365.2422 = 36,524.22 days. Thus, by adding a day every fourth year, after 100 years, the calendar days are more than the actual number of days taken by the Earth to go around the Sun. We have overcompensated. So, the number of days in 100 calendar years is, (100 × 365 + 100 4 × 1) = 36,525 days. 89 Math Talk Ganita Prakash | Grade 7 | Part-II So, the good people who designed calendars decided that they will not add 1 extra day in every hundredth year! Making Another Adjustment The year has 365 days Is the year divisible by 4? Is the year divisible by 100? The year has 366 days The year has 365 days Yes No Yes No No extra day in the 100th year. Chapter-4.indd 90 17-10-2025 16:57:09 Can you write an expression for the number of days in 100 calendar years with this new adjustment? We saw that there are 25 years divisible by 4 in 100 years. 100 is also divisible by 4, but we have to exclude it. So only 24 years have 366 days, and the rest (76 years) have 365 days. So the expression can be written as, This is close to 36524.22 days but is it close enough? What happens after 1000 years with this adjustment? 90 ( 100 4 – 100 100 ) × 366 + (100 – ( 100 4 – 100 100 ))× 365 = (24 × 366) + (76 × 365) = 36,524 days. Math Talk If we follow this new scheme of adding 1 day every four years, but not in the 100th year, the number of calendar days in 1000 years is 36524 × 10 = 3,65,240 days. So, there is a difference of 2.2 days. To bridge this gap, it was decided that every 400th year would be a leap year! The year has 366 days Is the year divisible by 100? The number of days the Earth takes to go around the Sun 1000 times is 1000 × 365.2422 = 3,65,242.2 days. Is the year divisible by 400? Yes No Add an extra day every 400th year. Another Peek Beyond the Point Chapter-4.indd 91 10/9/2025 2:44:46 PM With this scheme, let us calculate the number of days in 1000 calendar years. The year has 365 days The year has 366 days Is the year divisible by 4? Yes Yes No The year has 365 days No 91 Ganita Prakash | Grade 7 | Part-II In 1000 calendar years, how many years are divisible by 400? 2. In 1000 calendar years, how many years are divisible by 100 but not divisible by 400? 10 – 2 = 8. In 1000 calendar years, how many years are divisible by 4 but not divisible by 100 and 400? 250 – 10 = 240. The rest of the years are 1000 – (2 + 8 + 240) = 750. The Earth needs 3,65,242.2 days to go around the Sun 1000 times. Hence, in 1000 years, the calendar year is slightly shorter (by 0.2 days) than the actual number of days the Earth takes to go around the Sun. The calendar makers did not want to bother about a small difference that would happen in 1000 years! So, they left the scheme of leap years as is … Making Yet Another Adjustment So, the total number of days in 1000 calendar years is (750 × 365) + (240 × 366) + (8 × 365) + (2 × 366) = 3,65,242 days. Chapter-4.indd 92 10/9/2025 2:44:47 PM 92 With this final scheme of leap years can you calculate the number of calendar days in 10,000 years and the number of actual days the Earth will take to make 10,000 revolutions around the Sun? What is the difference? If there is a big difference, can you suggest a way to fix this problem? Try This Do you wonder how people figured out that the Earth completes one revolution around the Sun in exactly 364.2422 days? Investigate how traditional calendars in India managed to consistently align the days in the calendar with astronomical events like the Earth going around the Sun or even the positions of the stars in the sky accurately. Figure it Out 1. A 210 gram packet of peanut chikki costs ₹70.5, while a 110 gram packet of potato chips costs ₹33.25. Which is cheaper? 2. Write the decimal number at the arrow mark: 3. Shyamala bought 3 kg bananas at ₹30/- per kg. She counted 35 bananas in all. She sells each banana for ₹5/-. How much profit does she make selling all the bananas? 2.15 2.17 3.1 3.2 Another Peek Beyond the Point Try This Chapter-4.indd 93 10/9/2025 2:44:47 PM 4. A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much? 93 Ganita Prakash | Grade 7 | Part-II 5. Fill in the following blanks appropriately: 6. The following problem was set by Sridharacharya in his book, 7. Fill the boxes in at least 2 different ways: 8. Find the following quotients given that 756 ÷ 36 = 21: 9. Find the missing cells if each cell represents a÷b: Patiganita. “6 1 4 is divided by 2 1 2 , and 60 1 4 is divided by 3 1 2 . Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals? (a) × = 2.4 (b) × = 14.5 (a) 75.6 ÷ 3.6 (b) 7.56 ÷ 0.36 (c) 756 ÷ 0.36 (d) 75.6 ÷ 360 (e) 7560 ÷ 3.6 (f) 7.56 ÷ 0.36 1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m 5.5 km = _________ m 35 cm = ________ m 14.5 cm = _______ mm 68 g = ________ kg 9.02 m = ________ mm 125.5 ml = _______ l b a 1517 151.7 15.17 1.517 15170 1 kg = 1000 g 1 g = 1000 mg 1 l = 1000 ml Math Talk Chapter-4.indd 94 10/10/2025 3:31:03 PM 10. Using the digits 2, 4, 5, 8, and 0 fill the boxes . × . to get the: 94 (a) maximum product (b) minimum product 0.037 4100 0.37 370 3.7 4.1 37 41 11. Sort the following expressions in increasing order: • In this chapter, we learnt procedures to perform decimal multiplication and division. • For decimal multiplication, we first multiply the multiplier and multiplicand as counting numbers. The number of decimal digits in the product is the total number of decimal digits in the multiplier and multiplicand. • Division of decimals uses the same procedure, i.e., division using place value (long division), as with counting numbers. The regrouping continues after the Ones place to Tenths, Hundredths, Thousandths, and so on. When the Ones are regrouped to Tenths, a decimal point is placed in the quotient. • There are decimal divisions where the quotient never ends. After each regrouping and dividing there is always a remainder! (c) product greater than 150 (d) product nearest to 100 (e) product nearest to 5 (a) 245.05 × 0.942368 (b) 245.05 × 7.9682 (c) 245.05 ÷ 7.9682 (d) 245.05 ÷ 0.942368 (e) 245.05 (f) 7.9682 SUMMARY Another Peek Beyond the Point Chapter-4.indd 95 10/17/2025 5:22:32 PM 27 9 Puzzle 33 35 26 13 11 24 22 21 18 40 7 5 1 Hidato 30 25 23 21 12 39 39 26 20 13 11 27 28 14 19 9 10 32 33 35 36 37 31 34 24 22 38 Solution 15 16 18 8 2 17 7 16 3 40 5 4 1 95 In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it is made of only one piece. Usually, in every Hidato puzzle the lowest and the highest numbers are given on the grid. Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any one of the adjacent cells, including diagonally adjacent cells. The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution. Try solving the following Hidato puzzles. 25 1 20 4 14 8 15 9 7 10 9 39 36 2 26 23 21 34 12 31 1 30 4 6 13 17 Chapter-4.indd 96 10/9/2025 2:44:49 PM 56 44 49 28 54 42 50 29 26 5 17 15 39 35 34 25 23 1" class_7,13,connecting the dots...,ncert_books/class_7/gegp1dd2/gegp205.pdf,"5 Jemimah’s batting has been very consistent over the past year. We can expect a century from her in tomorrow’s match. 5.1 Of Questions and Statements Your teacher tells you that they are meeting two of their childhood friends this evening. One is 5 feet tall and the other is 6 feet tall. What is your guess as to each friend’s gender based on this information? You might have guessed that the 5-foot-tall person is a woman and the 6-foot-tall person is a man. There is a chance that you are wrong. But experience tells us that 5-foot-tall men and 6-foot-tall women are rare. We have seen that, more often, men are taller than women. The above is a simple example of statistical thinking. We regularly come across statements like — CONNECTING THE DOTS… I take about 15 minutes to cycle from school to home. I think my pen might last for 2 more weeks; it is time to get a new one soon. Chapter-5.indd 97 10/9/2025 3:45:40 PM The population of their village has reduced by about 100 in the last decade. Since I started to eat fruits and vegetables more frequently, I am able to run 2 km more each day. David spends about 7 hours daily in the school. Ganita Prakash | Grade 7 | Part-II We call these statistical statements. Simply put, a statistical statement is a claim or summary about some phenomenon, expressed in terms of numerical values, proportions, probabilities, or predictions. A statistical question is a question that can be answered by collecting data. For example, “How tall are Grade 7 students in our school?” is a statistical question. We expect that not all Grade 7 students have the same height, but we can collect data, analyse it, and make conclusions about the heights that do occur. The question “Typically, are onions costlier in Yahapur or Wahapur?” is also a statistical question. Prices can vary over time. Therefore, answering this question requires us to look at data, analyse it, and come to conclusions making suitable statistical statements. Which of the following are statistical questions? (a) What is the price of a tennis ball in India? (b) How old are the dogs that live on this street? (c) What fraction of the students in your class like walking up a hill? (d) Do you like reading? (e) Approximately how many bricks are in this wall? (f) Who was the best bowler in the match yesterday? (g) What was the rainfall pattern in Barmer last year? The term statistics refers to the study of collecting, organising, analysing, interpreting, and presenting data. In this chapter, we shall encounter some statistical questions and learn how analysing data and graphs can help answer them. 5.2 Representative Values Math Talk Chapter-5.indd 98 10/9/2025 3:45:41 PM The runs scored by Shubman and Yashasvi in a cricket series are given in the table below. Who do you think performed better? 98 Shubman 0 17 21 90 Yashasvi 67 55 18 35 Shreyas says, “Both their performances are similar since Yashasvi scored more in the first and second matches, whereas Shubman scored more in the third and fourth”. Match 1 Match 2 Match 3 Match 4 Vaishnavi says, “I think Shubman performed better because he scored the highest number of run in a match — 90!”. Shreyas says, “No! Yashasvi batted better since the total number of runs he made is 175, while Shubman made only 128”. Vaishnavi says, “Oh! Also, Yashasvi’s batting is more consistent — the difference between his maximum score and minimum score is lower”. The table below shows the runs scored by these two players in another series. Who do you think performed better in this series? Vaishnavi says, “Here, Shubman performed better since his total is 110 runs, while Yashasvi’s total is 96 runs”. What do you think of Vaishnavi’s statement? Shreyas says, “But Yashasvi made 96 runs in 4 matches and Shubman made 110 runs in 5 matches”. So, how do we say who performed better? It is often not simple to compare two groups of numbers and clearly say that one is better than the other. Can a single number act as a representative of a group of numbers? For example, can we represent Shubman’s or Yashasvi’s batting in this series with one number? Discuss. We saw one way already — the total of the values in the group! But, if the group sizes are different, then the total may not be an appropriate measure to compare. In some matches, a player could have scored more and in other matches less. A representative number for the group can be found by balancing out these highs and lows. For example, we can add up the runs scored in all the matches and divide the total by the number of matches played. We call this value the ‘average’ or ‘arithmetic mean’ of the given data. Here, the average number of runs scored by a player in a match = [Total runs scored by the player in all the matches] ÷ [Number of matches played]. Average number of runs scored by Shubman in a match = 110 ÷ 5 = 21 runs. Average number of runs scored by Yashasvi in a match = 96 ÷ 4 = 24 runs. In this series, Yashasvi’s average number of runs is higher than Shubman’s. Math Talk Shubman 23 07 10 52 18 Yashasvi 26 53 02 - 15 Match 1 Match 2 Match 3 Match 4 Match 5 Connecting the Dots… Chapter-5.indd 99 10/9/2025 3:45:41 PM 99 Ganita Prakash | Grade 7 | Part-II The Average or Arithmetic Mean (A.M.), or simply Mean, is calculated as follows: Average as Fair-Share The average can also be understood as fair-share or equal-share. Shreyas and 4 of his friends have collected the following numbers of guavas: 3, 8, 10, 5, and 4. Parag and 5 of his friends have collected the following numbers of guavas: 5, 4, 6, 3, 4, and 8. Each group will share their guavas equally amongst themselves. In which group will each member get a bigger share of guavas? To find this out, we first find out how many guavas each group has collected. Then we divide this total by the number of people in the group to get each member’s share. Shreyas’s group has collected 3 + 8 + 10 + 5 + 4 = 30 guavas. Each member of Shreyas’s group gets 30 ÷ 5 = 6 guavas. Parag’s group has collected 5 + 4 + 6 + 3 + 4 + 8 = 30 guavas. Each member of Parag’s group gets 30 ÷ 6 = 5 guavas. So, the members of Shreyas’s group get 1 more guava each than the members of Parag’s group. Mean = Sum of all the values in the data Number of values in the data . Chapter-5.indd 100 10/17/2025 4:31:15 PM 3 + 8 + 10 + 5 + 4 = 30 6 × 5 = 30 5 + 4 + 6 + 3 + 4 + 8 = 30 6 × 5 = 30 3 8 10 5 4 6 6 6 6 6 5 4 6 3 4 8 5 5 5 5 5 5 Vaishnavi tracks the number of Hibiscus flowers blooming in her garden each day. The data for the last few days’ is 2, 7, 9, 4, 3. What is the average number of Hibiscus flowers blooming per day in Vaishnavi’s garden? 100 The average = (the total number of Hibiscus flowers bloomed) ÷ (number of days) = (2 + 7 + 9 + 4 + 3) ÷ 5 = 5. On an average, 5 Hibiscus flowers bloom daily. In this case the average tells us the number of flowers blooming each day, if an equal number of flowers bloomed daily. One of the terms used for the Arithmetic Mean in ancient Indian mathematics is samamiti (mean measure): ‘sama’ means equal. Some terms used for the Arithmetic Mean in Indian texts include — samarajju (mean measure of a line segment) by Brahmagupta (628 CE), samīkaraṇa (levelling, equalising) by Mahāvīrācārya (850 CE) sāmya (equality, impartiality, equability towards) by Śrīpati (1039 CE) and samamiti (mean measure) by Bhāskarācārya (1150 CE) and Gaṇeṣa (1545 CE). The terminology shows that ancient Indian scholars perceived the Arithmetic Mean as the ‘common’ or ‘equalising’ value that is a representative measure of a collection of values. Figure it Out 1. Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat. 2. Try the activity above on your own. Collect data for 7 or more attempts and find the average. 3. Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day? Connecting the Dots… Try This Chapter-5.indd 101 10/9/2025 3:45:42 PM 4. Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker? 5. The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period. Know Your Onions! The table shows the monthly price of onions, in rupees per kilogram (kg), at two towns. Where are onions costlier, according to you? 101 Math Talk Ganita Prakash | Grade 7 | Part-II Khushboo: ‘I think Wahapur is costlier because it has the highest price of ₹60.’ Nafisa: ‘I added the prices of all months in each location - Yahapur’s total is 458, whereas Wahapur’s total is 450.’ Vishal: ‘Wahapur is costlier since it has 3 numbers in the 50s’. Sampat: ‘I compared the prices in each month in both locations. Prices in Yahapur are higher for 6 months, prices in Wahapur are higher for 5 months, and the prices are the same for 1 month. So, I feel Yahapur is costlier.’ Jithin: ‘I noticed that the difference between the highest and lowest prices in Yahapur is 59 – 24 = 35, and in Wahapur it is 60 – 17 = 43.’ Data can be described and compared by referring to its minimum value, maximum value, the average value, the sum total of all its values, and the difference between the maximum and minimum values. September 49 September 53 November 59 November 52 December 44 December 42 February 24 February 17 January 25 January 19 October 56 October 60 Month Yahapur Month Wahapur August 43 August 39 March 26 March 23 April 28 April 30 June 35 June 35 May 30 May 38 July 39 July 42 Chapter-5.indd 102 10/9/2025 3:45:42 PM 102 17 Can you think of any other ways to compare the data? To study data, we can visualise it in multiple ways. One way is shown below — it is called a dot plot. Dot plots show data points as dots on a line, helping us visualise variability and patterns in data. In the following figure, each dot represents the monthly price of onions. 10 20 30 40 50 60 10 20 30 40 50 60 Wahapur Yahapur The prices in Yahapur are in green and those in Wahapur are in purple. The horizontal line shows the prices from 10 to 60 (instead of starting from 0 as there are no values from 0 – 10 or above 60). The dots on the vertical line give the number of occurrences of a data value. Notice the equal spacing between the units along the horizontal as well as the vertical lines. Does this visualisation capture all the data presented in the tables earlier? One of the values is 19. Here it refers to one of the monthly prices of Wahapur Two of the values are 39. One belonging to each of Yahapur and Wahapur Two of the values are 42. Both belonging to Wahapur Connecting the Dots… Chapter-5.indd 103 10/9/2025 3:45:42 PM Looking at it, can we tell the price of onions in Yahapur in the month of January? This method of presentation orders or sorts the data, but it loses the original (month-wise) sequence of the values. However, it allows us to group the data however we wish, just as Vishal did. For instance, there are 2 data values between 11 – 20 for Wahapur, while Yahapur has none. This representation makes it easier to observe the variation in the data — where and how the data is clustered or spread out. We can easily see that the prices in Wahapur are more spread out than those in Yahapur. It is also easy to spot the highest and lowest values. We can also use the average as one of the ways to compare the prices at these two places. 103 Ganita Prakash | Grade 7 | Part-II Find the average price of onions at Yahapur and Wahapur. A statement such as, “The price of onions is ₹35 per kilo”, may not trigger any further questions. But looking at variations in data, like the prices of onions over a year in Yahapur and Wahapur, can spark one’s curiosity. For example, one might be curious to know more about the two locations. You might wonder — Do the seasons affect the price of onions? How much do onion prices vary across shops in the same area? Where are these two locations? Are they close to each other or far apart? What other commodities might have similar patterns? What are the factors that determine the price of onions? How do the price fluctuations impact farmers, consumers, and the industry? Chapter-5.indd 104 10/9/2025 3:45:43 PM 13 104 What else do you wonder about? You can discuss questions that you are curious about with your peers, teachers, or family members to find answers. Averages Around Us The Arithmetic Mean is frequently used in statistics, mathematics, experimental sciences, economics, sociology, sports, biology and diverse Observing and trying to make sense of data can reveal interesting things. It can also trigger our curiosity in different directions. Math Talk Smartphone users check their phone 58 times a day on average. The average rainfall per day in Jharkhand in the month of July is 37.2mm. disciplines as a representative of data. It is popular partly because the definition of the arithmetic mean is simple and easy to understand. Some statements involving averages in different scenarios are shown below: An average Indian citizen generates 0.45 kg of waste per day. My scooty's average mileage this year is about 45 kilometers per liter. Wheat yield averages 4.7 tonnes per hectare in Punjab vs. 2.9 tonnes per hectare in Bihar. Connecting the Dots… 3126 is the average number of Indian long films released annually between 2017 – 2024. Chapter-5.indd 105 10/9/2025 3:45:43 PM Outliers and Medians Does the average always give a reasonable summary of the values in a collection? If not, what is an alternative? Let us find out. Height of a Family The heights of the family members of Yaangba and Poovizhi are as follows: Yaangba’s family: 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm. Poovizhi’s family: 170 cm, 173 cm, 165 cm, 118 cm, 175 cm. Find the average height of each family. Can we say that Yaangba’s family is taller than Poovizhi’s family? 105 Math Talk 115 125 135 145 155 165 175 115 125 135 145 155 165 175 Ganita Prakash | Grade 7 | Part-II The average height of Poovizhi’s family (160.2 cm) is less than that of Yaangba’s family (164.3 cm). Although most members in Poovizhi’s family are taller, their family’s average height is less because one child is much younger and not as tall as the rest of the family. Their average height, 160.2 cm, is less than the heights of 4 out of 5 members. Here, the average doesn’t seem to represent the data very well. Can you think of any other number that can represent the data better? One way is to sort the data and pick the number in the middle. This number is called the Median. To find the median height of Poovizhi’s family, we first sort the heights — 118, 165, 170, 173, 175. The middle number in this sorted data is 170. Therefore, the median height is 170 cm. Let us find the median height of Yaangba’s family. Sorting the heights, we get 155, 160, 164, 165, 169, 173. Since the median is the number in the middle, it will have an equal number of values less than it and greater than it. This data does not have a single middle number because it has an even number of values (6). In such cases, we take the average of the two middle numbers in the sorted data. Therefore, the median height of Yaangba’s family is (164 + 165) ÷ 2 = 164.5 cm. Mean = 160.2 –– Mean = 164.3 –– Chapter-5.indd 106 10/9/2025 3:45:43 PM 115 125 135 145 155 165 175 115 125 135 145 155 165 175 In this case, does the median represent the heights of the families better than the average? 53 106 Mean = 164.3 –– Median = 164.5 - - - Mean = 160.2 –– Median = 170 - - - In Poovizhi’s family, the height of the youngest child is quite different from the heights of the rest of the family. We call such a value an outlier. Outliers are values which significantly deviate from the rest of the values in the data. Notice how the mean and the median are close to each other in Yaangba’s data, in the absence of any outlier. In Poovizhi’s data, because of the outlier, the mean is much lower than the median. Find the mean and median in Poovizhi’s data without the outlier value 118. What change do you notice? Are you a bookworm? After the summer vacation, a class teacher asked his class how many short stories they had read. Each student answered the number of stories read on a piece of paper, as shown below. Find the mean and median number of short stories read. Before calculating them, can you guess whether the mean will be less than or greater than the median? Connecting the Dots… Chapter-5.indd 107 10/17/2025 4:48:13 PM The median value 6 means that half of the class members have read 6 or more stories. 0 5 10 15 20 25 30 35 40 Mark the data, the mean, and the median on the dot plot below. 107 Ganita Prakash | Grade 7 | Part-II Which of the values would you consider an outlier? Find the mean and median in the absence of the outlier. What change do you notice? The average may not always be an appropriate representative of data that has outliers. A very high or a very low outlier can significantly impact the sum, thus affecting the average. For example, the 118 cm height in Poovizhi’s family is an outlier at the lower end of the data. And the count of 40 short stories read is an outlier at the higher end of the data. In these cases, we saw that the median was not affected much by the outliers. Are We on the Same Page? Do you read newspapers? Have you noticed how many pages a newspaper has on different days of the week — is it the same or different? The list below shows the number of pages for a particular newspaper from Monday to Sunday: 16, 18, 20, 22, 26, 16, 10. Mark the data, the mean, and the median on the dot plot below. In the three examples we considered — the heights, short-stories, and newspaper pages — observe the variability in data when: 0 5 10 15 20 25 Math Talk Chapter-5.indd 108 10/9/2025 3:45:45 PM (a) the mean and median are close to each other (b) the mean and median are comparatively far apart, with mean < median (c) the mean and median are comparatively far apart, with mean > median When the data is more balanced or uniformly spread out the mean, and the median appear to be close to each other. When the outlier is on the lower end, the mean appears to shift in that direction, i.e., mean < median. When the outlier is on the higher end, the mean appears to shift in that direction, i.e., mean > median. 108 Mean and Median are called measures of central tendency, i.e., the tendency of the values to pile up around a particular value. In other words, they represent the ‘centre’ of the data. Of Ends and the Essence As we have just seen, the mean and the median can give different perspectives on the data. As part of analysing data, it can also be valuable to look at the variability in the given data, i.e., its extremes (minimum and maximum values). How Tall is Your Class? Suppose you are asked the question, “How tall is your class?” What would you say? The table below shows the heights of students in a Grade 5 class in centimeters. We can visualise the data using a dot plot, identify the ends and patterns, and look at the variability. We can also find the measures of central tendency. The dot plot for the whole class, followed by the dot plots for boys and girls, respectively, are shown. The mean and the median are also shown for each collection. Discuss the effect on the mean and median when outliers are present on both sides. You may take some example data to examine and explain this. Girls 143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150 Boys 147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150 Connecting the Dots… Math Talk Chapter-5.indd 109 10/9/2025 3:45:45 PM Whole class Girls Boys 125 125 125 130 130 130 135 135 135 140 140 140 145 145 145 150 150 150 155 155 155 Mean = 142.94 Median = 144 Mean = 144.4 Median = 145 Mean = 146.9 Median = 148 109 Ganita Prakash | Grade 7 | Part-II What can we infer from the dot plots and the central tendency measures? The following points can help answer the question of how tall the class is. • The boys’ heights are more spread out and are between 128 and 158. The girls’ heights lie between 136 and 156. Both the tallest and shortest in the class are boys. How many students are taller than the class’ average height? How many boys are taller than the class’ average height? How long is a minute? Two groups of children were asked to estimate the length of 1 minute. They start by closing their eyes and then open when they think 1 minute has passed. Of course, they are not supposed to count while their eyes are closed. The dot plots below show after how many seconds the children opened their eyes. Discuss how well both the groups fared at this activity. Describe and compare the variability in data and their central tendency. • Yet, the boys’ average height is less than the whole class average, and also less than the girls’ average height. We can say girls are taller than boys in this class. Of course, this doesn’t mean every girl is taller than every boy! • For boys’ heights, mean < median (142.94 < 144) indicating a small influence of values on the lower side. For girls’ heights too, mean < median (146.9 < 148) indicating a small influence of values on the lower side. Math Talk Chapter-5.indd 110 10/9/2025 3:45:46 PM 11 110 40 40 45 45 50 50 55 55 60 60 65 65 70 70 Group B Mean = 59.28 Median = 59.5 Group A Mean = 58.21 Median = 60 Zero vs. No value Accounting for the extras, the average runs scored by a player in this innings is (407 – 19) ÷ 11 = 35.27. Zero Median Runs Scored! In a cricket match, can a team’s median runs scored by a player be 0 but the team’s total score be 407/10? In yesterday’s match, the median runs scored by England’s players was 0, and yet the team scored 407 for the loss of 10 wickets. Connecting the Dots… Chapter-5.indd 111 10/16/2025 3:30:40 PM Suppose a player scores 57, 13, 0, 84, — , 51, 27 in a series. Notice that the player played Match 3 and scored 0 runs whereas the player did not play Match 5. So, we consider the total number of matches to be 6 and not 7. We calculate their average runs scored per match as (57 + 13 + 0 + 84 + 51 + 27) ÷ 6. Sita has a mango tree in her backyard. The number of mangoes the tree gave every month over the last year, from January to December, is 0, 0, 8, 24, 41, 16, 5, 0, 0, 0, 0, 0 respectively. If we want to find the mean or median number of mangoes per month, it would be appropriate to consider only the (summer) months when mangoes are expected to grow. 111 Ganita Prakash | Grade 7 | Part-II Figure it Out 1. Find the median of onion prices in Yahapur and Wahapur. 2. Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, — , 2, 4. Find the mean and median. How would you describe this data? 3. Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height? A Mean Foot In the early 1500s in Europe, the basic unit of land measurement was the rod, defined as 16 feet long. At that time, a foot meant the length of a human foot! But foot sizes vary, so whose foot could they measure? To solve this, 16 adult males were asked to stand in a line, toe to heel, and the length of that line was considered the 16-foot rod. After the rod was determined, it was split into 16 equal sections, each representing the measure of a single foot. In essence, this was the arithmetic mean of the 16 individual feet, even though the term ‘mean’ was not mentioned anywhere. Jacob Kobel’s depiction of the determination of 1 foot Chapter-5.indd 112 10/9/2025 3:45:46 PM 4. The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4. 5. The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data. 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 112 (a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median. (b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data? 6. The dot plots of heights of another section of Grade 5 students of the Whole class Boys Girls 125 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures? Girls 4.0 3.1 3.4 3.7 2.5 3.4 125 125 Boys 3.5 4.1 2.6 3.2 3.4 3.8 130 130 130 135 135 135 140 140 140 145 145 145 150 150 150 155 155 155 Mean = 141.21 Median = 142.5 Mean = 142.05 Median = 143 Mean = 140.14 Median = 140 Connecting the Dots… Chapter-5.indd 113 10/9/2025 3:45:47 PM Compare the heights of the two sections. Share your observations. 7. The weights of some sumo wrestlers and ballet dancers are: Sumo wrestlers: 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers: 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximately how many times heavier is a sumo wrestler compared to a ballet dancer? 113 Math Talk Ganita Prakash | Grade 7 | Part-II 5.3 Visualising Data We can often understand data more clearly if it is presented as a picture. This is called data visualisation. Last year, we saw how to visualise data using graphs. Let us explore visualisation further. Clubbing the Columns Earlier, we looked at the monthly onion prices in Yahapur and Wahapur. Wahapur 19 17 23 30 38 35 42 39 53 60 52 42 Yahapur 25 24 26 28 30 35 39 43 49 56 59 44 Two column graphs for this data are given below. 60 Monthly onion prices in Yahapur 50 40 30 20 10 0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Chapter-5.indd 114 10/9/2025 3:45:47 PM 114 60 50 40 30 20 10 0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Monthly onion prices in Wahapur Price (in ₹) 60 50 40 30 20 10 The two graphs can also be combined into a single graph. We just draw the bars side by side! Verify if the data in the table matches the graph below. 0 We use different colours to clearly separate the data from the two places. This is called a clustered column graph. Since it has two columns in each cluster, we also call it a double column graph. What is the scale used in this graph? The relative heights of the bars tell us where onions are costlier in each month. We can also visually estimate the difference by referring to the markings along the vertical line. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Monthly onion prices in Yahapur and Wahapur Yahapur Wahapur The dots and slanted lines within the bars help people who find difficulty in distinguishing colours. It is also useful when things are printed in greyscale (black-and-white). Connecting the Dots… Chapter-5.indd 115 10/9/2025 3:45:47 PM Is it now easier to compare month-wise prices in both places? 10…9…8…7…6…5…4…3…2…1…Take Off! You might have heard about scientific probes (like Chandrayaan-3 launched in 2023 by ISRO or the Voyager-1 launched in 1977 by NASA), observational satellites (like Aryabhata launched in 1975 by ISRO or Sputnik-1 launched in 1957 by the Soviet Space program), or about human spaceflights to the International Space Station. All space missions are 115 Ganita Prakash | Grade 7 | Part-II launched using rockets. Look at the graph below showing the number of worldwide rocket launches by different organisations. Share your observations (you may take the teacher's help to identify the countries these organisations belong to). Chapter-5.indd 116 10/10/2025 3:52:55 PM Often there is a lot of information in graphs and it may be difficult to understand. We can follow a 2-step process to simplify making sense of the data in graphs. Step 1: Identify what is given Notice how the graph is organised, what scale is used, and what patterns the data shows. 116 Source: https://www.statista.com/chart/29410/number-of-worldwide-rocket-launches-bycompanies-and-space-agencies/I wish they had shown guidelines for 5, 10, and 15 along the horizontal line. It would help in estimating the smaller numbers. Step 2: Infer from what is given Analyse and interpret each of your observations. • For each organisation, the numbers of rocket launches for the years 2021, 2022, and 2023 are shown as three adjacent bars. The scale used is 1 unit length = 20 rockets. Notice the numbers at the bottom. • The ‘Others’ category indicates multiple organisations worldwide that are clubbed together to keep the graph short. • Note that in the double bar graph of onion prices, the months are shown in order, i.e., January to December, to observe the change over time, whereas in this case, a change in the order of organisations does not affect the meaning. Galactic Energy United Launch Alliance Arianespace Rocket Lab Roscosmos SpaceX Expace CASC ISRO 0 20 40 60 80 100 Perhaps 61 Decreasing year on year, bar length of 2023 is almost half of 2022 Increasing year on year, bar length of 2022 is almost double that of 2021 Connecting the Dots… The 2023 count of objects in this category is around 25 Increase and then decrease Chapter-5.indd 117 10/9/2025 3:45:48 PM • We can say that the USA, China, and Russia are the leading rocket launching countries in the given time period. • SpaceX launched about twice the number of rockets in 2022 compared to 2021. And it launched about 35 more rockets in 2023 compared to 2022. • The number of rockets launched by Arianespace decreased every year. • United Launch Alliance launched more rockets in 2022 than in 2021. They launched fewer rockets in 2023 than in both the years 2022 and 2021. • Other organisations launched about 25 rockets in 2023. 117 Ganita Prakash | Grade 7 | Part-II Identify which of the following statements can be justified using this data. (a) All organisations launched more rockets than the previous years. (b) Only an organisation from the USA launched more than 50 rockets in a single year. (c) The total number of rockets launched by France in all 3 years is less than 40. (d) The average number of rockets launched by CASC in these 3 years is around 40. (e) ISRO launched more rockets than Galactic Energy in these 3 years. (f) Russia launched more than 60 rockets in these 3 years. List the organisations that have consistently launched more rockets every year. Estimate the total number of rockets launched worldwide in 2023. We may have many questions after looking at the graph. We might wonder why USA launches so many more rockets than other countries. Or we might be eager to look at the data from previous and later years. What are you curious to know after looking at this graph? Summer and Winter at the Same Time The tables below show data related to weather in two cities in different countries. The numbers given are in hours. Can you guess what the data might be related to? (a) less than 200 (b) 200 to 400 (c) 400 to 600 (d) more than 600 Chapter-5.indd 118 10/9/2025 3:45:48 PM City 1 City 2 118 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 210 257 372 441 536 564 555 465 394 310 222 186 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 459 384 381 327 304 276 295 318 369 409 435 468 The data shows the monthly hours of daylight (i.e., the Sun is at least partly above the horizon) in these two cities over the year. Based on this data, a clustered bar graph showing the average daylight hours per day in each month is given below. This average is obtained by dividing the monthly daylight hours by the number of days in the month. Let us follow the 2-step process to identify and interpret the information presented. Step 1: Identify what is given Notice how the graph is organised, what scale is used, and what patterns the data shows. Nu mbe r of Ho urs p er day 10 15 20 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 0 5 Average daily sunshine hours in two cities City 1 City 2 Connecting the Dots… Chapter-5.indd 119 10/9/2025 3:45:49 PM Step 2: Infer from what is given Analyse and interpret each of your observations. Share appropriate summary and conclusion statements. • The average number of daylight hours per day in City 1 increases from January, reaching a maximum of about 17 – 18 hours in June. It then decreases, reaching a minimum of about 6 hours in December. • The horizontal line shows the months of the year. The vertical line shows the average daylight hours per day, using the scale 1 unit = 5 hours. The month of June has the maximum value for City 1 and the minimum value for City 2. 119 Ganita Prakash | Grade 7 | Part-II Does this give some idea of where these two cities are located? City 1 and City 2 are located away from the Equator in the Northern and Southern hemispheres, respectively. City 1 is Helsinki, Finland, and City 2 is Wellington, New Zealand. These are also shown on the map. In June, the Northern Hemisphere is tilted towards the Sun, resulting in longer daylight hours; it is summertime here. Meanwhile, the Southern Hemisphere is tilted away from the Sun, leading to shorter days; it is winter time here. The inverted seasonal daylight pattern is due to the cities’ location in opposite hemispheres. The large variation in the data is because they are away from the Equator. • The average number of daylight hours per day in City 2 decreases from January, reaching a minimum of about 9 hours in June. It then increases, reaching a maximum of about 15 hours in December. • The maximum and minimum values in City 1 are more extreme than those of City 2. That is, the maximum number of daylight hours per day of City 1 is more than that of City 2, and the minimum number of daylight hours per day of City 1 is less than that of City 2. • In June, City 1 experiences daylight for about 3 4 th of the full day (24 hours), whereas during December – January, it only experiences daylight for about 1 4 th of the full day. Chapter-5.indd 120 10/9/2025 3:45:52 PM 120 Is there anything more that you wish to explore? I got to know that during summer near the poles one can see the Sun even at midnight!! Very interesting! If we are there at that time, we can go sightseeing even at midnight! Connecting the Dots… Math Talk Chapter-5.indd 121 10/9/2025 3:45:52 PM All it Takes is a Minute Have you ever missed watching a cricket match? You can catch up in a minute by looking at a graph. You might have seen graphs like the following one. The Midnight Sun at Andøya, Norway. ©timeanddate.com/Brendan Goodenough 121 Ganita Prakash | Grade 7 | Part-II The horizontal line lists the overs starting from 1, and the vertical line indicates the runs scored in each over. The graph shows the number of runs scored per over as a double bar graph — each bar corresponding to a team. Let us call them the blue team (denoted by blue) and the red team (denoted by red). The scale used for the runs per over is 1 unit = 5 runs. The circles shown on top of the bars indicate that a wicket fell in that over. Answer the following questions based on the graph: 1. Can we tell who batted first? Who won the match? 2. How many runs did the blue team score in over 12? 3. In which over did the red team score the least number of runs? 4. Is it easy to tell the target set by the team batting first? Figure it Out Runs p er ove r 10 15 5 0 6 7 8 9 10 1 2 3 4 5 Overs 11 12 13 14 15 16 17 18 19 20 Chapter-5.indd 122 10/17/2025 4:35:11 PM 1. The following infographic shows the speeds of a few animals in air, on land, and in water. Can we call this graph a bar graph? 122 (a) What is the scale used in this graph? (b) What did you find interesting in this infographic? What do you want to explore further? (c) Identify a pair of creatures where one’s speed is about twice that of the other. (d) Can we say that a sailfish is about 4 times faster than a humpback whale? Can we say that a sailfish is the fastest aquatic animal in the world? Math Talk The peregrine falcon is the world's fastest animal. It hunts by diving at high speed and striking a pigeon or other bird in midair. The Australian tiger beetle moves at a blistering pace– about 120 body lengths per second. It's eyes can't process images fast enough to keep up, so at top speed this beetle is running blind. Chapter-5.indd 123 10/10/2025 3:55:39 PM Ganita Prakash | Grade 7 | Part-II 2. Preyashi asked her students ‘If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne which one would you choose?’. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale. 3. The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = 4°C. Can you guess which two months these days might belong to? 4. The following clustered-bar graph shows the number of electric vehicles registered in some states every year from 2022 to 2024. 100000 75000 Day 1 20°C 18°C 16°C 20°C 26°C 34°C 30°C 24°C Day 2 37°C 34°C 30°C 33°C 37°C 43°C 42°C 39°C Grade 5 w, a, a, a, w, n, s, a, n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n Grade 9 n, w, s, a, s, w, s, s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a 12 am 3 am 6 am 9 am 12 pm 3 pm 6 pm 9 pm 2022 Number of Electric Vehicles Registered Per Year 2023 2024 Chapter-5.indd 124 10/9/2025 3:45:54 PM 124 25000 50000 0 Uttarakhand West Bengal Andhra Pradesh Odisha Assam Gujarat Delhi 5.4 Data Detective We put well-formed sentences one after the other to make a beautiful story. In the same way, well-organised and well-presented data can tell interesting stories, and can also expose new mysteries or help solve mysteries! (a) The data (rounded-off to thousands) for the states of Gujarat and Delhi are given in the table below. Mark the corresponding bars on the bar graph. (It is enough if you place the top of the bars between the two appropriate vertical guidelines.) (b) Notice how the graph is organised, what scale is used, and what patterns the data shows. (c) How would you describe the change for various states between 2022 and 2024? (d) Approximately how many more registrations did Assam get in 2023 compared to 2022? (e) How many times more did the registrations in West Bengal increase from 2022 to 2024? (f) Is this statement correct — ‘There were very few new registrations in Uttarakhand in 2023 and 2024, as the increase in the bar lengths is minimal’? Gujarat 69000 89000 78000 Delhi 62000 74000 81000 2022 2023 2024 Connecting the Dots… Chapter-5.indd 125 10/9/2025 3:45:54 PM Telling Tall Tales Earlier, we saw data of two Grade 5 classrooms with heights of boys and girls in each class. There, the average height of girls was more than boys in one class and vice versa in the other class. Following are the dot plots of heights of boys (in blue) and girls (in orange) of Grades 6, 7 and 8 (in that order) of two different schools. What do you notice? Share your observations. 125 Math Talk 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 120 130 140 150 160 170 Ganita Prakash | Grade 7 | Part-II Looking at this data, you might wonder: across these two schools?” “Why is there a considerable difference in heights in the same grades School A School B Mean — = 134.8 Mean — = 149.84 Mean — = 137.78 Mean — = 150.2 Mean — = 141.8 Mean — = 156.14 Mean — = 141.83 Mean — = 155.41 Mean — = 149.35 Mean — = 156.14 Mean — = 147.81 Mean — = 156.83 Chapter-5.indd 126 10/17/2025 4:41:06 PM “Where are these schools located?” “How tall are students in Grades 6 to 8 in my school?” “What is the average height of all Grade 6 boys and girls?” We see that men are taller than women in general. But what about the heights of boys and girls? Are boys taller than girls? Well, just by looking at the data of one or two schools, we cannot generalise for all children in our country, or around the world. Let us look at some data (based on a survey) of the heights of boys and girls of different ages in India over time. The following table shows the average heights of boys and girls (in centimeters) across ages 5 to 19 in the years 1989, 1999, 2009, and 2019. In each year, the first column shows boys’ heights and the second column shows girls’ heights. 126 Spend sufficient time observing the data presented in this table. Share your findings with the class. These are some prompts for you to probe — • Changes in the heights of boys or girls of a certain age from 1989 to 2019. • The heights of boys vs. girls at different ages in a particular year. • Changes in height between successive ages in boys and girls in 2019. Age 1989 1999 2009 2019 10 127.5 127.3 128.3 127.8 129.4 129.9 132.6 132.8 11 132.2 133.4 132.8 133.6 133.7 135.7 137 138.6 12 137.7 139 138 139.1 138.9 141.1 142.2 143.8 13 144.2 143.2 144.3 143.1 145.2 145.1 148.4 147.7 14 150.6 146.2 150.5 146.1 151.5 148 154.4 150.4 15 155.4 148.5 155.2 148.4 156.3 150.1 159 152.4 16 158.9 150.1 158.7 150.1 159.9 151.6 162.3 153.8 17 161.3 151.2 161.4 151.3 162.6 152.6 164.6 154.7 18 162.9 151.8 163.2 152.1 164.3 153 166 155.2 19 163.5 151.9 164.2 152.4 165.1 153 166.5 155.2 5 101.3 100 102.4 101.7 105.1 104 107.1 107.2 6 107.5 106 108.7 107.5 111 109.7 113.1 112.9 7 113 111.4 114.2 112.6 116.2 114.8 118.6 118 8 118.1 116.5 119.2 117.5 120.9 119.6 123.5 122.7 9 122.9 121.7 123.9 122.4 125.2 124.5 128.1 127.6 Connecting the Dots… Math Talk Chapter-5.indd 127 10/9/2025 3:45:55 PM Which of the following statements can be justified using the data? 1. The average heights of both boys and girls at every age increased from 1989 to 2019. 2. The average height of 13-year-old girls in 1989 is more than the average height of 14-year-old girls in 2009. 3. The average height of 15-year-old boys in 2019 is more than the average height of 16-year-old boys in 1989. 4. All girls aged 13 are taller than all girls aged 11. 5. Throughout the age period 5 to 19, the average boy's height is more than the average girl's height. 6. Boys keep growing even beyond age 19. 127 Ganita Prakash | Grade 7 | Part-II In 2019, between which two successive ages from 5 to 19 did boys grow the most? Between which two successive ages from 5 to 19 did girls grow most? Suppose the average height of a newborn is 50 cm. Estimate the average height of young children of ages 1 to 4. Based on the trend observed in the table, write your estimates of the heights of boys and girls for ages 5 to 19 in the year 2029. You may want to look at the data of weights. Or you might be curious to see if such patterns are present in other countries as well. You might also wonder if humans were much shorter a few centuries ago! Also, are people in some countries taller than others? The following visualisation shows the change in average heights of 19-year-old boys and girls of different countries from 1989 to 2019. How is the graph organised? What information is presented? What do you find interesting? Notice that the vertical line starts from 145 cm — this helps give a closer (zoomed in) look at the heights. Whenever you see data or some graph, look closely to know the story it has to say and the mysteries it may hold. Math Talk Math Talk Chapter-5.indd 128 10/9/2025 3:45:56 PM He igh t (in c m) 128 A Mean Decision! Figure it Out 1. The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls. 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 Based on the dot plots, which of the following statements are true? (a) The data varies more for the boys than for the girls. (b) The median number of pockets for the boys is more than that for the girls. Number of Pockets Number of Pockets Boys Girls Perhaps using the average height of the family to make the door was not a good idea! Look at this picture of the current world’s tallest and shortest humans together. Connecting the Dots… Chapter-5.indd 129 10/9/2025 3:45:57 PM 2. The following table shows the points scored by each player in four games: (c) The mean number of pockets for the girls is more than that for the boys. (d) The maximum number of pockets for boys is greater than that for the girls. Player Game 1 Game 2 Game 3 Game 4 A 14 16 10 10 B 0 8 6 4 C 8 11 Did not play 13 129 Ganita Prakash | Grade 7 | Part-II 3. The marks (out of 100) obtained by a group of students in a General Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another group’s scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and 86. Compare and describe both the groups performance using, mean and median. 4. Consider this data collected from a survey of a colony. 5. Consider a group of 17 students with the following heights (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the ‘Telling Tall Tales’ section? Now answer the following questions: (a) Find the average number of points scored per game by A. (b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B? (c) Who is the best performer? Choose an appropriate scale and draw a double-bar graph. Write down your observations. Participating 620 320 320 250 105 Favourite Sport Cricket Basket Ball Swimming Hockey Athletics Watching 1240 470 510 430 250 Chapter-5.indd 130 10/9/2025 3:45:57 PM 6. Describe the mean and median of heights of your class. You can visualise the heights on a dot plot. 7. There are two 7th grade sections at a school. Each section has 15 boys and 15 girls. In one section, the mean height of students is 154.2 cm. From this information, what must be true about the mean height of students in the other section? 130 (a) The mean height of students in the other section is 154.2 cm. (b) The mean height of students in the other section is less than 154.2 cm. 8. Standing tall in the storm. (c) The mean height of students in the other section is more than 154.2 cm. (d) The mean height of students in the other section cannot be determined. Connecting the Dots… Chapter-5.indd 131 10/9/2025 3:45:58 PM 9. Estimate and then measure the objects listed in the following table. Draw a double bar graph based on the data. How accurate were your estimates? Find the average difference between the estimated and measured values. (a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London. (b) Are the following statements valid? (i) Only 12 cities have more skyscrapers than Mumbai. (ii) Only 7 cities have fewer skyscrapers than Mumbai. (iii) The tallest building in the world is in Hong Kong. 131 200 Ganita Prakash | Grade 7 | Part-II 10. Aditi likes solving puzzles. She recently started attempting the ‘Easy’ level Sudoku puzzles. The time she took (in seconds) to solve these puzzles are — 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220, 240. The first nine values correspond to Week 1 and the rest to Week 2. 11. Individual Project: Pick at least one of the following: (a) Construct a dot plot below showing the data for both weeks. (b) Describe the mean, median, and any observations you may have about the data. Length of a pen Length of an eraser Length of your plam Length of your geometry box Length of your math notebook 220 240 260 280 300 320 340 360 380 400 Object Estimate (in cm) Measure (in cm) Positive Difference Week 1 Week 2 Chapter-5.indd 132 10/9/2025 3:45:58 PM 132 (a) How Long is a Sentence? Pick any two textbooks from different subjects. Choose any page with a lot of text from each book. (b) What is in a Name? Write down the names of all of your classmates. The following are some interesting things you can do with this data! (i) Use a dot plot to describe how many words the sentences have on each page. (ii) Compare the data of both the pages using mean and median. (i) Find the mean and median name length (number of letters in a name). (ii) Visualise the data and describe its variability and central tendency. 12. Individual project (long term): This requires collecting data over 2 weeks or more. In and Out: Track how many times you step out of your house in a day. Do this for a month. 13. Small-group project: Pick at least one of the following. Make groups of 8 to 10. Collect data individually as needed. Put together everyone’s data and do the appropriate analysis and visualisation. (a) Our heights vs. our family’s heights: Collect the heights of your family members. (iii) Which starting letters are more popular? Which are less popular? (iv) What is the median starting letter? What does this say about the number of names starting with the letters A – M and N – Z? (v) Plot a double-bar graph showing the number of boys’ names and girls’ names that: • start and end with vowels, • start with vowels and end with consonants, (i) Describe the variability and central tendency of this data. Make a dot plot. (ii) Do you find anything interesting about this data? Share your observations. (iii) You can ask any of your family members or friends to do this as well. • start with consonants and end with vowels, • start and end with consonants. Connecting the Dots… Try This Chapter-5.indd 133 10/10/2025 3:59:02 PM (b) Estimating time: Check the time and close your eyes. Open them when you think 1 minute has passed (no counting). Note down after how m-any seconds you opened your eyes. Collect this data for yourself and for your family members. Repeat this activity to estimate 3 minutes. (i) Make a dot plot showing heights of just your family members. Describe its variability and central tendency. (ii) Make a double-bar graph showing each student’s height next to their family’s mean height. (iii) Look at everyone’s data and share your observations. (i) Make two dot plots (for 1 minute and 3 minutes) showing estimates of just your family members. 133 Ganita Prakash | Grade 7 | Part-II • Dot plots help us get a quick glimpse of the variability of the data — minimum, maximum, range, and how data is clustered or spread out. • The Arithmetic Mean = Sum of all the values in the data Number of values in the data . • The Median is the number in the middle of any sorted data. If there are an even number of values, then the median is the average of the two middle numbers. • We can describe and compare data in several ways including by referring to the minimum, maximum, total, range, arithmetic mean, and median. • We learnt how to read and make clustered bar graphs. These graphs can be used to compare and visualise values across categories and across time. • Examining data can lead to new questions and directions to probe further. (ii) Mark these on the respective dot plots. Describe its variability and central tendency. (iii) Make a double bar graph showing each family’s mean 1 minute estimate and mean 3 minute estimate. (iv) Look at everyone’s data and share your observations. SUMMARY Chapter-5.indd 134 10/9/2025 3:45:58 PM 134 A number lock has a 3-digit code. Find the code using the hints below. One digit is correct and well placed. One digit is correct but wrongly placed. Nothing is correct. One digit is correct but wrongly placed. Connect the Dots… Two digits are correct but wrongly placed. Chapter-5.indd 135 10/10/2025 4:00:55 PM" class_7,14,constructions and tilings,ncert_books/class_7/gegp1dd2/gegp206.pdf,"6 6.1 Geometric Constructions Eyes Do you recall the ‘Eyes’ construction we did in Grade 6? Of course, eyes can be drawn freehand, but we wanted to construct them so that the lower arc and upper arc of each eye look symmetrical. We relied on our spatial estimation to determine the two centres, A and B (see the figure), from which we drew the lower arc and upper arc respectively. The arcs define a line XY that ‘supports’ the drawing though it is not part of the final figure. We can start with this supporting line and systematically find the centres A and B. For the eye to be symmetrical, or for the supporting line to be the line of symmetry, the upper and lower arcs should have the same radius. In other words, we must have AX = BX. Since AX = AY and BX = BY, this means AX = AY = BX = BY CONSTRUCTIONS AND TILINGS Chapter-6.indd 136 10/17/2025 3:19:36 PM X Y A B How do we find such A and B? From X and Y, draw arcs above and below XY, with the same radii. The two points at which the arcs meet, above and below XY, give us A and B, respectively. Use this to construct an eye. In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the angle formed between them? We observe that AB passes through the midpoint of XY, and is also perpendicular to it. A division of a line, or any geometrical object, into two identical parts is called bisection. A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector. Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length? This can be answered through congruence. Let us consider a line segment XY. Find points A and B such that AX = AY = BX = BY. Draw the lines AB, AX, AY, BX and BY. Let O be the point of intersection between AB and XY. X Y O X Y Constructions and Tilings Fig. 6.1 A B A B Chapter-6.indd 137 10/10/2025 4:02:17 PM Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)? If we show that ∆AOX ≅ ΔAOY, then OX = OY, and ∠AOX = ∠AOY because they are corresponding parts of congruent triangles. Since ∠AOX and ∠AOY together form a straight angle, we have ∠AOX + ∠AOY = 180°. Thus, ∠AOX = ∠AOY = 90°. This establishes that O is the midpoint of XY and AB is perpendicular to XY. X Y O A B 137 Common side Ganita Prakash | Grade 7 | Part-II In ∆AOX and ΔAOY, we already know that AX = AY, and AO is common to both triangles. If we can show that ∠XAO = ∠YAO then, by the SAS congruence condition, we can conclude that the triangles are congruent. To show this, we observe that ∆ABX ≅ ΔABY. This is so because AX = AY, BX = BY, and AB is common to both the triangles. Thus, we have ∠XAB = ∠YAB, or ∠XAO = ∠YAO because they are corresponding parts of congruent triangles. Hence, AB is the perpendicular bisector of XY. We can have eyes of different shapes. X Y O A B Common side Chapter-6.indd 138 10/9/2025 3:05:55 PM 138 How do we get these different shapes? Try! One way is to choose two other points C and D such that CX = CY = DX = DY. An eye of a different shape can be drawn using these points. Will C and D lie on the perpendicular bisector AB? The points C and D are at the same distance from both X and Y. We have just seen that joining any two such points gives the perpendicular bisector of XY. Since XY has only one perpendicular bisector, which is the line AB, the points C and D must lie on the line AB. Justify the following statement using the facts that we have established. Any point that has the same distance from X and Y lies on the perpendicular bisector of XY. Thus, eyes of different shapes can be drawn by suitably choosing different pairs of points on the perpendicular bisector as centres to construct the upper and lower arcs of the eyes. Construction of Perpendicular Bisector Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass? We have seen that joining any two points — one above XY and one below — that are at equal distances from X and Y, gives the perpendicular bisector of XY. This gives a method to construct the perpendicular bisector. Constructions and Tilings X Y A B D C Chapter-6.indd 139 10/9/2025 3:05:56 PM 1. Taking some fixed radius, from X and then Y, construct two sufficiently long arcs above XY. Name the point where the arcs meet as A. X Y A 139 Ganita Prakash | Grade 7 | Part-II Thus, the perpendicular bisector can be constructed using the simplest geometric tools — an unmarked ruler and a compass. We will use only these two tools for all the other geometric constructions in this chapter, unless there is a need for drawing lines of specific lengths in standard units. Figure it Out 1. When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer. 2. Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer. 2. Using the same radius, from X and then Y, construct two sufficiently long arcs below XY. Name the point where the arcs meet as B. 3. AB is the required perpendicular bisector. [Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector. Hint 2: We can draw the whole line if any two of its points are known.] X Y A B Chapter-6.indd 140 10/9/2025 3:05:56 PM 3. While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them ? Explore this through construction, and then justify your answer. 4. Recreate this design using only a ruler and compass — 140 After completing the above design, you can use a colour pencil with a ruler or compass to trace its boundary. This will make the design stand out from the supporting lines and arcs. This method of constructing the perpendicular bisector is not only geometrically exact but also a practical way to construct it accurately. This method to find the midpoint of a line segment is more accurate than measuring the length using a marked scale. Construction of a 90° Angle at a Given Point Can we extend the method of constructing the perpendicular bisector to construct a 90° angle at any point on a line? Draw a line and mark a point O on it. Construct a 90° angle at point O. Find a segment of this line for which O is the midpoint. Using a compass, mark two points X and Y at equal distance from O, so that O is the midpoint of XY. The perpendicular bisector of XY will pass through O and is perpendicular to the given line. In this case, do we need to draw two pairs of intersecting arcs to get the perpendicular bisector of XY? No, we don’t. We already have one point, O, lying on the perpendicular bisector. Figures 6.2 and 6.3 describe the steps to construct a 90° angle at a given point on a line. Extend the line on either side of O. X O Y ( ) Fig. 6.2 X O Y ( ) O Constructions and Tilings A Chapter-6.indd 141 10/16/2025 3:54:39 PM Construction Methods in Śulba-Sūtras Ancient mathematicians from different civilizations, including India, knew exact procedures to construct perpendiculars and perpendicular bisectors. In India, the earliest known texts containing these methods are the Śulba-Sūtras. These are geometric texts of Vedic period dealing with the construction of fire altars for rituals. The Śulba-Sūtras are part of one of the six Vedāṅgas (a term that literally means ‘limbs of the Vedas’). The Śulbas contain the methods that we developed earlier to construct a perpendicular and the perpendicular bisector. All the construction Fig. 6.3 141 Ganita Prakash | Grade 7 | Part-II methods in the Śulba-Sūtras make use of a different kind of compass from what you would have used — a rope. A rope can be used to draw circles or arcs. It can also be stretched to form a straight line. In addition, the Śulba-Sūtras also contain other methods to construct perpendicular lines. Here is an interesting construction of the perpendicular bisector using a rope (Kātyāyana-Śulbasūtra 1.2). Let XY be the given line segment, drawn on the ground, for which we need to construct a perpendicular bisector. Fix a small pole or peg vertically into the ground at each point X and Y. Fasten the two loops at the ends of the rope to the poles at X and Y. Pull the midpoint of the rope above XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as A. Now similarly pull the midpoint of the rope below XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as B. Take a sufficiently long rope. Make two loops at its ends. Without taking into account the parts of the rope that has gone into the loops, fold the rope into half to find and mark its midpoint. X Y A A Chapter-6.indd 142 10/17/2025 3:20:34 PM Figure it Out 1. Justify why AB in Fig. 6.4 is the perpendicular bisector. 2. Can you think of different methods to construct a 90° angle at a given point on a line using a rope? 142 AB is the required perpendicular bisector. X Y Fig. 6.4 B Math Talk Angle Bisection for a Design What is the angle between two adjacent lines? We need the angle between every pair of adjacent lines to be equal. Since 360° is equally divided into 8 parts, every angle is 45°. How do we construct a 45° angle using only a ruler and a compass? We know how to construct a 90° angle. If we can divide it into two equal parts, or bisect it, then we get a 45° angle. How do we construct this figure? The supporting lines for this figure will look like this. Constructions and Tilings Fig. 6.5 Chapter-6.indd 143 10/9/2025 3:05:57 PM We will now develop a general method to bisect any angle. Consider an angle ∠XOY. We can bisect it if we can draw two congruent triangles ΔOBC and ΔOAC as shown in the figure. Then ∠BOC = ∠AOC. Y O A Y O B C X X 143 Ganita Prakash | Grade 7 | Part-II How do we construct these congruent triangles, given the angle? If A and B are marked such that OA = OB, and if C is chosen such that BC = AC, then by the SSS congruence condition, ΔOBC ≅ ΔOAC. So we can bisect an angle as follows. Steps for Angle Bisection 1. Mark points A and B such that OA = OB. 2. Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C. 3. OC bisects ∠AOB. So, a 45° angle can be constructed by first constructing a 90° angle and then bisecting it. Figure it Out 1. Construct at least 4 different angles. Draw their bisectors. 2. Construct the 8-petalled figure shown in Fig. 6.5. Y O Y O B B A A X X C Chapter-6.indd 144 10/9/2025 3:05:57 PM 3. In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer. 4. What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle? 5. Come up with a method to construct the angle bisector using a rope. 144 C A O Y B X Math Talk 6. Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square? Repeating Units and Repeating Angles Construct the following figure. In this figure, there is a single unit repeating itself. To construct this figure, we need to make exact copies of this unit in two different orientations. Fig. 6.6 Constructions and Tilings Chapter-6.indd 145 10/9/2025 3:05:57 PM In order to make exact copies, all the units must have the same arm lengths and the same angle between the arms. We can ensure equal arm lengths using a compass, but how do we ensure equal angles? Let us develop a method to create an exact copy of a given angle. Draw an angle. Create a copy of this angle using only a ruler and compass. A 145 Math Talk Ganita Prakash | Grade 7 | Part-II You might have developed your own method. Here is one simple approach. Steps of Construction to Copy an Angle 1. 2. 3. Draw an arc from A. This gives us three points that form the isosceles triangle ∆ABC. A X A C A C Draw an arc of the same radius from X. B B X Z X Chapter-6.indd 146 10/9/2025 3:05:57 PM 146 4. Measure BC using a compass. Transfer this length on the arc from Z to get YZ = BC. A C B X Z Y By the SSS congruence condition, ∆ABC ≅ XYZ. So, ∠A = ∠X. Figure it Out 1. Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles. 2. Construct the Fig. 6.6. This procedure to copy an angle finds an important application in constructing parallel lines using only a ruler and compass. Construction of a Line Parallel to the Given Line Recall that in the construction using a ruler and a set square, we constructed equal corresponding angles to get parallel lines. 5. A C B X Z Constructions and Tilings Y Chapter-6.indd 147 10/10/2025 4:12:52 PM How do we implement this idea using a ruler and a compass? Suppose there is a line m to which we need to draw a parallel line. We construct a line l that intersects m. Line l will serve as a transversal to line m and to the line parallel to m that we are going to construct. Let us choose a point B on l through which we are going to draw the parallel line. This parallel line must make the same corresponding angle, as shown in the figure. This can be done by copying the angle between m and l. Fig. 6.7 a A m A l a B 147 a m l Ganita Prakash | Grade 7 | Part-II Here is a step-by-step procedure for carrying this out. Figures 6.7–6.10 describe a method to construct a line parallel to the given line. Figure it Out 1. Construct 4 pairs of parallel lines in different orientations. Construct arcs of equal radius from A and B Fig. 6.8 A D C B A Fig. 6.10 m l D E n C m || n B F Transfer the length CD to the arc from F m l Fig. 6.9 A D E C B F m l Chapter-6.indd 148 10/9/2025 3:05:57 PM 2. Construct the following figure. 148 Z Y G H S T F D X W A E B C U V Arch Designs Trefoil Arch Have you seen this kind of beautiful arch? How did they make these arches? The first step is to be able to draw them on a plane surface such as paper or stone. Construct this arch shape on a piece of paper. Let us think about the support lines this figure will need. For symmetry, we should have AB = CD, and ∠BAD = ∠CDA. How would you construct these support lines? https://commons.wikimedia.org/wiki/ File:Central_Park_A.jpg Central park, New York City https://commons.wikimedia.org/w/index. php?curid=28374748 Diwan-i-Aam, Red fort B C A D Constructions and Tilings B C Chapter-6.indd 149 10/10/2025 4:21:56 PM Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing. A D A D Construct equal angles at A and D. Mark B and C such that AB = CD. 149 Ganita Prakash | Grade 7 | Part-II A Pointed Arch Some arches look like this. How do we construct this shape? What supporting lines will you use to draw this arch? https://en.m.wikipedia.org/wiki/File:Diwan-i-Aam,_Red_Fort,_Delhi_-_2.jpg Diwan-i-Aam, Red Fort Chapter-6.indd 150 10/10/2025 4:22:56 PM Remember ‘Wavy Wave’ from the Grade 6 Textbook? The supporting lines are just two line segments of equal length. If their midpoints are marked, will you be able to construct a pointed arch? 150 Fig. 6.11 Figure it Out 1. Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs. 2. Make your own arch designs. Regular Hexagons Recall that a regular polygon has equal sides and equal angles. A regular polygon with 3 sides is an equilateral triangle, and a regular polygon with 4 sides is a square. We have constructed these figures earlier. How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths. To construct a regular pentagon, we first need to have a better understanding of triangles and pentagons. We will discuss this in later years. However, constructing a regular hexagon is within our reach! Can we break a regular hexagon into smaller pieces that can be constructed? Constructions and Tilings Try This Chapter-6.indd 151 10/9/2025 3:06:11 PM Regular Hexagon and Equilateral Triangles What happens when we join the ‘opposite’ points of a regular hexagon? Since a regular hexagon has equal sides and angles, can we expect a figure like this? Will all the triangles in the figure be equilateral triangles? To answer these questions, we will reverse our approach. Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon? E F Fig. 6.12 A D O 151 B C Ganita Prakash | Grade 7 | Part-II If six congruent equilateral triangles can indeed be placed as shown in Fig. 6.12, then the sides of the resulting hexagon are equal, and their angles are 60 + 60 = 120° (how?). So what we really need to examine is whether 6 congruent equilateral triangles can fit this way without overlapping and without leaving any gaps around the centre. We have defined a degree by taking the complete angle around a point to be 360°. So all the angles around the centre should add up to 360°. Consider this figure. Will the 70° angle fit into the gap? What is the gap angle ∠AOI? We have, 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°. Use this to determine whether the 70°angle fits the gap. Thus, if there are angles that add up to 360°, their vertices can be joined together at a single point such that (a) the angles do not overlap, and (b) they completely cover the region around the point. Since each angle in an equilateral triangle is 60°, six such angles add up to 360°. Therefore, six congruent triangles can be arranged as shown in Fig. 6.12. In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines? Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass. ? A B C 70° 40° D E 60° I O 50°30° 90° 40° F H G Chapter-6.indd 152 10/9/2025 3:06:11 PM We can construct a regular hexagon more directly if we can construct a 120° angle using a ruler and a compass. How do we do it? This can be done if we can construct a 60° angle. 152 If we construct this, 60° 120° we also get this. Construction of a 60° angle How do we construct a 60° angle? We get a 60° angle if we construct an equilateral triangle! We can use the following steps for this. Suppose we need a 60° angle at point A on a line segment AX. Step 1 Step 2 We have ∠CAX = 60°. Why is ∠CAX = 60°? Is there an equilateral triangle here? With the same radius, cut another arc from B that meets the first arc. Let C be the point at which the arcs meet. Construct an arc with centre A and any radius. A B X A B X C Constructions and Tilings Chapter-6.indd 153 10/9/2025 3:06:11 PM We can use these ideas to construct a regular hexagon — Construct a regular hexagon of sidelength 5 cm. 153 Ganita Prakash | Grade 7 | Part-II Related Constructions Construction of 30° and 15° angles How will you construct 30° and 15° angles? 6-Pointed Star Construct the following 6-pointed star. Note that it has a rotational symmetry. Hint: Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG — equilateral? Why? [Hint: Find the angles.] Figure it Out G F K J E L I D C B H A Do you see a hexagon here? Math Talk Chapter-6.indd 154 10/17/2025 3:22:32 PM 1. Construct the following figures: 154 An Inflexed Arc (a) The fun part about this figure is that it can also be constructed using only a compass! Can you do it? (b) 2. Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook. 3. Construct this figure. 4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P. [Hint: Find a line segment on l whose perpendicular bisector passes through P.] [Hint: Find the angles in this figure.] (c) (d) (e) Constructions and Tilings Try This Chapter-6.indd 155 10/17/2025 3:40:29 PM 6.2 Tiling Tangrams are puzzles that originated in China. They make use of 7 pieces obtained by dividing a square as shown. For the problems ahead, we need these 7 tangram pieces. These are provided at the end of the book. Or, by looking at the figure, you could make cardboard cutouts of the pieces. A G C B D F 155 E Ganita Prakash | Grade 7 | Part-II We can form interesting pieces by rearranging the tangram pieces. Here is an arrow. Figure it Out How can the tangram pieces be rearranged to form each of the following figures? Chapter-6.indd 156 10/9/2025 3:06:12 PM 156 Covering a region using a set of shapes, without gaps or overlaps, is called tiling. Consider a rectangular grid made of unit squares — We call this a 4 × 6 grid, since it has 4 rows and 6 columns. Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles? We are allowed to rotate a 2 × 1 tile and use it. Here is one way. To see that there is no way to tile a 5 × 7 grid using 2 × 1 tiles, observe that this grid has 35 unit squares. Each tile covers exactly 2 unit squares. Math Talk Vertical tile Horizontal tile Constructions and Tilings Chapter-6.indd 157 10/9/2025 3:06:12 PM Obviously, this is not the only tiling possible. Can a 4 × 7 grid be tiled using 2 × 1 tiles? What about a 5 × 7 grid? 157 Ganita Prakash | Grade 7 | Part-II Complete the justification. Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it. One general strategy for this case is to cover each column with vertical tiles. This is possible because the number of rows is even. Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it. Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons. Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles? Even no. of tiles Math Talk Chapter-6.indd 158 10/9/2025 3:06:12 PM Is the following region tileable with 2 × 1 tiles? 158 Math Talk What about this one? Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable? There is an interesting way to look at these questions. For any tiling problem of this kind, we can create a similar problem with the unit squares coloured black and white so that black squares have only white neighbours and white squares have only black neighbours. For the tiling problem in Fig. 6.13, we get the following. Region to be tiled Plain grid Black and white grid Fig. 6.13 Constructions and Tilings Math Talk Chapter-6.indd 159 10/9/2025 3:06:12 PM In the black-and-white region, the problem is to tile the region with the 2 × 1 black-and-white tiles so that each black square of a tile sits on a black square of the grid, and each white square sits on a white square. If the plain grid is tileable, is the black-and-white-grid tileable? If the black-and-white grid is tileable, is the plain grid tileable? It can be seen that the answer to both the questions is yes. Tiles Fig. 6.14 159 Math Talk Ganita Prakash | Grade 7 | Part-II Is the black-and-white region in Fig. 6.14 tileable? Any region tiled with black-and-white-tiles must have an equal number of black tiles and white tiles. Since the black-and-white region in Fig. 6.14 has 8 white squares and 6 black squares, it can never be tiled with these tiles! Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable? Isn’t it surprising how, by introducing colours and making the problem more complicated, it becomes easier to tackle? What a creative way of looking at the problem! Figure it Out Are the following tilings possible? 1. 2. Region to be tiled Tile Region to be tiled Tile Math Talk Chapter-6.indd 160 10/9/2025 3:06:13 PM Tiling the Entire Plane So far we have seen how to tile a given region. What about tiling the entire plane? Can you think of a shape whose copies can tile the entire plane? Clearly, squares can. 160 Are there other regular polygons that can tile the plane? What about equilateral triangles? Tiling with equilateral triangles shows the possibility of tiling with another regular polygon. A plane can be tiled using regular hexagons as well. A plane can also be tiled using more than one shape, and using non-regular polygons. People such as the great Dutch artist Escher (1898 – 1972) — whose works explored mathematical themes such as tiling — have come up with creative ways of tiling a plane with animal shapes! Here are some examples. Constructions and Tilings Chapter-6.indd 161 10/9/2025 3:06:13 PM (a) (b) 161 Ganita Prakash | Grade 7 | Part-II Mathematicians are still exploring various ways of tiling the plane! Tiling (b) was found as recently as 2023. Have you seen tilings in daily life? They are often used in buildings and in designs. Tilings are found in nature too. The front face of bee hives and some wasp nests are tiled using hexagonal cells! (c) (d) Chapter-6.indd 162 10/9/2025 3:06:14 PM These cells are used by the insects to keep their eggs, larvae and pupae safe, as well as to store food. Because the region is tiled, no space is wasted. Scientists still wonder how bees and wasps are able to make hexagonal cells. Next time you see any tiling, pay closer attention to it! Tiling is still one of the most exciting and active areas of research in geometry. 162 • A division of a line segment, or any geometrical quantity, into two identical parts is called bisection. • Any point that is of equal distance from the two endpoints of a given line segment lies on its perpendicular bisector. This property can be used to construct the perpendicular bisector using a ruler and compass. • The method of constructing the perpendicular bisector can be modified to draw a 90° angle at any point on a line using only a ruler and compass. • An angle can be bisected and copied using the congruence properties of triangles. • A 60° angle can be constructed using a ruler and compass by constructing an equilateral triangle. • Covering a region using a set of shapes, without gaps or overlaps, is called tiling. SUMMARY Chapter-6.indd 163 10/17/2025 5:26:04 PM" class_7,15,finding the unknown,ncert_books/class_7/gegp1dd2/gegp207.pdf,"7 7.1 Find the Unknowns Unknown Weights We have a weighing scale that behaves as follows. The numbers represent same units of weight: Find the unknown weights in the following cases: FINDING THE UNKNOWN Chapter-7.indd 164 10/9/2025 3:19:51 PM Fig. 7.1 Fig. 7.2 Fig. 7.3 Discuss the answers with your classmates. Give reasons why you think your answer is right. Fig. 7.6 Fig. 7.7 Fig. 7.8 Fig. 7.4 Fig. 7.5 Finding the Unknown Math Talk Chapter-7.indd 165 10/9/2025 3:19:54 PM Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the sacks have the same weight. [Hint: If we remove equal weights from both the plates, will the weighing scale still be balanced? Remove one sack from each plate for Fig. 7.10.] Fig. 7.9 Fig. 7.10 165 Ganita Prakash | Grade 7 | Part-II Note to the Teacher: Encourage your students to find solutions to these problems using different strategies and methods, and ask them to compare and contrast their methods. [Hint: Can you remove objects so that the sacks are only on one plate?] Fig. 7.11 Fig. 7.12 Chapter-7.indd 166 10/10/2025 4:26:07 PM Let us find an unknown value in a different setting. Matchstick Pattern Consider this sequence of matchstick arrangements. Recall that we have studied this sequence in an earlier chapter. 166 1 2 3 4 The figure shows the first 4 matchstick arrangements along with their position numbers in the sequence. Jasmine decides to make a matchstick arrangement that appears in this sequence, using exactly 99 sticks. What will be the position number of this arrangement in the sequence? To answer this question, it is useful to know the number of matchsticks in each position. We see that the arrangement at position 1 has 2 × (1) + 1 = 3 matchsticks, the arrangement at position 2 has 2 × (2) + 1 = 5 matchsticks, the arrangement at position 3 has 2 × (3) + 1 = 7 matchsticks, and so on. So, the nth position will have 2n+1 matchsticks. To answer Jasmine’s question, we have to find the value of n such that 2n + 1 has the value 99, or, 2n + 1 = 99. Can you find ways to get the value of n, such that 2n + 1 = 99? Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks? A statement of equality between two algebraic expressions is called an equation. Nowadays, when using symbols, we write an equation as two algebraic expressions with an equal sign ‘=’ between them. Here are some more examples of equations: 3x + 4 = 7, 20 = y – 3, Finding the Unknown Math Talk Chapter-7.indd 167 10/9/2025 3:19:55 PM The process of finding the value(s) of the letter−numbers for which the equality holds, or for which the value of the Left Hand Side (LHS) of the equation becomes equal to the Right Hand Side (RHS) of the equation, is called solving the equation. As we saw in the matchstick problem, framing an equation using an unknown quantity as a letter−number can help us find its value. 2z + 4 = 5z – 14 etc. a 3 = 50, Left Hand Side (LHS) 2n + 1 = 99 Right Hand Side (RHS) 167 Ganita Prakash | Grade 7 | Part-II For the weighing scale problems in figures 7.6, 7.7, 7.8, 7.9, 7.10, and 7.11, frame equations by using letter-numbers to denote the unknown weight. For the problem in Fig. 7.6, let us denote the weight of one fried egg as e. Since each slice of bread is 2, we have 2 + 2 + 2 = 6 on one side and e + e on the other side. Since they are equal, we have 6 = e + e, or For the problem in Fig. 7.7, = 4, and we can denote the weight of one  as y. So, we have 16 on one side, and 4 + 2y on the other side. Thus, we have the equation Solve the equations that you frame and check if you get the same value for the unknown weight as you got previously. Frame 5 equations. Find methods to solve them. 7.2 Solving Equations Systematically How did you solve the various equations framed in the previous section? One way to solve an equation is to substitute different values in place of the letter−number and to check which value makes LHS = RHS. For example, consider the equation 2n + 1 = 99. If we substitute n = 5, we get LHS = 2 × (5) + 1 = 11. It is far away from 99, the RHS. We can try n = 10. The LHS is 21. It is still not equal to 99. Can we try n = 30? The LHS is now 61, still much lower than 99. Let us try n = 40. The LHS is 81. We are getting closer to 99. When n = 50, the LHS is 101. This is just a bit too high. When n = 49, the LHS is 99! Therefore, the solution to the equation 2n + 1 = 99 is n = 49. 4 + 2y = 16. 2e = 6. Math Talk Chapter-7.indd 168 10/9/2025 3:19:56 PM Can this equation have any other solution? This method is called the trial and error method. The trial and error method can be inefficient. Try solving 5x – 4 = 7 using trial and error. 168 Recall that in the case of the weighing scale, we didn’t use the trial and error method! For some problems, finding the solution was straightforward. For others, we used the fact that when we remove equal weights from both the plates of a balanced weighing scale, it remains balanced. Do equations have a similar property? Consider an equation 15 + 8 = 23. If we add, subtract, multiply or divide the same number on both sides, will it still preserve the equality of LHS and RHS? For example, you can check by adding 10 to both sides. Since the LHS and the RHS of an equation have the same value, performing the same operation on both sides will clearly not change their equality. In the weighing scale problems, we removed equal weights so that all the unknown weights were on only one plate of the scale. This made the problem easier to solve. We can use this same strategy to solve an equation as well! Let us first solve some arithmetic problems. Example 1: It is known that 14593 – 1459 + 145 – 14 + 88 = 13353. What is the value of 14593 – 1459 + 145 – 14? Why can we do this? We can do this because addition and subtraction are inverse operations. So, the value of the required expression can be found by subtracting 88 from both sides (LHS and RHS), which removes the term 88 and leaves only the expression to be evaluated on the LHS. To find the value, do we need to evaluate 14593 – 1459 + 145 – 14? No, we can get it by subtracting 88 from 13353. Finding the Unknown Chapter-7.indd 169 10/9/2025 3:19:56 PM Example 2: It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is the value of the expression 23 × 41 × 11 × 8? Using the fact that multiplication and division are inverse operations, we can simply divide 5,80,888 by 7 to get the value of the expression. Is this the same as dividing both sides by 7, which removes the factor 7 and leaves only the expression to be evaluated on the LHS? Example 3: It is known that 12345 – 5432 + 135 – 24 – (–67) = 7091. What is the value of the expression 12345 – 5432 + 135 – 24? 14593 – 1459 + 145 – 14 + 88 – 88 = 13353 – 88 169 Math Talk Ganita Prakash | Grade 7 | Part-II This problem also can be solved using the fact that addition and subtraction are inverse operations. However, we can use the following method that is perhaps easier to visualise. To retain only the expression to be evaluated on the LHS, we need to remove the term –(–67). We can remove this term by adding (–67) to both sides of the equation. 12345 – 5432 + 132 – 24 – (–67) + (–67) = 7091 + (–67) 12345 – 5432 + 132 – 24 – (–67) + (–67) = 7091 + (–67) Thus, 12345 – 5432 + 132 – 24 = 7024. Example 4: It is known that ( 35 113) × 24 × 14 × ( 8 9) = 94080 1017 . What is the value of the expression ( 35 113) × 24 × 14? Solution: To retain only the expression to be evaluated on the LHS we need to remove the factor ( 8 9). We can do that by dividing both sides by ( 8 9). Thus, Let us use these ideas to solve the equation 5x – 4 = 7. What can we do so that 5x is on one side and the equality between the LHS and the RHS still holds? To retain only 5x on the LHS, we need to remove the term – 4. This can be done by adding 4 to both sides. [( 35 113) × 24 × 14 × ( 8 9)] ÷ ( 8 9) = 94080 1017 ÷ ( 8 9) ( 35 113) × 24 × 14 = 11760 113 . ( 35 113) × 24 × 14 × ( 8 9)× ( 9 8)= 94080 1017 × ( 9 8) Chapter-7.indd 170 10/9/2025 3:19:56 PM Thus, 5x – 4 + 4 = 7 + 4. Hence, 5x = 11. To retain only the unknown x on the LHS, we need to remove the factor 5. This can be done by dividing both sides by 5. Thus, 5x 5 = 11 5 . So, x = 11 5 . 170 Can we check that x = 11 5 is the correct solution to the equation? We can check this by substituting x with the value 11 5 in the equation 5x – 4 = 7, and checking if the LHS = RHS. Substituting x with 11 5 in the LHS we get, This is the same as the RHS. So, 11 5 is indeed the correct solution to the equation. Example 5: Solve the equation 11y + (–5) = 61. To retain only the unknown term 11y on one side, we need to remove the term –5. This can be done by subtracting (–5) from both sides. 11y + (–5) – (–5) = 61 – (–5). That is, We can directly find the value of y, seeing that 11 × 6 = 66. We can also divide both sides by 11 to find y. 11y ÷11 = 66 ÷ 11. LHS = 5 ( 11 5 ) – 4 11y = 66. = 5 ( 11 5 ) – 4 = 11 – 4 = 7 Finding the Unknown Chapter-7.indd 171 10/9/2025 3:19:56 PM So y = 6 is the solution to the equation 11y + (–5) = 61. Can you check that this solution is correct? Example 6: Solve 6y + 7 = 4y + 21. In this equation, expressions with an unknown are on both sides. We have seen how to solve equations when the unknown term is on one side. What can be done to bring the unknown terms to the same side? Subtracting 4y from both sides, we get 6y + 7 – 4y = 4y + 21 – 4y. So, 2y + 7 = 21. 171 Ganita Prakash | Grade 7 | Part-II We can directly find the value of y, seeing that 2 × (7) = 14. We can also divide both sides by 2 to find y. Figure it Out [Hint: 4 more than a number, and 5 more than a number can never be equal!] The procedure to systematically solve an equation can be made efficient if we make an observation. Consider the equation 11y + (–5) = 61, which we solved. As the first step, we subtracted – 5 from both sides to remove the term –5. 1. Solve these equations and check the solutions. 2. Frame an equation that has no solution. Subtracting 7 from both sides we get 2y + 7 – 7 = 21 – 7, which gives (a) 3x – 10 = 35 (b) 5s = 3s (c) 3u – 7 = 2u + 3 (d) 4 (m + 6) – 8 = 2m – 4 (e) u 15 = 6 Remember, it is always good to check your solution. 2y ÷ 2 = 14 ÷ 2, which gives 2y = 14. y = 7. Math Talk Chapter-7.indd 172 10/10/2025 4:27:06 PM Since this action removes the term – 5 from the LHS, we could have written this step as: 11y = 61 – (–5). Note that we could also have arrived at this step by seeing addition and subtraction as inverse operations, as in the case of Example 1. Similarly, when we had 11y = 66, we divided both sides by 11 to remove the factor 11 from the LHS. 172 11y + (– 5) – (–5) = 61 – (– 5). 11y ÷ 11 = 66 ÷ 11. Since this action removes the factor 11 in the LHS, we can write this step as: y = 66 ÷ 11. Again, we could have arrived at this by seeing multiplication and division as inverse operations, as in the case of Example 2. Let us write down both these ways of solving an equation. What happens in cases like u 15 = 6? Multiplying both sides by 15 leaves only u in the LHS — u = 6 × 15 11y + (– 5) = 61 11y + (– 5) – (– 5) = 61 – (– 5). 11y = 66 11y ÷ 11 = 66 ÷ 11 y = 6 6y + 7 = 4y + 21 6y + 7 – 4y = 4y + 21 – 4y 2y + 7 = 21 2y + 7 – 7 = 21 – 7 2y = 14. 2y ÷ 2 = 14 ÷ 2 y = 7 Let us consider another equation that we solved earlier. u = 90. 11y + (– 5) = 61 11y = 61 – (– 5) 11y = 66 y = 66 ÷ 11 y = 6 6y + 7 = 4y + 21 6y + 7 – 4y = 21 2y + 7 = 21 2y = 21 – 7 2y = 14. y = 14 ÷ 2 y = 7 Finding the Unknown Chapter-7.indd 173 10/9/2025 3:19:56 PM Thus, we can make the following observations — (a) When a term that is added or subtracted on one side of an equation is removed from that side, its additive inverse should appear as a term on the other side for the equality to hold. For example, 2y + 7 = 21 becomes 2y = 21 – 7. (b) If one side of an equation is the product of two or more numbers or expressions, and we remove one of the factors, then the other side should be divided by this factor for the equality to hold. For example, 2y = 14 becomes y = 14 ÷ 2. (c) If one side of an equation is the quotient of two numbers or expressions, and we remove the divisor, then the other side should be multiplied by this divisor for the equality to hold. For example, u 15 = 6 becomes u = 6 × 15. 173 3 + 3 + 3 + 1 = 10 1 + 1 + 1 + 1 = 4 2 + 2 + 2 + 1 = 7 Step 2 Step 3 Step 4 Step k Method 2 Ganita Prakash | Grade 7 | Part-II Solving Problems Forming and solving equations gives us the ability to find solutions to many problems in our lives. Let us see a few such examples. Example 7: Ranjana creates a sequence of arrangements with square tiles as shown below. Can she extend the sequence and make an arrangement using 100 tiles? If yes, which step in the sequence will it be? She can look at the pattern in different ways. They are shown below. Step 2 Step 3 Step 4 Step k Method 1 Step 1 1 1 1 2 1 2 3 1 3 4 1 4 1 2 3 4 Step 1 Step 2 Step 3 4 + 4 + 4 + 1 = 13 k + k + k + 1 = 3k + 1 Chapter-7.indd 174 10/10/2025 4:29:05 PM 1 + 3 = 4 2 + 5 = 7 3 + 7 = 10 4 + 9 = 4 + (4 × 2 + 1) = 13 174 Step 1 3 5 7 9 = (4 × 2 + 1) 1 2 3 4 k + (2k + 1) = 3k + 1 We have the expression 3k + 1 which gives the number of tiles needed to make an arrangement in Step k. To check whether an arrangement is possible using 100 tiles at some Step k, we can solve the equation: 3k + 1 = 100. Find the value of k. Example 8: Madhubanti wants to organise a party. She decides to buy snacks for the party from the chaat shop in town. Each plate of snacks costs ₹25. The shop charges an additional fixed amount of ₹50 to deliver the snacks to Madhubanti’s house. There are 5 members in Madhubanti’s family, including herself. Her parents tell her she can spend ₹500 on this party. How many friends can she invite to the party if she wants to give a plate of snacks to each person, including her family and friends? Fatima’s method of solving this problem is shown below. She represented the situation using a rough diagram. Snack cost Delivery cost 25 × + 50 Finding the Unknown Chapter-7.indd 175 10/10/2025 4:29:05 PM Out of ₹500, if we subtract the fixed delivery charge, then Madhubanti is left with ₹450. So the question becomes “How many plates of snacks, each costing ₹25, can be bought for ₹450?”. The answer to this is 450 ÷ 25 = 18. 18 plates of snacks can be bought for ₹450. Considering her 5 family members, she can invite 18 – 5 = 13 friends to her party. 175 Mahesh represented the unknown quantity of the total number of people who can attend the party, including Madhubanthi and her family members, as p. Cost incurred = 25p + 50. Since this cost should be 500, we have the equation Subtracting 50 from both sides, we get Dividing both sides by 25, we get p = 450 ÷ 25 = 18. 18 people can be at her party, including her 5 family members. That means 13 friends can be invited. Ganita Prakash | Grade 7 | Part-II Example 9: Two friends want to save money. Jahnavi starts with an initial amount of ₹4000, and in addition, saves ₹650 per month. Sunita starts with ₹5050 and saves ₹500 per month. After how many months will they have the same amount of money? Let m denote the number of months after which their savings are equal. What are Jahnavi’s savings after m months? Jahnavi’s savings = 4000 + 650 m. 25p + 50 = 500. 25p = 500 – 50. 25p = 450. Srikanth decided to represent the unknown quantity of the total number of friends Madhubanthi can invite as f. What will be the cost in this case? Cost incurred = 25 (f + 5). Since Madhubanthi has ₹450 for snacks, we have the equation Dividing both sides by 25, f + 5 = 18. Subtracting 5 from both sides, 25 (f + 5) = 450. f = 13. Chapter-7.indd 176 10/10/2025 4:29:05 PM What are Sunita’s savings after m months? Sunita’s savings = 5050 + 500 m. As these amounts are equal after m months, we get the following equation: 4000 + 650 m = 5050 + 500 m 176 4000 + 650 m – 500 m = 5050 (subtracting 500 m from both sides) 150 m = 5050 – 4000 (subtracting 4000 from both sides) 4000 + 150 m = 5050 Jahnavi Sunita Equal 4000 5050 + + 650 × m 500 × m Subtracting 300 from both sides, we get Dividing both sides by 28, we get So x = 25 – 4, which gives x = 21. 28 (x + 4) = 1000 – 300 So, after 7 months, both will have the same amount of money. Check the answer. Let us solve a few equations. Example 10: Solve 28 (x + 4) + 300 = 1000. Here are some ways to solve this equation. 28 (x + 4) = 700. x + 4 = 700 ÷ 28 x + 4 = 25. m = 1050 ÷150 (dividing both sides by 150) 28 (x + 4) + 300 = 1000 Since 28, 300, and 1000 are divisible by 4, we get a simpler equation if we divide both sides by 4. Using the rules of fraction addition, we get 7 (x + 4) + 75 = 250 28 (x + 4) 28 (x + 4) + 300 4 = 1000 4 4 + 300 4 = 1000 4 150 m = 1050 m = 7. 28 (x + 4) + 300 = 1000 Simplifying the LHS, 28x + 112 + 300 = 1000 28x + 412 = 1000 Subtracting 412 from both sides, we get 28x = 1000 – 412 28x = 588 Dividing both sides by 28, x = 588 ÷ 28 Finding the Unknown Chapter-7.indd 177 10/10/2025 4:29:05 PM Subtracting 75 from both sides, 7 (x + 4) = 175 7x + 28 = 175 Subtracting 28 from both sides, 7x = 147 x = 147 7 = 21. x = 21. 177 Ganita Prakash | Grade 7 | Part-II Example 11: Riyaz created a math trick, which he tries out on his friend Akash. Riyaz asked Akash to perform the following steps without revealing the answer to any of the intermediate steps. The final answer revealed by Akash was 24. Using this, Riyaz correctly figured out the starting number that Akash had thought of. Find this number. Try the steps using different numbers as the starting number. Do you see any relation between the starting number and final answer? The answer can be found by algebraically modeling this scenario. In other words, we can form an equation using an unknown. Let the unknown starting number be x. What are the expressions we get after each step? 1. Think of a number. 2. Subtract 3 from the number. 3. Multiply the result by 4. 4. Add 8 to the product. 5. Reveal the final answer. Subtract 3 from the number. x – 3 Think of a number. x Steps Expression Chapter-7.indd 178 10/9/2025 3:19:57 PM Since Akash’s final answer was 24, we have the equation: Thus, Akash thought of the number 7. Can you think of a simple rule that you can use to get the starting number from the final answer? 178 Multiply the result by 4. 4 (x – 3) = 4x – 12 Add 8 to the product. 4x – 12 + 8 = 4x – 4 4x – 4 = 24 4 (x – 1) = 24 x – 1 = 6 (Dividing both sides by 4) Example 12: Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. How many marbles does each boy have? In this problem, we have two unknowns. If we denote the number of marbles with Ramesh as x and the number of marbles with Suresh as y, then what are the different equations that we have? How do we find the unknowns using these equations? So far, we have only developed a strategy to solve an equation with one unknown! To get such an equation, we can do the following. Denote the number of marbles with Suresh as y, and the number of marbles with Ramesh as y + 30. Since the total number of marbles is 60, we have the equation: y + (y + 30) = 60 1. The total number of marbles is 60. x + y = 60. 2. Ramesh has 30 marbles more than Suresh. x = y + 30. Suresh 1 2 3 30 Ramesh Finding the Unknown Chapter-7.indd 179 10/9/2025 3:19:57 PM Use this to find both the unknowns. Generating Equations So far, we have solved a given equation to find the value of the letter−number. Can we do the reverse — write equations using a given value of the letter−number? Write equations whose solution is y = 5. Share the equations you made with each other and discuss the methods used. Math Talk 2y + 30 = 60. 179 Ganita Prakash | Grade 7 | Part-II Two such equations are y + 1 = 6 and 3y = 15. Consider the following chains of equations, where one is obtained from the previous one by performing the same operation on both sides. Can you form a chain going from the bottom equation to the top? Compare the operations used when going from the top to the bottom and from the bottom to the top. Without calculating, can you find the value of the unknown in each equation in the chains above? [Hint: We have seen that the value that satisfies an equation also satisfies the new equation obtained by performing the same operation on both sides of the original equation.] –1 = – 6 + y – y – 1 = – 6 y + 1 = 6 5 = y Multiplying by – 1 Adding y Adding 6 3y + 6 = 21 y + 2 = 7 3y = 15 y = 5 Adding 6 Dividing by 3 Subtracting 2 Math Talk Chapter-7.indd 180 10/9/2025 3:19:57 PM Example 13: Can you give a real-life situation that can be modelled as the equation, 100x + 75 = 250? Solution: If we think of these numbers as representing money, we can see that the total money is ₹250. There are two terms that are adding to ₹250. The second term in the LHS, ₹75, is fixed and does not change. The value of the first term would change based on ‘x’. So we can think of 100 as the number of units and ‘x’ as the cost per unit. For instance, x could represent the cost of a plate of snacks and 75 could be the delivery charge. 180 Figure it Out 7.3 Mind the Mistake, Mend the Mistake The following are some equations along with the steps used to solve them to find the value of the letter−number. Go through each solution and decide whether the steps are correct. If there is a mistake, describe the mistake, correct it and solve the equation. 1. Write 5 equations whose solution is x = – 2. 2. Find the value of each unknown: 3. I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. My ten’s digit is 3 less than my unit’s digit. The sum of all the three digits is 15. Who am I? 4. The weight of a brick is 1 kg more than half its weight. What is the weight of the brick? 5. One quarter of a number increased by 9 gives the same number. What is the number? 6. Given 4k + 1 = 13, find the values of: (a) 2y = 60 (b) – 8 = 5x – 3 (c) – 53w = –15 (d) 13 – z = 8 (e) k + 8 = 12 – k (f) 7m = m – 3 (g) 3n = 10 + n (a) 8k + 2 (b) 4k (c) k (d) 4k – 1 (e) – k – 2 Finding the Unknown Chapter-7.indd 181 10/17/2025 4:11:50 PM 1 7 – 8z = 5 8z = 7 – 5 8z = 2 z = 4 4 15w – 4w = 26 15w = 26 + 4w 15w = 30 w = 2 4x + 6 = 10 4x = 10 + 6 4x = 16 x = 4 5z + 2 = 3z – 4 5z + 3z = – 4 + 2 8z = – 2 z = – 2 8 2 2v – 4 = 6 v – 4 = 6 – 2 v – 4 = 4 v = 8 5 3x + 1 = – 12 6 3 x + 1 = – 12 3 x + 1 = – 4 x = – 5 181 Ganita Prakash | Grade 7 | Part-II 7.4 A Pinch of History Forming expressions using symbols and solving equations with such expressions was an important component of mathematical explorations in ancient India. This area of mathematics was termed bījagaṇita, also now known as algebra. The word bīja means seed. Just as a tree is hidden inside a seed, the answer to a problem is hidden inside an unknown number. Solving the problem is like helping the tree grow — step by step, we discover what is hidden. We have seen Brahmagupta’s contributions to different areas of mathematics like integers and fractions. In Chapter 18 of his book Brāhmasphuṭasiddhānta (628 CE), he also explained how to add, subtract, and multiply unknown numbers using letters — similar to how we use x or y today. This chapter was one of the earliest known works in algebra in history. As renowned mathematician and Fields Medalist David Mumford remarked, ‘Brahmagupta is the key person in the creation of algebra as we know it’. In the 8th century, Indian mathematical ideas were translated into Arabic. They influenced a well-known mathematician named Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE, he wrote a book called Hisab al-jabr wal-muqabala, which means ‘calculation by restoring and balancing’. These ideas spread even further. By the 12th century, Al-Khwarizmi’s book was translated into Latin and brought to Europe, where it became very popular. The word al-jabr from his book gave us the word algebra, which we also still use today. Similar to how we use letters from the alphabet today to represent unknowns, ancient Indian mathematicians from the time of Brahmagupta used distinct symbols like yā, kā, nī, pī, lo, etc., for different unknowns. The symbol yā was short for yāvat-tāvat (meaning ‘as much as needed’). Others like kā and nī referred to as the first letters 7 8 9 4 (4q + 2) = 50 4 (4q) = 50 – 2 16q = 48 q = 3 – 2 (3 – 4x) = 14 –6v – 8x = 14 – 8x = 14 + 6 – 8x = 20 x = – 20 8 3 (7y + 4) = 9 + 5y 7y + 4 = 9 3 + 5y 7y + 4 = 3 + 5y 7y – 5y + 4 = 3 2y = 4 – 3 y = 1 2 Chapter-7.indd 182 10/10/2025 4:30:26 PM 182 of the names of colours — kālaka (black), nīlaka (blue), and so on. In contrast to these unknowns, the known quantities in an expression had a specific form (rūpa) and were denoted by rū. Here are some examples of how algebraic expressions in modern notation were written in ancient Indian notation: Example 16: Bījgaṇita by Bhāskarāchārya (1150 CE) mentions this problem. One man has ₹300 rupees and 6 horses. Another man has 10 horses and a debt of ₹100. If they are equally rich and the price of each horse is the same, tell me the price of one horse. Solution: Modern Notation Ancient Indian Notation 3x + 4 = 2x + 8 yā 3 rū 4 yā 2 rū 8 (the two sides of an equation were presented one below the other) 2x + 1 yā 2 rū 1 (in each term the indication of unknown/known came first) 2x – 8 yā 2 rū 8 . (a dot over the number indicated that it was negative) Finding the Unknown Chapter-7.indd 183 10/9/2025 3:19:58 PM Let price of one horse = ₹ x According to the problem Therefore, the price of one horse = ₹ 100. 300 + 6x = 10x – 100 300 + 6x + (100) = 10x 400 + 6x = 10x 400 = 10x – 6x 400 = 4x 400 ÷ 4 = x 100 = x 183 Ganita Prakash | Grade 7 | Part-II Such problems and solutions were well understood in ancient India. In fact, a very systematic way to solve problems with single unknowns was first proposed by Aryabhata (499 CE) and Brahmagupta has outlined it in his Brāhmasphuṭasiddhānta (628 CE). Let us understand his method. Let us look at a few equations of the following form: Can we come up with a formula to solve these equations? That is, for the first equation, can we perform some operations using 5, 4, 3, and 8 that will directly give us the solution? Using a similar method, can you solve the second equation using the numbers 3, – 6, 2 and 4? To get a formula, we can generalise equations of this form by taking the four numbers as A, B, C, and D. That is, Brahmagupta gives the solution to equations of this form with this formula: Using this approach, we can find the solution to the equation: simply by calculating, 5x + 4 = 3x + 8, or 3x – 6 = 2x + 4. 650m + 4000 = 500m + 5050 m = 5050 – 4000 650 – 500 . Ax + B = Cx + D. x = D – B A – C . Chapter-7.indd 184 10/10/2025 4:31:06 PM Using this formula can you solve this equation 2x + 3 = 4x + 5? Ancient Indian mathematicians were excellent at converting complex mathematical ideas into simple procedures so that everyone could use these procedures to solve problems! Bījagaṇita or algebra is the branch of mathematics that uses letter symbols to solve mathematical problems. We have seen some examples in previous pages. By studying algebra, we learn to generalise patterns— in numbers, shapes, and situations. We express these generalisations using the language of algebra, which is both precise and powerful. Bījagaṇita also gives us a way to justify mathematical claims (like why the sum of two odd numbers is always even) and to solve problems of many kinds. 184 The power of algebra was well recognised by ancient Indian mathematicians. We hope you recognise it too — and enjoy using it! Figure it Out 1. Fill in the blanks with integers. 2. Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer pays her in 50 and 100 rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have? 4. Here are machines that take an input, perform an operation on it and send out the result as an output. 3. In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem. (a) 5 × ___ – 8 = 37 (b) 37 – (33 – ____ ) = 35 (c) – 3 × (– 11 + ____ ) = 45 (a) Finding the Unknown 25 dots Chapter-7.indd 185 10/9/2025 3:19:58 PM Find the inputs in the following cases: 185 Ganita Prakash | Grade 7 | Part-II (b) 5. What are the inputs to these machines? Find the inputs in the following cases: Chapter-7.indd 186 10/9/2025 3:20:00 PM 186 6. A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each kilometer traveled. If the total cost for a taxi ride is ₹2200, determine the number of kilometres traveled. 10. Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38. 11. The steps to solve three equations are shown below. Identify and correct any mistakes. 7. The sum of two numbers is 76. One number is three times the other number. What are the numbers? 8. The figure shows the diagram for a window with a grill. What is the gap between two rods in the grill? 9. In a restaurant, a fruit juice costs ₹15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹600, find the cost of the fruit juice and milkshake. (a) 6x + 9 = 66 x + 9 = 11 x = 11 – 9 x = 2 11 (b) 14y + 24 = 36 7y + 12 = 18 7y = 6 2 cm 34 cm y = 6 7 3 cm (c) 4x – 5 = 9x + 8 4x = 9x + 8 – 5 4x = 9x + 3 4x – 9x = 3 –5x = 3 Finding the Unknown Chapter-7.indd 187 10/9/2025 3:20:00 PM 12. Find the measures of the angles of these triangles. y y + 15 x – 10 x + 10 x = – 5 3 x 187 Ganita Prakash | Grade 7 | Part-II 13. Write 4 equations whose solution is u = 6. 14. The Bakhśhāli Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second; and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person? 15. The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe? 16. Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure: (a) How many squares are in position number 11 of the sequence? (b) How many sticks are needed to make the arrangement in position number 11 of the sequence? (c) Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to? (d) Can an arrangement in this sequence be made using exactly 150 sticks? If yes, which position number will it correspond to? Chapter-7.indd 188 10/9/2025 3:20:01 PM 17. A number increased by 36 is equal to ten times itself. What is the number? 18. Solve these equations: 19. Solve the equations to find a path from Start to the End. Show your work in the given boxes provided and colour your path as you proceed. 188 (a) 5(r + 2) = 10 (b) – 3(u + 2) = 2(u – 1) (c) 2(7 – 2n) = – 6 (d) 2(x – 4) = – 16 (e) 6(x – 1) = 2(x – 1) – 4 (f) 3 – 7s = 7 – 3s (g) 2x + 1 = 6 – (2x – 3) (h) 10 – 5x = 3(x – 4) – 2(x – 7) 8x = 20 + 3x 8x – 3x = 20 5x = 20 x = 20 5 x = 4 8m + 8 = – 72 2x – 9 = – 3 –10 –10 – 5 – 4 4 5 6 8 5 – 8 2 3 + 4 4 –2 13 8 – 4 – 1 2 (x + 1) – 10 = 18 2x + 5 = 3 (x – 1) – 7 = 11 – 3x – 2x = – 42 – 44 21 1 2x + 3 = x + 5 15 = 19 – 4x Finding the Unknown Chapter-7.indd 189 10/9/2025 3:20:02 PM 20. There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there? – 4 = 16 – 5k 2x – 9 = 3 – x 30 = 4 – 50n 189 Try This Ganita Prakash | Grade 7 | Part-II I predict that you now have 8. Am I correct? Try the trick on your friends and family! Can you explain why the trick works? • An algebraic equation is a mathematical statement that indicates the equality of two algebraic expressions. • When the same operation is performed on both sides of an equation, equality is maintained. • Finding a solution to an equation means finding the values of the unknowns in the expressions such that the LHS is equal to the RHS. • Equations can often be solved by performing the same operation on both sides so that the value of the unknown becomes evident. Think of any number. Now multiply it by 2. Add 10. Divide by 2. Now subtract the original number you thought of. Finally, add 3. SUMMARY A Magic Trick Chapter-7.indd 190 10/9/2025 3:20:05 PM 190 [Hint: Denote the first number thought of by x.] Can you make your own such tricks? Note: Cut each shape along the white border. A tANGRAM B D E Chapter-7.indd 191 10/10/2025 4:41:18 PM F G C Chapter-7.indd 192 10/10/2025 4:41:18 PM Chapter-7.indd 193 10/10/2025 4:41:18 PM Chapter-7.indd 194 10/10/2025 4:41:18 PM Chapter-7.indd 195 10/10/2025 4:41:19 PM Chapter-7.indd 196 10/10/2025 4:41:19 PM Chapter-7.indd 197 10/10/2025 4:41:19 PM Chapter-7.indd 198 10/10/2025 4:41:19 PM" class_8,1,A square and a cube,ncert_books/class_8/hegp1dd/hegp101.pdf,"1 A SQUARE AND A CUBE Queen Ratnamanjuri had a will written that described her fortune of ratnas (precious stones) and also included a puzzle. Her son Khoisnam and their 99 relatives were invited to the reading of her will. She wanted to leave all of her ratnas to her son, but she knew that if she did so, all their relatives would pester Khoisnam forever. She hoped that she had taught him everything he needed to know about solving puzzles. She left the following note in her will— “I have created a puzzle. If all 100 of you answer it at the same time, you will share the ratnas equally. However, if you are the first one to solve the problem, you will get to keep the entire inheritance to yourself. Good luck.” The minister took Khoisnam and his 99 relatives to a secret room in the mansion containing 100 lockers. The minister explained— “Each person is assigned a number from 1 to 100. • Person 1 opens every locker. • Person 2 toggles every 2nd locker (i.e., closes it if it is open, opens it if it is closed). Chapter 1.indd 1 10-07-2025 14:06:40 This continues until all 100 get their turn. In the end, only some lockers remain open. The open lockers reveal the code to the fortune in the safe.” Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer? Hint: Find out how many times each locker is toggled. • Person 3 toggles every 3rd locker (3rd, 6th, 9th, … and so on). • Person 4 toggles every 4th locker (4th, 8th, 12th, … and so on). Ganita Prakash | Grade 8 If a locker is toggled an odd number of times, it will be open. Otherwise, it will be closed. The number of times a locker is toggled is the same as the number of factors of the locker number. For example, for locker #6, Person 1 opens it, Person 2 closes it, Person 3 opens it and Person 6 closes it. The numbers 1, 2, 3, and 6 are factors of 6. If the number of factors is even, the locker will be toggled by an even number of people and it will eventually be closed. Note that each factor of a number has a ‛partner factor’ so that the product of the pair of factors yields the given number. Here, 1 and 6 form a pair of partner factors of 6, and 2 and 3 form another pair. Does every number have an even number of factors? We see in some cases, like 2 × 2, that the numbers in the pair are the same. Can you use this insight to find more numbers with an odd number of factors? For instance, 36 has a factor pair 6 × 6 where both numbers are 6. Does this number have an odd number of factors? If every factor of 36 other than 6 has a different factor as its partner, then we can be sure that 36 has an odd number of factors. Check if this is true. 1: 1 × 1 The only factor is 1. 4: 1 × 4 2 × 2 Factors are 1, 2 and 4. 6: 1 × 6 2 × 3 Factors are 1, 2, 3 and 6. 9: 1 × 9 3 × 3 Factors are 1, 3 and 9. Chapter 1.indd 2 10-07-2025 14:06:40 2 Hence all the following numbers have an odd number of factors — 1 × 1, 2 × 2, 3 × 3, 4 × 4, ... A number that can be expressed as the product of a number with itself is called a square number, or simply a square. The only numbers that have an odd number of factors are the squares, because they each have one factor which, when multiplied by itself, equals the number. Therefore, every locker whose number is a square will remain open. Write the locker numbers that remain open. 1.1 Square Numbers Why are the numbers, 1, 4, 9, 16, …, called squares? We know that the number of unit squares in a square (the area of a square) is the product of its sides. The table below gives the areas of squares with different sides. Khoisnam immediately collects word clues from these 10 lockers and reads, “The passcode consists of the first five locker numbers that were touched exactly twice.” Which are these five lockers? The lockers that are toggled twice are the prime numbers, since each prime number has 1 and the number itself as factors. So, the code is 2-3-5-7-11. Sidelength (in units) 1 1 × 1 = 1 sq. unit 2 2 × 2 = 4 sq. units 3 3 × 3 = 9 sq. units 4 4 × 4 = 16 sq. units 5 5 × 5 = 25 sq. units (in sq units) Area A Square and A Cube Chapter 1.indd 3 10-07-2025 14:06:40 We use the following notation for squares. 1 × 1 = 12 = 1 2 × 2 = 22 = 4 3 × 3 = 32 = 9, 4 × 4 = 42 = 16 5 × 5 = 52 = 25. . . . In general, for any number n, we write n × n = n2 , which is read as ‛n squared’. Can we have a square of sidelength 3 5 or 2.5 units? Yes, there area in square units are ( 3 5 ) 2 = ( 3 5 ) × ( 3 5 ) = ( 9 25), and (2.5) 2 = (2.5) × (2.5) = 6.25. 10 10 × 10 = 100 sq. units 3 Ganita Prakash | Grade 8 The squares of natural numbers are called perfect squares. For example, 1, 4, 9, 16, 25, … are all perfect squares. Patterns and Properties of Perfect Squares Find the squares of the first 30 natural numbers and fill in the table below. What patterns do you notice? Share your observations and make conjectures. Study the squares in the table above. What are the digits in the units places of these numbers? All these numbers end with 0, 1, 4, 5, 6 or 9. None of them end with 2, 3, 7 or 8. 12 = 1 112 = 121 212 = 441 22 = 4 122 = 222 = 32 = 9 132 = 42 = 16 142 = 52 = 25 152 = 62 = 162 = 72 = 172 = 82 = 182 = 92 = 192 = 102 = 202 = Math Talk Math Talk Chapter 1.indd 4 10-07-2025 14:06:40 4 If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square? The numbers 16 and 36 are both squares with 6 in the units place. However, 26, whose units digit is also 6, is not a square. Therefore, we cannot determine if a number is a square just by looking at the digit in the units place. But, the units digit can tell us when a number is not a square. If a number ends with 2, 3, 7, or 8, then we can definitely say that it is not a square. Write 5 numbers such that you can determine by looking at their units digit that they are not squares. The squares, 12 , 92 , 112 , 192 , 212 , and 292 , all have 1 in their units place. Write the next two squares. Notice that if a number has 1 or 9 in the units place, then its square ends in 1. Let us consider square numbers ending in 6: 16 = 42 , 36 = 62 , 196 = 142 , 256 = 162 , 576 = 242 , and 676 = 262 . Which of the following numbers have the digit 6 in the units place? Find more such patterns by observing the numbers and their squares from the table you filled earlier. Consider the following numbers and their squares. If a number contains 3 zeros at the end, how many zeros will its square have at the end? What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end? What can you say about the parity of a number and its square? Perfect Squares and Odd Numbers Let us explore the differences between consecutive squares. What do you notice? (i) 382 (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822 We have two zeroes. We have one zero. 1002 = 10000 2002 = 40000 7002 = 49000 9002 = 81000 102 = 100 202 = 400 402 = 800 But we have two zeroes. But we have four zeroes. A Square and A Cube Chapter 1.indd 5 10-07-2025 14:06:40 4 – 1 = 3 9 – 4 = 5 16 – 9 = 7 25 – 16 = 9 See if this pattern continues for the next few square numbers. From this we observe that adding consecutive odd numbers starting from 1 gives consecutive square numbers, as shown below. 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36. 5 Ganita Prakash | Grade 8 Do you remember this pattern from Grade 6? The picture below explains why each subsequent inverted L gives the next odd number: We see that the sum of the first n odd numbers is n2 . Alternatively, every square is a sum of successive odd numbers starting from 1. Also, we can find out whether a number is a perfect square by successively subtracting odd numbers. Consider the number 25, successively subtract 1, 3, 5, ... until you get or cross over 0, This means 25 = 1 + 3 + 5 + 7 + 9 and is thus a perfect square. Since we subtracted the first five odd numbers, 25 = 52 . Using the pattern above, find 362 , given that 352 = 1225. From the question we know that 1225 is the sum of the first 35 odd numbers. To find 362 , we need to add the 36th odd number to 1225. 1 + 3 1 + 3 + (3 + 2) 1 + 3 + 5 1 + 3 + 5 + (5 + 2) 1 + 3 + 5 + 7 25 – 1 = 24 24 – 3 = 21 21 – 5 = 16 16 – 7 = 9 9 – 9 = 0 In mathematics, sometimes arguments and reasoning can be presented without any words. Visual proofs can be complete by themselves. Chapter 1.indd 6 10-07-2025 14:06:40 6 How do we find the 36th odd number? The 1st odd number is 1, 2nd odd number is 3, 3rd number is 5, …, 6th odd number is 11 and so on. What is the nth odd number? The nth odd number is 2n–1. Therefore, the 36th odd number is 71. By adding 71 to 1225, we get 1296, which is 362 . Consider a number such as 38 that is not a square and subtract consecutive odd numbers starting from 1. This shows that 38 cannot be expressed as a sum of consecutive odd numbers starting with 1. 38 – 1 = 37 37 – 3 = 34 34 – 5 = 29 29 – 7 = 22 22 – 9 = 13 13 – 11 = 2 2 – 13 = – 11 Thus, we can say that a natural number is not a perfect square if it cannot be expressed as a sum of successive odd natural numbers starting from 1. We can use this result to find out whether a natural number is a perfect square. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern? How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000? Perfect Squares and Triangular Numbers Do you remember triangular numbers? 1 –100 101 – 200 201 – 300 301 – 400 401 – 500 501 – 600 601 – 700 701 – 800 801 – 900 901 – 1000 1 3 6 10 15 A Square and A Cube Chapter 1.indd 7 10-07-2025 14:06:40 Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term. Square Roots The area of a square is 49 sq. cm. What is the length of its side? We know that 7 × 7 = 49, or 72 = 49. 1 + 3 = 4 = 22 3 + 6 = 9 = 32 6 + 10 = 16 = 42 7 Ganita Prakash | Grade 8 So, the length of the side of a square with an area of 49 sq. cm is 7 cm. We call 7 the square root of 49. In general, if y= x2 then x is the square root of y. What is the square root of 64? We know that 8 × 8 is 64. So, 8 is the square root of 64. What about – 8 × – 8 ? That is 64 too! 82 = 64, and (– 8)2 = 64. So, the square roots of 64 are + 8 and – 8. Every perfect square has two integer square roots. One is positive and the other is negative. The square root of a number is denoted by Thus, 64 8 =± and 100 10 =± . Note that 8 8 2 =± and 10 10 2 =± . In general, n n 2 =± . In this chapter, we shall only consider the positive square root. Given a number, such as 576 or 327, how do we find out if it is a perfect square? If it is a perfect square, how can we find its square root? We know that perfect squares end in 1, 4, 9, 6, 5, or an even number of zeros. But, it is not certain that a number that satisfies this condition is a square. We can clearly say that 327 is not a perfect square. However, we cannot be sure that 576 is a perfect square. 1. We can list all the square numbers in sequence and find out whether 576 occurs among them. We know that 202 = 400, we can find squares of 21, 22, 23, … and so on until we get 576 or a number greater than 576. 202 = 400 212 = 441 222 = 484 232 = 529 242 = 576 Math Talk Chapter 1.indd 8 10-07-2025 14:06:42 8 2. Recall that every square can be expressed as a sum of consecutive odd numbers starting from 1. Consider 81 . From 81, we successively subtracted consecutive odd numbers starting from 1 until we obtained 0 at the 9th step. Therefore 81 = 9. Can we find the square root of 729 using this method? Yes, but it will be time-consuming. However, this process becomes inefficient for larger numbers. 81 – 1 = 80 80 – 3 = 77 77 – 5 = 72 72 – 7 = 65 65 – 9 = 56 56 – 11 = 45 45 – 13 = 32 32 – 15 = 17 17 – 17 = 0 3. We know that a perfect square is obtained by multiplying an integer by itself. Will looking at a number’s prime factorisation help in determining whether it is a perfect square? Is 324 a perfect square? Is 156 a perfect square? The prime factorisation of 156 is 2 × 2 × 3 × 13. We cannot pair up these factors. Therefore, 156 is not a perfect square. Find whether 1156 and 2800 are perfect squares using prime factorisation. We can estimate the square root of larger perfect squares by looking at the closest perfect squares we are familiar with and then narrowing down the interval to search. For example, to find 1936 , we can reason as follows: 324 = 2 × 2 × 3 × 3 × 3 × 3. These can be grouped as 324 = (2 × 3 × 3) × (2 × 3 × 3). = (2 × 3 × 3)2 = 182 . We can also write the prime factors in pairs. That is, 324 = (2 × 2) × (3 × 3) × (3 × 3), which shows that 324 is a perfect square. Thus, 324 = (2×3×3)2 = 182 . Therefore, 324 =18. Yes, if we can divide the prime factors of a number into two equal groups, then the product of the prime factors in either group combine to form the square root. A Square and A Cube Chapter 1.indd 9 10-07-2025 14:06:43 Consider the following situations — Aribam and Bijou play a game. One says a number and the other replies with its square root. Aribam starts. He says 25, and Bijou quickly (i) 1936 is between 1600 (402 ) and 2500 (502 ), so 40 < 1936 < 50. (ii) The last digit of 1936 is 6. So, the last digit of the square root must either be 4 or 6. It can be 44 or 46. (iii) If we calculate 452 , we can compare it with 1936 to halve the interval to search from 40–50 to either 40–45 or 45–50. We can write 452 as (40 + 5) (40 + 5) = 402 + 2 × 40 × 5 + 52 = 1600 + 400 + 25 = 2025. (v) From the observation in point b we can guess and then verify that 1936 is 44. (iv) 2025 > 1936. So, 40 < 1936 < 45 9 Ganita Prakash | Grade 8 responds with 5. Then Bijou says 81, and Aribam answers 9. The game goes on till Aribam says 250. Bijou is not able to answer because 250 is not a perfect square. Aribam asks Bijou if he can at least provide a number that is close to the square root of 250. For this, Bijou needs to estimate the square root of 250. We know that 100 < 250 < 400 and 100 = 10 and 400 = 20. So, 10 < 250 < 20. But, we are still not very close to the number whose square is 250. We know that 152 = 225 and 162 = 256. Therefore, 15 < 250 < 16. Since 256 is much closer to 250 than 225, 250 is approximately 16. We also know it is less than 16. Here is another problem that requires estimating square roots. Akhil has a square piece of cloth of area 125 cm2 . He wants to know if he can cut out a square handkerchief of side 15 cm. If not, he wants to know the maximum size handkerchief that can be cut out from this piece of cloth with an integer side length. 125 is not a perfect square. The nearest perfect squares are 112 = 121 and 122 = 144. So the largest square handkerchief with integer side length that can be cut out from this piece of cloth has side length 11 cm. Figure it Out 1. Which of the following numbers are not perfect squares? 2. Which one among 642 , 1082 , 2922 , 362 has last digit 4? 3. Given 1252 = 15625, what is the value of 1262 ? (i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 512 4. Find the length of the side of a square whose area is 441 m2 . (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089 Chapter 1.indd 10 10-07-2025 14:06:44 10 5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10. 6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product. 7. How many numbers lie between the squares of the following numbers? 8. In the following pattern, fill in the missing numbers: (i) 16 and 17 (ii) 99 and 100 92 + 102 + (___)2 = (___)2 9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares. 32 + 42 + 122 = 132 42 + 52 + 202 = (___)2 12 + 22 + 22 = 32 22 + 32 + 62 = 72 A Square and A Cube Chapter 1.indd 11 10-07-2025 14:06:44 1.2 Cubic Numbers You know the word cube from geometry. A cube is a solid figure all of whose all sides meet at right angles and are equal. How many cubes of side 1 cm make a cube of side 2 cm? How many cubes of side 1 cm will make a cube of side 3 cm? 11 Ganita Prakash | Grade 8 Consider the numbers 1, 8, 27, ... These numbers are called perfect cubes. Can you see why they are named so? Each of them is obtained by multiplying a number by itself three times. We note that Is 9 a cube? We see that 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27. This shows that 9 is not a perfect cube. Nor is any number from 10 to 26. Can you estimate the number of unit cubes in a cube with an edge length of 4 units? It has 64 unit cubes! If you notice carefully, each layer of this cube has 4 × 4 unit cubes. Each square layer has 16 unit cubes (4 × 4), and there are 4 such layers, so the total number of unit cubes is 4 × 4 × 4 = 64. Since 53 = 5 × 5 × 5 = 125, 125 is a cube. In general, for any number n, we write the cube n × n × n as n3 . Complete the table below. 13 = 1 113 = 1331 23 = 8 123 = 33 = 27 133 = 2197 27 = 3 × 3 × 3 1 = 1 × 1 × 1 8 = 2 × 2 × 2 Chapter 1.indd 12 10-07-2025 14:06:44 12 What patterns do you notice in the table above? We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for 43 = 64 143 = 2744 53 = 125 153 = 63 = 163 = 73 = 173 = 4913 83 = 183 = 5832 93 = 193 = 6859 103 = 203 = Math Talk squares. What are the possible last digits of cubes? Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe? Can a cube end with exactly two zeroes (00)? Explain. (– 6)2 — we also can compute cubes of such numbers — (4 6 ) 3 , (13.08)3 , and (– 6)3. Taxicab Numbers Once when Srinivasa Ramanujan was working with G. H. Hardy at the University of Cambridge, Hardy had come to visit Ramanujan at a hospital when he was ill. Hardy had ridden in a taxicab numbered 1729 and he remarked that 1729 was ‘rather a dull number,’ adding that he hoped that this was not a bad sign. Ramanujan immediately replied, “No, Hardy, it is a very interesting number. It is the smallest number that can be expressed as the sum of two cubes in two different ways”. = 93 + 103 . Because of this story, 1729 has since been known as the Hardy– Ramanujan Number. And numbers that can be expressed as the sum of two cubes in two different ways are called taxicab numbers. Just as we can take squares of fractions/decimals — ( 4 6 ) 2 , (13.08)2 , and ( 4 6 ) 3 = ( 4 6 ) × ( 4 6 ) × ( 4 6 ) = ( 64 216) (13.08)3 = 13.08 × 13.08 × 13.08 = 2237.810112 (– 6)3 = – 6 × – 6 × – 6 = – 216. 1729 = 13 + 123 A Square and A Cube Chapter 1.indd 13 10-07-2025 14:06:44 The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes. How did Ramanujan know this? Well, he loved numbers. All through his life, he tinkered with numbers. During Ramanujan’s time in Cambridge, his colleagues often marveled at his ability to see deep patterns in numbers that seemed arbitrary to others. His colleague, John Littlewood, once said, “Every positive integer was one of his [Ramanujan's] personal friends”. Try This 13 Ganita Prakash | Grade 8 Perfect Cubes and Consecutive Odd Numbers Consecutive odd numbers have a role to play with cubes too. Look at the following pattern: Later in this series, we get the following set of consecutive numbers: Can you tell what this sum is without doing the calculation? Cube Roots We know that 8 = 23 . We call 2 the cube root of 8 and denote this by 2 = 83 . More generally, if y = x3 , then x is the cube root of y. This is denoted by x = y 3 . So, 83 = 23 3 = 2. Similarly, 27 3 3 3 3 3 = = and 1000 10 3 3 3 = =3. In general, n 3 3 = n . How do we find out if a number is a cube? Taking inspiration from the case of squares, let us see if we can use prime factorisations. 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. 31 + 33 + 35 + 37 + 39 + 41 = 216 = 63 . 21 + 23 + 25 + 27 + 29 = 125 = 53 13 + 15 + 17 + 19 = 64 = 43 7 + 9 + 11 = 27 = 33 3 + 5 = 8 = 23 1 = 1 = 13 Chapter 1.indd 14 10-07-2025 14:06:45 14 Let us check if 3375 is a perfect cube. Can the factors be split into three identical groups? For 3375, we can form three groups of (3 × 5). So, 3375 = (3 × 5) × (3 × 5) × (3 × 5) = (3 × 5)3 = 153 . Another way is to check if the factors can be grouped into triplet(s): 3375 = (3 × 3 × 3) × (5 × 5 × 5) = 33 × 53 . This means 3375 15 3 = . Is 500 a perfect cube? 500 = 2 × 2 × 5 × 5 × 5. We see that the factors cannot be split into three identical groups. Therefore, 500 is not a perfect cube. 3375 = 3 × 3 × 3 × 5 × 5 × 5. Observe that each prime factor of a number appears three times in the prime factorisation of its cube. Find the cube roots of these numbers: Successive Differences We know that the differences between consecutive perfect squares gives the sequence of odd numbers. Observe the figure below where the differences are computed successively for perfect squares. After two levels, all the differences are the same. Prime Factorisation of a Number Prime Factorisation of its Cube (i) 64 3 = (ii) 512 3 = (iii) 729 3 = 12 = 2 × 2 × 3 123 = 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 23 × 23 × 33 15 = 3 × 5 153 = 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 33 × 53 4 = 2 × 2 43 = 64 = 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 6 = 2 × 3 63 = 216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 2 2 2 2 ... Level 1 3 5 7 9 11 ... 1 4 9 16 25 36 ... Perfect Squares A Square and A Cube Chapter 1.indd 15 11-07-2025 14:27:18 Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice? 1.3 A Pinch of History The first known list of perfect squares and perfect cubes was compiled by the Babylonians as far back as 1700 BCE. These lists, found on clay tablets, were used to quickly find square roots and cube Level 2 1 8 27 64 125 216 ... Perfect Cubes 15 Ganita Prakash | Grade 8 roots in problems involving land measurement, architectural design, and other areas where geometric calculations were necessary. In ancient Sanskrit works the term varga was used both for the square figure or its area, as well as the square power, and the term ghana was used both for the solid cube as well as the product of a number with itself three times. The fourth power was called varga-varga . These terms were used in India at least from the third century BCE. Aryabhata (499 CE) states “A square figure of four equal sides and the number representing its area are called varga. The product of two equal quantities is also called varga.” Thus, the term varga for square power has its origin in the graphical representation of a square figure. Why is the word ‘root’ (the root of a plant) used for the mathematical operation √ (square root, cube root, etc.)? It is because, in ancient India, the Sanskrit word mula, meaning root of a plant, basis, cause, origin, etc., was used for the mathematical operations of taking roots. In Sanskrit, varga-mula (the basis, cause, origin of the square) was used for square-root and ghana-mula was used for cube-root. This use of mula for the mathematical concept of root was subsequently emulated in Arabic and Latin through their corresponding words for the root of a plant — jidhr and radix respectively. The term mula for root has been used in India at least from the first century BCE. Another term used was pada (foot, basis, cause, origin). Brahmagupta (628 CE) explains, ‘The pada (root) of a krti (square) is that of which it is a square.’ Figure it Out 1. Find the cube roots of 27000 and 10648. Chapter 1.indd 16 10-07-2025 14:06:46 16 2. What number will you multiply by 1323 to make it a cube number? 3. State true or false. Explain your reasoning. 4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (v) Cube numbers have an odd number of factors. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. 5. Which of the following is the greatest? Explain your reasoning. (i) 673 – 663 (ii) 433 – 423 (iii) 672 – 662 (iv) 432 – 422 A number obtained by multiplying a number by itself is called a square number. Squares of natural numbers are called perfect squares. All perfect squares end with 0, 1, 4, 5, 6 or 9. Squares can only have an even number of zeros at the end. Square root is the inverse operation of square. Every perfect square has two integral square roots. The positive square root of a number is denoted by the symbol √ . For example, √9 = 3. A number obtained by multiplying a number by itself three times is called a cube. For example 1, 8, 27, ... ,etc., are cubes. A number is a perfect square if its prime factors can be split into two identical groups. A number is a perfect cube if its prime factors can be split into three identical groups The symbol 3 denotes cube root. For example, 27 3 = 3. SUMMARY A Square and A Cube Chapter 1.indd 17 10-07-2025 14:06:47 17 Look at the following numbers: 3 6 10 15 1 They are arranged such that each pair of adjacent numbers adds up to a square. Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square. Can you arrange them in more than one way? If not, can you explain why? Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle? 3 + 6 = 9, 6 + 10 = 16, 10 + 15 = 25, 15 + 1 = 16. Square Pairs! Chapter 1.indd 18 10-07-2025 14:06:49" class_8,2,power play,ncert_books/class_8/hegp1dd/hegp102.pdf,"2 POWER PLAY 2.1 Experiencing the Power Play ... An Impossible Venture! Take a sheet of paper, as large a sheet as you can find. Fold it once. Fold it again, and again. How many times can you fold it over and over? Estu says “I heard that a sheet of paper can’t be folded more than 7 times”. Roxie replies “What if we use a thinner paper, like a newspaper or a tissue paper?” Try it with different types of paper and see what happens. If you can fold a paper 46 times, it will be so thick that it can reach the Moon! What! That’s crazy! Just 46 times!? You must have ignored several zeros after 46. Well, why don’t you find out yourselves. Chapter 2.indd 19 7/10/2025 3:29:53 PM Say you can fold a sheet of paper as many times as you wish. What would its thickness be after 30 folds? Make a guess. Let us find out how thick a sheet of paper will be after 46 folds. Assume that the thickness of the sheet is 0.001 cm. Ganita Prakash | Grade 8 The following table lists the thickness after each fold. Observe that the thickness doubles after each fold. (We use the sign ‘≈’ to indicate ‘approximately equal to’.) After 10 folds, the thickness is just above 1 cm (1.024 cm). After 17 folds, the thickness is about 131 cm (a little more than 4 feet). Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess. Fill the table below. After 26 folds, the thickness is approximately 670 m. Burj Khalifa in Dubai, the tallest building in the world, is 830 m tall. Fold Thickness Fold Thickness Fold Thickness Fold Thickness Fold Thickness Fold Thickness Fold Thickness Fold Thickness 18 ≈ 262 cm 21 24 19 ≈ 524 cm 22 25 20 ≈ 10.4 m 23 26 1 0.002 cm 7 0.128 cm 13 8.192 cm 2 0.004 cm 8 0.256 cm 14 16.384 cm 3 0.008 cm 9 0.512 cm 15 32.768 cm 4 0.016 cm 10 1.024 cm 16 65.536 cm 5 0.032 cm 11 2.048 cm 17 ≈ 131 cm 6 0.064 cm 12 4.096 cm Math Talk Chapter 2.indd 20 7/10/2025 3:29:55 PM 20 28 30 After 30 folds, the thickness of the paper is about 10.7 km, the typical height at which planes fly. The deepest point discovered in the oceans is the Mariana Trench, with a depth of 11 km. Fold Thickness Fold Thickness Fold Thickness 31 36 41 32 37 42 33 38 43 34 39 44 35 40 45 27 ≈ 1.3 km 29 It might be hard to digest the fact that after just 46 folds, the thickness is more than 7,00,000 km. This is the power of multiplicative growth, also called exponential growth. Let us analyse the growth. We have seen that the thickness doubles after every fold. Notice the change in thickness after two folds. By how much does it increase? After any 3 folds, the thickness increases 8 times (= 2 × 2 × 2). Check if that is true. Similarly, from any point, the thickness after 10 folds increases by 1024 times (= 2 multiplied by itself 10 times), as shown in the table below. 10 to 20 10.485 m – 1.024 cm ≈ 10.474 m 10.485 m ÷ 1.024 cm = 1024 20 to 30 10.737 km – 10.485 m ≈ 10.726 km 10.737 km ÷ 10.485 m = 1024 30 to 40 10995 km – 10.737 km ≈ 10984.2 km 10995 km ÷ 10.737 km = 1024 0 to 10 1.024 cm – 0.001 cm = 1.023 cm 1.024 ÷ 0.001 = 1024 Fold Thickness Times increased by Fold 4 0.016 cm Fold 5 0.032 cm Fold 10 1.024 cm Fold 9 0.512 cm Fold 4 0.016 cm Fold 6 0.064 cm Power Play Chapter 2.indd 21 7/10/2025 3:29:55 PM 2.2 Exponential Notation and Operations The initial thickness of the paper was 0.001 cm. Upon folding once, its thickness became 0.001 cm × 2 = 0.002 cm. Folding it twice, its thickness became — 0.001 cm × 2 × 2 = 0.004 cm, or 0.001 cm × 22 = 0.004 cm (in shorthand). Upon folding it thrice, its thickness became — 0.001 cm × 2 × 2 × 2, or 0.001 cm × 23 = 0.008 cm. When folded four times, its thickness became — 0.001 cm × 2 × 2 × 2 × 2, or 0.001 cm × 24 = 0.016 cm. Similarly, the expression for the thickness of the paper when folded 7 times will be 0.001 cm × 2 × 2 × 2 × 2 × 2 × 2 × 2, or 0.001 cm × 27 = 0.128 cm. We have seen that square numbers can be expressed as n2 and cube numbers as n3 . n × n = n2 (read as ‘n squared’ or ‘n raised to the power 2’) 21 Ganita Prakash | Grade 8 n × n × n = n3 (read as ‘n cubed’ or ‘n raised to the power 3’) n × n × n × n = n4 (read as ‘n raised to the power 4’ or ‘the 4th power of n’) n × n × n × n × n × n × n = n7 (read as ‘n raised to the power 7’ or ‘the 7th power of n’) and so on. In general, we write na to denote n multiplied by itself a times. 54 = 5 × 5 × 5 × 5 = 625. 54 is the exponential form of 625. Here, 4 is the exponent/power, and 5 is the base. Exponents of the form 5n are called powers of 5: 51 , 52 , 53 , 54 , etc. 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210 = 1024. Remember the 1024 from earlier? There, it meant that after every 10 folds, the thickness increased 1024 times. Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v. (i) 10v (ii) 10 + v (iii) 2 × 10 × v (iv) 210 (v) 210v (vi) 102 v Some more examples of exponential notation: 4 × 4 × 4 = 43 = 64. (– 4) × (– 4) × (– 4) = (– 4)3 = – 64. Similarly, a × a × a × b × b can be expressed as a3 b2 (read as a cubed b squared). a × a × b × b × b × b can be expressed as a2 b4 (read as a squared b raised to the power 4). Remember that 4 + 4 + 4 = 3 × 4 = 12, whereas 4 × 4 × 4 = 43 = 64. Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form. What is 02 , 05 ? What is 0n ? 54 is read as ‘5 raised to the power 4’ or ‘5 to the power 4’ or ‘5 power 4’ or ‘4th power of 5’ 2 32400 2 16200 2 8100 2 4050 5 2025 5 405 3 81 3 27 3 9 Chapter 2.indd 22 7/10/2025 3:29:55 PM 22 What is (– 1)5 ? Is it positive or negative? What about (– 1)56? Is (– 2)4 = 16? Verify. Figure it Out 1. Express the following in exponential form: In exponential form, this would be (i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b (iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d 32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3. 32400 = 24 × 52 × 34 . 3 3 1 2. Express each of the following as a product of powers of their prime factors in exponential form. 3. Write the numerical value of each of the following: The Stones that Shine ... Three daughters with curious eyes, Each got three baskets—a kingly prize. Each basket had three silver keys, Each opens three big rooms with ease. Each room had tables—one, two, three, With three bright necklaces on each, you see. Each necklace had three diamonds so fine… Can you count these stones that shine? Hint: Find out the number of baskets and rooms. How many rooms were there altogether? The information given can be visualised as shown below. (i) 2 × 103 (ii) 72 × 23 (iii) 3 × 44 (iv) (– 3)2 × (– 5)2 (v) 32 × 104 (vi) (– 2)5 × (– 10)6 (i) 648   (ii) 405   (iii) 540   (iv) 3600 King Power Play daughters baskets Chapter 2.indd 23 7/10/2025 3:29:55 PM From the diagram, the number of rooms is 34 . This can be computed by repeatedly multiplying 3 by itself, 3 × 3 = 9. 9 × 3 = 27. 27 × 3 = 81. 81 × 3 = 243. How many diamonds were there in total? Can we find out by just one multiplication using the products above? The number of diamonds is 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37 . keys 23 rooms Ganita Prakash | Grade 8 We had computed till 34 . To find 37 , we can just multiply 34 (= 81) with 33 (= 27). = 34 × 33 = 81 × 27 = 2187 37 can also be written as 32 × 35 . Can you reason out why? This can be easily extended to products where exponents are the same letter-numbers. Write the product p4 × p6 in exponential form. Use this observation to compute the following. (i) 29 (ii) 57 (iii) 46 46 can be evaluated in these two ways, 210 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) = (22 ) × (22 ) × (22 ) × (22 ) × (22 ) = (22 ) 5 . (4 × 4 × 4) × (4 × 4 × 4) = 43 × 43                      = 64 × 64                      = 4096. 43 × 43 is the square of 43 , i.e., 43 × 43 can also be written as (43 ) 2 . We can write 37 = (3 × 3 × 3 × 3) × (3 × 3 × 3) p4 × p6 = (p × p × p × p) × (p × p × p × p × p × p) = p10. We can generalise this to — Similarly, 74 = (7 × 7) × (7 × 7) = 72 × 72 = (72 ) 2 , and (i) 86    (ii) 715   (iii) 914   (iv) 58 3 × 3 × 3 × 3 × 3 × 3 × 3 na × nb = na+b , where a and b are counting numbers. (4 × 4) × (4 × 4) × (4 × 4) = 42 × 42 × 42                       = 16 × 16 × 16                       = 4096. 42 × 42 × 42 is the cube of 42 , i.e., 42 × 42 × 42 can also be written as (42 ) 3 . 34 33 Math Talk Chapter 2.indd 24 7/10/2025 3:29:55 PM 24 Is 210 also equal to (25 ) 2 ? Write it as a product. 2 10 = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) = (25 ) × (25 ) = (25 ) 2 . In general, Write the following expressions as a power of a power in at least two different ways: (na ) b =(nb ) a = na × b =n ab, where a and b are counting numbers. Magical Pond In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full? If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day? Since the number of lotuses doubles every day, the pond should be half covered on the 29th day. Write the number of lotuses (in exponential form) when the pond was — (i) fully covered (ii) half covered There is another pond in which the number of lotuses triples every day. When both the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4 days, she took all the lotuses from there and put them in the tripling pond. How many lotuses will be in the tripling pond after 4 more days? After the first 4 days, the number of lotuses is 1 × 2 × 2 × 2 × 2 = 24 . After the next 4 days, the number of lotuses is 24 × 3 × 3 × 3 × 3 = 24 × 34 . What if Damayanti had changed the order in which she placed the flowers in the lakes? How many lotuses would be there? Can this product be expressed as an exponent mn , where m and n are some counting numbers? By regrouping the numbers, = (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2) = (3 × 2)4 = 64 . In general form, 1 × 34 × 24 = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2). Power Play Chapter 2.indd 25 7/10/2025 3:29:56 PM ma × na = (mn) a , where a is a counting number. Use this observation to compute the value of 25 × 55 . Simplify 104 54 and write it in exponential form. In general, we can show that ma na = ( m n ) a . 25 Ganita Prakash | Grade 8 How Many Combinations Estu has 4 dresses and 3 caps. How many different ways can Estu combine the dresses and caps? For each cap, he can choose any of the 4 dresses, so for 3 caps, 4 + 4 + 4 = 4 × 3 = 12 combinations are possible. We can also look at it as—for each dress, Estu can choose any of the 3 caps, so for 4 outfits, 3 + 3 + 3 + 3 = 3 × 4 = 12 combinations are possible. Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up? Hint: Try drawing a diagram like the one above. Estu and Roxie came across a safe containing old stamps and coins that their great-grandfather had collected. It was secured with a 5-digit password. Since nobody knew the password, they had no option except to try every password until it opened. They were unlucky and the lock only opened with the last password, after they had tried all possible combinations. How many passwords did they end up checking? Chapter 2.indd 26 7/10/2025 3:29:57 PM 26 Instead of a 5-digit lock, let us assume we have a 2-digit lock and try to find out how many passwords are possible. There are 10 options for the first digit (0 to 9). For each of these, there are 10 options for the second digit (If 0 is the first digit then 00, 01, 02, 03, …, 09 are possible). Therefore the total number of combinations for a 2-digit lock is 10 × 10 = 100. Now, suppose we have a 3-digit lock. For each of the earlier 100 (2-digit) passwords there are 10 choices for the third digit. So, there are 100 × 10 = 1000 combinations for a 3-digit lock. You can list them all: 000, 001, 002, ….., 997, 997, 999. If you can’t solve a problem, try to find a simpler version of the problem that you can solve. This technique can come in handy many times. How many passwords are possible with such a lock? Think about how many combinations are possible in different contexts. Some examples are— (i) Pincodes of places in India—The Pincode of Vidisha in Madhya Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017. (ii) Mobile numbers. (iii) Vehicle registration numbers. Try to find out how these numbers or codes are allotted/generated. 2.3 The Other Side of Powers Imagine a line of length 16 units. Erasing half of it would result in How many 5-digit passwords are possible? Each digit has 10 choices, so a 5-digit lock will have: 10 × 10 × 10 × 10 × 10 = 105 = 1,00,000 passwords. This is the same as writing numbers till 99,999 with all 5 digits, i.e., 00000, 00001, 00002, …00010, 00011, …, 00100, 00101, …, 00999, …, 30456, …, 99998, 99999. Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.” 24 ÷ 2 = 2 × 2 × 2 × 2 2 = 2 × 2 × 2 = 23 = 8 units. Erasing half one more time would result in, Oh no! Maybe my entire vacation will be gone trying to unlock this… 0 1 2 3 4 5 6 7 8 9 Power Play 0 1 2 3 4 5 6 7 8 9 Try This Chapter 2.indd 27 7/10/2025 3:29:57 PM What is 2100 ÷ 225 in powers of 2? (24 ÷ 2) ÷ 2 = 24 ÷ 22 = 2 × 2 × 2 × 2 2 × 2 = 2 × 2 = 22 = 4 units. Halving 16 cm three times may be written as, 24 ÷ 23 = 2 × 2 × 2 × 2 2 × 2 × 2 = 2 = 21 = 2 units. From this we can see that In a generalised form, where n ≠ 0 and a and b are counting numbers and a > b. 24 ÷ 23 = 24–3 = 21 . na ÷ nb = na – b , 27 Ganita Prakash | Grade 8 Why can’t n be 0? We have not covered the case when the exponent is 0; for example, what is 20 ? When Zero is in Power! Let us define 20 in a way that the generalised form above holds true. 20 = 24–4 = 24 ÷ 24 = 2 × 2 × 2 × 2 2 × 2 × 2 × 2 = 1. In fact for any letter number a 20 = 2a – a = 2a ÷ 2a = 1. In general, xa ÷ xa = xa – a = x0 , and so 1 = x0 , where x ≠ 0 and a is a counting number. Math Talk Chapter 2.indd 28 7/10/2025 3:29:57 PM 28 24 ÷ 25 = 2 × 2 × 2 × 2 2 × 2 × 2 × 2 × 2 = 1 2 units. Using the generalised form, we get 24 ÷ 25 = 2(4 – 5) = 2– 1. So, 2– 1 = 1 2 . When a line of length 24 units is halved 10 times, we get 24 ÷ 210= 2(4 – 10) = 2– 6 units. When expanded, 24 ÷ 210 = 2 × 2 × 2 × 2 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1 26 = 1 64, which is also written as 2– 6. When a line of length 24 units is halved 5 times, Similarly, 10–3 = 1 103, 7–2 = 1 72 , etc. Can we write 103 = 1 10–3 ? We can write, Similarly, 72 = 1 7– 2 and 4a = 1 4– a . In a generalised form, n– a = 1 na and na = 1 n–a , where n ≠ 0. Consider the following general forms we have identified. We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true? Write equivalent forms of the following. (i) 2– 4 (ii) 10–5 (iii) (– 7)–2 (iv) (– 5)– 3 (v) 10–100 Simplify and write the answers in exponential form. (i) 2–4 × 27 (ii) 32 × 3–5 × 36 (iii) p3 × p–10 (iv) 24 × (– 4)–2 (v) 8 p × 8q Power Lines Let us arrange the powers of 4 along a line. na × nb = na + b (na ) b = (nb ) a = na × b na ÷ nb = na – b Math Talk 10–3 = 1 1/103 = 1 ÷ 1 10 3 = 1×103 = 103 . 1 Power Play Chapter 2.indd 29 7/10/2025 3:29:57 PM 47 × 4–2 48 ÷ 43 43 × 42 ÷ 16 ÷ 4 48 47 46 45 44 43 42 41 40 4–1 4–2 1 4 1 16 65536 16384 4096 1024 256 64 16 4 1 × 4 × 16 65536 ÷ 64 64 × 16 16384 × 1 16 29 Ganita Prakash | Grade 8 Can we say that 16384 (47 ) is 16 (42 ) times larger than 1,024 (45 )? Yes, since 47 ÷ 45 = 42 . How many times larger than 4–2 is 42 ? Use the power line for 7 to answer the following questions. 2.4 Powers of 10 We have used numbers like 10, 100, 1000, and so on when writing Indian numerals in an expanded form. For example, 77 76 75 74 73 72 71 70 7–1 7–2 7–3 7–4 823543 117649 16807 2401 1 343 1 2401 343 1 49 1 7 49 7 1 2,401 × 49 = 493 = 343 × 2,401 = 16,807 49 = 16,807 8,23,543 = 1,17,649 × 1 343 = 7 343 = 1 343 × 1 343 = Chapter 2.indd 30 7/10/2025 3:29:58 PM 30 Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374. How can we write 561.903? 561.903 = (5 × 102 ) + (6 × 101 ) + (1 × 100 ) + (9 × 10–1) + (0 × 10–2) + (3 × 10–3). This can be written using powers of 10 as 561.903 = (5 × 100) + (6 × 10) + 1 + (9 × 1 10) + (0 × 1 100) + (3 × 1 1000). Writing it using powers of 10, we have 47561 = (4 × 10000) + (7 × 1000) + (5 × 100) + (6 × 10) + 1. (4 × 104 ) + (7 × 103 ) + (5 × 102 ) + (6 × 101 ) + (1 × 100 ). Scientific Notation Let’s look at some facts involving large numbers— (i) The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy. (ii) The number of stars in our galaxy is 1,00,00,00,00,000. (iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg. As the number of digits increases, it becomes difficult to read the numbers correctly. We may miscount the number of zeroes or place commas incorrectly. We will then read the wrong value. It is like getting ₹5,000 when you were supposed to get ₹50,000. The number of zeroes is more important than the initial digits in several cases. Can we use the exponential notation to simplify and read these very large numbers correctly? For example, the number 5900 can be expressed as— 5900 = 590 × 10 = 590 × 101 = 59 × 100 = 59 × 102    = 5.9 × 1000 = 5.9 × 103    = 0.59 × 10000 = 0.59 × 104 . Any number can be written as the product of a number between 1 and 10 and a power of 10. For example, Write the large-number facts we read just before in this form. In scientific notation or scientific form (also called standard form), we write numbers as x × 10y , where x ≥ 1 and x < 10 is the coefficient and y, the exponent, is any integer. Often, the exponent y is more important than the coefficient x. When we write the 2 crore population of Mumbai as 2 × 107 , the 7 is more important than the 2. Indeed, if the 2 is changed to 3, the population increases by one-half, i.e., 2 crore to 3 crore, whereas if the 7 is changed to 8, the change in population is 10 times, i.e., 2 crores to 20 crores. Therefore, the standard form explicitly mentions the exponent, which indicates the number of digits. If we say that the population of Kohima is 1,42,395, then it gives the impression that we are quite sure about this number up to the units place. When we use large numbers, in most cases, we are more concerned about how big a quantity or measure is, rather than the exact value. If we are only sure that the population is around 1 lakh 42 thousand, we can write it as 1.42 × 105 . If we can only be certain that it is around 1 lakh 40 thousand, we write it as 1.4 × 105 . The number of digits in the coefficient reflects how well we know the number. The most important part of any 5900 = 5.9 × 103 20800 = 2.08 × 104 80,00,000 = 8 × 106 Power Play Chapter 2.indd 31 7/10/2025 3:29:58 PM 31 Ganita Prakash | Grade 8 number written in scientific form is the exponent, and then the first digit of the coefficient. The digits following the coefficient are small corrections to the first digit. These values are rounded-off estimates, averages, or approximations; most of the time, they serve the purpose at hand. The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 1012 m. The distance between Saturn and Uranus is 14,39,00,00,00,000 m = 1.439 × 1012 m. The distance between the Sun and Earth is 1,49,60,00,00,000 m = 1.496 × 1011 m. Can you say which of the three distances is the smallest? How old is this dinosaur’s skeleton? It is 70 million and 15 years old… ...when I started working here, it was 70 million years old. Chapter 2.indd 32 7/10/2025 3:29:58 PM 32 The number line below shows the distance between the Sun and Saturn (1.4335 × 1012 m). On the number line below, mark the relative position of the Earth. The distance between the Sun and the Earth is 1.496 × 1011 m. Express the following numbers in standard form. (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000 Sun Saturn 2.5 Did You Ever Wonder? Last year, we looked at interesting thought experiments in the chapter on Large Numbers. Let us continue this journey. Nanjundappa wants to donate jaggery equal to Roxie’s weight and wheat equal to Estu’s weight. He is wondering how much it would cost. What would be the worth (in rupees) of the donated jaggery? What would be the worth (in rupees) of the donated wheat? In order to find out, let us first describe the relationships among the quantities present. Worth of jaggery (in rupees) = Roxie’s weight in kg × cost of 1 kg jaggery. Worth of wheat (in rupees) = Estu’s weight in kg × cost of 1 kg wheat. Make necessary and reasonable assumptions for the unknowns and find the answers. Remember, Roxie is 13 years old and Estu is 11 years old. Assuming Roxie’s weight to be 45 kg and the cost of 1 kg of jaggery to be ₹70, the worth of donated jaggery is 45 × 70 = ₹3150. Assuming Estu’s weight to be 50 kg and the cost of 1 kg of wheat to be ₹50, the worth of donated wheat is 50 × 50 = ₹2500. Power Play Chapter 2.indd 33 7/10/2025 3:29:59 PM Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many coins are needed to equal my weight?”. How can we find out? For questions like these, you can consider following the steps suggested below. 1. Guessing: Make an instinctive (quick) guess of what the answer could be, without any calculations. The practice of offering goods equal to the weight of a person, called Tulābhāra or Tulābhāram, is quite old and is still followed in many places in Southern India. It is a symbol of bhakti (surrendering oneself), a token of gratitude; it also supports the community. 33 Ganita Prakash | Grade 8 2. Calculating using estimation and approximation — (i) Describe the relationships among the quantities that are needed to find the answer. (ii) Make reasonable assumptions and approximations if the required information is not available. (iii) Compute and find the answer (and check how close your guess was). Would the number of coins be in hundreds, thousands, lakhs, crores, or even more? Make an instinctive guess. Find the answer by making necessary and reasonable assumptions and approximations for the unknowns. Remember, we are not looking for an exact answer but a reasonably close estimate. Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead? How much money could it be?” Make an instinctive guess first. Then find out (make necessary and reasonable assumptions about the unknown details and find the answers). Estu says, “When I become an adult, I would like to donate notebooks worth my weight every year”. Roxie says, “When I grow up, I would like to do annadāna (offering grains or meals) worth my weight every year”. Initially, your guesses may be very far off from the answer and it is perfectly fine! You will get better at it like as you do it often and in different situations. Guessing and estimating can build intuition about numbers and various quantities. How about measuring to find out the weight of a 1-rupee coin? Chapter 2.indd 34 7/10/2025 3:29:59 PM 34 How many people might benefit from each of these offerings in a year? Again, guess first before finding out. Roxie and Estu overheard someone saying—“We did pādayātra for about 400 km to reach this place! We arrived early this morning.” How long ago would they have started their journey? Find answers by making necessary assumptions and approximations. Do guess first before calculating to check how close your guess was! Note to the Teacher: Assumptions can vary greatly at times, and as a result the answers computed using these you can also vary. This is perfectly alright. Modelling the situation properly is crucial, which can also be done in different ways sometimes. The accuracy of the assumed numbers or quantities can get better with exposure and practice. Math Talk Before the rise of modern transport, people moved from one place to another by walking — sometimes merchants, sages, and scholars walked thousands of kilometres to different parts of the world across deserts, mountains, and rivers. How many times can a person circumnavigate (go around the world) the Earth in their lifetime if they walk non-stop? Consider the distance around the Earth as 40,000 km. Linear Growth vs. Exponential Growth Roxie tells Estu about a science-fiction novel she is reading where they build a ladder to reach the moon, “... I wonder if we actually had a ladder like that, how many steps would it have?”. Pādayātra, is the traditional practice of walking long distances as part of a religious or spiritual pursuit. People across religions in our country observe similar forms of pilgrimage or spiritual walking, although they may have different names or purposes. Some of the pilgrimages are Ajmer Sharif Dargah Ziyarat, Pandharpur Wari, Kānwar Yatra, Sabarimala Yatra, Sammed Shikharji Yatra, Lumbini to Sarnath Yatra. Power Play Chapter 2.indd 35 7/10/2025 3:30:00 PM What do you think? Make an instinctive guess first. Would the number of steps be in thousands, lakhs, crores, or even more? To find out, we would need to know the gap between consecutive steps of the ladder. Let’s assume a reasonable distance of 20 cm. Visualising the problem as shown, 35 Ganita Prakash | Grade 8 We have to find out how many 20 cm make 3,84,400 km. If we calculate the value, we get the result as 1,92,20,00,000 steps, which is 192 crore and 20 lakh steps or 1 billion 922 million steps. The fixed increase in the distance from the earth with each step (a 20 cm gain after each step) is called linear growth. To cover the distance between the Earth and the Moon, it takes 1,92,20,00,000 steps with linear growth whereas it takes just 46 folds of a piece of paper with exponential growth! Linear growth is additive, whereas exponential growth is multiplicative. Some examples of exponential growth we have seen earlier in this chapter are ‘The Stones that Shine’, ‘Magical Pond’, ‘How Many Combinations’. We shall explore more such interesting examples in a later chapter and also in the next grade. Can you come up with some examples of linear growth and of exponential growth? 20 + 20 + 20 + ………… 1,92,20,00,000 times 0.001 × 2 × 2 × 2 ………. 46 times Chapter 2.indd 36 7/10/2025 3:30:00 PM 36 Getting a Sense for Large Numbers Last year, we learnt about lakhs and crores, as well as millions and billions. A lakh is 105 (1,00,000), a crore is 107 (1,00,00,000), and an arab is 109 (1,00,00,00,000), whereas a million is 106 (1,000,000) and a billion is 109 (1,000,000,000). You might know the size of the world’s human population. Have you ever wondered how many ants there might be in the world or how long ago humans emerged? In this section, we shall explore numbers significantly larger than arabs and billions. We shall use powers of 10 to represent and compare these numbers in each case. 100 As of mid-2025, there are only two northern white rhinos remaining in the world, both females, and they reside at the Ol Pejeta Conservancy in Kenya (= 2 × 100 ). 101 As of early 2024, the total population of Hainan gibbons is a meagre 42 ( ≈ 4 × 101 ). 102 There are just 242 Kakapo alive as of mid-2025 (≈ 2 × 102 ). 103 There are fewer than 3000 Komodo dragons in the world, all based in Indonesia (≈ 3 × 103 ). 104 A 2005 estimate of the maned wolf population showed that there are more than 17000 of them; most are located in Brazil (1.7 × 104 ). 105 As of 2018, there are around 4.15 lakh African elephants (≈ 4 × 105 ). 106 There are an estimated 50 lakh / 5 million American alligators as of 2025 (5 × 106 ). 107 The global camel population is estimated to be over 3.5 crore/ 35 million (3.5 × 107 ). India has only about 2.5 lakhs of them. The global horse population is around 5.8 crore / 58 million (5.8 × 107 ), with about half of them in America. 108 More than 20 crore / 200 million (2 × 108 ) water buffaloes are estimated worldwide, with a vast majority of them in Asia. 109 The estimated global population of starlings is around 1.3 arab/1.3 billion (_________). The global human population as of 2025 is 8.2 arab/8.2 billion (8.2 × 109 ). Power Play Chapter 2.indd 37 7/10/2025 3:30:03 PM A picture of a starling murmuration over a farm in the UK. Starling murmuration is a mesmerising aerial display of thousands of starlings flying in synchronised, swirling patterns. It is often described as a ‘choreographed dance’. 37 Ganita Prakash | Grade 8 With a global human population of about 8 × 109 and about 4 × 105 African elephants, can we say that there are nearly 20,000 people for every African elephant? 1016 The estimated population of ants globally is 20 padma/20 quadrillion (2 × 1016). Ants alone outweigh all wild birds and wild mammals combined. 1021 is supposed to be the number of grains of sand on all beaches and deserts on Earth. This is enough sand to give every ant 10 little sand castles to live in. 1010 The global chicken population living at any time is estimated at ≈ 33 billion (3.3 × 1010). 1012 The estimated number of trees (2023) globally stands at 30 kharab/3 trillion (3 × 1012). One kharab is 100 arab, and one trillion is 1000 billion. 1014 The estimated mosquito population worldwide (2023) is 11 neel/110 trillion (________). A derived estimate of the population of the Antarctic krill stands at 50 neel/500 trillion (5 × 1014). 1015 An estimate of the beetle population stands at 1 padma/1 quadrillion (1 × 1015). The estimate of the earthworm population is also at 1 padma/1 quadrillion. Chapter 2.indd 38 7/10/2025 3:30:04 PM 38 Calculate and write the answer using scientific notation: (i) How many ants are there for every human in the world? (ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world? 1023 The estimated number of stars in the observable universe is 2 × 1023. 1025 There are an estimated 2 × 1025 drops of water on Earth (assuming 16 drops per millilitre). 100 = 1 second - Time taken for a ball thrown up to fall back on the ground (typically a few seconds). 101 = 10 seconds - Time blood takes to complete one full circulation through the body: 10–20 seconds (1 × 101 – 2 × 101 seconds). Time in seconds Comparison to real-world events/phenomena (iii) If each tree had about 104 leaves, find the total number of leaves on all the trees in the world. (iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon? A different way to say your age! “How old are you?” asked Estu. “I completed 13 years a few weeks ago!” said Roxie. “How old are you?” asked Estu again. “I’m 4840 days old today!” said Roxie. “How old are you?” asked Estu again. “I’m ______ hours old!” said Roxie. Make an estimate before finding this number. Estu: “I am 4070 days old today. Can you find out my date of birth?” If you have lived for a million seconds, how old would you be? We shall look at approximate times and timelines of some events and phenomena, and use powers of 10 to represent and compare these quantities. What could this number mean? Find out! I am 69,70,710 … old Power Play Chapter 2.indd 39 7/10/2025 3:30:04 PM 102 seconds ≈1.6 minutes - Time needed to make a cup of tea: 5–10 minutes (≈ 4 × 102 – 8 × 102 seconds). Isn’t it quite amazing how someone is able to estimate things like the number of ants in the world or the time blood takes to fully circulate? You may carry this wonder whenever you encounter such facts. You will come across such facts in subjects like Science and Social Science, where such estimates are made frequently. - Time for light to reach the Earth from the Sun: about 8 minutes (≈ 5 × 102 seconds). - Typical waiting time at a traffic signal. 39 103 seconds ≈ 16.6 minutes - Satellites in low Earth orbits take between 90 minutes (≈ 5.5 × 103   seconds) to 2 hours to complete one full revolution around the Earth. 104 seconds ≈ 2.7 hours - The time needed to digest a meal: about 2–4 hours to pass through the stomach. 107 seconds ≈ 115.7 days / ≈ 3.8 months Ganita Prakash | Grade 8 105 seconds ≈ 1.16 days and 106 seconds ≈ 11.57 days. Think of some events or phenomena whose time is of the order of (i) 105 seconds and (ii) 106 seconds. Write them in scientific notation. - These liars!! It never gets cooked before 10 minutes… - Lifespan of an adult mayfly: about a day (≈ 9 × 104 seconds). Time spent sleeping in a year: about 4 months. They should approximate it as order of 102 seconds noodles  — that way whether it takes 120 seconds or 900 seconds, the claim will be true. Chapter 2.indd 40 7/10/2025 3:30:06 PM 40 108 seconds ≈ 3.17 years - The typical lifespan of most dogs is 3 to 15 years. 109 seconds ≈ 31.7 years - The orbital period of Halley’s comet is 75–79 years; the next expected return is in the year 2061 (≈ 2.4 × 109 seconds). - - Time taken by Mars for one full revolution around the Sun: 687 Earth-days/1.88 Earth-years (≈ 6 × 107 seconds). - Duration of one full revolution of Neptune around the Sun: 60,190 Earth-days/~165 Earth-years or 89,666 Neptunian days/1 Neptunian-year (≈ 5.2 × 109 seconds). A day on Neptune is about 16.1 hours Time taken by Mangalyaan mission to reach Mars: 298 days (≈ 2.65 × 107 seconds). Notice how rapid exponential growth is—106 seconds is less than a fortnight, but 109 seconds is a whopping 31 years (about half the life expectancy of a human)! 1010 seconds ≈ 317 years - The Chola dynasty ruled for more than 900 years (≈ 3 × 1010 seconds) between the 3rd Century BCE and 12th Century CE. 1011 seconds ≈ 3,170 years - Age of the oldest known living tree: about 5000 years (≈ 1.57 × 1011 seconds). 1012 seconds ≈ 31,700 years 1013 seconds ≈ 3.17 lakh years 1014 seconds ≈ 3.17 million years - Time since the last peak ice age: 19,000 – 26,000 years ago (≈ 6 × 1011 seconds – 8.2 × 1011 seconds). - Early Homo sapiens first appeared 2–3 lakh years ago (≈7 × 1012 – 9 × 1012 seconds). The entire population around that time could fit in a large cricket stadium. - The Steppe Mammoth is estimated to have appeared around 8–18 lakh years ago. - A fossil of Kelenken Guillermoi, a type of terror bird, is dated to 15 million years ago ( ≈_______________ seconds). Power Play Chapter 2.indd 41 7/10/2025 3:30:06 PM 1015 seconds ≈ 3.17 crore years - Age of Himalayas: 5.5 crore years/55 million years (≈ 1.7 × 1015 seconds); they continue to grow a few mm every year. - Dinosaurs went extinct 6.6 crore years ago/66 million years ago (≈ 2 × 1015 seconds). - Dinosaurs first appeared more than 20 crore/200 million years ago (≈ 6 × 1015 seconds). - It takes about 23 crore years for the Sun to make one complete trip around the Milky Way (≈ 7 × 1015 seconds). 41 Ganita Prakash | Grade 8 Notice that 109 seconds is of the order of the lifespan of a human, whereas 1018 seconds ago the universe did not exist according to modern physics!! The exponential notation can capture very large quantities in a concise manner. Calculate and write the answer using scientific notation: (i) If one star is counted every second, how long would it take to count all the stars in the universe? Answer in terms of the number of seconds using scientific notation. (ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take to finish the entire volume of water on Earth? 1016 seconds ≈ 31.7 crore years - Plants on land started 47 crore/470 million years ago ( ≈ _______________ seconds). 1017 seconds ≈ 3.17 billion years The oldest fossil evidence suggests that bacteria first appeared about 3.7 billion years ago. Very large quantities are often beyond our experience and comprehension. To put them into perspective, we can relate and compare them with quantities we are familiar with. This can give an essence of how large a number or a measure is! - The Earth is 4.5 billion years old. - The Milky Way galaxy was formed 13.6 billion years ago, and the Universe was formed 13.8 billion years ago. Try This Chapter 2.indd 42 7/10/2025 3:30:06 PM 42 2.5 A Pinch of History In the Lalitavistara, a Buddhist treatise from the first century BCE, we see number -names for odd powers of ten up to 1053. The following occurs as part of the dialogue between the mathematician Arjuna and Prince Gautama, the Bodhisattva. “Hundred kotis are called an ayuta (109 ), hundred ayutas a niyuta (1011), hundred niyutas a kankara (1013), …, hundred sarva-balas a visamjnagati (1047), hundred visamjna-gatis a sarvajna (1049), hundred sarvajnas a vibhutangama (1051), a hundred vibhutangamas is a tallakshana (1053).” Mahaviracharya gives a list of 24 terms (i.e., up to 1023) in his treatise Ganita-sara-sangraha. An anonymous Jaina treatise Amalasiddhi gives a list with a name for each power of ten up to 1096 (dasha-ananta). A Pali grammar treatise of Kāccāyana lists number-names up to 10140, named asaṅkhyeya. For expressing high powers of ten, Jaina and Buddhist texts use bases like sahassa (thousand) and koṭi (ten million); for instance, prayuta (106 ) would be dasa sata sahassa (ten hundred thousand). The modern naming is similar to this, where we say, Continuing this, a hundred kharab is a neel (1013), a hundred neel is a padma (1015), a hundred padma is a shankh (1017) and a hundred shankh is a maha shankh (1019). In the American/International system, we say Continuing this, a thousand trillion is a quadrillion (1015). This pattern continues. Observe the names million (106 ), billion (109 ), trillion (1012), quadrillion (1015), quintillion (1018), sextillion (1021), septillion (1024), octillion (1027), nonillion (1030), decillion (1033). A hundred thousand is a lakh 100 × 1000 = 1,00,000 102 × 103 = 105 A hundred arab is a kharab 100 × 1,00,00,00,000 = 1,00,00,00,00,000 102 × 109 = 1011 A thousand thousand is a million 1000 × 1000 = 1,000,000 103 × 103 = 106 A thousand million is a billion 1000 × 1,000,000 = 1,000,000,000 103 × 106 = 109 A hundred crores is an arab 100 × 1,00,00,000 = 1,00,00,00,000 102 × 107 = 109 A thousand billion is an trillion 1000 × 1,000,000,000 = 1,00,000,000,000 103 × 109 = 1012 A hundred lakhs is a crore 100 × 1,00,000 = 1,00,00,000 102 × 105 = 107 Power Play Chapter 2.indd 43 7/10/2025 3:30:07 PM What does the first part of each name denote? The number 10100 is also called a googol. The estimated number of atoms in the universe is 1078 to 1082. The number 10googol is called a googolplex. It is hard to imagine how large this number is! The currency note with the highest denomination in India currently is 2000 rupees. Guess what is the highest denomination of a currency note ever, across the world. The highest numerical value banknote ever printed was a special note valued 1 sextillion pengő (1021 or 1 milliard bilpengő) printed in Hungary in 1946, but it was never issued. In 2009, Zimbabwe printed a 100 trillion (1014) Zimbabwean dollar note, which at the time of printing was worth about $30. 43 Ganita Prakash | Grade 8 Figure it Out 1. Find out the units digit in the value of 2224 ÷ 432? [Hint: 4 = 22 ] 2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days? 3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers. 4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning. (i) Cube numbers are also square numbers. (ii) Fourth powers are also square numbers. (iii) The fifth power of a number is divisible by the cube of that number. (iv) The product of two cube numbers is a cube number. (v) q46 is both a 4th power and a 6th power (q is a prime number). 5. Simplify and write these in the exponential form. (i) 10– 2 × 10– 5 (ii) 57 ÷ 54 (iii) 9– 7 ÷ 94 (iv) (13– 2) – 3 (i) 643 (ii) 1928 (iii) 32–5 Chapter 2.indd 44 7/10/2025 3:30:07 PM 44 6. If 122 = 144 what is (i) (1.2)2 (ii) (0.12)2 (iii) (0.012)2 (iv) 1202 (v) m5 n12(mn) 9 10. 64 is a square number (82 ) and a cube number (43 ). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general? 11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible? 12. The worldwide population of sheep (2024) is about 109 , and that of goats is also about the same. What is the total population of sheep and goats? 13. Calculate and write the answer in scientific notation: 7. Circle the numbers that are the same— 24 × 36     64 × 32     610    182 × 62     624 8. Identify the greater number in each of the following— 9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of? (i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing. (ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees. (iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world. (iv) Total time spent eating in a lifetime in seconds. (ii) 209 (ii) 1011 (iii) 1010 (iv) 1018 (v) 2 × 109 (vi) 109 + 109 (i) 43 or 34 (ii) 28 or 82 (iii) 1002 or 2100 Power Play Math Talk Chapter 2.indd 45 7/10/2025 3:30:07 PM 14. What was the date 1 arab/1 billion seconds ago? Try This 45 Ganita Prakash | Grade 8 � We analysed some situations, asked questions, and found answers by first guessing, then modelling the problem statement, followed by making assumptions and approximations to carry out the calculations. We experienced how rapid exponential growth, also called multiplicative growth, can be compared to additive growth. na is n × n × n × n ×…× n (n multiplied by itself a times) and n–a = 1 na. Operations with exponents satisfy The scientific notation for the number 308100000 is 3.081 × 108 . The standard form of the scientific notation of any number is x × 10y , where x ≥ 1 and x < 10, and y is an integer. Engaging in interesting thought experiments can be used as means to understand how large a number or a quantity is. y na × nb = na+b y (na ) b = (nb ) a = na × b y na ÷ nb = na – b (n ǂ 0) y na × ma = (n × m) a y na ÷ ma = (n ÷ m) a (m ǂ 0) y n0 = 1 (n ǂ 0) SUMMARY Chapter 2.indd 46 7/10/2025 3:30:07 PM 46 Find a partner to play this game with. In 10 seconds, the person who writes a number or an expression, using only the digits 0-9 and arithmetic operations, that gives a number that is the larger between the two wins the round. In Round 1, Roxie wrote 10000000000000 and Estu wrote 999999 × 999999. Between these two, Roxie’s number is greater. Can you see why? Roxie’s number is 1013, whereas Estu’s number is less than (106 ) 2 . In Round 2, Roxie wrote 101000 + 101000 + 101000 + 101000 and Estu wrote (101000000) × 9000. Can you say which is greater? Below are some conditions that you may consider for different rounds. (i) Exponents are not allowed. Only addition is allowed. (ii) Exponents are not allowed. Only addition and multiplication are allowed. (iii) Exponents are allowed. Only addition is allowed. (iv) Exponents are allowed. Any arithmetic operation is allowed. You can create your own conditions and/or involve more people to play together. Tremendous in Ten! Chapter 2.indd 47 7/10/2025 3:30:10 PM" class_8,3,a story of numbers,ncert_books/class_8/hegp1dd/hegp103.pdf,"3 A STORY OF NUMBERS 3.1 Reema’s Curiosity One lazy afternoon, Reema was flipping through an old book when— whoosh!—a piece of paper slipped out and floated to the floor. She picked it up and stared at the strange symbols all over it. “What is this?” she wondered. She ran to her father, holding the paper as if it were a secret treasure. He looked at it and smiled. “Around 4000 years ago, there flourished a civilisation in a region called Mesopotamia, in the western part of Asia, containing a major part of the present-day Iraq and a few other neighbouring countries. This is one of the ways they wrote their numbers!” Reema’s eyes lit up, “Seriously? These strange symbols were numbers?” Her curiosity was sparked, and questions started swirling in her head. Chapter 3.indd 48 10-07-2025 17:44:17 Since when have people been writing numbers in the modern form? Since when have humans been counting? What was their need for counting? What were they counting? How would the Mesopotamians have written 20? 50? 100? Sensing her curiosity, her father started telling her how the idea of number and number representation evolved over the course of time, across geographies, to finally reach its modern efficient form. Get ready to travel back in time with them! Humans had the need to count even as early as the Stone Age. They were counting to determine the quantity of food they had, the number of animals in their livestock, details regarding trades of goods, the number of offerings given in rituals, etc. They also wanted to keep track of the passing days, e.g., to know and predict when important events such as the new moon, full moon, or onset of a season would occur. However, when they said or wrote down such numbers, they didn’t make use of the numbers that we use today. The structure of the modern oral and written numbers that we use today had its origin thousands of years ago in India. Ancient Indian texts, such as the Yajurveda Samhita, mentioned names of numbers based on powers of 10, almost as we say them orally today. For example, they listed names for the numbers one (eka), ten (dasha), hundred (shata), thousand (sahasra), ten thousand (āyuta), etc., all the way up to 1012 and beyond. The way we write our numbers today — using the digits 0 through 9 — also originated and were developed in India, around 2000 years ago. The first known instance of numbers being written using ten digits, including the digit 0 (which was then notated as a dot), occurs in the Bakhshali manuscript (c. 3rd century CE). Aryabhata (c. 499 CE) was the first mathematician to fully explain, and do elaborate scientific computations with the Indian system of 10 symbols. The Indian number system was transmitted to the Arab world by around 800 CE. It was popularised in the Arab world by the great Persian mathematician Al-Khwārizmī (after whom the word ‘algorithm’ is named) through his book On the Calculation with Hindu Numerals (c.825) and by the noted philosopher Al-Kindi through his work On the Use of the Hindu Numerals (c.830). From the Arab world, the Hindu numerals were transmitted to Europe and to parts of Africa by around 1100 CE. Though Al-Khwārizmī’s work on calculation with Hindu numerals was translated into Latin, it was the Italian mathematician Fibonacci who around the year 1200 really made the case to Europe to adopt the Indian numerals. However, the Roman numerals were so ingrained in European thinking and writing at the time that the Indian numerals did not gain widespread use for several more centuries. But eventually, during the European Renaissance and by the 17th century, not adopting them became impossible or it would impede scientific progress. Zero in the Bakhshali manuscript A Story of Numbers Chapter 3.indd 49 10-07-2025 17:44:18 49 Ganita Prakash | Grade 8 Their use then spread to every continent, and are now used in every corner of the world. Because European scholars learned the Indian numerals from the Arab world, they called them ‛Arabic numerals’ to reflect their European perspective. On the other hand, as noted above, Arab scholars, such as Al-Khwārizmī and Al-Kindi, called them ‛Hindu numerals’. During the period of European colonisation, the European term Arabic numbers became widely used. However, in recent years, this mistake is being corrected in many textbooks and documents around the world, including in Europe. The most commonly used terminologies for the numbers we use today are ‛Hindu numerals’, ‛Indian numerals’, and the transitional ‛Hindu-Arabic numerals’. It is worth noting that the word ‛Hindu’ here does not refer to a religion, but rather a geography/people from whom these numbers came. The shape of the digits 0, 1, 2, ..., 9 used to write numbers in the Indian number system today evolved over a period of time, as shown below: “The ingenious method of expressing every possible number using a set of ten symbols (each symbol having a place value and an absolute value) emerged in India. The idea seems so simple nowadays that its significance and profound importance is no longer appreciated. Its simplicity lies in the way it facilitated calculations and placed arithmetic foremost among useful inventions.” — Pierre-Simon Laplace (1749–1827) Chapter 3.indd 50 10-07-2025 17:44:18 50 Prior to the global adoption of the Indian system of numerals, different groups of people used different methods of representing numbers. We Evolution of the digits used in the Indian number system shall take a glimpse of some of them. We will not be looking at different systems in a chronological order, but rather an order that shows us the main stages in the development of the idea of number representation. But first, let us explore some of the foundational ideas needed to count and to determine the number of objects in a given collection. The Mechanism of Counting Imagine that we are living in the Stone Age, say, around ten thousand years ago. Suppose we have a herd of cows. Here are some natural questions that we might ask about our herd— Q1. How do we ensure that all cows have returned safely after grazing? Q2. Do we have fewer cows than our neighbour? Q3. If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour? We need to tackle these questions without the use of the number names or written numbers of the Hindu number system. How do we do it? Here are some possible methods. Method 1: We could tackle the questions by using pebbles, sticks or any object that is available in abundance. Let us choose sticks. For every cow in the herd, we could keep a stick. The final collection of sticks tells us the number of cows, which can be used to check if any cows have gone missing. A Story of Numbers Math Talk Chapter 3.indd 51 11-07-2025 14:29:33 51 Ganita Prakash | Grade 8 This way of associating each cow with a stick, such that no two cows are associated or mapped to the same stick is called a one-to-one mapping. This mapping can then be used to come up with a way to represent numbers, as shown in the table. How will you use such sticks to answer the other two questions (Q2 and Q3)? Method 2: Instead of objects, we could use a standard sequence of sounds or names. For example, we could use the sounds of the letters of any language. While counting, we could make a one-to-one mapping between the objects and the letters: that is, associate each object to be counted with a letter, following the letter-order. This mapping can then be used to come up with a way of verbally representing numbers. For example, we get the following number representation if we use English letters ‘a’ to ‘z’. Number Its representation (using sticks) 1 2 3 4 5 . . . . . . Chapter 3.indd 52 10-07-2025 17:44:18 52 Number Its representation (using sounds or names) 26 1 2 3 4 5 . . . . . . a b d c e z Representation using symbols I II III IV V VI VII VIII IX X Representation using symbols XI XII XIII XIV XV XVI XVII XVIII XIX XX An obvious limitation of using only the letters of the English alphabet in this form is that it cannot be used to count collections having more than 26 objects. How many numbers can you represent in this way using the sounds of the letters of your language? Do you see a way of extending this method to represent bigger numbers as well? How? From the discussion above, we see that for counting and finding the size of a collection, we need a standard sequence of objects, or names, or written symbols, that has a fixed order. Let us call this standard sequence a number system. A collection of objects can be counted by making a one-to-one mapping between them and the standard sequence, following the sequence order. Since there is no end to numbers, the challenge is to come up with an unending standard sequence/number system that is easy to count with. Using sticks gives an unending standard sequence/number system. However, it is not convenient to count larger collections, as we will need as many sticks as the number of objects being counted. Using the sounds of the letters of a language, as in Method 2, is convenient for the counting process but is not an unending standard sequence/number system. The standard sequence/number system given in Method 3 was actually the system used in Europe before it got replaced by the Hindu number system. It is called the Roman number system. It was widely used in Europe for centuries, and was convenient for many purposes, but had the similar drawback that one cannot write arbitrarily large numbers without introducing more and more symbols. We will learn more about this system of writing numbers later on. Number 1 2 3 4 5 6 7 8 9 10 Number 11 12 13 14 15 16 17 18 19 20 Method 3: We could use a sequence of written symbols as follows. Table 1 A Story of Numbers Math Talk Chapter 3.indd 53 10-07-2025 17:44:18 53 Ganita Prakash | Grade 8 As illustrated by the three methods, history gives us examples of number systems formed using physical objects (such as sticks, pebbles, body parts, etc.), names, and written symbols. Some groups of people had numbers represented both by physical objects as well as by names, while others like the Chinese had all three forms of representation. The symbols occurring in a written number system are called numerals. For example, 0, 1, 5, 36, 193, etc., are some of the numerals occurring in the Hindu number system. Numerals representing ‘smaller’ numbers always had names, and so a number system composed of written symbols always went hand in hand with a number system composed of names, as is the case with the modern-day Hindu system. Figure it Out 1. Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks. 2. One way of extending the number system in Method 2 is by using strings with more than one letter—for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it! Math Talk Math Talk Chapter 3.indd 54 10-07-2025 17:44:19 54 3. Try making your own number system. 3.2 Some Early Number Systems I. Use of Body Parts Many groups of people across the world have used their hands and body parts for counting. Here is how a group of people in Papua New Guinea used and still use their body parts as the standard sequence/number system. Try This II. Tally Marks on Bones and Other Surfaces One of the oldest methods of number representation is by making notches—marks cut on a surface such as a bone or a wall of a cave. These marks are also called tally marks. In this method, a mark is made for each object that is being counted. So the final collection of marks represents the total number of objects. This method is very similar to the method of using sticks to count (Method 1), except for the fact that a mark is made instead of adding a stick. Archaeologists have discovered bones dating back more than 20,000 years that seem to have tally marks. The oldest known such bones with markings that are thought to represent numbers are the Ishango bone and the Lebombo bone. The Ishango bone, dating back 20,000 to 35,000 years, was discovered in the Democratic Republic of Congo. It features notches arranged in columns, possibly indicating calendrical systems. The Lebombo bone, discovered in South Africa, is an even older tally stick with 29 notches, estimated to be around 44,000 years old. It is considered one of the oldest mathematical artefacts, and may have served as a tally stick or lunar calendar. A Story of Numbers Chapter 3.indd 55 10-07-2025 17:44:20 Lebombo bone Ishango bone 55 Ganita Prakash | Grade 8 III. Number Names Obtained by Counting in Twos A group of indigenous people in Australia called the Gumulgal had the following words for their numbers. Can you see how their number names are formed? The number name for 3 is composed of number names of 2 and 1. The number name for 4 is composed of two occurrences of the number name for 2. Can you see how the names of the other numbers are formed? The numbers are counted in 2s, using which the number names are formed: 3 = 2 + 1, 4 = 2 + 2, 5 = 2 + 2 + 1, 6 = 2 + 2 + 2. Gumulgal called any number greater than 6 ras. There is a very interesting and puzzling historical phenomenon associated with this number system. Look at the following number systems of a group of indigenous people in South America, and the Bushmen of South Africa: Gumulgal (Australia) 1. urapon 2. ukasar 3. ukasar-urapon 4. ukasar-ukasar 5. ukasar-ukasar-urapon 6. ukasar-ukasar-ukasar Chapter 3.indd 56 10-07-2025 17:44:21 56 Gumulgal Bushmen Bakairi 1 - tokale 2 - ahage 3 - ahage tokale (or ahawao) 4 - ahage ahage tokale 5 - ahage ahage ahage 1 - xa 2 - t'oa 3 - 'quo 4 - t'oa-t'oa 5 - t'oa-t'oa-t'a 6 - t'oa-t'oa-t'oa Map not to scale India 1 - urapon 2 - ukasar 3 - ukasar-urapon 4 - ukasar-ukasar 5 - ukasar-ukasar-urapon 6 - ukasar-ukasar-ukasar Despite being so far apart geographically, and with no trace of contact between them, these three groups have developed equivalent number systems! Historians have wondered how this happened. One theory is that these three groups of people may have had common ancestors, who used this number system. In course of time, their descendants migrated to these places. Even though the number system of Gumulgal had number names for numbers only till 6, we can see the emergence of an idea here. Counting in 2s is more efficient for representing numbers than, for example, a tally system. A general form which this idea has taken in different number systems is as follows: count in groups of a certain number (like 2 in the case of Gumulgal’s system), and use the word or symbol associated with this group size to represent bigger numbers. Some of the commonly used group sizes in different number systems have been 2, 5, 10 and 20. You can find the idea of counting by 5s in the Roman system (Table 1). One of the phenomena that could have led people to this idea might be the human limit for immediately knowing the size of a collection at a glance. Let us try out the following activity. Quickly count the number of objects in each of the following boxes: This idea of counting in a certain group size and using it to represent numbers is an important idea in the history of the evolution of number systems. A Story of Numbers Chapter 3.indd 57 10-07-2025 17:44:21 Up to what group size could you immediately see the number of objects without counting? Most humans find it difficult to count groups having 5 or more objects in a single glance. This limit of perception could have prompted people using tally marks to replace every group of, say, 5 marks, with a new symbol, as seen in the system shown in Table 1. Math Talk 57 Ganita Prakash | Grade 8 What could be the difficulties with using a number system that counts only in groups of a single particular size? How would you represent a number like 1345 in a system that counts only by 5s? Even though counting in groups of a particular size and using it for number representation is more efficient than the tally system, this method can still become cumbersome for larger numbers. The next system shows a refinement of this idea. IV. The Roman Numerals We have already seen the Roman number system till 20 (Table 1). We have seen that it uses I for 1, V for 5, X for 10. To get the Roman numeral for any number till 39, it is first grouped into as many 10s as possible, the remaining is grouped into as many 5s as possible, and finally the remaining is grouped into 1s. Example: Let us take the number 27. 27 = 10 + 10 + 5 + 1 + 1 Chapter 3.indd 58 10-07-2025 17:44:22 58 So, 27 in Roman numerals is XXVII. Instead of representing 50 as XXXXX, a new symbol is given to it: L. Following the way the number 4 is represented as 1 less than 5 — that is, as IV — 40 is represented as 10 less than 50 — that is, as XL. However, people using this system were not always consistent with this practice. Sometimes, 40 was also represented as XXXX. The Roman number system introduces newer symbols to represent certain bigger numbers. Let us call all these numbers that have a new basic symbol as landmark numbers. Here are some of the landmark numbers of the Roman system and their associated numerals. 1 5 10 50 100 500 1,000 I V X L C D M These symbols are used to denote other numbers as well. For example, consider the number 2367. Writing it as a sum of landmark numbers starting from 1000 such that we take as many 1000s as possible, 500s as possible, and so on, we get So in Roman numerals, this number is MMCCCLXII. Figure it Out 1. Represent the following numbers in the Roman system. We see how vastly efficient this system is compared to some of the previous number systems that we have seen. This system seems to have evolved out of the ancient Greek number system in around the 8th century BCE in Rome, and evolved over time. It spread throughout Europe with the expansion of the Roman empire. Despite the relative efficiency of the Roman system, it doesn’t lend itself to an easy performance of arithmetic operations, particularly multiplication and division. Example: Try adding the following numbers without converting them to Hindu numerals: The efficiency of this system is due to the grouping of a given number by not just one group size, but a sequence of group sizes that we call landmark numbers, and then using these landmark numbers to represent the given number. This idea is the next important breakthrough in the history of the evolution of number systems. (i) 1222   (ii) 2999   (iii) 302   (iv) 715 2367 = 1000 + 1000 + 100 + 100 + 100 + 50 + 10 + 5 + 1 + 1 A Story of Numbers Chapter 3.indd 59 10-07-2025 17:44:22 (a) CCXXXII + CCCCXIII Let us find the total number of Is, Xs, and Cs, and group them starting from the largest landmark number. Apparently, it looks like the largest landmark number is C, but note that 5 Cs (100s) make a D (500). So the sum is 59 Ganita Prakash | Grade 8 Do it yourself now: (b) LXXXVII + LXXVIII How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals? Try to find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX. People using the Roman system made use of a calculating tool called the abacus to perform their arithmetic operations. We will see what it is in a later section. However, only specially trained people used this tool for calculation. While going through the number systems discussed above, it should not be understood that, historically, one system developed as an improvement over the previous system. This point should be kept in mind when studying the upcoming number systems too. The actual story of how each of the number systems developed is much more complex, and many times not clearly known, and so we will not try to trace this in the chapter. CCXXXI and MDCCCLII Multiply Try This Chapter 3.indd 60 10-07-2025 17:44:22 60 Figure it Out 1. A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this? 2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, –, ×, ÷) for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following: Math Talk Math Talk 3. Identify the features of the Hindu number system that make it efficient when compared to the Roman number system. 4. Using the ideas discussed in this section, try refining the number system you might have made earlier. 3.3 The Idea of a Base I. The Egyptian Number System (iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar) (iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar) (ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasarukasar)(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasarukasar-urapon)A Story of Numbers Math Talk Try This Chapter 3.indd 61 10-07-2025 17:44:22 We are now going to see a written number system that the Egyptians developed around 3000 BCE. In this system, we see the use of landmark numbers to group and represent a given number. However, what makes this system special is its sequence of landmark numbers. Imagine making collections of pebbles. The first landmark number is 1. Group together 10 collections of the previous landmark number (1). Its size is the second landmark number which is 10. Group together 10 collections of the previous landmark number (10). Its size is the third landmark number which is 10 × 10 = 100, and so on. 61 Ganita Prakash | Grade 8 Each landmark number is 10 times the previous one. Since 1 is the first landmark number, they are all powers of 10. The following are the symbols given to these numbers — As in the case of Roman numbers, a given number is counted in groups of the landmark numbers, starting from the largest landmark number less than the given number. This is then used to assign the numeral. For example 324 which equals 100 + 100 + 100 + 10 + 10 + 4 is written as . Figure it Out 1. Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707. 2. What numbers do these numerals stand for? II. Variations on the Egyptian System and the Notion of Base (i) (ii) Chapter 3.indd 62 10-07-2025 17:44:23 62 Instead of grouping together 10 collections of size equal to the previous landmark number (as in the case of the Egyptian system), can we get a number system by grouping together 5 collections of size equal to the previous landmark number? Can this 5 be replaced by any positive integer? Let us examine this possibility. Let 1 be the first landmark number. Group together 5 collections of size equal to the previous landmark number (1). Its size is the second landmark number which is 5. Group together 5 collections of size equal to the previous landmark number (5). Its size is the third landmark number which is 5 × 5 = 25. Group together 5 collections of size equal to the previous landmark number (5). Its size is the fourth landmark number which is 5 × 25 = 125. Thus, we have a new number system where each landmark number is 5 times the previous one. Since 1 is the first landmark number, they are all powers of 5. Express the number 143 in this new system. Let us start grouping, starting with the size 53 = 125, as this is the largest landmark number smaller than 143. We get— So the number 143 in the new system is . Number systems having landmark numbers in which the (b) every next landmark number is obtained by multiplying the current landmark number by some fixed number n is said to be a base-n number system. The Egyptian number system is a base-10 system, and the number system that we created is a base-5 system. A base-10 number system is also called a decimal number system. (a) first landmark number is 1, and 50 = 1 51 = 5 52 = 25 53 = 125 54 = 625 55 = 3125 143 = 125 + 5 + 5 + 5 + 1 + 1 + 1. A Story of Numbers Chapter 3.indd 63 10-07-2025 17:44:23 Figure it Out 1. Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651. 2. Is there a number that cannot be represented in our base-5 system above? Why or why not? 3. Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system? The landmark numbers of a base-n number system are the powers of n starting from n0 = 1, n, n2 , n3 ,... 63 Ganita Prakash | Grade 8 Advantages of a Base-n System What is the advantage of having landmark numbers that are all the powers of a number? To understand this, let us perform some arithmetic operations using them. Example: Add the following Egyptian numerals: Let us find the total number of | and and group them starting from the largest possible landmark number. It has a total of— 15 and 15 |. Since 10 gives the next landmark number , the sum can be regrouped as— Chapter 3.indd 64 10-07-2025 17:44:23 64 Sum = Since 10| gives a , we have Sum = Figure it Out 1. Add the following Egyptian numerals: (i) (ii) 2. Add the following numerals that are in the base-5 system that we created: A Story of Numbers Chapter 3.indd 65 10-07-2025 17:44:23 Remember that in this system, 5 times a landmark number gives the next one! + 65 Ganita Prakash | Grade 8 Contrast the addition done in a base-n number system with that done in the Roman system. In the Roman system, the grouping and rearranging has to be done carefully as it is not always by the same size that each landmark number has to be grouped to get the next one. The advantage of a number system with a base becomes more evident when we consider multiplication. How to multiply two numbers in Egyptian numerals? I see a similarity in the method of adding numbers in the Egyptian and the Hindu system! + + 4 7 5 6 10 3 + 1 Chapter 3.indd 66 10-07-2025 17:44:23 66 Let us first consider the product of two landmark numbers. 1. What is any landmark number multiplied by (that is 10)? Find the following products— 2. What is any landmark number multiplied by (102 )? Find the following products— Each landmark number is a power of 10 and so multiplying it with 10 increases the power by 1, which is the next landmark number. Find the following products— Thus, the product of any two landmark numbers is another landmark number! Does this property hold true in the base-5 system that we created? Does this hold for any number system with a base? What can we conclude about the product of a number and (10), in the Egyptian system? A Story of Numbers Math Talk Chapter 3.indd 67 10-07-2025 17:44:24 (ii) | × | is the same as + + |. Thus, | × = ( + + |) × Will the distributive property hold here? For the same reason that it holds for (a + b) × n, it also holds when one of the numbers has more than 2 terms. For example, (a + b + c) × n = an + bn + cn. So, ( + + |) × = ( × ) + ( × ) + (| × )   = = + + 67 Ganita Prakash | Grade 8 Now find the following products— (i) (ii) What would be a simple rule to multiply a number with ? As has been seen, a process of multiplying two numbers involves the multiplication of landmark numbers. When the landmark numbers are powers of a number, then their product is another landmark number. This fact simplifies the process of multiplication. However, this is not the case with the Roman numerals, which is why multiplication using them is difficult. Thus, a number system whose landmark numbers are powers of a number, i.e., a number system with a base, is efficient not only in number representation but also in its utility in carrying out arithmetic operations. The idea of a number system with a base was a turning point in the history of the evolution of number systems. Our modern Hindu number system is built on this structure. Abacus that Makes Use of the Decimal System In around the 11th century, even the people still using the Roman numerals started using a calculating device—the abacus—constructed using a decimal system. It was a board with lines, as shown in the Fig. 3.1. Starting from the line that stood for 1, each successive line stood for a successive power of 10. Chapter 3.indd 68 10-07-2025 17:44:25 68 Numbers were represented in it as follows: the given number was first grouped into the landmark numbers (powers of 10), in exactly the same way we have been grouping them so far. For each power of 10, as many counters were placed on its line as the number of times it occurred Fig. 3.1: Abacus in the grouping. The presence of a counter above a line contributed a value of 5. For example let us take the number 3426. It can be grouped as 3426 = 1000 + 1000 + 1000 + 100 + 100 + 100 + 100 + 10 + 10+ 1 + 1 + 1 + 1 + 1 + 1 This number was represented as shown in Fig. 3.1. Notice how the 6 ones are represented. To get an idea of how the abacus was used for calculations, let us consider a simple addition problem: 2907 + 43. The two numbers were taken on either side of the vertical partition. How would you use this to find the sum? The counters along each line were brought together. What is to be done if the total in a line exceeded 10? Hint: In this problem, the 7 ones and the 3 ones together make 10 ones which contributes a counter to the line representing 10s. III. Shortcomings of the Egyptian System A Story of Numbers Chapter 3.indd 69 10-07-2025 17:44:25 Despite being a number system that enabled relatively efficient number representations for numbers till a crore (107 ), and relatively easy computations, the Egyptian system had a drawback. If larger and larger numbers needed to be represented, then there was a need for inventing an unending sequence of symbols for higher and higher powers of 10. Here we see the original challenge of number representation resurfacing in a different form! The next and the final idea in the history of the evolution of number systems not only solves this problem but also remarkably simplifies number representation and computations! Figure it Out 1. Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not? Math Talk 69 Ganita Prakash | Grade 8 2. Create your own number system of base 4, and represent numbers from 1 to 16. 3. Give a simple rule to multiply a given number by 5 in the base-5 system that we created. 3.4. Place Value Representation I. The Mesopotamian Number System In the beginning, the number system used in ancient Mesopotamia had different symbols for different landmark numbers. In later times, it became a base-60 system, also called the sexagesimal system, with a very efficient number representation. It has puzzled many why they chose base 60. Different theories exist to explain this, ranging from the connection between 60 and the periods of some important events (like the length of their lunar month which had 30 days, or the time taken for the Sun to complete one revolution around the Earth when Earth is taken to be stationary), the ease of representing fractions (we will not go into this here), their earlier sequence of landmark numbers — 1, 10, 60, 600, 3600, 36000,… — getting reduced to only the powers of 60, and so on. The influence of the Mesopotamian sexagesimal system, also known as the Babylonian number system, can be seen even now in our units of time measurements—1 hour = 60 minutes and 1 minute = 60 seconds. This system used the symbol for 1 and for 10. Let us now briefly pause on the study of their number system, and ideate on how one can build an efficient number system using the Mesopotamian features seen so far. Math Talk Chapter 3.indd 70 10-07-2025 17:44:25 70 Let us give our own symbols to their landmark numbers— Note that we have actually used Indian numerals in creating these symbols. We could have invented our own symbols but for the sake of easy recall and use, we have chosen to take help of the familiar numerals 1, 2, 3, ... Using and , numbers from 1 to 59 can be represented— 10 11 12 20 30 40 50 59 1 2 3 4 5 6 7 8 9 1 60 602 = 3600   603 = 216000 ... 1 2 3 ... A Story of Numbers Chapter 3.indd 71 10-07-2025 17:44:26 Example: Let us represent the number 640 in this system. Grouping it into landmark numbers, we see that 10 1 s, and 40 would be represented using 4 s. If we use the Egyptian idea, this number would be represented using Reproduction of a Mesopotamian Tablet 640 = (10) × 60 + 40. 71 Ganita Prakash | Grade 8 Can we represent this more compactly? We can simply represent this number as which can be read as ten 60s and one 40, just as we have written in the equation. Example: Let us try another number — 7530. So, its representation would be Note that when a number is grouped into powers of 60 for its representation, no power of 60 can occur 60 or more times. If this happens, then 60 of them can be grouped to form the next power of 60. For example, consider the expression— (1) × 3600 + (70) × 60 + 2 = (1) × 602 + (60 + 10) × 60 + 2 = (1) × 602 + 602 + (10) × 60 + 2 = (2) × 602 + (10) × 60 + 2 Therefore, any number can be represented using the numerals from 1–59, along with the numerals for landmark numbers. Now, what if we make the representation even more compact by dropping the symbols for the different powers of 60 altogether? 7530 = (2) ×3600 + (5) × 60 + 30 Chapter 3.indd 72 10-07-2025 17:44:26 72 This is exactly what the Mesopotamians did! In their numeral, the rightmost set of symbols showed the number of 1s, the set of symbols to its left showed the number of 60s, the next showed the number of 3600s and so on. Whenever there was no occurrence of a power of 60, a blank space was given in that position. It does not seem that the Mesopotamians arrived at this idea in the same way we did. Some scholars suggest that the similarity of symbols given to the landmark numbers 1 and 60 in their earlier number system, and an accidental usage of them, might have made them stumble upon this idea. Figure it Out 1. Represent the following numbers in the Mesopotamian system — Thus, we can see how the Mesopotamian system removes the need for generating an unending sequence of symbols for the landmark numbers by making use of the positions where the symbols are written. Such a number system (having a base) that makes use of the position of each symbol in determining the landmark number that it is associated with is called a positional number system or a place value system. This idea of place value marks the highest point in the history of evolution of number systems, and gives a very elegant solution to the problem of representing the unending sequence of numbers using only a finite number of different symbols! The Mesopotamian system however cannot be considered a fully developed place value system. It has certain defects that lead to confusion while reading a number. Look at the representation of 60. What will be the representation for 3,600? While writing the numerals, the spacing between symbols was not given the way we are giving it here. It was also difficult to maintain a consistent spacing for blanks across different manuscripts written by different people. These created ambiguities. For example, consider the representations of the following numbers — (i) 63  (ii) 132 (iii) 200 (iv) 60 (v) 3605 A Story of Numbers Chapter 3.indd 73 10-07-2025 17:44:26 73 Ganita Prakash | Grade 8 Because of the ambiguity in finding which symbols correspond to which powers of 60, the same numeral can be read in different ways. Even in our representation which uses uniform spacing between symbols for different powers of 60, it is difficult to know the number of blanks between two sets of symbols, as in the representation of 36002. To address the issue arising out of blank spaces, the later Mesopotamians used a brilliant idea of assigning a ‘placeholder’ symbol to denote a blank space. This is like the 0 (zero) we use in our system. Thus, zero—the symbol that shows nothingness—is indispensable as a placeholder in a place value system in which numbers are written in an unambiguous manner. Even with the problem arising out of blank spaces solved, other ambiguities still remained in the system. For example, the placeholder symbol was primarily used in the middle of numbers and not at the end; so they would not use it to represent a number like (what we would write as) 3600. II. The Mayan Number System Chapter 3.indd 74 10-07-2025 17:44:26 74 In Central America, there flourished a civilisation known as the Mayan civilisation that made great intellectual and cultural progress between the 3rd and 10th centuries CE. Among their intellectual achievements stands their place value system designed independently of those in Asia. They also made use of a placeholder symbol, for the modern-day ‘0’, that looked like a seashell. Chapter 3.indd 75 10-07-2025 17:44:27 Ganita Prakash | Grade 8 Here we find a puzzling phenomenon. Why was their third landmark number 360 rather than 400? Some scholars feel that this might have something to do with their calendars. They used a dot for 1, and a bar for 5. These were used to denote numbers from 1 to 19. The symbols associated with different landmark numbers were written one below the other with the lowermost set of symbols corresponding to the number of 1s, the set above corresponding to the number of 20s, the set above to the number of 360s and so on. Represent the following numbers using the Mayan system: Because the Mayan system is not an actual base-20 system, it lacks the advantages that a system with a base has for computations. Nevertheless, their place value notation and their use of a placeholder symbol for zero is considered an important advance in the history of number systems. A curious fact is that we can still find the use of base-20 in the number names of some European languages. III. The Chinese Number System (i) 77  (ii) 100 (iii) 361 (iv) 721 Chapter 3.indd 76 10-07-2025 17:44:27 76 The Chinese used two number systems—a written system for recording quantities, and a system making use of rods for performing computations. The numerals in the rod-based number system are called rod numerals. Here we discuss the rod numerals, which are more efficient in writing and computing with numbers than the written system of the Chinese. The rod numerals developed in China by at least by 3rd century AD and were used till the 17th century. It was a decimal system (base-10). The symbols for 1 to 9 were as follows: Note: The zongs represent units, hundreds, tens of thousands, etc., and the hengs tens, thousands, hundreds of thousands, etc. How is this to be read? A numeral Landmark number positions 103 102 10 1 CHINESE NUMBER SYSTEM 2 (Heng) 6 (Zong) 3 (Heng) 4 (Zong) Base-10 or Decimal A Story of Numbers Chapter 3.indd 77 10-07-2025 17:44:27 Like the Mesopotamians, the rod numerals used a blank space to indicate the skipping of a place value. However, because of the slightly more uniform sizes of the symbols for one through nine, the blank spaces were easier to locate than in the Mesopotamian system. = (2) × 103 + (6) × 10² + (3) × 10 + (4) × 1 = 2634 77 Ganita Prakash | Grade 8 Notice how similar the rod numerals are to the Hindu system. The Chinese system, with a symbol for zero, would be a fully developed place value system. IV. The Hindu Number System Where does the Hindu/Indian number system figure in the evolution of ideas of number representation? What are its landmark numbers? And does it use a place value system? Hindu Number System Ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 A numeral 375 Base-10 or Decimal Chapter 3.indd 78 10-07-2025 17:44:28 78 As can be seen, the Hindu number system is a place value system. The Hindu number system has had a symbol for 0 at least as early as 200 BCE. Because of the use of 0 as a digit, and the use of a single digit in each position, this system does not lead to any kind of ambiguity when How is this to be read? 3 7 5 Landmark number positions = (3) × 10² + (7) × 10 + (5) × 1 = 375 102 10 1 reading or writing numerals. It is for this reason that the Hindu number system is now used throughout the world. The use of 0 as a digit, and indeed as a number, was a breakthrough that truly changed the world of mathematics and science. In Indian mathematics, indeed, zero was not just used as a placeholder in the place value system, but was also given the status of a number in its own right, on par with other numbers. The arithmetic properties of the number 0 (e.g., that 0 plus any number is the same number, and that 0 times any number is zero) were explicitly used by Aryabhata in his Āryabhaṭīya in 499 CE to compute with and do elaborate scientific computations using Hindu numerals. The use of 0 as a number like any other number, on which one can perform the basic arithmetic operations, was codified by Brahmagupta in his work Brāhmasphuṭasiddhānta in 628 CE, as we learned in an earlier grade. By introducing 0 as a number, along with the negative numbers, Brahmagupta created what in modern terms is called a ring, i.e., a set of numbers that is closed under addition, subtraction, and multiplication (i.e., any two numbers in the set can be added, subtracted, or multiplied to get another number in that set). These new ideas laid the foundations for modern mathematics, and particularly for the areas of algebra and analysis. Hopefully, this gives you a sense of all the ideas that went into writing and computing with numbers in the way that we do today. The discovery of 0 and the resulting Indian number system is truly one of the greatest, most creative, and most influential inventions of all time—appearing constantly in our daily lives and forming the basis of much of modern science, technology, computing, accounting, surveying, and more. The next time you are writing numbers, think about the incredible history behind them and all the deep ideas that went into their discovery! A Story of Numbers Chapter 3.indd 79 10-07-2025 17:44:28 1. Count in groups of a single number. ukasar-ukasar-urapon 2. Group using landmark numbers. I V X L C M 3. Choosing powers of a number as landmark numbers—the idea of a base. 1 101 102 103 104 … 4. Using positions to denote the landmark numbers—the idea of place value system. 1 7 2 9 5. The idea of 0 as a positional digit and as a number. Evolution of Ideas in Number Representation 79 Ganita Prakash | Grade 8 Figure it Out 1. Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions? 2. Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s. 3. Where in your daily lives, and in which professions, do the Hindu numerals, and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of? 4. The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2? Math Talk Math Talk Chapter 3.indd 80 10-07-2025 17:44:29 80 The map shows the locations of the different civilisations. They existed in different time periods. Map not to scale To represent numbers, we need a standard sequence of objects, names, or written symbols that have a fixed order. This standard sequence is called a number system. The symbols representing numbers in a written number system are called numerals. In a number system, landmark numbers are numbers that are easily recognisable and used as reference points for understanding and working with other numbers. They serve as anchors within the number system, helping people to orient themselves and make sense of quantities, particularly larger ones. A number system whose landmark numbers are the powers of a number n is referred to as a base-n number system. Number systems having a base that make use of the position of a symbol in determining the landmark number that it is associated with are called positional number systems or place value systems. Place value representations were used in the Mesopotamian (Babylonian), Mayan, Chinese and Indian civilisations. The system of numerals that we use throughout the world today is the Hindu number system (also sometimes called the Indian number system, or the Hindu-Arabic number system). It is a place value system with (usually) 10 digits, including the digit 0 which is treated on par with other digits. Due to its use of 0 as a number, the system enables the writing of all numbers unambiguously using just finitely many symbols, and also enables efficient computation. The system originated in India around 2000 years ago, and then spread across the world, and is considered one of human history’s greatest inventions. SUMMARY A Story of Numbers Chapter 3.indd 81 10-07-2025 17:44:30 81" class_8,4,quadrilaterals,ncert_books/class_8/hegp1dd/hegp104.pdf,"4 QUADRILATERALS In this chapter, we will study some interesting types of four-sided figures and solve problems based on them. Such figures are commonly known as quadrilaterals. The word ‘quadrilateral’ is derived from Latin words — quadri meaning four, and latus referring to sides. Observe the following figures. (i) (ii) (iii) Chapter 4 Quadrilaterals 06-07-2025.indd 82 7/10/2025 3:19:32 PM Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why? The angles of a quadrilateral are the angles between its sides, as marked in Figs. (i), (ii), and (iii). We will start with the most familiar quadrilaterals—rectangles and squares. (iv) (v) 4.1 Rectangles and Squares We know what rectangles are. Let us define them. Rectangle: A rectangle is a quadrilateral in which— The definition precisely states the conditions a quadrilateral has to satisfy to be called a rectangle. Are there other ways to define a rectangle? Let us consider the following problem related to the construction of rectangles. A Carpenter’s Problem A carpenter needs to put together two thin strips of wood, as shown in Fig. 1, so that when a thread is passed through their endpoints, it forms a rectangle. She already has one 8 cm long strip. What should be the length of the other strip? Where should they both be joined? Let us first model the structure that the carpenter has to make. The strips can be modelled as line segments. They are the diagonals of the quadrilateral formed by their endpoints. For the quadrilateral to be a rectangle, we need to answer the following questions — 1. What is the length of the other diagonal? (i) The angles are all right angles (90°), and (ii) The opposite sides are of equal length. A B Fig. 1 O Quadrilaterals D C Chapter 4 Quadrilaterals 06-07-2025.indd 83 7/10/2025 3:19:32 PM 2. What is the point of intersection of the two diagonals? 3. What should the angle be between the diagonals? Let us answer these questions using geometric reasoning (deduction). If that is challenging, try to construct/ measure some rectangles. To find the answers to these questions, let us suppose that we have placed the diagonals such that their endpoints form the vertices of a rectangle, as shown in Fig. 2. A D B C Fig. 2 AC = 8 cm 83 Ganita Prakash | Grade 8 A B Deduction 1— What is the length of the other diagonal? This can be deduced using congruence as follows— Since ABCD is a rectangle, we have AB = CD ∠BAD = ∠CDA = 90° AD is common to both triangles. So, ∆ADC ≅ ∆DAB by the SAS congruence condition. Therefore, AC = BD, since they are corresponding parts of congruent triangles. This shows that the diagonals of a rectangle always have the same length. So the other diagonal must also be 8 cm long. You can verify this property by constructing/measuring some rectangles. Deduction 2—What is the point of intersection of the two diagonals? This can also be found using congruence. Since we need to know the relation between OA and OC, and OB and OD, which two triangles of the rectangle ABCD should we consider? O D C A B O 1 Common side A B D C D C Chapter 4 Quadrilaterals 06-07-2025.indd 84 7/10/2025 3:19:33 PM 84 A B The blue angles are equal since they are vertically opposite angles. 1 3 Since ∠B = 90°, ∠3 + ∠1 = 90°. In ∆BCD, since ∠3 + ∠2 + 90 = 180, we have ∠3 + ∠2 = 90°. O D C A B In order to show congruence, consider ∠1 and ∠2. Are they equal? 3 O 2 2 D C So, ∠1 = ∠2 (= 90° – ∠3). Thus, by the AAS condition for congruence, ∆AOB ≅ ∆COD. Hence OA = OC and OB = OD, since they are corresponding parts of congruent triangles. So, O is the midpoint of AC and BD. This shows that the diagonals of a rectangle always intersect at their midpoints. Therefore, to get a rectangle, the diagonals must be drawn so that they are equal and intersect at their midpoints. When the diagonals cross at their midpoints, we say that the diagonals bisect each other. Bisecting a quantity means dividing it into two equal parts. Verify this property by constructing some rectangles and measuring their diagonals and the points of intersection. Can the following equalities be used to establish that ∆AOD ≅ ∆COB? Deduction 3— What are the angles between the diagonals? Let us check what quadrilateral we get if we draw the two diagonals such that their lengths are equal, they bisect each other and have an arbitrary angle, say 60°, between them as shown in the figure to the right. Can you find all the remaining angles? A B AO = CO (proved above) ∠AOB = ∠COD (vertically opposite angles) AD = CB A B Quadrilaterals 60° O Math Talk Chapter 4 Quadrilaterals 06-07-2025.indd 85 7/10/2025 3:19:33 PM D C In ∆AOB, since OA = OB, the angles opposite them are equal, say a. Can you find the value of a? 60° 120° 120° O 60° We can find the remaining angles between the diagonals using our understanding of vertically opposite angles and linear pairs. a a A B D C D C 60° 120° 120° 60° O 85 Ganita Prakash | Grade 8 In ∆AOB, we have, a + a + 60 = 180 (interior angles of a triangle). Therefore 2a = 120. Thus a = 60. Similarly, we can find the values of all the other angles. Can we now identify what type of quadrilateral ABCD is? Notice that its angles all add up to 90° (30° + 60°). What can we say about its sides? A B 60° O 60° A B D C 30° 30° 30° 30° 60° 120° 120° 60° 60° 60° 60° O 60° A B 120° 120° O Chapter 4 Quadrilaterals 06-07-2025.indd 86 7/10/2025 3:19:33 PM 86 We can see that ∆AOB ≅ ∆COD and ∆AOD ≅ ∆COB. Hence, AB = CD, and AD = CB, since they are corresponding parts of congruent triangles. Therefore, ABCD is a rectangle since it satisfies the definition of a rectangle. Will ABCD remain a rectangle if the angles between the diagonals are changed? Can we generalise this? Take one of the angles between the diagonals as x. D C D C D C A x O B A B D C What is the value of a (in degrees) in terms of x? We have, Similarly, in the isosceles ∆AOD, let the base angles be b. b + b + 180 – x = 180 Since we know that ∆AOB is isosceles, we can denote the measures of both of its base angles by a. x 180° – x 180° – x a + a + x = 180 (sum of the interior angles of a triangle) x a = (180 – x) 2 = 90 – x 2 . 2b = 180 – (180 – x) 2b = 180 – 180 + x 2a = 180 – x We can compute the four angles between the diagonals to be x, x, 180 – x, and 180 – x. Can you find the other angles? A B A B D C x 2 a a b b x 180° – x 180° – x O 90°– x 2 90°– x 2 90°– x 2 Quadrilaterals x x 2 Chapter 4 Quadrilaterals 06-07-2025.indd 87 7/10/2025 3:19:33 PM All the angles of the quadrilateral are a + b, which is Thus, all four angles of the quadrilateral ABCD are 90°. What can we say about AB and CD, and AD and BC? We have ∆AOB ≅ ∆COD and ∆AOD ≅ ∆COB. Hence, AB = CD, and AD = CB, since they are the corresponding parts of congruent triangles. 90 – x 2 + x 2 = 90. 2b = x b = x 2 . D C 180° – x 180° – x x 2 O x x 90°– x 2 x 2 87 Ganita Prakash | Grade 8 Hence, no matter what the angles between the diagonals are, if the diagonals are equal and they bisect each other, then the angles of the quadrilateral formed are 90° each, and the opposite sides are equal. Thus, the quadrilateral is a rectangle. Now we know how the wooden strips have to be put together to form the vertices of a rectangle! They should be equal and connected at their midpoints. As we have been seeing from lower grades, properties of geometric objects such as parallel lines, angles, and triangles can be deduced through geometric reasoning. We will continue to deduce properties of special types of quadrilaterals in this chapter. Once you have deduced a property of a quadrilateral, it is good to verify it with a real-world quadrilateral, either the quadrilateral constructed on paper or simply a surface having the shape of the quadrilateral. If you are not able to figure out the property using deduction, you could experiment by taking real-world quadrilaterals and observing the property through measurement. Note that these observations give useful insights about the property, but with them, we can only form a conjecture, that is, a statement about which we are highly confident, but not yet sure if it always holds true. For example, constructing a few rectangles and observing through measurement that their diagonals bisect each other does not necessarily mean that this will always be the case—can we be sure that the 1000th rectangle we construct will also have this property? The only way we can be sure of this property is by justifying or proving the statement, just as we did in Deduction 2. A O C O C O 4 cm 4 cm 4 cm 4 cm 4 cm 4 cm This method is actually used in practice to make rectangles. Carpenters in Europe use this method to get a rectangular frame. It is also known that farmers in Mozambique, a country in Africa, use this method while constructing houses to get the base of the house in a rectangular shape. The Process of Finding Properties A C 4 cm D 4 cm 4 cm 4 cm B A D B Chapter 4 Quadrilaterals 06-07-2025.indd 88 7/10/2025 3:19:33 PM 88 Note to the Teacher: Gently encourage students to deduce or justify properties. Whenever students face challenges in doing it, encourage them to experiment and observe, and use their intuition to figure out the properties. The Carpenter’s Problem shows that rectangles can also be defined as follows— Rectangle: A rectangle is a quadrilateral whose diagonals are equal and bisect each other. Observe how different this definition is from the earlier one. Yet, both capture the same class of quadrilaterals. Further, it turns out that the first definition can be simplified. In the earlier definition, we stated that a rectangle has (a) opposite sides of equal length, and (b) all angles equal to 90°. Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are 90°? If you think that this definition is incomplete, try constructing a quadrilateral in which the angles are all 90° but the opposite sides are not equal. Are you able to construct such a quadrilateral? Let us prove why this is impossible. Deduction 4—What is the shape of a quadrilateral with all the angles equal to 90°? Consider a quadrilateral ABCD with all angles measuring 90°. What can we say about the opposite sides of such a quadrilateral? Join BD. ∆BAD and ∆DCB seem congruent. Can we justify this claim? A B 1 Common side D C Two equalities can be directly seen in the triangles. What can we say about ∠1 and ∠2? Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 89 7/10/2025 3:19:33 PM A B Recall that we tackled a very similar problem in Deduction 2. We can use the same reasoning here. 1 Since ∠B = 90°, ∠3 + ∠1 = 90°. In ∆BCD, since ∠3 + ∠2 + 90° = 180°, ∠3 + ∠2 = 90°. 3 2 D C A B 3 2 D C 89 Ganita Prakash | Grade 8 So, ∠1 = ∠2. Thus, by the AAS congruence condition, ∆BAD ≅ ∆DCB. Therefore, AD = CB, and DC = BA, since these are corresponding sides of congruent triangles. Is it wrong to write ∆BAD ≅ ∆CDB? Why? Thus, we have established that if all the angles of a quadrilateral are right angles, then the opposite sides have equal lengths. Therefore, the quadrilateral is a rectangle. Thus, a rectangle can simply be defined as follows — Rectangle: A rectangle is a quadrilateral in which the angles are all 90°. Let us list the properties of a rectangle. Property 1: All the angles of a rectangle are 90°. Property 2: The opposite sides of a rectangle are equal. Are the opposite sides of a rectangle parallel? They definitely seem so. This fact can be justified using one of the transversal properties. Notice that AB acts as a transversal to AD and BC, and that ∠A + ∠B = 90° + 90° = 180°. When the sum of the internal angles on the same side of the transversal is 180°, the lines are parallel. We can use this fact to conclude that the lines AD and BC are parallel, which we represent as AD || BC. Can you similarly show that AB is parallel to DC (AB||DC)? Property 3: The opposite sides of a rectangle are parallel to each other. A B D C Chapter 4 Quadrilaterals 06-07-2025.indd 90 7/10/2025 3:19:33 PM 90 Property 4: The diagonals of a rectangle are of equal length and they bisect each other. A Special Rectangle In the quadrilaterals below, are there any non-rectangles? 5 cm 2 cm 6 cm (i) (ii) (iii) (iv) 5 cm 5 cm 1 cm 2 c m 3.6 cm 3.6 cm 6 c m 5 cm 1 cm 4 cm 4 c m 4 c m 4 cm All these quadrilaterals are rectangles, including (iv). Quadrilateral (iv) is a rectangle because all its angles are 90°. However, it is a special kind of rectangle with all sides of equal length. We know that this quadrilateral is also called a square. Square: A square is a quadrilateral in which all the angles are equal to 90°, and all the sides are of equal length. Thus, every square is also a rectangle, but clearly every rectangle is not a square. This relation can be pictorially represented using a Venn diagram. We have seen these diagrams before. In a Venn diagram, a set of objects is represented as points inside a closed curve. Typically, these closed curves are ovals or circles. For example, the set of all squares is represented as I am an Indian and I am a Malayali. Square Wait, which one of them are you — an Indian or a Malayali? How can you be both? Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 91 7/10/2025 3:19:33 PM Each point in the region represents a square, thereby covering all the possible squares. Since every square is a rectangle, the Venn diagram representation of these two sets would be as follows — Square Rectangle 91 Ganita Prakash | Grade 8 Let us consider the Carpenter’s Problem again. If the wooden strips have to be placed such that the thread passing through their endpoints forms a square, what must be done? What more needs to be done to get equal sidelengths as well? Can this be achieved by properly choosing the angle between the diagonals? See if you can reason and/or experiment to figure this out! Deduction 5— What should be the angle formed by the diagonals? The angle between the diagonals can be found using the notion of congruence! Suppose we join the equal diagonals such that they bisect each other and result in a square. Let us label the square ABCD. To find the angle formed by the diagonals, what are the two triangles we should consider for congruence? A A B O O Common Side D C C B As in the previous case, let us try to construct a square, one of whose diagonals is of length 8 cm. While solving the Carpenter’s Problem for the case of a rectangle, we have seen that to get a quadrilateral with all angles 90° (and opposite sides of equal length), the diagonals have to be drawn such that — (i) they are of equal lengths, and (ii) they bisect each other. Chapter 4 Quadrilaterals 06-07-2025.indd 92 7/10/2025 3:19:33 PM 92 Can this be used to find the angles ∠BOA and ∠BOC formed by the diagonals? Since these angles are corresponding parts of congruent triangles, they are equal. Further, these angles together form a straight angle. So ∠BOA + ∠BOC = 180°. Thus, these angles have to be 90° each. D By the SSS condition for congruence, ∆BOA ≅ ∆BOC This shows that the diagonals of a square bisect each other at right angles. This means that the diagonals have to be drawn such that they are of equal lengths and bisect each other at right angles. Since the endpoints of the diagonals uniquely determine the vertices of a quadrilateral, we will get a square when the diagonals are joined this way. Using this fact, construct a square with a diagonal of length 8 cm. Properties of a Square Since a square is a special type of rectangle, all the properties of a rectangle hold true for a square. Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well. Property 1: All the sides of a square are equal to each other. Property 2: The opposite sides of a square are parallel to each other. Property 3: The angles of a square are all 90°. Property 4: The diagonals of a square are of equal length and they bisect each other at 90°. There is one more special property of a square. What are the measures of ∠1, ∠2, ∠3, and ∠4? See if you can reason and/or experiment to figure this out! A B 1 2 In ∆ADC, we have, ∠1 + ∠3 + 90 = 180 Since AD = DC, we have ∠1 = ∠3. Thus, ∠1 = ∠3 = 45°. Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 93 7/10/2025 3:19:33 PM Similarly, find ∠2 and ∠4. Thus, we have another property of a square — Property 5: The diagonals of a square divide the angles of the square into equal halves. This can also be expressed as—The diagonals of a square bisect the angles of the square. D C 3 4 93 Ganita Prakash | Grade 8 Figure it Out 1. Find all the other angles inside the following rectangles. 2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of 3. Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out. 4. We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these? 5. We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle? 4.2 Angles in a Quadrilateral (i) 30° (ii) 40° (iii) 90° (iv) 140° P Q D C A B 30° (i) (ii) S R 110° Math Talk Try This Chapter 4 Quadrilaterals 06-07-2025.indd 94 7/10/2025 3:19:33 PM 94 Is it possible to construct a quadrilateral with three angles equal to 90° and the fourth angle not equal to 90°? You might have observed through constructions that this may not be possible. But why not? This is due to a general property of quadrilaterals related to their angles. We have seen that the sum of the angles of a triangle is 180°. There is a similar regularity in the sum of the angles of a quadrilateral. Consider a quadrilateral SOME. Draw a diagonal SM. We get two triangles ∆SEM and ∆SOM. In ∆SEM, we have ∠1 + ∠2 + ∠3 = 180°. And in ∆SOM, ∠4 + ∠5 + ∠6 = 180°. What do we get when we add all six angles? We will have ∠1+ ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180° = 360°. Or, (∠1+ ∠4) + (∠3 + ∠6) +∠2 + ∠5 = 360°. Since (∠1 + ∠4), (∠3 + ∠6), ∠2 and ∠5 are the angles of this quadrilateral, we have the following result— The sum of all angles in any quadrilateral is 360°. This explains why it is impossible for a quadrilateral to have three right angles, with the fourth angle not right angle. 4.3 More Quadrilaterals with Parallel Opposite Sides Rectangles (and therefore squares) have parallel opposite sides. Are there quadrilaterals that have parallel opposite sides that are not rectangles? Let us try constructing one. This can be easily done by drawing two pairs of parallel lines, ensuring that they do not meet at right angles. Construct such a figure by recalling how parallel lines can be constructed using a ruler and a set-square, or a compass and a ruler. E M 2 3 S 1 4 5 Quadrilaterals O 6 Chapter 4 Quadrilaterals 06-07-2025.indd 95 7/10/2025 3:19:34 PM Observe the quadrilateral ABCD. It has parallel opposite sides but is not a rectangle. Thus, a larger set of quadrilaterals exists in which the opposite sides are parallel. Such quadrilaterals are called parallelograms. A B D C 95 Ganita Prakash | Grade 8 Is a rectangle a parallelogram? A rectangle has opposite sides parallel. So, it satisfies the parallelogram’s definition. Hence, it is indeed a parallelogram. More specifically, a rectangle is a special kind of parallelogram with all its angles equal to 90°. Let us represent this relation using a Venn diagram. To understand the relations between the sides and the angles of a parallelogram, let us construct the following figure. Draw a parallelogram with adjacent sides of lengths 4 cm and 5 cm, and an angle of 30° between them. 30° Square Rectangle Parallelogram A 4 cm B D C 5 cm 5 cm 4 cm B D Chapter 4 Quadrilaterals 06-07-2025.indd 96 7/10/2025 3:19:34 PM 96 ABCD is the required parallelogram. What are the remaining angles of the parallelogram? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out. Step 1: Draw line segments AB = 4 cm and AD = 5 cm with an angle of 30° between them. Step 2: Draw a line parallel to AB through the point D, and a line parallel to AD through B. Mark the point at which these lines intersect as C. A 4 cm B D C 30° 5 cm A 30° Deduction 6— What can we say about the angles of a parallelogram? In the parallelogram ABCD, AB||CD, and AD is a transversal to them. ∠A + ∠D = 180° (sum of the internal angles on the same side of a transversal). Therefore, ∠D = 180 – ∠A = 180 – 30 = 150°. Similarly, AD||BC, and AB and CD are transversals to them. So, ∠C + ∠D = 180°. Using these equations, we get ∠B = 150° and ∠C = 30°. We see that in this parallelogram, the adjacent pairs of angles add up to 180° and opposite pairs of angles are equal. Thus, And, Since the adjacent angles are the interior angles on the same side of a transversal to a pair of parallel lines, they must add up to 180°. What about the opposite angles? Will they be equal in all parallelograms? If yes, how can we be sure? Let us take one of the angles to be x. What are the other angles? Since ∠P + ∠R = 180°, ∠R = 180 – ∠P = 180 – x. Similarly, since ∠A + ∠R = 180°, ∠A = 180 – ∠R = 180 – (180–x) = 180 – 180 + x = x. Thus, ∠P = ∠A = x. Similarly, we can deduce that ∠ R = ∠E = 180 – x. Therefore, this shows that the opposite angles of a parallelogram are always equal. ∠A + ∠B = 180°, ∠A + ∠D = 180°, ∠C + ∠D = 180°, and ∠B + ∠C = 180°. ∠A = ∠C, and ∠B = ∠D. So, ∠A + ∠B = 180°. A 4 cm B D C 30° 5 cm R A Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 97 7/10/2025 3:19:34 PM Deduction 7— What can we say about the sides of a parallelogram? By looking at a parallelogram, it appears that the opposite sides are equal. Can we again use congruence to show this? Which two triangles can be considered for this? P E x 97 Ganita Prakash | Grade 8 In ∆ABD and ∆CDB, the angles marked with a single arc are equal as they are the opposite angles of a parallelogram. Since AD||BC, and BD is a transversal to it, the angles marked with double arcs are equal as they are alternate angles. So, by the AAS condition, the triangles are congruent, that is, ∆ABD ≅ ∆CDB. Therefore, AD = CB, and AB = CD. Thus, the opposite sides of a parallelogram are equal. Is it wrong to write ∆ABD ≅ ∆CBD? Why? From these deductions we can find the remaining sides and angles of the parallelogram. Let us list the properties of a parallelogram. Property 1: The opposite sides of a parallelogram are equal. Property 2: The opposite sides of a parallelogram are parallel. Property 3: In a parallelogram, the adjacent angles add up to 180°, and the opposite angles are equal. A 4 cm B D C 30° 5 cm 150° 150° A B 4 cm 30° 5 cm D C Common Side Chapter 4 Quadrilaterals 06-07-2025.indd 98 7/10/2025 3:19:34 PM 98 Are the diagonals of a parallelogram always equal? Check with the parallelogram that you have constructed. We see that the diagonals of a parallelogram need not be equal. Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out. Deduction 8— What is the point of intersection of the two diagonals in a parallelogram? As in the case of a rectangle, we can find out if the diagonals bisect each other by examining the congruence of ∆AOE and ∆YOS in the parallelogram EASY. AE = YS (as they are the opposite sides of the parallelogram) The angles marked using a single arc are equal, and so are the angles marked using a double arc, since they are alternate angles of parallel lines. Thus, by the ASA condition, the triangles are congruent, that is, ∆AOE ≅ ∆YOS. Therefore, OA = OY, and OE = OS, since they are corresponding parts of congruent triangles. Thus, O is the midpoint of both diagonals. Is it wrong to write ∆AOE ≅ ∆SOY? Why? Property 4: The diagonals of a parallelogram bisect each other. Do the diagonals of a parallelogram intersect at a particular angle? 4.4 Quadrilaterals with Equal Sidelengths Are squares the only quadrilaterals that have equal sidelengths? Let us explore this question through construction. Draw two equal sides AD and AB, that are not perpendicular to each other. Can we complete this quadrilateral so that all its sides are of the same length? E Y A B A S 50° O D Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 99 7/10/2025 3:19:34 PM Mark a point C whose distance from B and D is equal to AB (or AD). To do this, measure AB using a compass. Keeping this length as the radius, cut arcs from B and D. Now we have a quadrilateral with equal sidelengths and one of its angles 50°. Note that we could have constructed such a quadrilateral by taking any angle less than 180° (in place of 50°). A quadrilateral in which all the sides have the same length is a rhombus. A B 50° D C 99 Ganita Prakash | Grade 8 What are the other angles of the rhombus ABCD that we have constructed? Reason and/or experiment to figure this out. Deduction 9— What can we say about the angles in a rhombus? Consider a rhombus GAME. In ∆GAE, since GE = GA, a = d. Similarly, in ∆MAE, since ME = MA, b = c. It can be seen that ∆GAE ≅ ∆MAE (How?) So, a = b, c = d and ∠G = ∠M (since they are corresponding parts of congruent triangles). Thus, we have, a = b = c = d. These facts hold for any rhombus. Let us apply them to the rhombus ABCD that we constructed earlier. Let the four equal angles formed by the diagonal be a, as shown in the figure Thus, the angles of the rhombus ABCD are 50°, 130°, 50°, and 130°. So, in a rhombus opposite angles are equal to each other. Interestingly, there is one more way by which we could have figured out the other angles of the rhombus ABCD. We have shown that in a general rhombus GAME, the four angles formed by a diagonal are equal to each other. In ∆ADB, we have a + a + 50 = 180°. So, a = 65°. A B a a 50° D C a a G A 50° E M a b d c D C 65° 65° 65° 65° 50° Chapter 4 Quadrilaterals 06-07-2025.indd 100 7/10/2025 3:19:34 PM 100 Consider the lines EM and GA and its transversal AE. Since the alternate angles are equal, EM||GA. G A E M a a Similarly consider the lines GE and AM and its transversal AE. Since the alternate angles are equal, GE||AM. G A A B 50° E M a a As opposite sides are parallel, GAME is also a parallelogram. Thus, every rhombus is a parallelogram, and the properties of a parallelogram hold true for a rhombus as well. Thus, the adjacent angles of a rhombus add up to 180°, and the opposite angles are equal (verify that the arguments in Deduction 6 can be applied to a rhombus as well!). Thus, in rhombus ABCD, So a rhombus is a parallelogram, and a rectangle is also a parallelogram. How can this be represented using a Venn diagram? Where will the set of squares occur in this diagram? We know that a square is a rectangle. Since the opposite sides of a square are parallel, a square is also a parallelogram. Further, since all the sides of a square have the same length, a square is also a rhombus. Thus, the Venn diagram will be as follows. Rectangle Square Rhombus Rectangle Rhombus ∠D = ∠B = 180 – 50 = 130°. ∠A = ∠C = 50°, and A B 50° Parallelogram D C Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 101 7/10/2025 3:19:34 PM Let us list the properties of a rhombus. Property 1: All the sides of a rhombus are equal to each other. Property 2: The opposite sides of a rhombus are parallel to each other. Property 3: In a rhombus, the adjacent angles add up to 180°, and the opposite angles are equal. Are the diagonals of a rhombus equal? Property 4: The diagonals of a rhombus bisect each other. Property 5: The diagonals of a rhombus bisect its angles. Square 101 Ganita Prakash | Grade 8 Do the diagonals of a rhombus intersect at any particular angle? Reason out and/or experiment to figure this out! Deduction 10—What can we say about the angles formed by the diagonals of a rhombus at their point of intersection? In the rhombus GAME, we have ∆GEO ≅ ∆MEO (why?). So, ∠GOE = ∠MOE, as they are corresponding parts of congruent triangles. As they add up to 180°, they should be 90° each. Property 6: Diagonals of a rhombus intersect each other at an angle of 90°. Figure it Out 1. Find the remaining angles in the following quadrilaterals. P E 40° (i) (ii) (iii) X W R A G A (iv) O I S R E M P Q 110° O Chapter 4 Quadrilaterals 06-07-2025.indd 102 7/10/2025 3:19:34 PM 102 2. Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°. 3. Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm. U V 30° A E 20° 4.5 Playing with Quadrilaterals Geoboard Activity Take a geoboard and some rubber bands. If you do not have these, you could just use the dot grid papers given at the end of the book for this activity. Place two rubber bands perpendicular to each other, forming diagonals of equal length. Join the ends. Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 103 7/10/2025 3:19:34 PM What is the quadrilateral that you get? Justify your answer. Extend one of the diagonals on both sides by 2 cm. What quadrilateral will you get now? Justify your answer. 103 Ganita Prakash | Grade 8 Joining Triangles 1. Take two cardboard cutouts of an equilateral triangle of sidelength 8 cm. 2. Take two cardboard cutouts of an isosceles triangle with sidelengths 8 cm, 8 cm, and 6 cm. Can you join them to get a quadrilateral? What type of a quadrilateral is this? Justify your answer. 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm 8 cm Chapter 4 Quadrilaterals 06-07-2025.indd 104 7/10/2025 3:19:34 PM 104 What are the different ways they can be joined to get a quadrilateral? What quadrilaterals are these? Justify your answers. Joining them in this way you get Joining them in this way you get 8 cm 8 cm 8 cm 8 cm 6 cm 6 cm 8 cm 8 cm 6 cm 6 cm 6 cm 8 cm 3. Take two cardboard cutouts of a scalene triangle with sides 6 cm, 9 cm, and 12 cm. 4.6 Kite and Trapezium Kite One of the ways the two triangles of sides 6 cm, 9 cm and 12 cm can be joined together is as follows — What are the different ways they can be joined to get a quadrilateral? Are you able to identify the different quadrilaterals that are obtained by joining the triangles? Justify your answer whenever you identify a quadrilateral. 6 cm 9 cm 12 cm 9 cm 6 cm 12 cm 6 cm 6 cm 9 cm 9 cm 12 cm Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 105 7/10/2025 3:19:34 PM This quadrilateral looks like a kite. Observe that the adjacent sides are of the same length. Kite: A kite is a quadrilateral that can be labelled ABCD such that AB = BC, and CD = DA. Property 1: In the kite, show that the diagonal BD (i) bisects ∠ABC and ∠ADC, (ii) bisects the diagonal AC, that is, AO = OC, and is perpendicular to it. Hint: Is ∆AOB ≅ ∆COB? A B D O 105 C Ganita Prakash | Grade 8 Trapezium Parallelograms are quadrilaterals that have parallel opposite sides. We get a new type of quadrilateral if we relax this condition. Trapezium: A trapezium is a quadrilateral with at least one pair of parallel opposite sides. Construct a trapezium. Measure the base angles (marked in the figure). Can you find the remaining angles without measuring them? Since PQ||SR, we have Property 1: ∠ S + ∠P = 180° and ∠R + ∠Q = 180°. Using these facts, the remaining angles can easily be found. Verify your answer after finding them. When the non-parallel sides of a trapezium have the same lengths, the trapezium is called an isosceles trapezium. How do we construct an isosceles trapezium? Construct an isosceles trapezium UVWX, with UV||XW. Measure ∠U. X W S R X W P Q Chapter 4 Quadrilaterals 06-07-2025.indd 106 7/10/2025 3:19:34 PM 106 Can you find the remaining angles without measuring them? Does it appear that the angles opposite to the equal sides—∠U and ∠V—are also equal? Can we find congruent triangles here? Consider line segments XY and WZ perpendicular to UV. What type of quadrilateral is XWZY? Since XW||UV, a = 180° – ∠XYZ = 90°, and b = 180° – ∠WZY = 90° (since the internal angles on the same side of a transversal add up to 180°) Hence, XWZY is a rectangle. U V Mark X and W such that UX = VW U V U Y Z X W a b V Now, it can be shown that ∆UXY ≅ ∆VWZ. (How?) Thus, ∠U = ∠V. Using this fact, the remaining angles of the isosceles trapezium can be determined. Verify the angles by measurement. Property 2: In an isosceles trapezium, the angles opposite to the equal sides are equal. Figure it Out 1. Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm. 2. Construct a kite whose diagonals are of lengths 6 cm and 8 cm. 3. Find the remaining angles in the following trapeziums— 4. Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions— (i) What is the quadrilateral that is both a kite and a parallelogram? 135° 105° 100° Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 107 7/10/2025 3:19:34 PM (iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals? 5. If PAIR and RODS are two rectangles, find ∠IOD. (ii) Can there be a quadrilateral that is both a kite and a rectangle? 5 cm P A I D R O 5 cm 30° S 107 Ganita Prakash | Grade 8 10. Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring. 6. Construct a square with diagonal 6 cm without using a protractor. 7. CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b). 8. If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement. 9. What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer. Hint: Draw a diagonal and check for congruent triangles. V C U A E S (a) (b) W X Try This Chapter 4 Quadrilaterals 06-07-2025.indd 108 7/10/2025 3:19:34 PM 108 11. State whether the following statements are true or false. Justify your answers. (i) A quadrilateral whose diagonals are equal and bisect each other must be a square. A B D C A rectangle is a quadrilateral in which the angles are all 90°. A square is a quadrilateral in which all the angles are 90°, and all the sides are of equal length. (vii) Isosceles trapeziums are parallelograms. (iii) A quadrilateral whose diagonals bisect each other must be a parallelogram. (iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus. (vi) A quadrilateral in which all the angles are equal is a rectangle. (ii) A quadrilateral having three right angles must be a rectangle. (v) A quadrilateral in which the opposite angles are equal must be a parallelogram. Properties of a rectangle — Properties of a square — y The opposite sides of a square are parallel to each other. y Opposite sides of a rectangle are equal. y Opposite sides of a rectangle are parallel to each other. y Diagonals of a rectangle are of equal length and they bisect each other. y The diagonals of a square are of equal lengths and they bisect each other at 90°. SUMMARY Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 109 7/10/2025 3:19:34 PM A parallelogram is a quadrilateral in which opposite sides are parallel. A rhombus is a quadrilateral in which all the sides have the same length. Properties of a parallelogram — y The diagonals of a square bisect the angles of the square. y The opposite sides of a parallelogram are equal. y In a parallelogram, the adjacent angles add up to 180°, and the opposite angles are equal. y The diagonals of a parallelogram bisect each other. 109 Ganita Prakash | Grade 8 A kite is a quadrilateral with two non-overlapping adjacent pairs of sides having the same length. A trapezium is a quadrilateral having at least one pair of parallel opposite sides. The sum of the angle measures in a quadrilateral is 360°. Properties of a rhombus — Kite y The opposite sides of a rhombus are parallel to each other. y In a rhombus, the adjacent angles add up to 180°, and the opposite angles are equal. y The diagonals of a rhombus bisect each other at right angles. y The diagonals of a rhombus bisect its angles. Trapezium Rectangle Parallelogram Chapter 4 Quadrilaterals 06-07-2025.indd 110 7/10/2025 3:19:34 PM 110 Square Rhombus Gameplay 1. Fold a sheet into half. 3. Make a triangular crease at the corner that is at the middle of the paper. 5. How would you fold the quarter paper to get the kinds of creases shown in the following image. 2. Now, fold it once more into a quarter. 4. Open the sheet. What is the shape formed by the creases? Which Quad? Quadrilaterals Chapter 4 Quadrilaterals 06-07-2025.indd 111 7/10/2025 3:19:37 PM 6. How would you fold the quarter paper such that a square is formed? 111" class_8,5,number play,ncert_books/class_8/hegp1dd/hegp105.pdf,"5 NUMBER PLAY 5.1 Is This a Multiple Of? Sum of Consecutive Numbers Anshu is exploring sums of consecutive numbers. He has written the following— Now, he is wondering— • “Can I write every natural number as a sum of consecutive numbers?” 7 = 3 + 4 10 = 1 + 2 + 3 + 4 12 = 3 + 4 + 5 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 Chapter 5 Letter Number Play 06-07-2025.indd 112 10-07-2025 15:07:56 Explore these questions and any others that may occur to you. Discuss them with the class. Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in between the numbers. How many different possibilities exist? Write all of them. • “Which numbers can I write as the sum of consecutive numbers in more than one way?” • “Ohh, I know all odd numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?” • “Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.” 3 + 4 – 5 + 6 3 – 4 – 5 – 6 Math Talk Eight such expressions are possible. You can use the diagram below to systematically list all the possibilities. Evaluate each expression and write the result next to it. Do you notice anything interesting? Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have done before. Find out the results of each expression. What do you observe? Repeat this for one more set of 4 consecutive numbers. Share your findings. 3 + 4 – 5 + 6 = 8 3 – 4 – 5 – 6 = – 12 . . . 3 + – 4 4 – + + – 5 5 5 5 5 + 6 – 7 + 8 = 12 5 – 6 – 7 – 8 = – 16 . . . – + + + + – – – 6 6 6 6 6 6 6 6 3 + 4 + 5 + 6 3 + 4 + 5 – 6 __ + __ – __ + __ = __ __ – __ – __ – __ = __ . . . Number Play Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 113 10-07-2025 15:07:57 Some sums appear always no matter which 4 consecutive numbers are chosen. Isn’t that interesting? Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning? Hint: Use algebra and describe the 8 expressions in a general form. You might have noticed that the results of all expressions are even numbers. Even numbers have a factor of 2. Negative numbers having a factor 2 are also even numbers, for example, – 2, – 4, – 6, and so on. Check if anyone in your class got an odd number. When 4 consecutive numbers are chosen, no matter how the ‘+’ and ‘–’ signs are placed between them, the resulting expressions always have even parity. 113 Ganita Prakash | Grade 8 Now take any 4 numbers, place ‘+’ and ‘–’ signs in the eight different ways, and evaluate the resulting expression. What do you observe about their parities? Repeat this with other sets of 4 numbers. Is there a way to explain why this happens? Hint: Think of the rules for parity of the sum or difference of two numbers. Explanation 1: Let us consider any of the 8 expressions formed by four numbers a, b, c, and d. When one of its signs is switched, its value always increases or decreases by an even number! Let us see why. Consider one of the expressions: a + b – c – d. Replacing +b by – b, we get = 2b (this is an even number). If the difference between two numbers is even, can they have different parities? No! So either both are even or both are odd. Now, let us see what happens when a negative sign is switched to a positive sign. Replace any negative sign in the expression a + b – c – d with a positive sign and find the difference between the two numbers. What do you conclude from this observation? = a + b – c – d – a + b + c + d (notice how the signs changed when we opened the second set of brackets) a – b – c – d. By how much has the number changed? It has changed by (a + b – c – d) – (a – b – c – d) Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 114 10-07-2025 15:07:57 114 Starting from any expression, we can get 7 expressions by switching one or more ‘+’ and ‘–’ signs. Thus, all the expressions have the same parity! Explanation 2: We know that odd ± odd = even We have seen that the parity of a + b and a – b is the same, regardless of the parities of a and b. In short, a ± b have the same parity. By the same argument, a ± b + c and a ± b – c have the same parity. Extending this further, we can say that all the expressions a ± b ± c ± d have the same parity. even ± even = even odd ± even = odd. Explanation 3: This can also be explained using the positive and negative token model you studied in the chapter on Integers. Try to think how. The number of ways to choose 4 numbers a, b, c, d and combine them using ‘+’ and ‘–’ signs is infinite. Mathematical reasoning allows us to prove that all the combinations a ± b ± c ± d always have the same parity, without having to go through them one by one. Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think? Breaking Even We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even. Several problems in mathematics can be thought about and solved in different ways. While the method you came up with may be dear to you, it can be amusing and enriching to know how others thought about it. Two tidbits: ‘share’ and ‘listen’. ‘What if …?’, ‘Will it always happen?’— Wondering and posing questions and conjectures is as much a part of mathematics as problem solving. a + – b b – + + – c c c c Number Play – – – – + + + + d d d d d d d d Chapter 5 Letter Number Play 06-07-2025.indd 115 10-07-2025 15:07:58 Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers. 13k – 5k 6m – 3n b2 x + 1 4k × 3j 2 + 2 809 + 214 119 × 303 5133 543 – 479 2a + 2b 3g + 5h 4m + 2n 2u – 4v 43 + 37 672 – 348 4 × 347 × 3 708 – 477 115 Ganita Prakash | Grade 8 The expression 4m + 2q will always evaluate to an even number for any integer values of m and q. We can justify this in two different ways— • We know 4m is even and 2q is even for any integers m and q. Therefore, their sum will also be even. • The expression 4m + 2q is equal to the expression 2(2m + q). Here, the expression 2(2m + q) means 2 times 2m + q. In other words, 2 is a factor of this expression. Therefore, this expression will always give an even number for any integers m and q. For example, if m = 4 and q = – 9, the expression 4m + 2q becomes 4 × 4 + 2 × (–9) = – 2, which is an even number. In the expression x2 + 2, x2 is even if x is even, and x2 is odd if x is odd. Therefore, the expression x2 + 2 will not always give an even number. An example and a non-example for when the expression evaluates to an even number — (i) if x = 6, then x2 + 2 = 38, and (ii) if x = 3, then x2 + 2 = 11. Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression. Write a few algebraic expressions which always give an even number. Pairs to Make Fours Take a pair of even numbers. Add them. Is the sum divisible by 4? Try this with different pairs of even numbers. When is the sum a multiple of 4, and when is it not? Is there a general rule or a pattern? Even numbers can be of two types based on the remainders they leave when divided by 4. Chapter 5 Letter Number Play 06-07-2025.indd 116 10-07-2025 15:07:58 116 Even numbers that are multiples of 4 leave a remainder of 0 when divided by 4. . . . . . . . Even numbers that are not multiples of 4 leave a remainder 2 when divided by 4. Adding two (even) numbers that are multiples of 4 will always give a multiple of 4. Adding two even numbers that are not multiples of 4 will always give a multiple of 4 because their remainders of 2 add up to 4. When will two even numbers add up to give a multiple of 4? This problem is similar to the question of identifying when adding two numbers will result in an even number. Can you see this? There are three cases to examine: Explanation with Algebra and Visualisation Examples (4p + 2) and (4q + 2). (4p + 2) + (4q + 2) = 4p + 4q + 4 = 4 (p + q + 1). 4p and 4q. 4p + 4q = 4 (p + q). p . ro w s . . . . . . = = p r o w s q r o w s + q . r o w s . . + . . . . . . . . . . . . . . . p ro w s 1 row q r o w s (p + q) r o w s Number Play (p + q + 1 ) r o w s 4, 12, 16, 24, 36. 12 + 16 = 4 (3 + 4) = 28. 16 + 28 = 4 (4 + 7) = 44. 2, 6, 10, 18, 22, 42. 2 + 6 = 8. 6 + 10 = 16. 22 + 6 = 28. Chapter 5 Letter Number Play 06-07-2025.indd 117 10-07-2025 15:07:58 What happens when we add a multiple of 4 to an even number that is not a multiple of 4? Is it similar to the case of the parity of the sum of an even and an odd number? Look at the following expressions and the visualisation. Write the corresponding explanation and examples. . . . 117 Ganita Prakash | Grade 8 Notice how we are able to generalise and prove properties of arithmetic using algebra and also using visualisation. Always, Sometimes, or Never We examine different statements about factors and multiples and determine whether a statement is ‘Always True’, ‘Sometimes True’, or ‘Never True’. We know that the sum of any two multiples of 2 is also a multiple of 2. 1. If 8 exactly divides two numbers separately, it must exactly divide their sum. Explanation with Algebra and Visualisation Examples Explanation with Algebra and Visualisation Examples 4p and (4q + 2) = 4p + (4q + 2) = 4p + 4q + 2 = 4 (p + q) + 2. . . . . = p r o w s q r o w s + . . . . . . . . . . a re m ai n d er 2 (p + q ) r o w s wi th Chapter 5 Letter Number Play 06-07-2025.indd 118 10-07-2025 15:07:58 118 The two numbers have 8 as a factor; in other words, the two numbers are multiples of 8. As multiples of 8 are obtained by repeatedly adding 8, the sum of two multiples of 8 will also be a multiple of 8. Statement 1 is always true. Determine if it is true with subtraction. 8a and 8b. 8 and 16. 16 and 56. 80 and 120. 8a + 8b = 8 (a + b). a ro w s . . . . . . . . b r o w s (a + b) r o w s 8 + 16 = 8(1 + 2) = 24. 16 + 56 = 72. 80 + 120 = 200. A number divisible by 8 is a multiple of 8. A number divisible by 8 can be expressed as a sum of two multiples of 8 or sum of two nonmultiples of 8. Numbers divisible by 7 will have 7 as a factor. 7j 14 = 7 × 2 (j = 2). 42 = 7 × 6 (j = 6). 98 = 7 × 14 (j = 14). In general, if a divides M and a divides N, then a divides M + N and a divides M – N. In other words, if M and N are multiples of a, then M + N and M – N will also be multiples of a. 2. If a number is divisible by 8, then 8 also divides any two numbers (separately) that add up to the number. So, statement 2 is sometimes true. 3. If a number is divisible by 7, then all multiples of that number will be divisible by 7. Explanation with Algebra and Visualisation Examples Explanation with Algebra and Visualisation Examples 8m = 8a + 8b 8m = p + q (p, q not multiples of 8) 8m 8, 16, 56, 72. . . . . . . . . a ro w s b r o w s m = ( a +b ) ro w s . . p . . . . . q = 72 = 48 + 24 (8×9 = 8×6 + 8×3). 72 = 50 + 22 Number Play Chapter 5 Letter Number Play 06-07-2025.indd 119 10-07-2025 15:07:58 This contains a total of mj rows. So this is also a multiple of 7. The number 7jm or (7 × j × m) has a factor of 7. We can see that Statement 3 is always true. (7j) × m . . . . . . . . . . . . . . . . . . . j r o w s j ro w s j ro w s m ti m es Some multiples of 14: 28 = (7 × 2) × 2. 70 = (7 × 2) × 5. 154 = (7 × 2) × 11 119 A number divisible by 12 is a multiple of 12. 12m 12, 24, 36, 48, 108, 132. Factors of multiples of 12 will include factors of 12. Numbers divisible by 7 are multiples of 7. 7k . . . . . Ganita Prakash | Grade 8 In general, if A is divisible by k, then all multiples of A are divisible by k. 4. If a number is divisible by 12, then the number is also divisible by all the factors of 12. In general, if A is divisible by k, then A is divisible by all the factors of k. Hence, Statement 4 is always true. 5. If a number is divisible by 7, then it is also divisible by any multiple of 7. Explanation with Algebra and Visualisation Examples Explanation with Algebra and Visualisation Examples 12m = 2 × 6 × m = 3 × 4 × m . . . . . A factor of 12 covers a row fully. Hence, it covers all multiples of 12 fully. . . . . . . . . 12 k ro w s m r o w s Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Chapter 5 Letter Number Play 06-07-2025.indd 120 10-07-2025 15:07:58 120 Multiples of 7. 7k will be divisible by 7m if and only if m is a factor of k. We can see that this statement is only sometimes true. If k = ym then 7k ÷ 7m = 7ym ÷ 7m = y 7m . . . . . . . . . . . . . . . . . . . m r o w s m r o w s m r o w s k ro w s 42 (7 × 6) is divisible by 7 but it is not divisible by 28 (7 × 4). 42 (7 × 6) is divisible by 7 and it is divisible by 14 (7 × 2). Examine each of the following statements, and determine whether it is ‛Always true’, ‛Sometimes true’, ‛Never true’. 6. If a number is divisible by both 9 and 4, it must be divisible by 36. 7. If a number is divisible by both 6 and 4, it must be divisible by 24. In general, if A is divisible by k and A is also divisible by m, then A is divisible by the LCM of k and m. This is because A is a multiple of k and also a multiple of m, so A’s prime factorisation should contain the prime factorisation of LCM (k, m). 8. When you add an odd number to an even number we get a multiple of 6. We know that multiples of 6 are all even numbers. The sum of an odd number and an even number will be an odd number. Therefore, this statement is never true. We can also explain this algebraically. Suppose, (2n) + (2m + 1) = 6j, where 2n is an even number, 2m + 1 is an odd number, and 6j is a multiple of 6. Then 2n + 2m = 6j – 1 2 (n + m) = 6j – 1 which means 2(n + m), which is an even number, should be equal to 6j – 1, which is an odd number. This is never true. What Remains? Find a number that has a remainder of 3 when divided by 5. Write more such numbers. Can I write an even and an odd number as 2n and 2n+1 instead? Number Play Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 121 10-07-2025 15:07:58 Which algebraic expression(s) capture all such numbers? (i) 3k + 5 (ii) 3k – 5 (iii) 3k 5 (iv) 5k + 3 (v) 5k – 2 (vi) 5k – 3 The numbers that leave a remainder of 0 when divided by 5 are the multiples of 5. But we want numbers that leave a remainder of 3 when divided by 5. These numbers are 3 more than multiples of 5. Multiples of 5 are of the form 5k. So, numbers that leave a remainder of 3 when divided by 5 are those of the form 5k + 3 5 . . . k r o w s 5k + 3 = 3 8 13 18 23 k = 0 1 2 3 4 121 Ganita Prakash | Grade 8 Let us consider another expression, 5k – 2, and see the values it takes for different values of k. Numbers that leave a remainder of 3 when divided by 5 can also be seen as 2 less than multiples of 5; 5k – 2, where k ≥ 1. Are there other expressions that generate numbers that are 3 more than a multiple of 5? Figure it Out 1. The sum of four consecutive numbers is 34. What are these numbers? 2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p. 3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. 4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9. (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. 5k – 2 = 3 8 13 18 23 k = 1 2 3 4 5 Chapter 5 Letter Number Play 06-07-2025.indd 122 10-07-2025 15:07:59 122 5. “I hold some pebbles, not too many, 6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true? When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?” 7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. 8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest? 5.2 Checking Divisibility Quickly Earlier, you have learnt shortcuts to check whether a given number, written in the Indian number system is divisible by 2, 4, 5, 8, and 10. Let us revisit them. Divisibility by 10, 5, and 2: If the units digit of a number is ‘0’, then it is divisible by 10. Let us understand why this works through algebra. We can write the general form of a number in the Indian system using a set of letter-numbers. For example, a 5-digit number can be expressed as, edcba denoting e × 10000 + d × 1000 + c × 100 + b × 10 + a. The letter-numbers e, d, c, b, and a denote each digit of a 5-digit number. Any number can be written in general as…dcba, where the letternumbers a, b, c and d represent the units, tens, hundreds and thousands digit, respectively, and so on. As a sum of place values, this number is — … +1000d + 100c + 10b + a. (For example, in the number 4075, d = 4, c = 0, b = 7, and a = 5.) We know that each place value, with the exception of the units place, is a multiple of 10. So, 10b, 100c, … all will be multiples of 10. Hence, the number will be divisible by 10 if and only if the units digit a is 0. (i) 4779 + 661 (ii) 4779 – 661 Number Play Chapter 5 Letter Number Play 06-07-2025.indd 123 10-07-2025 15:07:59 Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work. Let us now examine shortcuts to check divisibility by some other numbers and explain why they work! A Shortcut for Divisibility by 9 Can you say, without actually calculating, which of these numbers are divisible by 9: 999, 909, 900, 90, 990? All of them. 123 Ganita Prakash | Grade 8 Can we say that any number made up of only the digits ‘0’ and ‘9’, in any order, will always be divisible by 9? Yes, if each digit is either 0 or 9, then each term in its expanded form will be 9 × or 0 × (the ‘ ’ denotes a place value). This means each term will be a multiple of 9, for example, But this shortcut alone cannot identify all the multiples of 9. Unlike the numbers 2, 5, and 10, we cannot identify the multiples of 9 by just looking at the unit’s digit. 99 and 109 are two numbers with 9 as the units digit; but 99 is divisible by 9, while 109 is not. Is 10 divisible by 9? If not, what is the remainder? Check the divisibility of other multiples of 10 (10, 20, 30, ...) by 9. You will notice that for any multiple of 10, the remainder is the same as the number of tens. Similarly, look at the remainder when the multiples of 100 (100, 200, 300, … ) are divided by 9. What do you notice? The remainder is the same as the number of hundreds for any multiple of 100. Using this observation, find the remainder when 427 is divided by 9. 99009 = 9 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 9 × 1. 99 1 99 1 99 1 400 20 7 9 1 9 1 1 1 1 1 1 1 1 9 1 99 99 9 1 9 1 1 1 Chapter 5 Letter Number Play 06-07-2025.indd 124 10-07-2025 15:07:59 124 We see that 427 has 4 hundreds; thus, its corresponding remainder (upon division by 9) would be 4. 427 has 2 tens, and its corresponding remainder would be 2. We have 7 units also remaining. Adding all the remainders, we get 4 + 2 + 7 = 13. We can make one more group of 9 with 13, leaving a remainder of 4. Therefore, 427 ÷ 9 gives a remainder of 4. 99 1 99 1 99 1 99 1 99 1 (Remainder) 9 1 9 1 4 1 1 1 1 1 1 1 9 Will this work with bigger numbers? You can see that this is true for any place value: 1 = 0 + 1 10 = 9 + 1 100 = 99 + 1 1000 = 999 + 1 10000 = 9999 + 1, and so on. Each digit thus denotes the remainder when the corresponding place value is divided by 9. For example, to find the remainder of 7309 when divided by 9, we can just add all the digits—7 + 3 + 0 + 9—to get 19. This can be seen as follows: 7 × 1000 + 3 × 100 + 0 × 10 + 9 × 1 = 7 × (999 + 1) + 3 × (99 + 1) + 0 × (9 + 1) + 9 × (0+1) = (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 × 1 + 3 × 1 + 0 × 1 + 9 × 1) = (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 + 3 + 0 + 9). 7 × 1000 3 × 100 9 × 1 999 1 999 1 999 1 999 1 999 1 999 1 999 1 99 1 99 1 99 1 1 1 1 1 1 1 1 1 1 7 × 999 3 × 99 7 3 9 999 999 999 999 999 999 999 This is a multiple of 9 99 99 99 9999 1 99 9 1 1 999 1 So, we need to just consider this part (i.e., 7 + 3 + 9 = 19) 1 1 1 1 1 1 1 Number Play 1 1 1 1 1 1 1 1 1 1 1 1 Chapter 5 Letter Number Play 06-07-2025.indd 125 10-07-2025 15:07:59 This means that the number 7309 is 19 more than some multiple of 9. The digits 1 and 9 can further be added to get 1 + 9 = 10. Now, we can say that 7309 is 10 more than a multiple of 9. And repeating this step for the number 10, we get the remainder to be 1 + 0 = 1, meaning 7309 is 1 more than a multiple of 9. Therefore, 7309 ÷ 9 gives a remainder of 1. A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Also, we can add the digits of a number repeatedly till a single digit is obtained. This single digit is the remainder when the number is divided by 9. Look at each of the following statements. Which are correct and why? This is a multiple of 9 So, we need to just consider this part (i) If a number is divisible by 9, then the sum of its digits is divisible by 9. 125 Ganita Prakash | Grade 8 Figure it Out 1. Find, without dividing, whether the following numbers are divisible by 9. 2. Find the smallest multiple of 9 with no odd digits. 3. Find the multiple of 9 that is closest to the number 6000. 4. How many multiples of 9 are there between the numbers 4300 and 4400? A Shortcut for Divisibility by 3 We know that all the multiples of 9 are also multiples of 3. That is, if a number is divisible by 9, it will also be divisible by 3. However, there are other multiples of 3 that are not multiples of 9 for example— 15, 33, and 87. (ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. (iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9. (iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9. Learning maths is not just about knowing some shortcuts and following procedures but about understanding ‘why’ something works. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095 Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 126 10-07-2025 15:07:59 126 The shortcut to find the divisibility by 3 is similar to the method for 9. A number is divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when powers of 10 are divided by 3. Explain why this method works. A Shortcut for Divisibility by 11 Interestingly, the shortcut for 11 is also based on checking the remainders with place value. Let us see how. Figure it Out 1. Find, without dividing, whether the following numbers are divisible by 9. 2. Find the smallest multiple of 9 with no odd digits. 3. Find the multiple of 9 that is closest to the number 6000. 4. How many multiples of 9 are there between the numbers 4300 and 4400? A Shortcut for Divisibility by 3 We know that all the multiples of 9 are also multiples of 3. That is, if a number is divisible by 9, it will also be divisible by 3. However, there are other multiples of 3 that are not multiples of 9 for example— 15, 33, and 87. (ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. (iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9. (iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9. Learning maths is not just about knowing some shortcuts and following procedures but about understanding ‘why’ something works. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095 This alternating pattern of one more than 11 and one less than 11 continues for higher place values. Since 400 contains 4 hundreds, 400 is 4 more than a multiple of 11 (396 + 4). Since 60 contains 6 tens, 60 is 6 less than a multiple of 11 (66 – 6). Since 2 contains 2 units, 2 is 2 more than a multiple of 11, i.e., 2 = (0 + 2). Units place (1) 11 × 0 = 0 1 = 11 × 0 + 1 1 is one more than a multiple of 11. Thousands place (1000) Hundreds place (100) 11 × 9 = 99 100 = 11×9 + 1 100 is one more than a multiple of 11. Tens place (10) 11 × 1 = 11 10 = 11×1 – 1 10 is one less than a multiple of 11. 11 . . . . 11 × 91 = 1001 1000 = 11×91 – 1 . . . . 1000 is one less than a multiple of 11. . . . . 11 . . . 9 11 . . . . . . 91 . . . . Number Play Math Talk The shortcut to find the divisibility by 3 is similar to the method for 9. A number is divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when powers of 10 are divided by 3. Explain why this method works. A Shortcut for Divisibility by 11 Interestingly, the shortcut for 11 is also based on checking the remainders with place value. Let us see how. Chapter 5 Letter Number Play 06-07-2025.indd 127 10-07-2025 15:07:59 Using these observations, can you tell whether the number 462 is divisible by 11? What could be a general method or shortcut to check divisibility by 11? Math Talk Math Talk 127 1. Add the digits of place values which are 1 more (than a multiple of 11), i.e., place values corresponding to 1, 100, 10000, and so on. 2. Add the digits of place values which are 1 less (than a multiple of 11), i.e., place values corresponding to 10, 1000, 100000, and so on. Ganita Prakash | Grade 8 We saw that the place values alternate as 1 more and 1 less than a multiple of 11. Using this observation, Steps Purpose Example for the Number 320185 To know how much in excess we are with respect to a multiple of 11 for these place values. 11 . . . 9 To know how short we are with respect to a multiple of 11 for these place values. Total short, 3 + 0 + 8 = 11. 2 × 10,000 Total excess, 2 + 1 + 5 = 8. 11 . . . . . . . . 9091 3 × 10,000 0 × 100 8 × 10 11 . . . . . . . . 9091 11 . . . . . . . . 9091 11 . . . . . . 909 11 . . . . . . 909 320185 1 × 100 5 × 1 11 11 11 11 11 11 11 11 320185 Chapter 5 Letter Number Play 06-07-2025.indd 128 10-07-2025 15:07:59 128 3. Compute the difference between these two sums, i.e., (number in excess) – (number short). The difference between these two sums 8 – 11 = – 3, indicating that the number 3,28,105 is 3 short of or 8 more than a multiple of 11. If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11? Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11. (i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076 To know the remainder obtained when divided by 11. 8 – 11 = – 3. (3 short of a multiple of 11) Look at the following procedure— Is this method similar to or different from the method we saw just before? Fill in the following table. Find a quick way to do this? 1. Place alternating ‘+’ and ‘–’ signs before every digit starting from the unit’s digit. –3 + 2 – 8 + 1 – 0 + 5 2. Evaluate the expression. –3 + 2 – 8 + 1 – 0 + 5 = – 3 3. The result denotes the remainder obtained when the number is divided by 11. Number Divisible by 128 Yes No No No No Yes No No No 990 1586 275 6686 639210 Steps to follow Example for the number 328105 2 3 4 5 6 8 9 10 11 328105 is 3 less than or 8 more than a multiple of 11 Number Play Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 129 10-07-2025 15:07:59 More on Divisibility Shortcuts Divisibility Shortcuts for Other Numbers How can we find out if a number is divisible by 6? Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify— 38, 225, 186, 64. 429714 2856 3060 406839 129 Ganita Prakash | Grade 8 How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not? Determining divisibility by 24 by checking divisibility by 4 and by 6 does not work. For example, the number 12 is divisible by both 4 and 6, but not by 24. To check for the divisibility by 24, we can instead check for the divisibility by 3 and divisibility by 8. Explain using prime factorisation why checking divisibility by 3 and 8 works for checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for checking divisibility by 24. There are such shortcuts to check divisibility by every number until 100, and for some numbers beyond 100. You may try to understand how these work after learning certain concepts in higher grades. Digital Roots Take a number. Add its digits repeatedly till you get a single-digit number. This single-digit number is called the digital root of the number. For example, the digital root of the number 489710 will be What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9. Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3? Write the digital roots of any 12 consecutive numbers. What do you observe? We saw that the digital root of multiples of 9 is always 9. 2 (4 + 8 + 9 + 7 + 1 + 0 = 29, 2 + 9 = 11, 1 + 1 = 2). Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 130 10-07-2025 15:08:00 130 Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6. What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice? Try to explain the patterns noticed. I’m made of digits, each tiniest and odd, No shared ground with root #1—how odd! My digits count, their sum, my root— All point to one bold number’s pursuit— The largest odd single-digit I proudly claim. What’s my number? What’s my name? Aryabhata II’s (c. 950 CE) work Mahāsiddhānta, mentions the method of computing the digital root of a number by repeatedly adding the digits till a single-digit number is obtained. This method is known to have been used to perform checks on calculations of arithmetic operations. Figure it Out 1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number? 2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations. 3. What will be the digital root of the number 9a + 36b + 13? 4. Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root. 5.3 Digits in Disguise Last year, we saw cryptarithms—puzzles where each letter stands for a digit, each digit is represented by at most one letter, and the first digit of a number is never 0. Solve the cryptarithms given below. (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9. (i) A1 + 1B B0 (ii) AB + 37 6A (iii) ON ON + ON PO (iv) QR QR + QR PRR Number Play Math Talk Chapter 5 Letter Number Play 06-07-2025.indd 131 10-07-2025 15:08:00 Let us now try solving some cryptarithms involving multiplication. (v) PQ × 8 = RS. Guna says, “Oh, this means a 2-digit number multiplied by 8 should give another 2-digit number. I know that 10 × 8 = 80. But the units digits of 10 and 80 are the same, which we don’t want. For the same reason PQ cannot be 11 as P and Q correspond to different digits. 12 × 8 = 96 fits all the conditions”. Can PQ be 13? Think. It is not possible because 13 × 8 = 104. For all 2-digit numbers greater than 12, the product with 8 is a 3-digit number. 131 Ganita Prakash | Grade 8 (vi) Try this now: GH × H = 9K. This means a 2-digit number multiplied by a 1-digit number gives another 2-digit number in the 90s. Observe the letters corresponding to the units digits in this cryptarithm. Pick the solution to this question from the options given below: 11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93, 16 × 6 = 96. (vii) Here is one more: BYE × 6 = RAY. Anshu says, “Since the product is a 3-digit number, B can’t be 2 or more. If B = 2, i.e., 2 hundreds, the product will be more than 1200. So, B = 1.” What can you say about ‘Y’? What digits are possible/not possible? “Y cannot be 7 or more because, if Y = 7, then 170 × 6 = 1020; but we want a 3-digit product. Also, Y will be even”, Anshu explains. We can solve cryptarithms using patterns, properties, and reasoning related to numbers and operations. Solve the following: Figure it Out 1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem. 2. “I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion. (i) UT × 3 = PUT (ii) AB × 5 = BC (iii) L2N × 2 = 2NP (iv) XY × 4 = ZX (v) PP × QQ = PRP (vi) JK × 6 = KKK Chapter 5 Letter Number Play 06-07-2025.indd 132 10-07-2025 15:08:00 132 3. When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern. 4. Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”. 5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b. (i) Examine if her conjecture is true for any multiple of 9. (ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9? 7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur? 10. Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6. 11. Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion. 12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning. 6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q. 8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class. 9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers will be divisible by 6. (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. (iv) 8 (7b – 3) – 4 (11b + 1) is a multiple of 12. Number Play Math Talk Try This Chapter 5 Letter Number Play 06-07-2025.indd 133 10-07-2025 15:08:00 13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise. 16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32? 14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers? 15. Solve the cryptarithms — (i) EF × E = GGG (ii) WOW × 5 = MEOW 133 Ganita Prakash | Grade 8 (i) (iii) We explored and learnt various properties of divisibility— Multiples of 32 Multiples Multiples of 4 of 8 Multiples of 4 Multiples of 32 Multiples of 8 Multiples of 4 Multiples of 32 Multiples of 8 • If a is divisible by b, then all multiples of a are divisible by b. • If a is divisible by b, then a is divisible by all the factors of b. SUMMARY Multiples of 4 Multiples of 32 (ii) (iv) Multiples of 8 Chapter 5 Letter Number Play 06-07-2025.indd 134 10-07-2025 15:08:00 134 We learnt shortcuts to check divisibility by 3, 9 and 11, and why they work. Through all this we were exposed to the power of mathematical thinking and reasoning, using algebra, visualisation, examples and counterexamples. • If a divides m and a divides n, then a divides m + n and m – n. • If a is divisible by b and is also divisible by c, then a is divisible by the LCM of b and c. Navakankari, also known as Sālu Mane Āṭa, Chār-Pār, or Navkakri, is a traditional Indian board game that is the same as ‛Nine Men’s Morris’ or ‛Mills in the West’. It is a strategy game for two players where the goal is to form lines of three pawns to eliminate the opponent’s pawns or block their movement. Gameplay 1. Each player starts with 9 pawns. The players take turns in placing their pawns on the marked intersections. An intersection can have at most one pawn. Navakankari Chapter 5 Letter Number Play 06-07-2025.indd 135 10-07-2025 15:08:03 2. Once all the pawns are placed, the players take turns to move one of their pawns to adjacent empty intersections to form lines of three. The line can be horizontal or vertical. 3. Once a player makes a line with their pawns they can remove any one of the opponent’s pawns as long as it is not a part of one of their lines. A player wins if the opponent has less than 3 pawns or is unable to make a move. 135" class_8,6,"we distribute, yet things multiply",ncert_books/class_8/hegp1dd/hegp106.pdf,"6 We have seen how algebra makes use of letter symbols to write general statements about patterns and relations in a compact manner. Algebra can also be used to justify or prove claims and conjectures (like the many properties you saw in the previous chapter) and to solve problems of various kinds. Distributivity is a property relating multiplication and addition that is captured concisely using algebra. In this chapter, we explore different types of multiplication patterns and show how they can be described in the language of algebra by making use of distributivity. 6.1 Some Properties of Multiplication Increments in Products Consider the multiplication of two numbers, say, 23 × 27. 1. By how much does the product increase if the first number (23) is increased by 1? WE DISTRIBUTE, YET THINGS MULTIPLY Chapter 6 We Distribute, Yet Things Multiply.indd 136 10-07-2025 15:10:48 2. What if the second number (27) is increased by 1? 3. How about when both numbers are increased by 1? Do you see a pattern that could help generalise our observations to the product of any two numbers? Let us first consider a simpler problem—find the increase in the product when 27 is increased by 1. From the definition of multiplication (and the commutative property), it is clear that the product increases by 23. This can be seen from the distributive property of multiplication as well. If a, b and c are three numbers, then— This is called the distributive property of multiplication over addition. Using the identity a (b + c) = ab + ac with a = 23, b = 27, and c = 1, we have Remember that here, a (b + c) and 23 (27 + 1) mean a × (b + c), and 23 × (27 + 1), respectively. We usually skip writing the ‘×’ symbol before or after brackets, just as in the case of expressions like 5a, xy, etc. We can also similarly expand (a + b) c using the distributive property as follows— (a + b) c = c (a + b) (commutativity of multiplication) = ca + cb (distributivity) = ac + bc (commutativity of multiplication) We can use the distributive property to find, in general, how much a product increases if one or both the numbers in the product are increased by 1. Suppose the initial two numbers are a and b. If one of the numbers, say b, is increased by 1, then we have— This property can be visualised nicely using a diagram: a rows Increase 23 ( 27 + 1) = 23 × 27 + 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b columns c columns ab ac a (b + c) = ab + ac a (b + c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . We Distribute, Yet Things Multiply Chapter 6 We Distribute, Yet Things Multiply.indd 137 10-07-2025 15:10:48 Now let us see what happens if both numbers in a product are increased by 1. If in a product ab, both a and b are increased by 1, then we obtain (a + 1) (b + 1). Increase a ( b + 1) = ab × a 137 Ganita Prakash | Grade 8 (a + 1) (b – 1) = (a + 1) b – (a + 1) 1 = ab + b – (a + 1) = ab + b – a – 1 Increase How do we expand this? Let us consider (a + 1) as a single term. Then, by the distributive property, we have Thus, the product ab increases by a + b + 1 when each of a and b are increased by 1. What would we get if we had expanded (a + 1) (b + 1) by first taking (b + 1) as a single term? Try it? What happens when one of the numbers in a product is increased by 1 and the other is decreased by 1? Will there be any change in the product? Let us again take the product ab of two numbers a and b. If a is increased by 1 and b is decreased by 1, then their product will be (a + 1) (b – 1). Expanding this, we get (a + 1) (b + 1) = (a + 1) b + (a + 1) 1 = ab + (b + a + 1) Increase (a + 1) (b + 1) = (a + 1) b + (a + 1) 1 Again applying the distributive property, we obtain (23 + 1) (27 – 1) = (23 + 1) 27 – (23 + 1) 1 = 23 × 27 + 27 – (23 + 1) = 23 × 27 + 27 – 23 – 1 Increase If a = 23, and b = 27, we get (23 + 1) (27 + 1) = (23 + 1) 27 + (23 + 1) 1 = 23 × 27 + (27 + 23 + 1) Increase If a = 23, and, b = 27, we get Chapter 6 We Distribute, Yet Things Multiply.indd 138 10-07-2025 15:10:49 138 Will the product always increase? Find 3 examples where the product decreases. What happens when a and b are negative integers? Check by substituting different values for a and b in each of the above cases. For example, a = –5, b = 8; a = –4, b = –5; etc. We have seen that integers also satisfy the distributive property, that is, if x, y and z are any three integers, then x (y + z) = xy + xz. Thus, the expressions we have for increase of products hold when the letter-numbers take on negative integer values as well. Recall that two algebraic expressions are equal if they take on the same values when their letter-numbers are replaced by numbers. These numbers could be any integers. Mathematical statements that express the equality of two algebraic expressions, such as (a + 1) (b – 1) = ab + b – a – 1, etc., are called identities. By how much will the product of two numbers change if one of the numbers is increased by m and the other by n? If a and b are the initial numbers being multiplied, they become a + m and b + n. = ab + mb + an + mn The increase is an + bm + mn. Notice that the product is the sum of the product of each term of (a + m) with each term of (b + n). This identity can be visualised as follows— (a + m) (b + n) = ab + mb + an + mn Identity 1 (a + m) (b + n) = (a + m)b + (a + m)n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b columns n columns a (b + 8) = ab + 8a, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . We Distribute, Yet Things Multiply Chapter 6 We Distribute, Yet Things Multiply.indd 139 10-07-2025 15:10:49 a rows m rows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . mb ab (a + m) (b + n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . mn an 139 Ganita Prakash | Grade 8 This identity can be used to find how products change when the numbers being multiplied are increased or decreased by any amount. Can you see how this identity can be used when one or both numbers are decreased? For example, let us reconsider the case when one number is increased by 1 and the other decreased by 1. Let us write the product (a + 1) (b – 1) as (a + 1) (b + (–1). Taking m = 1 and n = –1 in Identity 1, we have which is the same expression that we obtain earlier. Use Identity 1 to find how the product changes when Verify the answers by finding the products without converting the subtractions to additions. = ab + ub – av – uv. Check that this is the same as taking m = u and n = –v in Identity 1. As in Identity 1, the product (a + u) (b – v) is the sum of the product of each term of a + u (a and u) with each term of b – v (b and (–v)). Notice that the signs of the terms in the products can be determined using the usual rules of integer multiplication. Generalising this, we can find the product (a + u) (b – v) as follows. (a + u) (b – v) = (a + u) b – (a + u) v See how the rules of integer multiplication allows us to handle multiple cases using a single identity! (ii) both numbers are decreased, one by 3 and the other by 4. (i) one number is decreased by 2 and the other increased by 3; ab + (1) × b + a × (–1) + (1) × (–1) = ab + b – a – 1, = ab + ub – (av + uv) Chapter 6 We Distribute, Yet Things Multiply.indd 140 10-07-2025 15:10:49 140 Expand (i) (a – u) (b + v), (ii) (a – u) (b – v). We get (a – u) (b – v) = ab – ub – av + uv. The distributive property is not restricted to two terms within a bracket. Example 1: Expand 3a 2 (a – b + 1 5). 3a 2 (a – b + 1 5) = (3a 2 × a) – (3a 2 × b) + (3a 2 × 1 5 ). The terms can be simplified as follows— 3a 2 × a = 3 2 × (a × a). (a – u) (b + v) = ab – ub + av – uv, and Using exponent notation, we can write 3 2 × (a × a) = 3 2 a2 . 3a 2 × b = 3 2 × (a × b) = 3 2 ab. 3a 2 × 1 5 = (3 2 × 1 5) a = 3 10 a So we get 3a 2 (a – b + 1 5) = 3 2a2 – 3 2 ab + 3 10 a. Can any two terms be added to get a single term? For example, can 3 2 a2 and 3 10 a be added to get a single term? We see that no two terms have exactly the same letter-numbers, which would have allowed them to be simplified into a single term. So, a further simplification of the expression is not possible. Recall that we call terms having the same letter-numbers like terms. Example 2: Expand (a + b) (a + b). = a2 + ba + ab + b2 Since ba = ab, we have two terms having the same letter-numbers ab (or, that are like terms), and so can be added — Example 3: Expand (a + b) (a2 + 2ab + b2 ). (a + b) (a2 + 2ab + b2 ) = (a + b)a2 + (a + b) × 2ab + (a + b)b2 = (a × a2 ) + ba2 + (a × 2ab) + (b × 2ab) + ab2 + (b × b2 ) The terms can be simplified as follows— We have (a + b) (a + b) = (a+b) a + (a + b)b = a × a + b × a + ab + b × b ba + ab = ab + ab = 2ab So we get (a + b) (a + b) = a2 + 2ab + b2 . We Distribute, Yet Things Multiply Chapter 6 We Distribute, Yet Things Multiply.indd 141 10-07-2025 15:10:49 So, (a + b)(a2 + 2ab + b2 ) = a3 + a2 b + 2a2 b + 2ab2 + ab2 + b3 . We see that a2 b and 2a2 b have the same letter-numbers (or, are like terms) and so can be added— a × 2ab = 2 × a × a × b = 2a2 b b × 2ab = 2 × a × b × b = 2ab2 a × a2 = a3 (why?) b × b2 = b3 ba2 = a2 b 141 Ganita Prakash | Grade 8 A Pinch of History The distributive property of multiplication over addition was implicit in the calculations of mathematicians in many ancient civilisations, particularly in ancient Egypt, Mesopotamia, Greece, China, and India. For example, the mathematicians Euclid (in geometric form) and Āryabhaṭa (in algebraic form) used the distributive law in an implicit manner extensively in their mathematical and scientific works. The first explicit statement of the distributive property was given by Brahmagupta in his work Brahmasphuṭasiddhānta (Verse 12.55), who referred to the use of the property for multiplication as khaṇḍa-guṇanam (multiplication by parts). His verse states, “The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added”. That is, if there are two parts, then using letter symbols this is equivalent to the identity (a + b) c = ac + bc. In the next verse (Verse 12.56), Brahmagupta further describes a method for doing fast multiplication using this distributive property, which we explore further in the next section. Figure it Out 1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid. a2 b + 2a2 b = (1 + 2) a2 b = 3a2 b. Similarly, ab2 and 2ab2 are like terms and so can be added— ab2 + 2ab2 = (1 + 2)ab2 = 3ab2 . Thus, we have (a + b) × (a2 + 2ab +b2 ) = a3 + 3a2 b + 3ab2 + b3 . Chapter 6 We Distribute, Yet Things Multiply.indd 142 10-07-2025 15:10:49 142 10 10 20 30 40 50 60 70 80 90 100 x 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 3 × 5 3 × 6 3 × 7 4 × 5 4 × 6 4 × 7 5 × 5 5 × 6 5 × 7 pq 2. Expand the following products. (i) (3 + u) (v – 3) (ii) 2 3 (15 + 6a) (iii) (10a + b) (10c + d) (iv) (3 – x) (x – 6) (v) (–5a + b) (c + d) (vi) (5 + z) (y + 9) 3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4. 4. Expand (i) (a + ab – 3b2 ) (4 + b), and (ii) (4y + 7) (y + 11z – 3). 5. Expand (i) (a – b) (a + b), (ii) (a – b) (a2 + ab + b2 ) and (iii) (a – b)(a3 + a2 b + ab2 + b3 ), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding? Fast Multiplications Using the Distributive Property The distributive property can be used to come up with quick methods of multiplication when certain types of numbers are multiplied. When one of the numbers is 11, 101, 1001, ... Use the following multiplications to find the product of a number with 11 in a single step. Let us take the first multiplication. 3874 × 11 = 3874 (10 + 1) = 38740 + 3874 (a) 3874 × 11 (b) 5678 × 11 We Distribute, Yet Things Multiply Math Talk Chapter 6 We Distribute, Yet Things Multiply.indd 143 10-07-2025 15:10:49 Notice how the digits are getting added. Let us take a 4-digit number dcba, that is, the number that has d in the thousands place, c in the hundreds place, b in the tens place and a in the units place. dcba × (10 + 1) = dcba × 10 + dcba. This becomes d  c    b    a    o +    d c b a d (c + d) (b + c) (a + b) a 38740 + 3874 143 Ganita Prakash | Grade 8 This can be used to obtain the product in one line. Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line. Evaluate (i) 94 × 11, (ii) 495 × 11, (iii) 3279 × 11, (iv) 4791256 × 11. Can we come up with a similar rule for multiplying a number by 101? Multiply 3874 by 101. Let us take a 4-digit number dcba. dcba × 101 = dcba × (100 + 1) = dcba × 100 + dcba. This becomes d  c    b    a    o    o +      d c b a d  c (b + d) (a + c) b a Use this to multiply 3874 × 101 in one line. What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, … 3874 × 11 4 Step 1 Step 3 1 3874 × 11 2614 3874 × 11 42614 1 1 1 1 1 1 + Step 4 Step 5 3874 × 11 14 + Step 2 1 1 3874 × 11 614 + Math Talk Chapter 6 We Distribute, Yet Things Multiply.indd 144 10-07-2025 15:10:49 144 Use this to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, (iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999. Such methods of applying the distributive property to easily multiply two numbers were discussed extensively in the ancient mathematical works of Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE). In his work Brahmasphuṭasiddhānta (Verse 12.56), Brahmagupta refers to such methods for fast multiplication using the distributive property as ista-gunana. Math Talk 6.2 Special Cases of the Distributive Property Square of the Sum/Difference of Two Numbers The area of a square of sidelength 60 units is 3600 sq. units (602 ) and that of a square of sidelength 5 units is 25 sq. units (52 ). Can we use this to find the area of a square of sidelength 65 units? A square of sidelength 65 can be split into 4 regions as shown in the figure—a square of sidelength 60, a square of sidelength 5, and two rectangles of sidelengths 60 and 5. The area of the square of sidelength 65 is the sum of the areas of all its constituent parts. Can you find the areas of the four parts in the figure above? We get 652 = (60 + 5)2 = 602 + 52 + 2 × (60 × 5). = 3600 + 25 + 600 = 4225 sq. units. Let us multiply (60 + 5) × (60 + 5) using the distributive property. What if we write 652 as (30 + 35)2 or (52 + 13)2 ? Draw the figures and check the area that you get. Let us look at the general expression for the square of sum of two numbers, (a + b) 2 . (60 + 5) × (60 + 5) = 60 × 60 + 5 × 60 + 60 × 5 + 5 × 5 = 602 + 2 × (60 × 5) + 52 . a b Using the distributive property, (a+b)2 can be expanded as We Distribute, Yet Things Multiply 60 5 5 5 × 60 2 602 60 5 60 × 5 Chapter 6 We Distribute, Yet Things Multiply.indd 145 10-07-2025 15:10:49 a b If a and b are any two integers, is (a + b) 2 always greater than a2 + b2 ? If not, when is it greater? Use Identity 1A to find the values of 1042 , 372 . (Hint: Decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute.) b2 b × a a × b a2 (a+b) × (a+b) = a×a + a×b + b×a + b×b = a2 + 2ab + b2 , as we had already seen in Example 2. Identity 1A (a+b)2 = a2 + 2ab + b2 Math Talk 145 Ganita Prakash | Grade 8 (6x + 5)2 = (6x + 5) (6x + 5) = (6x × 6x) + (5 × 6x) +(6x × 5) + 5 × 5 = (6x) 2 + 2 (6x × 5) + 52 = 36x2 + 60x + 25. Use Identity 1A to write the expressions for the following. Expand (6x + 5)2 . Expand (3j + 2k) 2 using both the identity and by applying the distributive property. Can we use 602 (=3600) and 52 (=25) to find the value of (60 – 5)2 or 552 ? Let us approach this through geometry by drawing a square of side length 55 sitting inside a square of sidelength 60. Area of a square of sidelength 55 is (60 – 5)2 = 552 . Using the Distributive Property Using the Identity If you have difficulty remembering or using the general rule, you can just apply the distributive property to multiply and get the desired result. (i) (m + 3)2 (ii) (6 + p) 2 55 5 We can get the area of the square of sidelength 55 by taking the area of the square of sidelength 60 and removing the areas of the two rectangles of sidelengths 60 and 5, i.e., 602 – (60 × 5) – (5 × 60). By doing this, we remove the area of the small square of sidelength 5 twice. What can we do with this expression to get the actual area? We can add back the area of the square of sidelength 5 to this expression. That way, we are only subtracting this area once. (6x + 5)2 = (6x) 2 + 52 + 2 × (6x × 5) = 36x2 + 25 + 60x. Chapter 6 We Distribute, Yet Things Multiply.indd 146 10-07-2025 15:10:49 146 55 60 60 × 60 –5 × 60 +5 × 5 –60 × 5 = 3600 – 300 – 300 + 25 = 3025. The area of the square of sidelength 55 is 3025 sq. units. We have seen what (a + b)2 gives when expanded. What is the expansion of (a – b) 2 ? Using the distributive property, = a2 – 2ab + b2 . We can also use the expansion of (a + b) 2 to find the expansion of (a – b) 2 . Think how. Hint: (a – b) 2 = (a + (–b))2 . We can now directly use the expansion of (a + b) 2 . Find the general expansion of (a – b) 2 using geometry, as we did for 552 . Use the identity (a – b) 2 to find the values of (a) 992 and (b) 582 . Expand the following using both Identity 1B and by applying the distributive property So, (60 – 5)2 = 602 – (60 × 5) – (5 × 60) + 52 (i) (b – 6)2 (ii) (–2a + 3)2 (iii) (7y – 3 4z) 2 (a + (–b) 2 = (a) 2 + (–b) 2 + 2 × (a) × (–b) Identity 1B (a – b) 2 = a2 + b2 – 2ab (a – b) 2 = (a – b) × (a – b) = (a) 2 – ba – ab + (b) 2 We Distribute, Yet Things Multiply Chapter 6 We Distribute, Yet Things Multiply.indd 147 10-07-2025 15:10:50 Investigating Patterns Pattern 1 Look at the following pattern. 2 (62 + 52 ) = 112 + 12 2 (52 + 32 ) = 82 + 22 . 2 (22 + 12 ) = 32 + 12 2 (32 + 12 ) = 42 + 22 147 Ganita Prakash | Grade 8 Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares? Try this with other pairs of numbers. Have you figured out a pattern? Notice that 2 (52 + 62 ) = (6 + 5)2 + (6 – 5)2 . Do the identities below help in explaining the observed pattern? (a + b) 2 + (a – b) 2 = (a2 + 2ab + b2 ) + (a2 – 2ab + b2 ) Adding the like terms a2 + a2 = 2a2 , b2 + b2 = 2b2 and 2ab – 2ab = 0, we get Pattern 2 Here is a related pattern. Try to describe the pattern using algebra to determine if the pattern always holds. The pattern here appears to be a2 – b2 = (a + b) × (a – b). Is this a true identity? Using the distributive property, we get (a + b) × (a – b) = a2 – ab + ba – a2 . Adding the like terms, ab + (–ab) = 0, we see that indeed 2 (a2 + b2 ) = (a + b) 2 + (a – b) 2 . 10 × 10 – 4 × 4 = 14 × 6 (a + b) 2 = a2 + 2ab + b2 (a – b) 2 = a2 – 2ab + b2 8 × 8 – 6 × 6 = 14 × 2 9 × 9 – 1 × 1= 10 × 8 7 × 7 – 2 × 2 = 9 × 5 Chapter 6 We Distribute, Yet Things Multiply.indd 148 10-07-2025 15:10:50 148 You had seen this identity earlier in Figure it Out 5 (i). Use Identity 1C to calculate 98 × 102, and 45 × 55. Show that (a + b) × (a – b) = a2 – b2 geometrically. Identity 1C (a + b) × (a – b) = a2 – b2 . Try This Hint: Sridharacharya (750 CE) gave an interesting method to quickly compute the squares of numbers using Identity 1C! Consider the following modified form of this identity — Why is this identity true? Now, for example, 312 can be found by taking a = 31 and b = 1. 312 = (31 + 1) (31 – 1) + 12 1972 can be found by taking a = 197, and b = 3. 1972 = (197 + 3) (197 – 3) + 32 a b a2 = (a + b) (a – b) + b2 = 32 × 30 + 1 a b = 961. We Distribute, Yet Things Multiply What do we get when this piece is moved? Chapter 6 We Distribute, Yet Things Multiply.indd 149 10-07-2025 15:10:50 Figure it Out 1. Which is greater: (a – b) 2 or (b – a) 2 ? Justify your answer. 2. Express 100 as the difference of two squares. 3. Find 4062 , 722 , 1452 , 10972 , and 1242 using the identities you have learnt so far. 4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer. = 200 × 194 + 9 = 38809. Math Talk 149 Ganita Prakash | Grade 8 6.3 Mind the Mistake, Mend the Mistake We have expanded and simplified some algebraic expressions below to their simplest forms. (iii) Then write the correct expression. (i) Check each of the simplifications and see if there is a mistake. (ii) If there is a mistake, try to explain what could have gone wrong. 10 11 12 1 2(x – 1) + 3(x + 4) = 2x – 1 + 3x + 4 = 5x + 3 4 7 (a + 2) (b + 4) = ab + 8 ab2 + a2 b + a2 b2 = ab (a + b + ab) (x + 2)(x + 5) = (x + 2)x + (x + 2)5 = x2 + 2x + 5x +10 = x2 + 7x + 10 –3p (–5p + 2q) = –3p + 5p – 2q = p – 2q 1 2 (10s – 6) + 3 = 5s – 3 + 3 = 5s (5m + 6n)2 = 25m2 + 36n2 2 y + 2(y + 2) = (y + 2)2 = y2 + 4y + 4 5 8 (–q + 2)2 = q2– 4q + 4 5w2 + 6w = 11w2 3 6 9 2a3 + 3a3 + 6a2 b + 6ab2 = 5a3 + 12a2 b2 3a (2b × 3c) = 6ab × 9ac = 54a2bc Chapter 6 We Distribute, Yet Things Multiply.indd 150 10-07-2025 15:10:50 150 6.4 This Way or That Way, All Ways Lead to the Bay Observe the pattern in the figure below. Draw the next figure in the sequence. How many circles does it have? How many total circles are there in Step 10? Write an expression for the number of circles in Step k. There are many ways of interpreting this pattern. Here are some possibilities: 1 2 3 . . . . . . Math Talk 22 – 1 = (1 + 1)2 – 1 32 – 1 = (2 + 1)2 – 1 42 – 1 = (3 + 1)2 – 1 52 – 1 = (4 + 1)2 – 1 (k + 1)2 . . . – 1 1 + 2 × 1 = 12 + 2 × 1 22 + 2 × 2 = 22 + 2 × 2 . . . Method 2 Step 1 Step 2 Step 3 Step 4 . . . Step k Method 1 Step 1 Step 2 Step 3 Step 4 . . . Step k 32 + 2 × 3 = 32 + 2 × 3 42 + 2 × 4 = 42 + 2 × 4 k2 + 2 × k We Distribute, Yet Things Multiply . . . . . . Chapter 6 We Distribute, Yet Things Multiply.indd 151 10-07-2025 15:10:50 1 × 2 + 1 = 1 × (1 + 1) + 1 Method 3 Step 1 Step 2 Step 3 Step 4 . . . Step k 2 × 3 + 2 = 2 × (2 + 1) + 2 3 × 4 + 3 = 3 × (3 + 1) + 3 4 × 5 + 4 = 4 × (4 + 1) + 4 k × (k + 1) + k . . . . . . 151 Ganita Prakash | Grade 8 1 × 3 = 1 × (1 + 2) . . . Does your method match any of these, or is it different? Each expression that we have identified appears different, but are they really different? Since they describe the same pattern, they should all be the same. Let us simplify each expression and find out. When carried out correctly, all methods lead to the same answer; k2 + 2k. The expression k2 + 2k gives the number of circles at Step k of this pattern. Method 4 (k + 1)2 – 1 = k2 + 1 + 2k – 1 = k2 + 2k Step 1 Step 2 Step 3 Step 4 . . . Step k In Mathematics, there are often multiple ways of looking at a pattern, and different ways of approaching and solving the same problem. Finding such ways often requires a great deal of creativity and imagination! While one or two of the ways might be your favourite(s), it can be amusing and enriching to explore other ways as well. 2 × 4 = 2 × (2 + 2) 3 × 5 = 3 × (3 + 2) 4 × 6 = 4 × (4 + 2) k × (k+2) k2 + 2 × k = k2 + 2k k × (k + 1) + k = k2 + k + k = k2 + 2k . . . k × (k + 2) = k2 + 2k Chapter 6 We Distribute, Yet Things Multiply.indd 152 10-07-2025 15:10:50 152 Use this formula to find the number of circles in Step 15. Consider the pattern made of square tiles in the picture below. 1 2 3 . . . . . . How many square tiles are there in each figure? How many are there in Step 4 of the sequence? What about Step 10? Write an algebraic expression for the number of tiles in Step n. Share your methods with the class. Can you find more than one method to arrive at the answer? Find the area of the (interior) shaded region in the figure below. All four rectangles have the same dimensions. Tadang’s method: The total region is a square of side (m + n) with an area (m + n) 2 . Subtracting the area of four rectangles from the total area will give the area of the interior shaded region. That is, (m + n) 2 – 4mn . Yusuf’s method: The shaded region is a square with sidelength (n – m). So, its area is (n – m) 2 . By expanding both expressions, check that (m + n)2 – 4mn = (n – m)2 . Find out the area of the region with slanting lines in the figure. All three rectangles have the same dimensions (Fig. 1). We Distribute, Yet Things Multiply m y Fig. 1 x Math Talk n Chapter 6 We Distribute, Yet Things Multiply.indd 153 10-07-2025 15:10:50 Anusha’s method: Required area = Area (ABCD) – Area (EFGH) Area of ABCD = x2 . Area of EFGH = xy. Required area = x2 – xy. Vaishnavi’s method: QS = y + x + y = x + 2y. Area of PQSR = x (x + 2y) Required area = Area of PQSR – (area of the three rectangles) = x (x + 2y) – 3xy. 153 Ganita Prakash | Grade 8 Aditya’s method: The required area is 2 times the area of JKLM. Area (JKML) = x ( x – y 2 ) Required area = 2 × Area of JKML = 2x (x – y 2 ) = x (x – y). By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region. Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 3.5, and s = 9, and calculate the area. Figure it Out 1. Compute these products using the suggested identity. JK = x – y 2 , KM = x r s r p Math Talk Chapter 6 We Distribute, Yet Things Multiply.indd 154 10-07-2025 15:10:50 154 (iv) 43 × 45 using Identity 1C for (a + b) (a – b) 2. Use either a suitable identity or the distributive property to find each of the following products. (i) (p – 1) (p + 11) (ii) (3a – 9b) (3a + 9b) (iii) –(2y + 5) (3y + 4) (iv) (6x + 5y) 2 (ii) 397 × 403 using Identity 1C for (a + b) (a – b) (iii) 912 using Identity 1B for (a – b) 2 (vi) (7p) × (3r) × (p + 2) (v) (2x – 1 2) 2 (i) 462 using Identity 1A for (a + b) 2 3. For each statement identify the appropriate algebraic expression(s). m2 + (m – 1)2 (m + (m + 1))2 (2m) 2 + (2m + 1)2 4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square as 2 + s (s + 2)2 s2 + 2 s2 + 4 2s2 22 s (ii) The sum of the squares of two consecutive numbers (i) Two more than a square number. m2 + n2 (m + n) 2 m2 + 1 m2 + (m + 1)2 Su M Tu W Th F Sa 16 17 18 19 20 21 22 23 24 25 26 27 28 2 3 4 5 6 7 8 9 10 11 12 13 14 15 February a + 7 (a + 8) a (a + 1) We Distribute, Yet Things Multiply 1 Math Talk Chapter 6 We Distribute, Yet Things Multiply.indd 155 10-07-2025 15:10:50 5. Verify which of the following statements are true. (iv) (6n + 2)2 – (4n + 3)2 is 5 less than a square number. 6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7? 7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other (i) (k + 1) (k + 2) – (k + 3) is always 2. (ii) (2q + 1) (2q – 3) is a multiple of 4. (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8. 155 Ganita Prakash | Grade 8 sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity. 10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g2 sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled. 11. For each pattern shown below, 8. What is the algebraic expression describing the following steps—add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers. 9. Which is larger? Find out without fully computing the product. (iii) Write an expression to describe the number of basic units in Step y. (ii) 25 × 75 or 26 × 74 (ii) How many basic units are there in Step 10? (i) 14 × 26 or 16 × 24 (i) Draw the next figure in the sequence. w ft. g2 sq. ft. g2 sq. ft. w ft. Chapter 6 We Distribute, Yet Things Multiply.indd 156 10-07-2025 15:10:50 156 Step 1 Step 2 Step 3 Step 1 Step 2 Step 3 � We extended the distributive property to find the product of two expressions each of which has two terms. The general form for the same is (a + b) × (c + d) = ac + ad + bc + bd. � We saw some special cases of this identity. � We considered different patterns, and explored how to understand them using algebra. We saw that, often, there are multiple ways to solve a problem and arrive at the same correct answer. Finding different methods to approach and solve the same problem is a creative process. y (a + b) 2 = a2 + 2ab + b2 y (a – b)2 = a2 – 2ab + b2 y (a + b) (a – b) = a2 – b2 . SUMMARY We Distribute, Yet Things Multiply Chapter 6 We Distribute, Yet Things Multiply.indd 157 10-07-2025 15:10:51 157 Arrange 10 coins in a triangle as shown in the figure below on the left. The task is to turn the triangle upside down by moving one coin at a time. How many moves are needed? What is the minimum number of moves? A triangle of 3 coins can be inverted (turned upside down) with a single move, and a triangle of 6 coins can be inverted by moving 2 coins. The 10-coin triangle can be flipped with just 3 moves; did you figure out how? Find out the minimum possible moves needed to flip the next bigger triangle having 15 coins. Try the same for bigger triangular numbers. Is there a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement? Coin Conjoin Chapter 6 We Distribute, Yet Things Multiply.indd 158 10-07-2025 15:10:54 158" class_8,7,propositional reasoning - 1,ncert_books/class_8/hegp1dd/hegp107.pdf,"7 7.1 Observing Similarity in Change We are all familiar with digital images. We often change the size and orientation of these images to suit our needs. Observe the set of images below— PROPORTIONAL REASONING-1 Image A Image B Image C Chapter 7.indd 159 10-07-2025 15:14:27 We can see that all the images are of different sizes. Which images look similar and which ones look different? Images (A, C, and D) look similar, even though they have different sizes. Image D Image E Ganita Prakash | Grade 8 Do images B and E look like the other three images? No, they are slightly distorted. The tiger appears elongated in B, and compressed and fatter in E! Why? You may notice that images A, C, and D are rectangular, but E is square. Maybe that is why E looks different. But B is also a rectangle! Why does it look different from the other rectangular images? Can we observe any pattern to answer this question? Perhaps by measuring the rectangles? What makes images A, C, and D appear similar, and B and E different? When we compare image A with C, we notice that the width of C is half that of A. The height is also half of A. Both the width and height have changed by the same factor (through multiplication), 1 2 in this case. Since the widths and heights have changed by the same factor, the images look similar. When we compare image A with image B, we notice that the width of B is 20 millimetre (mm) less than that of A. The height too is 20 mm less than the height of A. Even though the difference (through subtraction) is the same, the images look different. Have the width and height changed by the same factor? The height of B is half the height of A. But the width of B is not half the width of A. Since the width and height have not changed by the same factor, the images look different. Image A 60 40 Image D 90 60 Image B 40 20 Image C 30 20 Image E 60 60 Image Width (in mm) Height (in mm) Math Talk Chapter 7.indd 160 10-07-2025 15:14:27 160 Can you check by what factors the width and height of image D change as compared to image A? Are the factors the same? Images A, C, and D look similar because their widths and heights have changed by the same factor. We say that the changes to their widths and heights are proportional. 7.2 Ratios We use the notion of a ratio to represent such proportional relationships in mathematics. We can say that the ratio of width to height of image A is The numbers 60 and 40 are called the terms of the ratio. The ratio of width to height of image C is 30 : 20, and that of image D is 90 : 60. So, in image A, we can say that for every 60 mm of width, there are 40 mm of height. We can say that the ratios of width to height of images A, C, and D are proportional because the terms of these ratios change by the same factor. Let us see how. Multiplying both the terms by 1 2, we get 60 × 1 2 : 40 × 1 2 which is 30 : 20, the ratio of width to height in image C. By what factor should we multiply the ratio 60 : 40 (image A) to get 90 : 60 (image D)? A more systematic way to compare whether the ratios are proportional is to reduce them to their simplest form and see if these simplest forms are the same. In a ratio of the form a : b, we can say that for every ‘a’ units of the first quantity, there are ‘b’ units of the second quantity. Image A — 60 : 40 60 : 40. Proportional Reasoning-1 Chapter 7.indd 161 10-07-2025 15:14:28 7.3 Ratios in their Simplest Form We can reduce ratios to their simplest form by dividing the terms by their HCF. In image A, the terms are 60 and 40. What is the HCF of 60 and 40? It is 20. Dividing the terms by 20, we get the ratio of image A to be 3 : 2 in its simplest form. The ratio of image D is 90 : 60. Dividing both terms by 30 (HCF of 90 and 60), we get the simplest form to be 3 : 2. So the ratios of images A and D are proportional as well. What is the simplest form of the ratios of images B and E? The ratio of image B is 40 : 20; in its simplest form, it is 2 : 1. The ratio of image E is 60 : 60; in its simplest form, it is 1 : 1. 161 Ganita Prakash | Grade 8 These ratios are not the same as 3 : 2. So, we can say that the ratios of width to height of images B and E are not proportional to the ratios of images A, C, and D. Thus, 60 : 40 :: 30 : 20 and 60 : 40 :: 90 : 60. 7.4 Problem Solving with Proportional Reasoning Example 1: Are the ratios 3 : 4 and 72 : 96 proportional? 3 : 4 is already in its simplest form. To find the simplest form of 72 : 96, we need to divide both terms by their HCF. What is the HCF of 72 and 96? The HCF of 72 and 96 is 24. Dividing both terms by 24, we get 3 : 4. Since both ratios in their simplest form are the same, they are proportional. Example 2: Kesang wanted to make lemonade for a celebration. She made 6 glasses of lemonade in a vessel and added 10 spoons of sugar to the drink. Her father expected more people to join the celebration. So he asked her to make 18 more glasses of lemonade. To make the lemonade with the same sweetness, how many spoons of sugar should she add? To maintain the same sweetness, the ratio of the number of glasses of lemonade to the number of spoons of sugar should be proportional. For 6 glasses of lemonade, she added 10 spoons of sugar. The ratio of glasses of lemonade to spoons of sugar is 6 : 10. If she needs to make 18 more glasses of lemonade, how many spoons of sugar should she use? We can model this problem as — When two ratios are the same in their simplest forms, we say that the ratios are in proportion, or that the ratios are proportional. We use the ‘::’ symbol to indicate that they are proportional. So a : b :: c : d indicates that the ratios a : b and c : d are proportional. Chapter 7.indd 162 10-07-2025 15:14:28 162 We know that each term in the ratio must change by the same factor, for the ratios to be proportional. How can we find the factor of change in the ratio? The first term has increased from 6 to 18. To find the factor of change, we can divide 18 by 6 to get 3. 6 : 10 :: 18 : ? The second term should also change by the same factor. When 10 increases by a factor of 3, it becomes 30. Thus, So, she should use 30 spoons of sugar to make 18 glasses of lemonade with the same sweetness as earlier. Example 3: Nitin and Hari were constructing a compound wall around their house. Nitin was building the longer side, 60 ft in length, and Hari was building the shorter side, 40 ft in length. Nitin used 3 bags of cement but Hari used only 2 bags of cement. Nitin was worried that the wall Hari built would not be as strong as the wall he built because she used less cement. Is Nitin correct in his thinking? In Nitin and Hari’s case, we should compare the ratio of the length of the wall to the bags of cement used by each of them and see whether they are proportional. The ratio in Nitin’s case is 60 : 3, i.e., 20 : 1 (in its simplest form). The ratio in Hari’s case is 40 : 2, i.e., 20 : 1 (in its simplest form). Since both ratios are proportional, the walls are equally strong. Nitin should not worry! Example 4: In my school, there are 5 teachers and 170 students. The ratio of teachers to students in my school is 5 : 170. Count the number of teachers and students in your school. What is the ratio of teachers to students in your school? Write it below. Is the teacher-to-student ratio in your school proportional to the one in my school? 6 : 10 :: 18 : 30. ______ : ______ Proportional Reasoning-1 Chapter 7.indd 163 10-07-2025 15:14:28 Example 5: Measure the width and height (to the nearest cm) of the blackboard in your classroom. What is the ratio of width to height of the blackboard? Can you draw a rectangle in your notebook whose width and height are proportional to the ratio of the blackboard? Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same? Math Talk Note to the Teacher: Give more such examples that students can relate to and ask them to give reasons why they think they are correct. Engaging with these problems and finding solutions through a process of proportional reasoning should go along with learning procedures and methods to solve the problems. ______ : ______ 163 Ganita Prakash | Grade 8 Example 6: When Neelima was 3 years old, her mother’s age was 10 times her age. What is the ratio of Neelima’s age to her mother’s age? What would be the ratio of their ages when Neelima is 12 years old? Would it remain the same? The ratio of Neelima’s age to her mother’s age when Neelima is 3 years old is 3 : 30 (her mother’s age is 10 times Neelima’s age). In the simplest form, it is 1 : 10. When Neelima is 12 years old (i.e., 9 years later), the ratio of their ages will be 12 : 39 (9 years later, her mother would be 39 years old). In the simplest form, it is 4 : 13. When we add (or subtract) the same number from the terms of a ratio, the ratio changes and is not necessarily proportional to the original ratio. Example 7: Fill in the missing numbers for the following ratios that are proportional to 14 : 21. In the first ratio, we don’t know the first term. But the second term is 42. It is 2 times the second term of the ratio 14 : 21. So, the first term should also be 2 times 14 (the first term). Hence the proportional ratio is 28 : 42. For the second ratio, the first term is 6. What factor should we multiply 14 by to get 6? Can it be an integer? Or should it be a fraction? We can model this as 14y = 6. So, y = 6 14 = 3 7 . So, we need to multiply 21 (the second term of 14 : 21) also by the same factor 3 7 . 21 × 3 7 is 9. So, the ratio is 6 : 9. In the third ratio, the first term is 2. We can see that when we divide 14 (the first term of 14 : 21) by 7 (HCF of 14 and 21) we get 2. If we divide 21 also by 7, we get 3. So, the ratio is 2 : 3. ______ : 42 6 : ______ 2 : ______ × 2=? × 2 ______ :42 14:21 Chapter 7.indd 164 10-07-2025 15:14:28 164 Filter Coffee! Filter coffee is a beverage made by mixing coffee decoction with milk. Manjunath usually mixes 15 mL of coffee decoction with 35 mL of milk to make one cup of filter coffee in his coffee shop. In this case, we can say that the ratio of coffee decoction to milk is 15 : 35. If customers want ‘stronger’ filter coffee, Manjunath mixes 20 mL of decoction with 30 mL of milk. The ratio here is 20 : 30. Why is this coffee stronger? And when they want ‘lighter’ filter coffee, he mixes 10 mL of coffee and 40 mL of milk, making the ratio 10 : 40. Why is this coffee lighter? The following table shows the different ratios in which Manjunath mixes coffee decoction with milk. Write in the last column if the coffee is stronger or lighter than the regular coffee. Figure it Out Coffee Decoction (in mL) Milk (in mL) Regular/Strong/ Light 300 600 150 500 200 400 100 300 24 56 Proportional Reasoning-1 Math Talk Math Talk Chapter 7.indd 165 10-07-2025 15:14:29 1. Circle the following statements of proportion that are true. 2. Give 3 ratios that are proportional to 4 : 9. 3. Fill in the missing numbers for these ratios that are proportional to 18 : 24. (i) 4 : 7 :: 12 : 21 (ii) 8 : 3 :: 24 : 6 (iii) 7 : 12 :: 12 : 7 (iv) 21 : 6 :: 35 : 10 (v) 12 : 18 :: 28 : 12 (vi) 24 : 8 :: 9 : 3 3 : ______ 12 : ______ 20 : ______ 27 : ______ ______ : ______ ______ : ______ ______ : ______ 165 Ganita Prakash | Grade 8 4. Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios. 5. Look at the following rectangle. Can you draw a smaller rectangle and a bigger rectangle with the same width to height ratio in your notebooks? Compare your rectangles with your classmates’ drawings. 6. The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form. Are all of them the same? If they are different from yours, can you think why? Are they wrong? A B C D E Math Talk Chapter 7.indd 166 10-07-2025 15:14:29 166 (b) (a) 7. Let us draw some human figures. Measure your friend’s body—the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below— Does the drawing look more realistic if the ratios are proportional? Why? Why not? Trairasika—The Rule of Three Example 8: For the mid-day meal in a school with 120 students, the cook usually makes 15 kg of rice. On a rainy day, only 80 students came to school. How many kilograms of rice should the cook make so that the food is not wasted? 15 × 2 3 = 10. So, the cook should make 10 kg of rice on that day. The situation above is a typical example of a problem where we need to use proportional reasoning to find a solution. Four quantities are Math Talk Note to the Teacher: In all these activities, encourage students to reason why their drawings are proportional. Now, draw a figure with head, torso, arms, and legs with equivalent ratios as above. head : torso ______ : ______ torso : arms ______ : ______ torso : legs ______ : ______ Proportional Reasoning-1 Chapter 7.indd 167 10-07-2025 15:14:29 The ratio of the number of students to the amount of rice needs to be proportional. So, 120 : 15 :: 80 : ? What is the factor of change in the first term? We can find that by dividing the terms 80 120 = 2 3. The number of students is reduced by a factor of 2 3 . On multiplying the weight of rice by the same factor, we get, 167 Ganita Prakash | Grade 8 linked proportionally, out of which three are known and we must find the fourth, unknown, quantity. To solve such problems, we can model two proportional ratios using algebraic notation as — a : b :: c : d. For these two ratios to be proportional, we know that term c should be a multiple of term a by a factor, say f, and term d should be a multiple of term b by the same factor f. So, ab × c a = ab × d b bc = ad or ad = bc Thus, when a : b :: c : d, then ad=bc. This is known as cross multiplication of terms. Since ad = bc, we can show that Two ratios are proportional if their terms are equal when cross multiplied. The fourth unknown quantity can be found through such cross multiplication. d = f b ...(2) From (1) and (2), we can say that, Therefore, c a = d b. Multiplying both sides by ab, we get, f = c a and f = d b . c = f a ...(1) ab × c a = ab × d b d = bc a . Chapter 7.indd 168 10-07-2025 15:14:29 168 In ancient India, Āryabhaṭa (199 CE) and others called such problems of proportionality Rule of Three problems. There were 3 numbers given—the pramāṇa (measure—‘a’ in our case), the phala (fruit—‘b’ in our case), and the ichchhā (requisition—‘c’ in our case). To find the ichchhāphala (yield—‘d’ in our case), Āryabhaṭa says, “Multiply the phala by the ichchhā and divide the resulting product by the pramāṇa.” In other words, Āryabhaṭa says, “pramāṇa : phala :: ichchhā : ichchhāphala,” therefore, pramāṇa × ichchhāphala = phala × ichchhā. ichchhāphala = phala × ichchhā pramāṇa . Using the cross multiplication method proposed by Āryabhaṭa, ancient Indians solved complex problems that involved proportionality. Example 9: A car travels 90 km in 150 minutes. If it continues at the same speed, what distance will it cover in 4 hours? If it continues at the same speed, the ratio of the time taken should be proportional to the ratio of the distance covered. Is this the right way to formulate the question? No, because 150 is in minutes, but 4 is in hours. The second ratio should use the same units for time as the first ratio. Since 4 hours is 240 minutes, the right form is How can you find the distance covered in 240 minutes? Discuss with your classmates and find the answer using different strategies. Note to the Teacher: Instead of giving one ‘method’ to solve the problem of the distance, encourage students to reason out the answer through different strategies. They can use their understanding of equivalent fractions and equivalent ratios to find the answer. Thus, We can model this proportion as 150 : 90 :: 240 : ? 150 : 90 :: 4 : ? Proportional Reasoning-1 Chapter 7.indd 169 10-07-2025 15:14:29 Therefore, x = 240 × 90 150 . 48 3 = 240 × 90 150 = 144. 5 1 The distance covered by the car in 4 hours is 144 km. Example 10: A small farmer in Himachal Pradesh sells each 200 g packet of tea for ₹200. A large estate in Meghalaya sells each 1 kg packet of tea for ₹800. Are the weight-to-price ratios in both places proportional? Which tea is more expensive? 150 : 90 :: 240 : x. By cross multiplication, we get 150 × x = 240 × 90 169 Ganita Prakash | Grade 8 The ratio of weight to price of the Himachal tea is 200 : 200. What is the weight to price ratio of the Meghalaya tea? Is it 1 : 800? This would not be appropriate, because we considered the weight in grams in the case of Himachal. So, the weight to price ratio is 1000 : 800 in Meghalaya after we convert the weight to grams. To check if the ratios are proportional, we need to see if both ratios are the same in their simplest forms. The Himachal tea ratio in its simplest form is 1 : 1. The Meghalaya tea ratio in its simplest form is 5 : 4. So, the ratios are not proportional. Which tea is more expensive? Why? To answer the question as to which tea is more expensive, we should compare the price of tea for the same weight in both places. What is the price of 1 kg of tea from Meghalaya? It is ₹800. In Himachal, if 200 g of tea costs rupees 200, what is the cost of 1 kg of tea? Let us say that the price of 1 kg of tea is x rupees. 200 g is 1 5 of 1 kg. So, 1 5 × x = 200. Multiplying both sides by 5, we get 1 5 × x × 5 = 200 × 5 1 5 × x × 5 = 1000 x = 1000. So, the cost of 1 kg of tea is ₹800 in Meghalaya and ₹1,000 in Himachal Pradesh. Therefore, the tea from Himachal Pradesh is more expensive. Note to the Teacher: Encourage a discussion on the more expensive tea, how they came to their conclusions and what the reasons for that tea being more expensive could be. Chapter 7.indd 170 10-07-2025 15:14:29 170 Activity 1: Take your favourite dish. Find out all the ingredients and their respective quantities needed to make the dish for your family. Suppose you are celebrating a festival and you want to invite 15 guests. Find out the quantities of the ingredients required to cook the same dish for them. Figure it Out 1. The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week? 2. A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle at a speed of 50 km/h. If he drives at 75 km/h, how long will it take him to reach Kanpur? Can we form this problem as a proportion— 50 : 2 :: 75 : __ Would it take Puneeth’s father more time or less time to reach Kanpur? Think about it. Even though this problem looks similar to the previous problems, it cannot be solved using the Rule of Three! The time of travel would actually decrease when the speed increases. So this problem cannot be modelled as 50 : 2 :: 75 : __. separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness. 12 ft 9 ft 6 ft 9 ft 15 ft Proportional Reasoning-1 Math Talk Chapter 7.indd 171 10-07-2025 15:14:29 Activity 2: Go to the market and collect the prices of different sizes of shampoo containers of the same shampoo and create a table like the one given below. See if the volume of shampoo is proportional to the price. Medium Bottle 340 mL ₹276 Small Bottle 180 mL ₹154 Large Bottle 1000 mL ₹540 Container Volume Price Sachet 6 mL ₹2 171 Ganita Prakash | Grade 8 Let us compare the ratios for the sample table above. The ratio of the volume of a sachet to a small bottle is 6 : 180. The ratio of their prices is 2 : 154. Are these ratios proportional? Why do you think that the ratio of the prices is not proportional to the ratio of the volumes? Discuss the pros and cons of different size bottles for the company and for customers. For reducing ecological footprint, what would you recommend to the company and to the customer? Does the same occur for other products? Make similar tables for other products in the market, capturing different prices for different measures of the same product, e.g., rice or atta (flour). Observe the products for which the prices are proportional to the different measures. Discuss in class the proportionality of prices to measures of the same product. 7.5 Sharing, but Not Equally! Activity 3: Form a pair. Collect 12 countable objects or counters (it can be coins, seeds, or pebbles). Now, share them between the two of you in different ways. Note to the Teacher: Give a project to students. Start by dividing the class into groups. Each group should go to one shop and collect prices for different measures of the same product. For example, they should note down the prices of 500 g of rice, 1 kg of rice, and 10 kg of rice. They should make tables with measure sizes and prices, and present them to the rest of the class. They should discuss if the prices are proportional or not, and why. Math Talk Chapter 7.indd 172 10-07-2025 15:14:29 172 If you divide them equally, what is the ratio of the number of counters with each of you? Each of you will get 6 counters. So, the ratio is 6 : 6, or 1 : 1 in its simplest form. Now let us not share equally. If your partner gets 5 counters, how many objects will you get? What is the ratio of the counters? Sharing 12 in the ratio 3 : 1 The ratio of the counters of your partner to yours is 5 : 7. Now, if you want to share the counters between the two of you in the ratio of 3 : 1, how many counters would each of you get? Share the counters in different ways and see which combination is in the ratio 3 : 1. One way to share the counters in the ratio of 3 : 1 is as follows — 1. Your partner takes 3 counters and you take 1 counter. There are now 8 counters left. 2. Your partner takes 3 more counters and you take 1 more counter. There are now 4 counters left. 3. Your partner takes 3 more counters and you take 1 more counter. There are no more counters left. So, your partner gets 9 counters in total and you get 3 counters. When we divide 12 counters in the ratio of 3 : 1 between two people, one gets 9 counters and the other gets 3 counters. If your partner gets 4 groups and you get 3 groups, the total number of groups is 7. So, the size of each group is 42 ÷ 7 = 6. Multiplying the number of groups by the size of each group, your partner gets 24 counters and you get 18 counters, when you share 42 counters in the ratio of 4 : 3. 12 × 3 × 3 3 : 1 9 : 3 The ‛whole’ is 12 The ‛whole’ is 4 (= 3 + 1) and 12 is 3 times 4 The ‛parts’ are 3 times the ratio 3 : 1 Proportional Reasoning-1 Chapter 7.indd 173 10-07-2025 15:14:29 Now, if you want to share 42 counters between the two of you in the ratio of 4 : 3, how will you do it? Using the same procedure would take a long time! There is a simpler way to share the whole with parts in a specified ratio. You need to divide 42 into groups such that your partner gets 4 groups and you get 3 groups. What is the size of each group? 173 Ganita Prakash | Grade 8 In general, when we want to divide a quantity, say x, in the ratio m : n, we do the following: 1. We need to split x into groups such that it can be divided into two parts where the first part has m groups and the second part has n groups. 2. But what is the size of each group? This can be found out by dividing x by the number of groups. The number of groups are m + n. So, the size of each group is x m + n . 3. So, the first part has m × x m + n objects and the second part has n × x m + n objects. Thus, if we want to divide a quantity x in the ratio of m: n, then the parts will be m × x m + n and n × x m + n . We see that Sharing 42 in the ratio 4 : 3 × 6 × 6 24 : 18 4 : 3 42 The ‛whole’ is 42 The ‛whole’ is 7 (= 4 + 3) and 42 is 6 times 7 6 times the ratio 4 : 3 The ‛parts’ are Chapter 7.indd 174 10-07-2025 15:14:29 174 Example 11: Prashanti and Bhuvan started a food cart business near their school. Prashanti invested ₹75,000 and Bhuvan invested ₹25,000. At the end of the first month, they gained a profit of ₹4,000. They decided that they would share the profit in the same ratio as that of their investment. What is each person’s share of the profit? The ratio of their investment is 75000 : 25000. Reducing this ratio to its simplest form, we get 3 : 1. 3 + 1 is 4 and dividing the profit of 4000 by 4, we get 1000. So, Prashanti’s share is 3 × 1000 and Bhuvan’s share is 1 × 1000. So, Prashanti would get ₹3,000 and Bhuvan would get ₹1,000 of the profit. Example 12: A mixture of 40 kg contains sand and cement in the ratio of 3 : 1. How much cement should be added to the mixture to make the ratio of sand to cement 5 : 2? m × x m + n : n × x m + n :: m : n. Let us find the quantity of sand and cement in the original mixture. The ratio is 3 : 1 and the total weight is 40 kg. So, the weight of sand is 3 (3 + 1) × 40 = 30 kg. The weight of cement is 1 (3 + 1) × 40 = 10 kg. The weight of sand is the same in the new mixture. It remains 30. But the new ratio of sand to cement is 5 : 2. So the question is, 5 : 2 :: 30 : ? If the ratio is 5 : 2, then the second term is 2 5 times the first term. Since the new ratio is equivalent to 5 : 2, the second term in the new ratio should also be 2 5 times of 30. 2 5 × 30 = 12. The new mixture should have 12 kg of cement if the ratio of sand to cement is to be 5 : 2. There is 10 kg of cement already. So, we need to add 2 kg of cement to the original mixture. Figure it Out 1. Divide ₹4,500 into two parts in the ratio 2 : 3. 2. In a science lab, acid and water are mixed in the ratio of 1 : 5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain? 3. Blue and yellow paints are mixed in the ratio of 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of these two colours are needed? To make the paint a lighter shade of green, I added 20 mL of yellow to the mixture. What is the new ratio of blue and yellow in the paint? Proportional Reasoning-1 Chapter 7.indd 175 10-07-2025 15:14:30 4. To make soft idlis, you need to mix rice and urad dal in the ratio of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need? 5. I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3 : 5. I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture? 7.6 Unit Conversions We have noticed earlier that solving problems with proportionality often requires us to convert units from one system to another. Here are 175 Ganita Prakash | Grade 8 a few important unit conversions for your reference. Temperature Temperature conversion between Fahrenheit and Celsius is a bit more complicated. 0o C = 32 o F, and Celsius = 5 9 × (Fahrenheit–32) For example, 25 o C is 77 o F. Figure it Out 1. Anagh mixes 600 mL of orange juice with 900 mL of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in its simplest form. 2. Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers who went on that trip and all the buses were full. This year we have 204 students. How many buses will we need? Will all the buses be full? Length Area 1 hectare = 2.471 acres Volume 1 millilitre (mL) = 1 cubic centimetre (cc) 1 square metre = 10.764 square feet 1 acre = 43,560 square feet 1 hectare = 10,000 square metres 1 litre = 1,000 mL or 1,000 cc Fahrenheit = 9 5 × Celsius + 32 1 metre = 3.281 feet and Chapter 7.indd 176 10-07-2025 15:14:30 176 4. A crane of height 155 cm has its neck and the rest of its body in the ratio 4 : 6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be? 5. Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If 21 2 palas of saffron 3. The area of Delhi is 1,484 sq. km and the area of Mumbai is 550 sq. km. The population of Delhi is approximately 30 million and that of Mumbai is 20 million people. Which city is more crowded? Why do you say so? 12. The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel are mixed in a 3 : 1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the cost of copper is ₹906 per kg and the cost of nickel is ₹1,341 per kg, what is the cost of these metals in a ₹10 coin? 10. One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land? 11. A tractor can plough the same area of a field 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an acre of land. How much time would it take if the farmer used a pair of oxen to plough the field? How much time would it take him if he decides to use a tractor instead? costs 3 7 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?” 6. Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2? 7. The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold? 8. It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy? (Please refer to the section on Unit Conversions earlier in this chapter). 9. A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 mL. How much time does the same tap take to fill a bucket of water if the bucket has a 10-litre capacity? Proportional Reasoning-1 Try This Chapter 7.indd 177 10-07-2025 15:14:30 Ratios in the form of a : b indicate that for every ‘a’ unit of the first quantity, there are ‘b’ units of the second quantity. ‘a’ and ‘b’ are the terms in the ratio. Two ratios—a : b and c : d—are proportional (written a : b :: c : d) if their terms change by the same factor, i.e., if ad = bc. If x is divided into two parts in the ratio m : n, then the quantity of the first part is m × x m + n and the quantity of the second part is n × x m + n. SUMMARY 177 Binairo, also known as Takuzu, is a logic puzzle with simple rules. Binairo is generally played on a square grid with no particular size. Some cells start out filled with two symbols: here horizontal and vertical lines. The rest of the cells are empty. The task is to fill cells in such a way that: 1. Each row and each column must contain an equal number of horizontal and vertical lines. 2. More than two horizontal or vertical lines can’t be adjacent. 3. Each row is unique. Each column is unique. Solve the following Binairo puzzles: Puzzle Solution Binairo Chapter 7.indd 178 10-07-2025 15:14:33 Dot Grid Chapter 7.indd 179 10-07-2025 15:14:33 Dot Grid Chapter 7.indd 180 10-07-2025 15:14:33" class_8,8,fractions in disguise,ncert_books/class_8/hegp2dd/hegp201.pdf,"1 1.1 Fractions as Percentages You might have heard statements like, “Mega Sale — up to 50% off!” or “Hiya scored 83% in her board exams”. Do you know what the symbol ʻ%ʼ means? This symbol is read as per cent. The word ‘per cent’ is derived from the Latin phrase ‘per centum’, meaning ‘by the hundred’ or ‘out of hundred’. So, 25 per cent (25%) means 25 out of every 100 — like 25 people out of 100, 25 rupees out of 100 rupees, or 25 marks out of 100 marks. If we say 50% of some quantity s, it means = 50 100 × s = 1 2s. Thus, percentages are simply fractions where the denominator is 100. Examples: 20% = 20 100 = 2 10 = 1 5 , FRACTIONS IN DISGUISE 50% = 50 × 1 100 × s (50 times the unit fraction of s) 01_Chapter 1.indd 1 20-12-2025 16:28:46 33% = 33 100. We saw that percentages are just fractions. Given any fraction, can we express it as a percentage? Yes, let us see how. Expressing Fractions as Percentages Example 1: Surya wants to use a deep orange colour to capture the sunset. He mixes some red paint and yellow paint to make this colour. The red paint makes up 3 4 of this mixture. What percentage of the colour is made with red? 3 4 is 3 out of every 4. That is, 6 out of every 8 (equivalent fraction). That is, 30 out of every 40. That is, 75 out of every 100. This means 75%. 3 4 = 6 8 = 30 40 = 75 100 Ganita Prakash | Grade 8 | Part-II Explaining this in a different way: To express 3 4 as a percentage, we need to find its equivalent fraction with 100 as the denominator. Two ways of going ahead are shown below. So, 3 4 can be expressed as 75%. Observe the following bar model diagram showing the equivalence between 3 4 and 75%. 0 0 Method 1 Method 2 3 4 = 3 × 25 4 × 25 = 75 100 = 75% (25%) (50%) (75%) (100%) 25 100 50 100 75 100 100 100 1 4 3 4 2 4 = 1 2 4 4 = 1 3 4 × 100 = x 100 × 100 [multiply both sides by 100] 3 4 = x 100 . x = 3 4 × 100 = 75. 01_Chapter 1.indd 2 20-12-2025 16:28:47 2 Can you tell what percentage of the colour was made using yellow? Example 2: Surya won some prize money in a contest. He wants to save 2 5 of the money to purchase a new canvas. Express this quantity as a percentage. Try to understand the different methods for solving this problem, as shown below. 2 5 = 20 50 = 40 100 = 40%. Method 1 Method 2 2 5 = x 100 . x = 2 5 × 100 = 40. Try completing Method 3 by filling the boxes. Method 3 Example 3: Given a percentage, can you express it as a fraction? For example, express 24% as a fraction. Since a percentage is a fraction, 24% is the same as 24 100. We can find other equivalent forms of 24 100 = 12 50 = 6 25 = 48 200. In general, we can say that a percentage, z%, can be expressed by any of the fractions that are equivalent to z 100. Figure it Out A fraction is of a unit, while a percentage is per 100. Therefore, to express a fraction as a percentage, we can just multiply the fraction by 100. Several problems in mathematics can be approached and solved in different ways. While the method you came up with may be dear to you, it can be amusing and enriching to know how others thought about it. 0% 1 5 2 5 3 5 4 5 5 5 = 1 0 Savings for canvas 100% Total prize money Isn’t it the same as finding 2 5 th of 100? Fractions in Disguise 01_Chapter 1.indd 3 20-12-2025 16:28:48 1. Express the following fractions as percentages. 2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white? (iv) 72 150 (v) 1 3 (vi) 5 11 (iv) 60% (v) 40% (vi) None of these (i) 3 5 (ii) 7 14 (iii) 9 20 (i) 10% (ii) 15% (iii) 25% 3 Ganita Prakash | Grade 8 | Part-II 3. In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking? 4. A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed. 5. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations. (i) 50% ____ 5% (ii) 5 10 ____ 50% Well, if percentages are just a particular type of fraction, why do we need them? Why can’t we just continue using fractions? Let us consider an example. biscuits. Sugar makes up 9 34 of Variety 1 and 13 45 of Variety 2. Which variety is more sugary? It may not be clear at first glance and we may have to do some calculations. But, when the same information is presented as — Sugar makes up 26.47% of Variety 1 and 28.88% of Variety 2, it is immediately clear which variety is more sugary. If we want to have the same denominator, why choose 100 in particular? Why not 10, 50, 1000, or 43? Think. In principle, we could choose any number as the denominator. But with 100, there are some advantages. Since our number system is base 10, numbers like 10, 100, and 1000 fit easily with decimals. For example, A biscuit-making factory is experimenting with 2 new varieties of (iii) 3 11 _____ 61% (iv) 30% ____ 1 3 Start Finish 55% 20% 38% 72% 84% 93% A B C D 01_Chapter 1.indd 4 20-12-2025 16:28:48 4 31% = 31 100 = 0.31. Converting between fractions, decimals, and percentages becomes quick and intuitive. The number 100 is round, and easy to understand and work with. We could say “per 1000” or “per 100,000” (this usage is present in statistics like “per thousand people” or “per lakh”), but 100 hits the sweet spot — it’s large enough to give detail, yet simple enough to grasp mentally. Per 10 would be too small for many purposes. 9 34 = 26.47 percent (per 100) 9 34 = 2.647 per decem (per 10) 9 34 = 264.7 per mille (per 1000) Long before the decimal fraction was introduced, the need for it was felt in computations by tenths, twentieths, and hundredths. The idea of ‘per hundred’ can be found as early as the 4th century BCE in Kautilya’s Arthaśhāstra, “An interest of a pana and a quarter per month per cent is just. Five panas per month per cent is commercial interest. Ten panas per month per cent prevails among forests. Twenty panas per month per cent prevails among sea traders”. Around the same time, the Romans used taxes of 1 20, 1 100 in transactions related to trade and auctions. In the Italian manuscripts of the 15th century, expressions such as ‘xx p cento’, ‘x p cento’, ‘vii p cento’ can be found (equivalent to our 20%, 10%, and 7%). Percentages Around Us Percentages are widely used in a variety of contexts. Here are some interesting findings that involve percentages. The human body, on average, is about 60% water by weight. Ice cream is about 30 – 50% air by volume. 45% of the world̛ s population watched at least part of the 2022 FIFA World Cup. Fractions in Disguise 01_Chapter 1.indd 5 20-12-2025 16:28:49 Over 80% of the teenagers globally fail to meet the recommendation of at least one hour of daily physical actvity. About 99.86% of the Solar System̛s mass is contained in the Sun. An estimated 52% of the agricultural land worldwide is degraded. 5 25% sugar means — 25 g sugar per 100 g of biscuits, which means 5 g sugar per 20 g biscuits, so 30 g sugar per 120 g biscuits. Ganita Prakash | Grade 8 | Part-II 1.2 Percentage of Some Quantity Example 1: Madhu and Madhav each ate biscuits of a different variety. Madhu’s biscuits had 25% sugar, while Madhav’s had 35% sugar. Can you tell who ate more sugar? As we just saw, percentages represent fractional quantities or proportions. It would be inappropriate to compare just the percentages when they are referring to different quantities or wholes. That is, if they both had 100 g of biscuits, then clearly Madhav ate more sugar — 35 g (35% of 100 g is 35 g per 100 g) vs. Madhu’s 25 g (25% of 100 g is 25 g per 100 g). Suppose Madhu ate 120 g of biscuits and Madhav ate 95 g of biscuits. Who consumed more sugar? Try to find out. We know that the weight of sugar is proportional to the weight of the biscuits consumed. Madhu ate 120 g of biscuits having 25% sugar. The amount of sugar he ate is the value in the blank — 25 : 100 :: ____ : 120. A few ways of going forward are shown. The proportional relationship can be written as 25 100 = s 120. s = 25 100 × 120 = 30 . 0 0 Sugar 25% 50% 75% 100% 30 g 60 g 90 g 120 g 01_Chapter 1.indd 6 20-12-2025 16:28:49 6 Do any of the methods match your thinking? Were you able to understand all the methods? Now, let us find out how much sugar Madhav ate. He ate 95 g of biscuits with 35% sugar. Sometimes the numbers may not be convenient to calculate using different methods as we did just before. 100 g of the biscuits he ate has 35 g of sugar. relationship 35 100 = s 95. Therefore, Madhav ate more sugar. This means, 1 g of the biscuits has 35 100 g of sugar. We can say that 95 g of the biscuit will have 35 100 g × 95 = 33.25 g. This can also be solved by finding the value of s in the proportional 35 100 = x 1 → x = 35 100 Free-hand Computations We just calculated 25% of 120. Is 25% the same as 1 4th (a quarter)? Suppose we want to find 25% of 40. Is it the same as 1 4th of 40? Yes, since 25 is 1 4th (a quarter) of 100 ( 25 100 = 1 4). Therefore 25% of 40 = 25 100 × 40 is the same as 1 4 × 40. Try to calculate (without using pen and paper) the indicated percentages of the values shown in the table below. Write your answers in the table. More generally, y% of some value, say 80, is given by y 100 × 80. We can also say that 45% of some value, say z, is given by 45 100 × z. 25% 25 10% 20% 5% 100 200 50 80 10 35 287 Fractions in Disguise 01_Chapter 1.indd 7 20-12-2025 16:28:49 How did you find these values? Discuss the methods with the class. Do you find anything interesting in the table? You may have noticed that 20% of a value is double that of 10% of the same value. This will always happen as 20% (20 parts out of 100) is twice that of 10% (10 parts out of 100). Using this understanding, mentally calculate how much 40% of the values in the table above would be. What relationship do you observe among 20%, 5% and 25% of a value? It appears that (20% of y) + (5% of y) = 25% of y. We can verify that this property always holds: ( 20 100 × y) + ( 5 100 × y) = ( 25 100 × y). Math Talk 7 Ganita Prakash | Grade 8 | Part-II Using this observation, mentally calculate how much 15% of the values in the table would be. Suppose you have to mentally calculate the following percentages of some value: 75%, 90%, 70%, 55%. How would you do it? Discuss. The FDP Trio — Fractions, Decimals, and Percentages Example 2: We can find 50% of a value by multiplying 1 2 with the value. Will multiplying the value by 0.5 also give the answer for 50% of the value? Yes, since 1 2 = 0.5. Similarly, to find 10% of a quantity, what decimal value should be multiplied? Complete the following table: Per cent 50% 100% 25% 75% 10% 1% 5% 43% Fraction 50 100 Decimal 0.5 50% = 50 100 = 1 2 = 0.5 1 = 0.5. 50% of 24 = 12 0.5 × 24 = 12 Math Talk 01_Chapter 1.indd 8 20-12-2025 16:28:49 8 Example 3: The maximum marks in a test are 75. If students score 80% or above in the test, they get an A grade. How much should Zubin score at least to get an A grade? We can find 80% of 75 in different ways, using our understanding of fraction and decimal multiplication, as well as of proportionality. Make a pair. Each of you choose a number. Suppose, the numbers chosen are a and b. Share your numbers with each other. Both of you should estimate the percentage equivalent to the fraction a b (where a < b) and announce your answers by a fixed time, say, 5 seconds. The one whose estimate is the closest wins this round. Play this for 10 rounds. Activity: How Close Can You Get? Fraction Multiplication → 80 100 × 75 = 4 5 × 75 = 60. Decimal Multiplication → 0.8 × 75 = 60. Proportional Reasoning → Out of 100, the minimum mark is 80. Out of 75, it is Example 4: To prepare a particular millet kanji (porridge), suppose the ratio of millet to water to be mixed for boiling is 2:7. What percentage does the millet constitute in this mixture? If 500 ml of the mixture is to be made, how much millet should be used? This situation can be modelled as shown in the bar model on the right side. The ratio of millet to the volume of the mixture is 2:9. In other words, in one unit of the mixture, millet occupies 2 9 units and water occupies 7 9 units. Estimate first what percentage 2 9 would be. The percentage (i.e., in 100 such units) of millet in the mixture is The percentage of water in the mixture will be 100 – 22.22 = 77.78%. A mixture with 22.22% millet means 100 ml mixture will have 22.22 ml millet. Therefore, 500 ml with 22.22% millet will have 5 × 22.22 = 111.1 ml of millet. 2 9 × 100 = 22.22%. 75 × 80 100 = 60. 0 0 1.0 (100%) Total marks Minimum marks for A grade 0.2 (20%) 0.4 (40%) 0.6 (60%) 0.8 (80%) So, 2 9 is clearly less than 50%. Half of 4.5 is 2.25. So, 2 9 is less than 25%. 15 30 45 60 Half of 9 is 4.5. 2 7 9 Fractions in Disguise 10% of 9 is 0.9. 20% of 9 is 1.8. So, 2 9 could be between 20% – 25%. 75 01_Chapter 1.indd 9 20-12-2025 16:28:50 A given ratio can be converted to a fraction, and then to a percentage. Note to the Teacher: Incorporate the practice of estimating before calculating or solving as part of the problem-solving process. You may remind and encourage students as needed. The practice of estimating first before calculating can improve number sense and help reduce mistakes. Often, we may not need exact values. The ability to make quick estimates is useful at these times. 9 Ganita Prakash | Grade 8 | Part-II 40% is 92 km, therefore 20% is 46 km. This makes 60% to be 92 + 46 = 138 km. Example 5: A cyclist cycles from Delhi to Agra and completes 40% of the journey. If he has covered 92 km, how many more kilometres does he have to travel to reach Agra? Let us first try to model this situation by a bar model. Estimate first before solving further. A few ways of solving this problem are shown below. Does your method match any of the given ones? Do you like any of the other methods? It is given that 40% of the distance is 92 km. We have to find out how much the rest of the 60% distance is. If 40% is 92, 60% is ? 40 : 92 :: 60: ? 40 92 = 60 r (r is the remaining distance) r = 60 × 92 40 = 138. Method 1 Method 3 Method 4 Method 2 40 100 = 92 d (d is the total distance) d = 92 × 100 40 = 230. Remaining distance = 230 km – 92 km = 138 km. Delhi Agra ? 92 km 40% 100% If x is the remaining distance then the total distance from Delhi to Agra is x + 92. Since, we know that 92 is 40% of this total distance, 40 100 × (x + 92) = 92. (x + 92) = 92 × 100 40 x = 230 – 92 = 138. 01_Chapter 1.indd 10 24-12-2025 16:20:18 10 Percentages Greater than 100 Till now, we saw percentages with a value 100 or less than 100. Can there be percentages with a value more than 100? What could it mean when a percentage is greater than 100? Let us explore. Example 6: Kishanlal recently opened a garment shop. He aims to achieve a daily sales of at least ₹5000. The sales on the first 2 days were ₹2000 and ₹3500. What percentage of his target did he achieve? Drawing rough diagrams can help understand the given situation better and make it easier to think further about the problem. The percentage target achieved is visualised below. It is 40% on Day 1 and 70% on Day 2. Another way of saying it is — he was 60% short of his target on Day 1 and 30% short of his target on Day 2. In the next two days, he made ₹5000 and ₹6000 respectively. What percentage of his target are these values? His target is ₹5000, and he made ₹5000 on Day 3 — this is 100%. On Day 4, he made ₹6000, which is 1000 more than his target. What percentage of the target was achieved on Day 4? 1000 is 20% of 5000. Therefore, 6000, (5000 + 1000) is 100% + 20% 0% Day 3: 5000 0 0 Day 1: 2000 0 % of target: 5000 5000 × 100 = 100% 100% = 100 100 = 1 1 = 1.0 % of target: 2000 5000 × 100 = 40% 40% = 40 100 = 2 5 = 0.4 100% 5000 100% 5000 0 0 0% 0 % of target: 6000 5000 × 100 = 120% 120% = 120 100 = 6 5 = 1.2 % of target: 3500 5000 × 100 = 70% 70% = 70 100 = 7 10 = 0.7 5000 Day 2: 3500 Day 4: 6000 5000 Fractions in Disguise 100% 100% 20% 01_Chapter 1.indd 11 20-12-2025 16:28:51 = 120% of 5000. It can also be computed as 6000 5000 × 100 = 6 5 × 100 = 120%. This means he achieved 120% of his target, i.e., 20% more than his target. On Days 5 and 6 his sales were ₹7800 and ₹9550 respectively. Calculate the percentage of the target achieved on these days. 0% Day 5: 7800 0 0% 0 Day 6: 9550 100% 100% 5000 5000 2800 4550 200% 10,000 10,000 200% 11 Ganita Prakash | Grade 8 | Part-II (0) (1) (2) (3) (4) 0% 100% 200% 300% 400% On Day 7, he achieved 150% of his target. On Day 8, he achieved 210% of his target. Find the sales made on these days. Suppose on some day, he made ₹2500. This can be expressed as “He achieved 1 2 of his target” or “He achieved 50% of his target” or “He achieved 0.5 of his target”. On some other day, he made ₹10,000. We can say “He achieved twice/double/2 times his target” or “He achieved 200% of his target”. Complete the table below. Mark the approximate locations in the following diagram. Example 7: A farmer harvested 260 kg of wheat last year. This year, they harvested 650 kg of wheat. What percentage of last year’s harvest is this year’s harvest? Fraction Decimal Percent 90% 110% 200% 250% 15% 173% 358% 28.9% 305% 90% 01_Chapter 1.indd 12 20-12-2025 16:28:51 12 This year’s harvest = 650 260 × 100 = 250% of last year’s harvest. 250% indicates that it is 2.5 times the original value. Figure it Out Estimate first before making any computations to solve the following questions. Try different methods including mental computations. 1. Find the missing numbers. The first problem has been worked out. (ii) (i) 20% 75 ? 90 100% 100% ? 60 100% 100% 2. Find the value of the following and also draw their bar models. 3. Surya made 60 ml of deep orange paint, how much red paint did he 4. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the boxes. Visualising or estimating can help. Compute only if necessary or for verification. 5. Fill in the blanks appropriately: (i) 25% of 160 (ii) 16% of 250 (iii) 62% of 360 (iv) 140% of 40 (v) 1% of 1 hour (vi) 7% of 10 kg (iii) use if red paint made up 3 4 of the deep orange paint? (i) 50% of 510 50% of 515 (ii) 37% of 148 73% of 148 (iii) 29% of 43 92% of 110 (iv) 30% of 40 40% of 50 (v) 45% of 200 10% of 490 (vi) 30% of 80 24% of 64 (iii) 90% of n is 270, 9% of n is ______, 18% of n is _____, 100% of n is ______. (iv) Make 2 more such questions and challenge your peers. (ii) 100% of m is 215, 10% of m is _____, 1% of m is ______, 6% of m is ______. (i) 30% of k is 70, 60% of k is _____, 90% of k is _____, 120% of k is ______. ? 140 100% ? 100% Fractions in Disguise 01_Chapter 1.indd 13 20-12-2025 16:28:51 6. Fill in the blanks: 7. Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers. 8. Mariam’s farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day, she gave the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe? (ii) _____ is 40% of 4. (iii) 40 is 80% of _____. (i) 3 is ____ % of 300. Math Talk 13 Ganita Prakash | Grade 8 | Part-II 1.3 Using Percentages To Compare Proportions Example 1: Eesha scored 42 marks out of 50 on an English test and 70 marks out of 80 in a Science test. Since she lost only 8 marks in English but 10 marks in Science, she thinks she has done better at English. Reema does not agree! She argues that since Eesha has scored more marks in Science, she has done better at Science. Vishu thinks we cannot compare the scores because the maximum marks are different. Who do you think is correct? 10. The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10% : 80% : 10%. If he wants to conduct a training of 90 minutes. How long should each activity be done? 11. An estimated 90% of the world’s population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year’s worldwide population. 12. A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions — Rava: 40%, Sugar: 40%, and Ghee: 20%. 9. Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary? (ii) If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present? (i) If you want to make halwa for 8 people, what is the proportion of each of the above ingredients? 01_Chapter 1.indd 14 20-12-2025 16:28:51 14 If the maximum marks are the same, the comparison becomes easier, isn’t it? For these kinds of comparisons, we need to convert both values to percentages. English score as a percentage = 42 50 × 100 = 84% Science score as a percentage = 70 80 × 100 = 87.5%. The Science score (as a percentage) is higher than the English score (as a percentage). So, we can conclude that Eesha has scored better on the Science test. Zacni DEF 66% Know Your Contents (KYC) Example 2: Madhu and Madhav recently learnt about the importance of reading labels on processed food before purchase. They are at a shop to buy badam drink mix. They are looking at two products and wondering which has a larger share of badam. Can you figure it out? Which product uses a smaller proportion of food chemicals? It is easier to compare the proportions of the ingredients if we convert them into percentages. For example, DEF’s sugar content as a percentage of total weight = 99 150 × 100 = 66%. Complete this table by calculating the percentages to answer the questions: Sugar Milk Solids Badam Powder Food Chemicals Maybe they should call it ʻSugar drink mixʼ Fractions in Disguise Ya! It is very important to Know Your Contents. Do KYC! 01_Chapter 1.indd 15 20-12-2025 16:28:55 Check if the percentages of each product add up to 100. Percentage Increase or Decrease Percentages are often used to describe the rate of change of quantities. For example, 1. Suppose the price of 1 kg tomatoes 3 years ago was ₹30, and the price now is ₹42. The increase in the price is ₹12. = 12 30 × 100 = 40%. We say the price of tomatoes increased by 40% over the last 3 years. Percentage increase = amount of increase original amount or base × 100 15 Ganita Prakash | Grade 8 | Part-II 2. The average footfall in this theater before COVID was 160. Now it is just 100. The decrease in the footfall is 60. = 60 160 × 100 = 37.5%. We say the footfall in this theater post-COVID has decreased by 37.5%. Example 3: Do the following two statements mean the same thing? (ii) The population of this state has increased by 65% from 1961 to 1991. Yes, both mean the same. Suppose p is the population of the state in 1961 and q is the population of the state in 1991. In other words, the population of the state in 1991 is 1.65 times that in 1961. Profit and Loss You may have the experience of buying something — snacks, groceries, clothes, toys, etc. Very often, the shopkeeper quotes a price and after some bargaining the customer pays the negotiated amount and buys the item(s). We call the price quoted by the shopkeeper the marked price (sometimes this can be the MRP of an item). The price the customer pays after a discount is called the selling price. Also, the price the shopkeeper paid to purchase that item is called the cost price. Let us see how these labels are relative to the context through an example. The picture below shows the journey of a sweater from the manufacturer to the customer and what the cost price (CP), the marked price (MP), and the selling price (SP) mean in each step. (i) The population of this state in 1991 is 165% of that in 1961. Statement A implies, q = 165% of p Percentage decrease = amount of decrease original amount or base × 100 q = 165 100 × p = 1.65p Statement B implies, q = p + 65% of p q = p + 0.65 × p = 1.65p 01_Chapter 1.indd 16 20-12-2025 16:28:56 16 Kishanlal (retailer) buys sweaters from a wholesaler at a price of ₹300 per sweater. The marked price he quotes his customers is ₹480. After bargaining, he sells this sweater at ₹430. Notice that the selling price is greater than the cost price, resulting in a profit of ₹430 – ₹300 = ₹130. If the selling price is less than the cost price, it will result in a loss. Example 4: Find out the percentage profit Kishanlal made on this sweater. We shall consider the cost price to be 100% to find out the percentage profit made with reference to the cost price. The following rough diagram describes this situation. The profit amount is ₹130. The percentage profit is 130 300 × 100 = 43.3%. Find the profit percentage of the wholesaler and the manufacturer. Shambhavi owns a stationery shop. She procures 200 page notebooks at ₹36 per book. She sells them with a profit margin of 20%. Find the selling price. She sells crayon boxes at ₹50 per box with a profit margin of 25%. How much did Shambhavi buy them from the wholesaler? Example 5: The rice stock in Raghu’s provision store is getting old. He had purchased the rice at ₹35 per kg. To clear his stock, he sells 10 kg rice for ₹300. Find out the percentage loss. The amount Raghu had paid towards buying the 10 kg rice is ₹350. He sold it for ₹300. 0 0 130 Kishanlal’s buying price 0 0 Selling price Fractions in Disguise 100% 300 430 350 ? 01_Chapter 1.indd 17 20-12-2025 16:28:56 The loss is ₹350 – ₹300 = ₹50. The percentage loss is 50 350 × 100 = 14.28%. Could we have just calculated the loss percentage per kg instead? Would it be the same? Example 6: Shyamala had procured decorative vases at ₹2650 per piece. One of the pieces was slightly damaged. She decides to sell it at a loss of 18%. How much will she get by selling this piece? Try making an estimate. Draw a rough diagram depicting the given situation. 50 Selling price Cost price 100% 17 Ganita Prakash | Grade 8 | Part-II Two methods of solving this are shown. The sale amount of the damaged vase would be ₹2173. Due to heavy rains, Snehal could not transport strawberries to Hyderabad from his farm in Panchgani. He sells some of his stock at ₹80 per kg with a 12% loss. What is the cost price? You have probably seen percentages mentioned when shops offer discounts! Do you know what 30% off (or 30% discount) means? It means that the shop is willing to reduce the price of the item by 30%. A utensil store is offering a 35% discount on the cooker with an MRP ₹1800. What is the selling price? If the cost price was ₹900, what is the percentage profit made after the sale? With respect to the buying price being 100%, the selling price is 18% less than the buying price. That is, the selling price would be 82%. 82% of 2650 = 0.82 × 2650 = 2173. Suppose Kishanlal achieved a sales of ₹80,000 last month. Out of this, the amount he spent on buying these goods he sold last month was ₹48,000. The difference amount, ₹80,000 – ₹48,000 = ₹32,000, is called gross profit. After deducting the other expenses he incurred (such as transport cost, employee’s salary, electricity bill, etc.) amounting to ₹8,000, the amount remaining is called net profit, which is ₹32,000 – ₹8,000 = ₹24,000. Note: In this chapter, the term profit refers to gross profit. The loss amount is 18% of 2650. That is, 18 100 × 2650 = 477. Reducing this from the buying price, 2650 – 477 = 2173. 01_Chapter 1.indd 18 20-12-2025 16:28:59 18 Manisha sells fertiliser. She buys 50 kg bags at ₹500 per bag. She sells it at ₹750 per bag making a profit of 250 500 × 100 = 50%. Although, with respect to the ₹750 amount she has earned by selling a bag, the profit percentage is 250 750 × 100 = 33.33%. Profit percentage is calculated based on the price the goods bought when we want to know “How much profit did I make compared to what I invested in buying the goods?”. 0 0 0 0 100% 500 500 33.33% 50% 250 250 750 100% Taxes Tax rates, like the GST (Goods and Services Tax) rate or the Income Tax rates, are also specified as percentages. You may have noticed GST mentioned as part of bills. This means that the tax is part of the amount we pay and this amount goes to the government. Check if the calculations are correct in the bill shown. You may share any bills you have at home with the class. Observe the different elements present in the bills. Are there any similarities or differences in these bills? Figure it Out Profit percentage is calculated based on the revenue (sales amount) when we want to know “How much (net) profit did I make on my overall revenue?”. Suppose, in a month she made sales of ₹1,50,000. The cost of buying the goods was ₹1,00,000. So, the gross profit is ₹50,000. The monthly expenses amounted to ₹5,000. The net profit is ₹45,000. Out of monthly revenue (₹1,50,000), the net profit percentage is 45000 150000 × 100 = 3 10 × 100 = 30%. 1. If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost? Fractions in Disguise Try This 01_Chapter 1.indd 19 20-12-2025 16:29:00 2. I am a carpenter and I make chairs. The cost of materials for a chair is ₹475 and I want to have a profit margin of 50%. At what price should I sell a chair? 3. The total sales of a company (also called revenue) was ₹2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year? 4. A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar have to pay to buy this shirt? 5. The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol? (iv) 66.66% (v) 140% (vi) 160.66% (i) 50% (ii) 40% (iii) 60% 19 Ganita Prakash | Grade 8 | Part-II 3. Samson bought a car for ₹4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car? 4. 1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get? Can you guess the minimum number of candidates who stood for the election? 5. The price of 1 kg of rice was ₹38 in 2024. It is ₹42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.) 6. A number increased by 20% becomes 90. What is the number? 7. A milkman sold two buffaloes for ₹80,000 each. On one of them, he made a profit of 5% and on the other a loss of 10%. Find his overall profit or loss. 8. The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants last decade is p, the population now is 9. Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”? (iii) The demand now is 15% of the demand a decade ago. (i) p × 0.5 (ii) p × 0.05 (iii) p × 1.5 (iv) p × 1.05 (v) p + 1.50 (iv) The demand a decade ago was 15% of the demand now. (ii) The demand a decade ago was 85% of the demand now. (v) The demand a decade ago was 185% of the demand now. (i) The demand now is 85% of the demand a decade ago. 01_Chapter 1.indd 20 20-12-2025 16:29:00 20 Growth and Compounding You might have come across statements like, “1 year interest for FD (Fixed Deposit) in the bank @ 6% per annum” or “Savings account with interest @ 2.5% per annum”. Interest is the extra money paid by institutions like banks or post offices on money deposited (kept) with them. Interest is also paid by people or institutions when they borrow money. In a Fixed Deposit (FD), you deposit a specific amount of money for a fixed period at a predetermined interest rate. The money remains locked for the chosen duration, and the bank pays you interest on it. You cannot withdraw the amount before the maturity date without incurring a penalty. At the end of the term, you receive both your original deposit and the interest earned. (vi) The demand now is 185% of the demand a decade ago. For example, a bank says that the interest rate for fixed deposits is 10% p.a. What do you understand by this statement? What is the ‘p.a.’ next to the percentage? ‘p.a.’ is the short form of per annum, which means for every year. The specified interest rate indicates that if you invest ₹6000 as a deposit for a year with the bank, they will give you ₹600 as interest on this deposit. The 10% is the rate of interest and the ₹6000 on which the interest is calculated is the principal. The amount after 1 year in your bank account will be In other words, the amount deposited, ₹6000, will become 110% (or increase by 10%), This can also be expressed as Example 7: If one deposits ₹6000 in the bank, what is the amount after 3 years? That depends on the choice of FD. There are two possibilities: 1. Option 1: The interest is paid out regularly (for example, every year). The principal amount is returned after the maturity period. (principal) + (10% interest on principal) Amount after 1 year = Principal (P) + rate of interest (r) of P = 6000 + 600 = 6600. 6000 + (0.10 × 6000) 6000 × 110% = 6000 × 110 100 = 6000 × 1.1 = 6600. Interest returned (10% p.a.) Amount in the FD 0 0 Principal (initial deposit) Amount after 1 year Fractions in Disguise 100% 1000 ? 10% 01_Chapter 1.indd 21 20-12-2025 16:29:00 Year 1 Year 2 Year 3 Total amount received ₹1800 + ₹6000 = ₹7800 Beginning ₹6000 Beginning ₹6000 Beginning ₹6000 Ending ₹600 returned ₹6000 Ending ₹600 returned ₹6000 Ending ₹600 returned ₹6000 × 0.10 × 0.10 × 0.10 21 Ganita Prakash | Grade 8 | Part-II 2. Option 2: The interest gained every time (say after each year) is added back to the FD, thus increasing the principal amount for the subsequent period. After the maturity period, the entire amount is returned. This phenomenon is called compounding. Example 8: What percent is the total amount received with respect to the amount deposited in both the options? This can be calculated by finding total amount received amount deposited × 100. We can see that with compounding, the final amount is more. Year 1 Year 2 Year 3 Total amount received ₹7986 Beginning ₹6000 Beginning ₹6600 Beginning ₹7260 Ending ₹600 added back ₹6600 Ending ₹660 added back ₹7260 Ending ₹726 ₹7986 Interest added back (10% p.a.) Amount in the FD ₹7986 is returned × 0.10 × 0.10 × 1.10 × 0.10 × 1.10 × 1.10 01_Chapter 1.indd 22 20-12-2025 16:29:00 22 7800 6000 × 100 = 130% = 1.3. In other words, the total amount received = 6000 × (1 + 0.1 + 0.1 + 0.1) = 6000 × 1.3. The percentage gain over 3 years is 30%. Figure it Out 1. Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually. Without Compounding With Compounding 7986 6000 × 100 = 133.1% = 1.331. In other words, the total amount received = 6000 × 1.1 × 1.1 × 1.1. = 6000 × 1.331 The percentage gain over 3 years is 33.1%. The interest gained in 1 term is 6000 × 0.1 The interest gained in 3 terms is 6000 × 0.1 × 3 The total amount at the end of an FD of 3 years is 6000 + (6000 × 0.1 × 3). 2. Bank of Wahapur offers an interest of 5% p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually. 3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss. Let us try to generalise the pattern observed in each of the options. Example 9: What is the amount we get back if we invest ₹6000 at an interest rate of 10% p.a. for ‘t’ years? No Compounding Here, the interest gained every term is paid back. Therefore, the principal amount for every term shall remain the same, and as a result, the interest gained every term also shall be the same. With Compounding Here, the interest gained every term/year is added back to the FD. Therefore, the principal amount increases after every term, and as a result, the interest gained every term also increases proportionately. The interest gained in 1 term is p × r (p is the principal, r is the rate of interest in percentage) The interest gained in t terms is p × r × t The total amount at the end of an FD of t years is p + (p × r × t )= p + prt = p (1 + rt). Fractions in Disguise 01_Chapter 1.indd 23 20-12-2025 16:29:00 The total amount in the FD after Year 1 is (6000) × 1.1 (principal for Year 1) The total amount in the FD after Year 2 is (6000 × 1.1) × 1.1 (principal for Year 2) The total amount in the FD after Year 3 is (6000 × 1.1 × 1.1) × 1.1 (principal for Year 3) The total amount in the FD after t years is 6000 × (1.1 × 1.1 × 1.1…..× 1.1) t times 6000 × (1.1)t . The total amount in the FD after Year 1 is (p) × (1 + r) (principal for Year 1) The total amount in the FD after Year 2 is p × (1 + r) × (1 + r) (principal for Year 2) The total amount in the FD after Year 3 is p × (1 + r) × (1 + r) × (1 + r) (principal for Year 3) The total amount in the FD after t years is p × (1 + r) × (1 + r) × …..× (1 + r) t times p × (1 + r)t . 23 Ganita Prakash | Grade 8 | Part-II Suppose we want to know the expression/formula to find the total interest amount gained at the end of the maturity period. What would be the formula for each of the two options? Figure it Out 4. Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done? 5. The post office offers an interest of 7% p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded? 6. Giridhar borrows a loan of ₹12,500 at 12% per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much? 7. Consider an amount ₹1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth? 8. The population of a city is rising by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years? (i) p × 6 × 4 (ii) p × 0.6 × 4 (iii) p × 0.6 100 × 4 (iv) p × 0.06 100 × 4 (v) p × 1.6 × 4 (vi) p × 1.06 × 4 (vii) p + (p × 0.06 × 4) Math Talk Math Talk 01_Chapter 1.indd 24 20-12-2025 16:29:00 24 9. In a laboratory, the number of bacteria in a certain experiment increases at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000. Decline Several items or materials lose financial value over time. Suppose someone buys a bike at your home for ₹1,00,000 and after a few years they want to sell it. The value of the bike at that time will be less than ₹1,00,000. It could depend on various factors, including how many years have passed since the purchase, how many kilometres the vehicle has been used for, if there has been any damage, or if any parts have been replaced. This is called depreciation — reduction of value due to use and age of the item. The population 1 decade later will be 0.9 times the population of the current decade. Therefore, the population after 1 decade will be 1250 × 0.9. The population after 2 decades will be 1250 × 0.9 × 0.9. The population after 3 decades will be 1250 × 0.9 × 0.9 × 0.9 = 911.25. The amount of reduction in the value is 5% of 21,000 = 0.05 × 21,000 = 1050. The current value is 21,000 – 1050 = 19,950. Example 10: A TV is bought at a price of ₹21,000. After 1 year, the value of the TV depreciates by 5%. Find the value of the TV after one year. The value of the TV after 1 year will be ₹19,950. Example 11: The population of a village was observed to be reducing by about 10% every decade. If the current population is 1250, what is the expected population after 3 decades? Rounding off, we can say that the expected population after 3 decades will be around 910. First decade’s decrease = 0.1 × 1250 = 125. Population after 1 decade = 1250 – 125 = 1125. Second decade’s decrease = 0.1 × 1125 = 112.5 ≅ 112. Population after 2 decades = 1125 – 112 = 1013 Third decade’s population decrease = 0.1 × 1013 = 101.3 ≅ 101. Population after 3 decades = 1013 – 101 = 912. The value of the TV after 1 year will be 95% of the current value = 95% of 21,000 = 0.95 × 21,000 = 19,950. Fractions in Disguise 01_Chapter 1.indd 25 20-12-2025 16:29:01 Tricky Percentages Would You Rather? You have won a contest. The organisers offer you two options to choose from: Option A: You deposit ₹100 and you get back ₹300. Option B: You deposit ₹1000 and you get back ₹1500. What is the percentage gain each option gives? You can choose any option only once. Which option would you choose? Why? Math Talk 25 Ganita Prakash | Grade 8 | Part-II A provision store is offering a stock clearance sale. Customers can choose one of the two options — 20% discount or ₹50 discount—for any purchase above ₹150. Which option would you choose if you want to: (iii) buy items worth ₹300 Example 12: A bakery called Cakely is offering a 30% + 20% discount on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Would you rather choose Cakely or Cakify if you want the cheaper cost? (ii) buy items worth ₹225 (i) buy items worth ₹180 While comparing percentages, we have to be mindful that we are comparing fractions or proportions and not absolute values. 01_Chapter 1.indd 26 23-12-2025 17:20:08 26 It seems that both the options should give the same benefit. Although mathematically 30% + 20% is the same as 50%, the usage of 30% + 20% in shopping means compounding. Suppose you want to buy a cake worth ₹200. Cakely’s 30% + 20% → Applying the 30% discount → the price of cake is ₹200 – ₹60 = ₹140. Applying the 20% discount on ₹140 → the price of cake is ₹140 – ₹28 = ₹112. Cakify’s 50% → The 50% discount makes the price of the cake ₹100. A Mishap Example 13: After Surbhi bought cookware from the wholesaler, she kept a profit margin of 50% on all the products. To clear off the remaining stock, she thought she would offer a 50% discount and come out without any loss. (i) Do you think she didn’t make any loss? (iii) What should have been the percentage discount offered so that she sold the goods at the price she had bought (i.e., no profit or loss)? Let us model the situation first. Suppose the worth of the products she bought from the wholesaler is x. The worth corresponding to the selling price (with a 50% margin) is 1.5x. A 50% discount on this price will make the worth 0.75x. (ii) If she had sold goods worth ₹12,000, (iii) To sell the goods at the same price, the discount offered should be (i) This means the selling price is 3 4 of the price the goods were bought at, i.e., a 25% loss. (ii) If she had sold goods (originally) for ₹12,000 after discount, how much loss did she incur? What is the percentage loss? She lost ₹4000. 0.75x = 12,000 x = 16,000. After discount Price bought at 0.75x Selling price including profit x Fractions in Disguise 1.5x 01_Chapter 1.indd 27 20-12-2025 16:29:04 Ariba and Arun have some marbles. Ariba says, “The number of marbles with me is 120% of the marbles Arun has”. What would be an appropriate statement Arun could make comparing the number of marbles he has with Ariba’s? d = 1 3 = 0.33. The discount offered should have been 33.33%. 1.5x – d × (1.5x) = x Try This 27 Ganita Prakash | Grade 8 | Part-II Figure it Out 13% 8% 18% 10% 1% 35% 2% 2% 0.1% 1. The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025? 2. The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population. [Hint: Writing these numbers in the standard form and estimating can help]. 3. The price of a mobile phone is ₹8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the GST? 4. The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was +5%, Month 2 change was –2%, and Month 3 change was –3%. Which of the following statement(s) are true? The initial population is p . (i) 8250 + 18 (ii) 8250 + 1800 (iii) 8250 + 18 100 (iv) 8250 × 18 (v) 8250 × 1.18 (vi) 8250 + 8250 × 0.18 (vii) 1.8 × 8250 Germany 83 million India 1.46 billion Bangladesh 175 million USA 347 million 01_Chapter 1.indd 28 20-12-2025 16:29:04 28 5. A shopkeeper initially set the price of a product with a 35% profit margin. Due to poor sales, he decided to offer a 30% discount on the selling price. Will he make a profit or a loss? Give reasons for your answer. 6. What percentage of area is occupied by the region marked ‘E’ in the figure? (iii) The population after three months was p + 0.05 – 0.02 – 0.03. (iv) The population after three months was p. (vi) The population after three months was less than p. (i) The population after three months was p × 0.05 × 0.02 × 0.03. (ii) The population after three months was p × 1.05 × 0.98 × 0.97. (v) The population after three months was more than p. 10. The bus fares were increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years? 11. If the length of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) does the breadth decrease by? 12. The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet. 7. What is 5% of 40? What is 40% of 5? 8. A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls. [Hint: Drawing a rough diagram can help]. 9. A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage? (ii) For each shop, calculate the percentage discount on the items. [Hint: Compare the free items to the total items you receive.] (iii) Suppose you need 4 items. Which shop would you choose? Why? Try This What is 25% of 12? What is 12% of 25? What is 15% of 60? What is 60% of 15? What do you notice? Can you make a general statement and justify it using algebra, comparing x% of y and y% of x? (ii) If the total number of students going to the excursion is 160, how many of them are Grade 8 girls? (i) What percentage of the students going to the excursion are Grade 8 girls? C D A E Fractions in Disguise B 01_Chapter 1.indd 29 22-12-2025 12:48:23 13. Three shops sell the same items at the same price. The shops offer deals as follows: Shop A: “Buy 1 and get 1 free” Shop B: “Buy 2 and get 1 free” Shop C: “Buy 3 and get 1 free” Answer the following: (i) If the price of one item is ₹100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest. 29 Twenties The ability to use computers is highest among those in their twenties and teenagers. Ability to use computer by age and gender (2023) Children Teenage Thirties Seniors Source: NSS Round 79, Comprehensive Annual Modular Survey, National Statistics Office Forties Fifties Ganita Prakash | Grade 8 | Part-II 14. In a room of 100 people, 99% are left-handed. How many left-handed people have to leave the room to bring that percentage down to 98%? 15. Look at the following graph. 0% 2% 4% 4% 4% 7% 9% 5% 10% 15% 20% 25% 30% 35% 40% 14% 14% 26% 25% 24% 29% Female Male 37% Try This 01_Chapter 1.indd 30 22-12-2025 12:48:23 30 Based on the graph, which of the following statement(s) are valid? (iii) There are more people in their twenties than teenagers. (iv) More than a quarter of people in their thirties can use computers. (vi) Half of the people in their twenties can use computers. (ii) Women lag behind in the ability to use computers across age groups. (v) Less than 1 in 10 aged 60 and above can use computers. (i) People in their twenties are the most computer-literate among all age groups. y Percentages are widely used in our daily life. y Percentages are fractions with denominator 100. Percentages are y Fractions can be converted to percentages and vice versa. Decimals too can be converted to percentages and vice versa. For example, 4 10 = 0.4 = 40%. y We have learnt to find the exact number when a certain percentage of the total quantity is given. y When parts of a quantity are given to us as ratios, we have seen how to convert them to percentages. y The increase or decrease in a certain quantity can also be expressed as a percentage. y The profits or losses incurred in transactions, and tax rates, can be expressed in terms of percentages. y We have seen how a quantity or a number grows when compounded. Interest rates are a common example of compounding. If p is the principal, r is the rate of interest, and t is the number of terms, then the total amount after the maturity period is denoted using the symbol ʻ%ʼ, pronounced ʻper centʼ. x% = x 100. SUMMARY 0 0% 40% 100% 40 100 = 0.4 100 100 = 1 Fractions in Disguise 01_Chapter 1.indd 31 22-12-2025 12:48:23 y A situation or a problem can often be solved by describing it using a rough diagram. We have learnt to estimate and do mental computations to solve problems related to percentages. Without compounding, p (1 + rt) p remains the principal for all the terms. With compounding, p × (1 + r) × (1 + r) × … × (1 + r) = p (1 + r)t Principal for term 1 Principal for term 2 Principal for term 3 Principal for term t 31 Place 8 knights on the chess board so that no knight attacks another. A knight moves in an ‘L-shape’. It can move either (a) two steps vertically and one step horizontally, or (b) two steps horizontally and one step vertically. Possible moves of a knight are shown below. Peaceful Knights 01_Chapter 1.indd 32 20-12-2025 16:29:12" class_8,9,the baudhayana-pythagoras theorem,ncert_books/class_8/hegp2dd/hegp202.pdf,"2 2.1 Doubling a Square In Baudhāyana’s Śulba-Sūtra (c. 800 BCE), Baudhāyana considers the following question: How can one construct a square having double the area of a given square? A first guess might be to simply double the length of each side of the square. Will this new square have double the area of the original square? It’s not hard to see that the new square will have area 2 × 2 = 4 times the area of the original square: PYTHAGORAS THEOREM THE BAUDHĀYANA02_Chapter 2.indd 33 20-12-2025 5.05.37 PM So how can one make a square that has double the area? Baudhāyana in his Śulba-Sūtra (Verse 1.9) gave an elegant answer to this question: As Baudhāyana says, the key is to construct a square on the diagonal of the original square: The diagonal of a square produces a square of double the area of the original square. Ganita Prakash | Grade 8 | Part-II Why does the new dotted square have double the area of the original square? In many of the constructions in the Śulba-Sūtra, it is desirable to construct, where needed, what Baudhāyana calls ‘east-west’ and ‘north-south’ lines, i.e., horizontal and vertical lines that are perpendicular to each other. Can you draw some horizontal and vertical lines to see why the new square has double the area of the original square? You could draw some horizontal and vertical lines as shown on the right. Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square? [Hint: From the diagonal property of a square, the line that bisects an angle passes through the opposite vertex. Argue why the vertical and horizontal sides of the original square bisect the two angles of the dotted square.] So, the new square has double the area of the original square, because the original square is made up of two small triangles, while the new square is made up of four small triangles. Moreover, all these small triangles are congruent to each other. Can you explain why? Math Talk 02_Chapter 2.indd 34 20-12-2025 5.05.37 PM 34 Adding some more horizontal and vertical lines can make the situation even clearer: So we can make the following sequence of squares: Each square has double the area of the previous one, as they are made up of 2, 4, and 8 small triangles, respectively. Doubling a Square Using Paper Cut out two identical squares of paper. Draw, label, and cut as follows: The Baudhāyana-Pythagoras Theorem 02_Chapter 2.indd 35 20-12-2025 5.05.38 PM Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square with double the area. 1 3 Square 1 Identical Square 2 2 4 5 7 6 8 35 Ganita Prakash | Grade 8 | Part-II 2.2 Halving a Square Now suppose we are given a square, and we want to construct a square whose area is half that of the original square. How would you do it? One way to do it is to reverse the construction of the previous section. We draw a tilted smaller square inside the larger one: Why is the smaller inside square half the area of the larger square? Again, adding some east-west and north-south lines can explain it: Halving a Square Using Paper Cut out a square from a piece of paper. Now make a square whose area is half the area of the first square. Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square? 02_Chapter 2.indd 36 20-12-2025 5.05.38 PM 36 Fold the square paper inward, as shown, such that the crease lines pass through the midpoints of the sides. PQRS is the required square with half the area. S R P Q Why is PQRS a square? Why is its area half that of the original paper? Explain by connecting QS and PR, finding the different angles formed, and then using tringle congruence. 2.3 Hypotenuse of an Isosceles Right Triangle Recall that in a right triangle, the side opposite to the right angle is called the hypotenuse. Find the hypotenuse of this isosceles right triangle. We know that a square of side 1 unit is made of two such isosceles right triangles. We also know that the square constructed on the diagonal of this square has twice the area of the original square. We do not yet know the length of the hypotenuse, but we know the area of the square REST on it! R S P E 1 A 1 1 1 T The Baudhāyana-Pythagoras Theorem 1 1 ? 02_Chapter 2.indd 37 20-12-2025 5.05.38 PM We know the relation between the side and the area of a square. If c is the hypotenuse ER, then Therefore, c = 2. Thus, the hypotenuse is of length 2 units. 1 Area of REST = c × c = c2 . So, c2 = 2. Area of REST = 2 × Area of PEAR 1 2 1 = 2 × 1 = 2 sq. units. 1 2 37 Ganita Prakash | Grade 8 | Part-II In the rest of this chapter, we will assume that all the lengths are of a fixed unit unless specified otherwise. What is the value of 2? Decimal Representation of 2 The decimal representation of 2 can be found using the following argument. Is 2 less than or greater than 1? A square of sidelength 1 unit has an area of 1 sq. unit. A square of sidelength 2 has an area of 2 sq. units. So, 1 is less than 2. In other words, 12 = 1, and 22 = 2. Is 2 less than or greater than 2? A square of sidelength 2 units has an area of 4 sq. units. A square of sidelength 2 has an area of 2 sq. units. So, 2 is greater than 2. In other words, 22 = 2, and 22 = 4. We call 1 a lower bound on 2 and 2 an upper bound. Can we find closer bounds for 2? 1.12 = 1.21 1.22 = 1.44 1.32 = 1.69 1.42 = 1.96 1.412 = 1.9881 1.422 = 2.0164 Therefore, 1 < 2. Therefore, 2 < 2. Thus, 1< 2 < 2. 1.4112 = 1.990921 1.4122 = 1.993744 1.4132 = 1.996569 02_Chapter 2.indd 38 20-12-2025 5.05.38 PM 38 Will we ever get a number with a terminating decimal representation whose square is 2? If there is such a terminating decimal starting with 1.414… whose square is 2, then it must have a non-zero last digit. If this is the case, then the decimal representation of its square will also have a non-zero last digit after the decimal point. For example, if 2 is of the form 1.414…4, then its square will be of the form— 1.52 = 2.25 So, 1.4 < 2 < 1.5 . . . . 6 So, 1.41 < 2 < 1.42 1.4142 = 1.999396 1.4152 = 2.002225 So, 1.414 < 2 < 1.415 So a terminating decimal cannot have 2 or 2.000... as its square. Thus, the decimal expansion of 2 must go on forever, i.e., it has a non-terminating decimal representation. Can 2 be expressed as a fraction m n , where m and n are counting numbers? If 2 could be expressed as m n , then we would have 2 = m2 n2 2n2 = m2 . Recall that in the prime factorization of a square number, each prime occurs an even number of times. So in the equation 2n2 = m2 , the prime 2 would occur an odd number of times on the left side and an even number of times on the right side. This is impossible. Thus 2 cannot be expressed as a fraction m n . This beautiful proof that 2 cannot be expressed as m n where m, n are counting numbers was given by Euclid in his book Elements (c. 300 BCE). We will discuss it in more detail in a later class. Thus the number 2 cannot be expressed as a terminating decimal or a fraction. But we can think of it as a certain non-terminating decimal: 2 = 1.41421356… Figure it Out 1. Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way. 2 = m n The Baudhāyana-Pythagoras Theorem Math Talk Try This 02_Chapter 2.indd 39 20-12-2025 5.05.38 PM Can you arrange these pieces to create a square with double the area of either square? 2. The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point. (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9 1 2 3 4 39 Ganita Prakash | Grade 8 | Part-II 3. The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.] General Solution The relation between the areas of a square and the square on its diagonal can be used to find a general relation between the hypotenuse and the other two sides of an isosceles right triangle. So, c2 = 2a2 . This formula can be used to find c when a is known, or to find a when c is known. Let a be the length of the equal sides and c the length of the hypotenuse. Area of SQVU = 2 × Area of PQRS a S R V P a Q c U 02_Chapter 2.indd 40 20-12-2025 5.05.39 PM 40 Example 1: Find the hypotenuse of an isosceles right triangle whose equal sides have length 12. We have a = 12. Using the formula, we get We have 162 = 256, and 172 = 289. So, 288 lies between 16 and 17. The length of the hypotenuse of an isosceles right triangle, whose length of the equal sides is 12 units, is between 16 and 17 units. Example 2: If the hypotenuse of an isosceles right triangle is 72, find its other two sides. We have c = 72. Using the formula, we get c = 2×122 = 288 . c2 = 2a2 So, a = 36 = 6. Therefore, each of the other two sides has length 6. Use this formula to check your answers in the Figure it Out on page 39. 2.4 Combining Two Different Squares We have seen in the previous sections how to combine two copies of the same square to make a larger square whose area is the sum of the areas of the two smaller squares. The sidelength of the larger square is the length of the diagonal of either of the smaller squares. What if we wish to combine two squares of ‘different’ sizes to make a large square whose area is the sum of the two smaller squares? + → So, ( 72) 2 = 2a2 Thus, a2 = 72 2 = 36. 72 = 2a2 . The Baudhāyana-Pythagoras Theorem 02_Chapter 2.indd 41 20-12-2025 5.05.39 PM In his Śulba-Sūtra (Verse 1.12), Baudhāyana also gives a truly remarkable solution to this more general problem of combining two different sized squares. He writes: The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides. + → ? 41 Ganita Prakash | Grade 8 | Part-II That is, to combine two different squares, make a right-angled triangle whose perpendicular sides are the sidelengths of the two squares. The square whose sidelength is the hypotenuse of this right-angled triangle has an area that is the sum of the areas of the original two squares. Why does Baudhāyana’s method work? Can you see why the method works in the case where the two squares are the same size? Does it agree with the method we used earlier to combine two same sized squares into a bigger square? Subsequently in his Śulba-Sūtra (Verse 2.1), Baudhāyana provides another verse that helps in explaining why the method works in general: To combine different squares, mark a rectangular portion of the larger square using a side of the smaller square. The diagonal of this rectangle is the side of a square that has area equal to the sum of the areas of the smaller squares. + → Area A Area B Area A + Area B 02_Chapter 2.indd 42 20-12-2025 5.05.39 PM 42 Let us follow Baudhayan᾽s instructions. a • Join the two squares. a b b a a b b a • Mark a rectangular portion of the larger square using a side of the smaller one, and draw its diagonal. By doing this, we get a right triangle with perpendicular sides a and b. • Make a 4-sided figure over the hypotenuse by drawing three more of such right triangles: 1. 2. a a b – a a a b – a b a b b a b a The Baudhāyana-Pythagoras Theorem a b – a b b 02_Chapter 2.indd 43 20-12-2025 5.05.39 PM a a a 3. a b – a b – a b a b a a a b – a a b – a b a 43 b Ganita Prakash | Grade 8 | Part-II Why? • The 4-sided figure obtained (T + U + V) is in fact a square with an area equal to the sum of the areas of the two smaller squares! • As T, U, X and W are all congruent, the sides of this new 4-sided figure all have the same length. Notice the angles. 4. b a a a a b b – a b – a a 90 – x 90 – x b a a a x 90 – x b – a b W X a x x T b – a a V U b 02_Chapter 2.indd 44 20-12-2025 5.05.39 PM 44 Explain why all the angles of this new 4-sided figure are right angles and so it is a square. Notice that this new square has as its sidelength, the hypotenuse of the right triangle of sides a and b. The area of the square on the hypotenuse = the sum of the areas of T, U, and V = the sum of the areas of V, W, and X = the sum of the areas of the two given squares. • Baudhāyana’s assertion is now clear: a b – a x 90 – x a Combining Two Squares Using Paper Cut out and join two different sized squares as follows: Now make two cuts to make three pieces: Rearrange the three pieces into a larger square: T U a V a a a c c b b – a b The Baudhāyana-Pythagoras Theorem V W X b 02_Chapter 2.indd 45 20-12-2025 5.05.39 PM You have now combined two smaller squares into a larger square! Now make a right triangle using the two smaller squares. Draw a square on the hypotenuse. c c 45 Ganita Prakash | Grade 8 | Part-II Cover the square on the hypotenuse using your pieces. a a c c b 02_Chapter 2.indd 46 20-12-2025 5.05.39 PM 46 This shows that if a right triangle has shorter sides of length a and b, and hypotenuse of length c, then the areas of the two smaller squares, a2 and b2 , add up to the area of the larger square, c2 : a2 + b2 = c2 . This is the famous and fundamental theorem of Baudhāyana on right-angled triangles: Baudhāyana’s Theorem on Right-angled triangles: If a right-angled triangle has sidelengths a, b, and c, where c is the length of the hypotenuse, then a2 + b2 = c2 . b Baudhāyana was the first person in history to state this theorem in this generality and essentially modern form. The theorem is also known as the Pythagorean Theorem, after the Greek philosopher-mathematician Pythagoras (c. 500 BCE) who also admired and studied this theorem, and lived a couple hundred years after Baudhāyana. It is also often called the transitional name ‘Baudhāyana-Pythagoras Theorem’ so that everyone knows what theorem is being referred to. Using Baudhāyana’s Theorem Make a right-angled triangle in your notebook whose shorter sidelengths are 3 cm and 4 cm. Now, measure the length of the hypotenuse. It should read about 5cm. In fact, we could have used Baudhāyana’s Theorem to predict that the hypotenuse is exactly 5cm! Let a = 3 and b = 4, the lengths in cm of the two shorter sides. Then, by Baudhāyana’s Theorem, the length c of the hypotenuse satisfies the equation, Figure it Out 1. If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana’s Theorem. a2 + b2 = c2 32 + 42 = c2 9 + 16 = c2 25 = c2 So, c = 5 cm. The Baudhāyana-Pythagoras Theorem 3 cm ? 4 cm 02_Chapter 2.indd 47 20-12-2025 5.05.39 PM 2. If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana’s Theorem. 3. Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhāyana’s Śulba-Sūtra, Verse 1.10) 4. Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases: (i) a = 5, b = 7 (ii) a = 8, b = 12 (iii) a = 9, c = 15 (iv) a = 7, b = 12 (v) a = 1.5, b = 3.5 47 Ganita Prakash | Grade 8 | Part-II 2.5 Right–Triangles Having Integer Sidelengths In his Śulba-Sūtra (Verse 1.13) Baudhāyana lists a number of integer triples (a, b, c) that form the sidelengths of a right-triangle and therefore satisfy a2 + b2 = c2 . These include (3, 4, 5), (5, 12, 13) (8, 15, 17), (7, 24, 25), (12, 35, 37), and (15, 36, 39). For this reason, such triples (a, b, c) that form the sidelengths of a right triangle (equivalently, satisfy a2 + b2 = c2 ) are called Baudhāyana triples. They are also called Baudhāyana-Pythagoras triples, Pythagorean triples, and right-angled triangle triples. List down all the Baudhāyana triples with numbers less than or equal to 20. Is there an unending sequence of Baudhāyana triples? Mathematicians have answered this question and have found a method to generate all such triples! Let us take a few steps in this direction. We have seen that (3, 4, 5) is a Baudhāyana triple. Is (30, 40, 50) a Baudhāyana triple? Is (300, 400, 500) a Baudhāyana triple? The list of Baudhāyana triples having numbers less than or equal to 20 contains the following triples — (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20). Do you see any pattern among them? All these triples can be obtained by multiplying each term of (3, 4, 5) by a certain positive integer. Math Talk 02_Chapter 2.indd 48 20-12-2025 5.05.39 PM 48 Can we form a conjecture on Baudhāyana triples based on this observation? Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, where k is any positive integer. Is this true? We have to check if (3k) 2 + (4k) 2 = (5k) 2 . We have (3k) 2 + (4k) 2 = 9k2 + 16k2 = 25k2 , which is equal to (5k) 2 . So, (3k, 4k, 5k) is indeed a Baudhāyana triple. This shows that there are infinitely many Baudhāyana triples. Can we further generalise the conjecture? If (a, b, c) is a Baudhāyana triple, then (ka, kb, kc) is also a Baudhāyana triple where k is any positive integer. Is this statement true? This statement can be shown to be true in the same way. Since (a, b, c) is a Baudhāyana triple, we have a2 + b2 = c2 . We need to check whether (ka) 2 + (kb) 2 = (kc) 2 . So, (ka) 2 + (kb) 2 = k2 a2 + k2 b2 . Taking out the common factor, we have (ka) 2 +(kb) 2 = k2 (a2 + b2 ). Since a2 + b2 = c2 , we have (ka) 2 + (kb) 2 = k2 c2 = (kc) 2 . Thus, (ka, kb, kc) is a Baudhāyana triple if (a, b, c) is a Baudhāyana triple. We call (ka, kb, kc) a scaled version of (a, b, c). A Baudhāyana triple that does not have any common factor greater than 1 is called a primitive Baudhāyana triple. So, (3, 4, 5) is primitive, whereas (9,12,15) is not. Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive Baudhāyana triples with numbers less than or equal to 20? Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive? If (a, b, c) is non-primitive, and the integers have f — greater than 1 — as a common factor, then is ( a f , b f , c f) a Baudhāyana triple? Check this statement for (9, 12, 15). Justify this statement. If we can find all the primitive triples, we can find all the Baudhāyana triples. (ka) 2 = ka × ka = k2 a2 , and (kb) 2 = kb × kb = k2 b2 . The Baudhāyana-Pythagoras Theorem 02_Chapter 2.indd 49 20-12-2025 5.05.39 PM How do we generate more primitive triples? We know the relation between the sum of consecutive odd numbers and square numbers. 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 The sum of the first n odd numbers is n2 . Let us express this algebraically. For this, we need to know the nth odd number. What is it? The nth odd number is 2n – 1. So, 1 + 3 + 5 + … + (2n – 3) + (2n – 1) = n2 Sum of first (n – 1) odd numbers 49 Ganita Prakash | Grade 8 | Part-II What is the sum of the first (n – 1) odd numbers? Thus, (n – 1)2 + (2n – 1) = n2 . Note that this equation could also have been directly obtained by expanding (n – 1)2 and adding 2n – 1 to it. If the nth odd number, 2n – 1, is also a square number, then we have a sum of two square numbers equal to another square number. We will use this idea to generate Baudhāyana triples. 1. 9 is an odd square. It is the 5th odd number (9 = 2 × 5 – 1). So, we have Could we have obtained this triple using the equation (n – 1)2 + (2n – 1) = n2 ? Since we took the 5th odd number, the value of n is 5. Substituting n = 5 into the equation, we get (5 – 1)2 + 9 = 52 2. 25 is an odd square. It is the 13th odd number (25 = 2 × 13 – 1). So, Figure it Out 1 + 3 + 5 + … + 23 + 25 = 132 122 + 52 = 132 We have n = 13. Substituting this value in the equation, we get (13 – 1)2 + 25 = 132 122 + 52 = 132 1 + 3 + 5 + 7 + 9 = 52 42 + 32 = 52 . 42 + 32 = 52 02_Chapter 2.indd 50 24-12-2025 15:58:28 50 1. Find 5 more Baudhāyana triples using this idea. 2. Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.] 3. Are there primitive triples that cannot be obtained through this method? If yes, give examples. 2.6 A Long-Standing Open Problem The study of Baudhāyana triples inspired the great French mathematician Fermat — who lived during the 17th century — to make a general statement about the sum of powers of positive integers. Math Talk We have seen that there are an infinite number of square numbers that can be written as a sum of two square numbers. This made Fermat wonder if there is a perfect cube that can be written as a sum of two perfect cubes, a fourth power that can be written as a sum of two fourth powers, and so on. In other words, he wondered if there is a solution to the equation xn + yn = zn , where x, y, and z are natural numbers, and n > 2. In the margin of a book that dealt with properties and patterns of positive integers (like that of Baudhāyana triples), Fermat wrote that amongst the unending sequence of numbers, one cannot find a single perfect cube that is a sum of two perfect cubes, a fourth power that is a sum of two fourth powers, and so on. So, the equation has no solution for powers greater than 2. In addition to stating this, Fermat wrote, “I have found a truly marvellous proof of this statement, but the margin is too small to contain it”. The Baudhāyana-Pythagoras Theorem 02_Chapter 2.indd 51 20-12-2025 5.05.40 PM No one could ever find Fermat’s proof of this statement, which is called Fermat’s Last Theorem. After his death, many great mathematicians tried their hand at proving this theorem. There followed more than 300 years of failed attempts in proving it. In 1963, a 10-year-old boy named Andrew Wiles read a book (The Last Problem by Eric Bell) about Fermat’s Last Theorem and its history. Despite reading about the failures of so many great mathematicians, this young boy resolved to prove this theorem. He eventually did prove this theorem in 1994! 51 Ganita Prakash | Grade 8 | Part-II 2.7 Further Applications of the BaudhāyanaPythagoras Theorem The Baudhāyana-Pythagoras theorem is one of the fundamental theorems of geometry. Let us see some of its applications. A Problem from Bhāskarāchārya’s Līlāvatī The following is a translation of a problem from Bhāskarāchārya’s (Bhāskara II) Līlāvatī. Try to visualise what you read. “In a lake surrounded by chakra and krauñcha birds, there is a lotus flower peeping out of the water, with the tip of its stem 1 unit above the water. On being swayed by a gentle breeze, the tip touches the water 3 units away from its original position. Quickly tell the depth of the lake.” At first glance, it seems like there is insufficient data to find the solution. But the solution exists! Let x be the length of the stem inside the water. This is the required depth of the lake. Since the stem of the lotus is sticking 1 unit above the water, the total length of the stem is x + 1. We can make a (very reasonable) assumption that the lotus stem is perpendicular to the surface of the water in the first position. With this assumption, we get a right triangle of sides 3, x and x + 1. As this satisfies the Baudhāyana-Pythagoras theorem, we have 02_Chapter 2.indd 52 22-12-2025 1.46.59 PM 52 9 + x2 = x2 + 2x + 1. Subtracting x2 from both sides, we get 9 = 2x + 1. x = 4. Thus, the depth of the lake is 4 units. Figure it Out 1. Find the diagonal of a square with sidelength 5 cm. 2. Find the missing sidelengths in the following right triangles: 32 + x2 = (x + 1)2 3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units. 4. Is the hypotenuse the longest side of a right triangle? Justify your answer. 150 10 7 40 9 41 The Baudhāyana-Pythagoras Theorem 4 10 27 45 10 200 02_Chapter 2.indd 53 20-12-2025 5.05.40 PM 5. True or False—Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple. 6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers. 7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units. 8. (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq.units, and (d) 5 sq. unit? (ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner? Math Talk Try This 53 Ganita Prakash | Grade 8 | Part-II 9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.] y The Baudhāyana-Pythagoras Theorem is one of the most fundamental theorems in geometry. It expresses the relationship among the three sides of a right-angled triangle. y If a, b, c, are the sidelengths of a right-angled triangle, where c is the length of hypotenuse, then a2 + b2 = c2. y In an isosceles triangle with sidelengths a, a, c, we have the relation a2 + a2 = 2a2 = c2, i.e., c = a 2 . y The number 2 lies between 1.414 and 1.415. However, it cannot be expressed as a terminating decimal. It also cannot be expressed as a fraction m n with m, n positive integers. y A triple (a, b, c) of positive integers satisfying a2 + b2 = c2 is called a Baudhāyana-Pythagoras triple. Examples include (3, 4, 5), (6, 8, 10), and (5, 12, 13). Infinitely many such triples can be constructed. y The equation an + bn = cn has no solution in positive integers when n>2. This is known as ‘Fermat’s Last Theorem’. It was proven by Andrew Wiles in 1994. SUMMARY Try This 02_Chapter 2.indd 54 22-12-2025 12.58.44 PM There are 3 closed boxes—one containing only red balls, the second containing only blue balls and the third containing only green balls. The boxes are labelled RED, BLUE and GREEN such that ‘no’ box has the correct label. We need to find which label goes with which box. How can this be done if we are allowed to open only one box? Find the Colours!" class_8,10,proportional reasoning-2,ncert_books/class_8/hegp2dd/hegp203.pdf,"3 3.1 Proportionality—A Quick Recap In an earlier chapter, we explored proportional relationships between quantities and we used the ratio notation to represent such relationships. When two or more related quantities change by the same factor, we call that relationship a proportional relationship. For example, idli batter is made by mixing rice and urad dal. The proportion of these two can have regional variations. One of the proportions used is: for 2 cups of rice, we add 1 cup of urad dal. We represent this relationship using the ratio notation 2 : 1. Viswanath made idlis by mixing 6 cups of rice with 3 cups of urad dal, while Puneet made idlis by mixing 4 cups of rice with 2 cups of urad dal. If cooked in the same way, would their idlis taste the same? Viswanath’s mixture can be represented as 6 : 3 and Puneet’s mixture as 4 : 2. Recall that, to verify that these two ratios are proportional, we can use the cross-multiplication method: × PROPORTIONAL REASONING–2 03_Chapter 3.indd 55 20-12-2025 16:36:21 The two products are the same (12). So, the two ratios are proportional. It is likely that the idlis would taste the same, if all the other ingredients are proportional too! In general, we can say that two ratios a : b and c : d are proportional if a × d = b × c, or 6 : 3 4 : 2 a c = b d . × Ganita Prakash | Grade 8 | Part-II 3.2 Ratios in Maps Have you noticed that in many maps there is a ratio given, usually in the lower right corner of the map? It usually contains 1 and a very large number, such as 1 : 60,00,000. What does RF 1 : 60,00,000 mean? What does it indicate? Can you guess? A Representative Fraction (RF) is an expression that shows the ratio between a distance on the map and the corresponding actual distance on the ground. For example, if the ratio on a map is 1 : 60,00,000, that means a distance of 1 cm on the map is equivalent to a geographical distance of 60,00,000 cm. Remember, this is geographical distance and not road distance! Map not to scale 03_Chapter 3.indd 56 20-12-2025 16:36:24 56 Convert 60,00,000 cm to kilometres. It is 60 km. Verify this. Using the map above, can you find the geographical distance between Bengaluru and Chennai? Also, find the geographical distance between Mangaluru and Chennai. [Hint: Use a ruler to find the distance between the cities on the map. Then, use the ratio given on the map to find the actual geographical distance.] Try to find the distances between the same two pairs of cities with different maps that have different scales (ratios). Do they all give the same geographical distance, approximately? 3.3 Ratios with More than 2 Terms Viswanath is experimenting with a spice mix powder. He makes the powder by grinding 8 spoons of coriander seeds, 4 red chillies, 2 spoons of toor dal, and 1 spoon of fenugreek (methi) seeds. For his spice mix powder, the ratio of coriander seeds to red chillies to toor dal to fenugreek seeds is Note to the Teacher: Bring maps and atlases to the classroom and encourage students to observe the scale given as a ratio (usually in the lower right corner) on the map. Use the maps and atlases in the library and ask students to find the geographical distances between two locations on the map that are of local interest. Ask them to verify with each other if they get similar distances and to find the reasons if the distances are very different. Map Making Activity: Guide students to make a sketch of their classroom with an accurate scale (ratio of 1: 50). They should mark the location of various objects in the classroom like the teacher᾿s desk, blackboard, fans and lights, according to scale. Students can use appropriate symbols to represent different objects like fans, lights, tables, chairs, and so on. 8 : 4 : 2 : 1. Proportional Reasoning–2 03_Chapter 3.indd 57 20-12-2025 16:36:26 Notice that the ratio has 4 terms. Ratios can have many terms if each of the quantities change by the same factor to maintain the proportional relationship. Puneet has only 2 red chillies in his kitchen. But he wants to make spice mix powder that tastes the same as Viswanath’s spice mix powder. How much of the other ingredients should Puneet use to make his spice mix powder? For Puneet’s spice mix powder to be similar to Viswanath’s, the ratio of all the ingredients should be the same as Viswanath’s spice mix powder. Puneet has only 2 red chillies. He has half the number of chillies that Viswanath used in his mixture. So, the quantity of the other ingredients should also be reduced to half. Thus, Puneet should add 4 spoons of coriander seeds, 2 red chillies, 1 spoon of toor dal, and half a spoon of fenugreek seeds. The ratio is 4 : 2 : 1 : 0.5. 57 Ganita Prakash | Grade 8 | Part-II Both the ratios are proportional to each other. We denote this by In general, when two ratios with multiple terms are proportional then, a p = b q = c r = d s . Example 1: To make a special shade of purple, paint must be mixed in the ratio, Red : Blue : White :: 2 : 3 : 5. If Yasmin has 10 litres of white paint, how many litres of red and blue paint should she add to get the same shade of purple? In the ratio 2 : 3 : 5, the white paint corresponds to 5 parts. If 5 parts is 10 litres, 1 part is 10 ÷ 5 = 2 litres. So, the purple paint will have 4 litres of red, 6 litres of blue, and 10 litres of white paint. What is the total volume of this purple paint? The total volume of purple paint is 4 + 6 + 10 = 20 litres. Example 2: Cement concrete is a mixture of cement, sand, and gravel, and is widely used in construction. The ratio of the components in the mixture varies depending on how strong the structure needs to be. For structures that need greater strength like pillars, beams, and roofs, the ratio is 1 : 1.5 : 3, and the construction is also reinforced with steel rods. Using this ratio, if we have 3 bags of cement, how many bags of concrete mixture can we make? The concrete mixture is in the ratio Red = 2 parts = 2 × 2 = 4 litres. Blue = 3 parts = 3 × 2 = 6 litres. 8 : 4 : 2 : 1 :: 4 : 2 : 1 : 0.5. a : b : c : d :: p : q : r : s 03_Chapter 3.indd 58 20-12-2025 16:36:26 58 If we have 3 bags of cement, we have to multiply the other terms by 3. So, the ratio is cement : sand : gravel :: 3 : 4.5 : 9. In total, we have 3 + 4.5 + 9 = 16.5 bags of concrete. 3.4 Dividing a Whole in a Given Ratio In an earlier chapter, we learnt how to divide a whole in a ratio, e.g., 12 in the ratio 2 : 1. To do this, we add the terms (2 + 1 = 3), and divide the whole by this sum (12 ÷ 3 = 4). We multiply each term by this quotient : 2 × 4 = 8 and 1 × 4 = 4 . So, 12 divided in the ratio 2 : 1 is 8 : 4. Bags of cement : bags of sand : bags of gravel :: 1 : 1.5 : 3. We can extend this to ratios with multiple terms. Let us look at the earlier example of making a concrete mixture. We need to mix cement, sand, and gravel in the ratio of 1 : 1.5 : 3 to get the concrete. Example 3: For some construction, 110 units of concrete are needed. How many units of cement, sand, and gravel are needed if these are to be mixed in the ratio 1 : 1.5 : 3? For 1 unit of cement, we need to add 1.5 units of sand and 3 units of gravel. Together, they add up to 5.5 units of concrete. We need to do this 20 times (110 ÷ 5.5 = 20) to get 110 units of concrete. So each term has to be multiplied by 20. 1 × 20 = 20 units of cement, 1.5 × 20 = 30 units of sand, and 3 × 20 = 60 units of gravel. So, we need 20 units of cement, 30 units of sand, and 60 units of gravel to make the concrete. When we divide a quantity x in the ratio a : b : c : …, the terms in the ratio are — Proportional Reasoning–2 03_Chapter 3.indd 59 20-12-2025 16:36:29 x × a (a + b + c + …) , x × b (a + b + c + …) , x × c (a + b + c + …) , and so on. Example 4: You get a particular shade of purple paint by mixing red, blue, and white paint in the ratio 2 : 3 : 5. If you need 50 ml of purple paint, how many ml of red, blue, and white paint will you mix together? Red paint = 50 × 2 (2 + 3 + 5) = 50 × 2 10 = 10 ml. Blue paint = 50 × 3 (2 + 3 + 5) = 50 × 3 10 = 15 ml. White paint = 50 × 5 (2 + 3 + 5) = 50 × 5 10 = 25 ml. 59 Ganita Prakash | Grade 8 | Part-II Example 5: Construct a triangle with angles in the ratio 1 : 3 : 5. We know that the sum of the angles in a triangle is 180°. So the angles are ∠A = 180° × 1 (1 + 3 + 5) = 180° × 1 9 = 20°. ∠B = 180° × 3 (1 + 3 + 5) = 180° × 3 9 = 60°. ∠C = 180° × 5 (1 + 3 + 5) = 180° × 5 9 = 100°. Figure it Out 1. A cricket coach schedules practice sessions that include different activities in a specific ratio — time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5. If each session is 150 minutes long, how much time is spent on each activity? 2. A school library has books in different languages in the following ratio — no. of Odiya books : no. of Hindi books : no. of English books :: 3 : 2 : 1. If the library has 288 Odiya books, how many Hindi and English books does it have? 3. I have 100 coins in the ratio — no. of ₹10 coins : no. of ₹5 coins : no. of ₹2 coins : no. of ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in coins? 4. Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not? 5. Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not? A B 60° 20° Math Talk 100° C 03_Chapter 3.indd 60 20-12-2025 16:36:29 60 3.5 A Slice of the Pie Have you seen pie charts like the one shown in the figure? Pie charts show different proportions of a whole. This one shows the proportions of students that have scored each grade in an assessment. These kinds of visualisations help us quickly interpret data. Let us try to create this pie chart. Here is a table showing the grades scored by students: Students 12 10 8 6 4 Grade A B C D E E Grade Distribution (40 students) A D 12 B 10 8 4 6 C How do we mark the different slices of the pie chart? To mark a slice in the pie chart, the angle corresponding to a grade should be proportional to the number of students who have scored that grade. The total angle in a circle is 360°. So, we need to divide 360 in the ratio of 12 : 10 : 8 : 6 : 4. Can we reduce this ratio to its simplest form? We can use the same procedure we used to reduce a ratio with two terms to its simplest form. We divide all the terms by their HCF to get the simplest form. In this example, 2 is the HCF of the terms. We get the simplest form by dividing all the terms by 2. The ratio becomes 6 : 5 : 4 : 3 : 2. So, the angles are: Now let us construct a pie chart using these angles. Step 1: Draw a circle and mark the radius AB as shown below: Step 2: To draw the slice to represent the proportion of students who got an A grade, measure 108° from the segment AB on A (anti-clockwise), and mark the new radius AC. Grade A = 6 (6 + 5 + 4 + 3 + 2) × 360° = 6 20 × 360° = 6 × 18 = 108°. Grade B = 5 (6 + 5 + 4 + 3 + 2) × 360° = 5 × 18 = 90°. Grade C = 4 × 18 = 72°. Grade D = 3 × 18 = 54°. Grade E = 2 × 18 = 36°. Proportional Reasoning–2 03_Chapter 3.indd 61 20-12-2025 16:36:29 A B 108° A B C 61 Ganita Prakash | Grade 8 | Part-II Step 3: Measure 90° from AC and draw AD. Step 4: Measure 72° from AD and draw AE. Steps 5,6: Similarly, we can complete the rest of the pie chart. D D 108° 90° A B C C 108° 90° A B 72° 54° 36° Step 7: You can colour and label the different slices of the pie chart appropriately. D Grade B C 108° 90° A B 72° E Grade A 03_Chapter 3.indd 62 20-12-2025 16:36:30 62 Figure it Out 1. A group of 360 people were asked to vote for their favourite season from the three seasons — rainy, winter and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information. 2. Draw a pie chart based on the following information about viewers᾿ favourite type of TV channel: Entertainment — 50%, Sports — 25%, News — 15%, Information — 10%. 3. Prepare a pie chart that shows the favourite subjects of the students in your class. You can collect the data of the number of students for E F Number of Students Subject Language Arts Education Vocational Education Social Science Physical Education Maths Science 3.6 Inverse Proportions Do you recall the rule of three? When two ratios are proportional, i.e., when a : b :: c : d, then d = bc a . We call such proportions direct proportions. We use this understanding to find the value of the fourth quantity (d), when the value of three quantities (a, b, and c) are given. Example 1: If 5 workers can move 4500 bricks in a day, how many workers are needed to move 18000 bricks in a day? This can be represented as a statement of proportionality — 4500 : 18000 :: 5 : x. We can find the value of x by x = 18000 × 5 4500 = 20 Thus, the number of workers needed are 20. Example 2: Puneeth’s father went from Lucknow to Kanpur in 3 hours by riding his motorcycle at a speed of 30 km/h. If he takes a car instead and drives at 60 km/h, how long will it take him to reach Kanpur? each subject shown in the table (each student should choose only one subject). Then write these numbers in the table and construct a pie chart: Proportional Reasoning–2 03_Chapter 3.indd 63 20-12-2025 16:36:32 Can we represent this problem with the following statement of proportionality — 30 : 60 :: 3 : x ? Will the travel time increase or decrease as the speed of the motorcycle increases? The following table shows the time taken to travel from Lucknow to Kanpur using different modes of transport: From this table, we notice that when the speed increases, the time taken to travel the same distance decreases. Time (in hours) 18 6 3 1.5 Speed (km/h) 5 15 30 60 Walk Bicycle Motorcycle Car Math Talk 63 Ganita Prakash | Grade 8 | Part-II Does it decrease by the same rate (or factor)? Going by bicycle is 3 times faster than walking (15 ÷ 5). The speed has increased 3 times. The travel time has decreased 3 times too (18 ÷ 6). The speed has increased by the same factor by which the travel time has decreased. Check if this is the case for the other modes of transport. Since both quantities, speed and time, change by the same factor, they are proportional. But they change in opposite directions, or inversely. Such proportions are called inverse proportions. From the table, we can see that the product of speed and time are the same for all modes of transport, namely 90 km. Two quantities x and y vary in inverse proportion if there exists a relation of the type xy = k, where k is a constant. In the previous example, if x represents the speed and y represents the time taken, then k is the distance between Lucknow and Kanpur, which remains constant. Time (in hours) 18 6 3 1.5 Speed (km/h) 5 15 30 60 × 1 3 × 1 2 × 4 × 3 × 2 × 1 4 03_Chapter 3.indd 64 20-12-2025 16:36:32 64 Thus, if quantities x and y are inversely proportional, and x1 and x2 are values of x that have corresponding y values y1 and y2, then x1 y1 = x2 y2 = k (some constant). From these we can see that x1 x2 = y2 y1 . Let us check if this is true for other modes of transport, such as walking and by car. Let us use x to represent speed and y to represent time. x1 = 5, x2 = 60, y1 = 18, y2 = 1.5. x1 x2 = 5 60 = 0.083333... What is the value of 1.5 18 y2 y1  ? It is 0.083333... too! Figure it Out 1. Which of these are in inverse proportion? 2. Fill in the empty cells if x and y are in inverse proportion. Example 3: 20 workers take 4 days to complete laying a road. How many days will 10 workers take to complete laying the same length of road? If we decrease the number of workers, then the number of days to complete the work will increase by the same factor. So these quantities are inversely proportional. So, x1 y1 = x2 y2. (iii) x 30 90 150 10 (ii) x 40 80 25 16 (i) x 40 80 25 16 x 16 12 36 y 9 48 y 20 10 32 50 y 20 10 12.5 8 y 15 5 3 45 Proportional Reasoning–2 03_Chapter 3.indd 65 20-12-2025 16:36:32 Thus, 20 × 4 = 10 × y2 It will take 8 days for 10 workers to complete the work. We notice that when the number of workers halved, the number of days to complete the work doubled. Quantities are inversely proportional if, when one quantity changes by a factor n, the other quantity changes by the inverse 1 n . y2 = 20 × 4 10 = 8. 65 Ganita Prakash | Grade 8 | Part-II Example 4: 2 pumps can fill a tank in 18 hours. How much time will it take to fill the tank if we add 2 more pumps of the same kind? If we add 2 more pumps, we will have 4 pumps. Let us denote the time taken to fill the tank with 4 pumps by x. If we increase the number of pumps by n, then the time taken to fill the tank will decrease by the same factor n, so the quantities are inversely proportional. Since the quantities are inversely proportional, the product remains constant. So, It will take 9 hours to fill the tank. Example 5: A school has food provisions to feed 80 students for 15 days. If 20 more students join the school, for how many days will the provisions last? More students → fewer days the provisions will last. The quantities are inversely proportional. If x is the number of days, The provisions will last for only 12 days. Example 6: If Ram takes 1 hour to cut a given quantity of vegetables and Shyam takes 1.5 hours to cut the same quantity of vegetables, how much time will they take to cut the vegetables if they do it together? Consider the work done to cut the given quantity of vegetables as 1 unit of work. Let us figure out the work done by each person in 1 hour. • Ram finishes the work in 1 hour, so in 1 hour he does 1 unit of work. • Shyam finishes the work in 1.5 hours, so in 1 hour he does 1 1.5 = 2 3 units of work. 80 × 15 = 100 × x. x = 80 × 15 100 = 12. 2 × 18 = 4 × x. x = 2 × 18 4 = 9. 03_Chapter 3.indd 66 20-12-2025 16:36:32 66 So, the work done by both in 1 hour is 1 + 2 3 = 5 3 units of work. Therefore, to complete 5 3 units of work, it takes them 1 hour if they work together. How much time will it take them to complete 1 unit of work? Is the quantity of work and time taken to complete it directly or inversely proportional? It is directly proportional. So, this can be represented as 5 3 : 1 :: 1 : x where x is the time taken. Math Talk Since it is a direct proportion, we know that 5 3 1×1 If they cut the vegetables together, they will finish in 3 5 hours. Figure it Out 1. Which of the following pairs of quantities are in inverse proportion? 2. If 24 pencils cost ₹120, how much will 20 such pencils cost? 3. A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem? (iii) The distance a car can travel and the amount of petrol in the tank. (iv) The speed of a cyclist and the time taken to cover a fixed route. (v) The length of cloth bought and the price paid at a fixed rate per metre. (vi) The number of pages in a book and the time required to read it at a fixed reading speed. (ii) The number of painters hired and the days needed to paint a wall of fixed size. (i) The number of taps filling a water tank and the time taken to fill it. x . = = = × x 1×1 5 3 = 1 5 3 5 3 Proportional Reasoning–2 Math Talk 03_Chapter 3.indd 67 20-12-2025 16:36:33 4. Fill in the average number of hours each living being sleeps in a day by looking at the charts. Select the appropriate hours from this list : 15, 2.5, 20, 8, 3.5, 13, 10.5, 18. 67 Ganita Prakash | Grade 8 | Part-II 10. A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill? 5. The pie chart on the right shows the result of a survey carried out to find the modes of transport used by children to go to school. Study the pie chart and answer the following questions. 6. Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make? 7. It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump? 8. A given set of chairs are arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have? 9. A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same? (iii) If 18 children travel by car, how many children took part in the survey? How many children use taxis to travel to school? (iv) By which two modes of transport are equal numbers of children travelling? (ii) What fraction of children travel by car? (i) What is the most common mode of transport? Cycle Walk 120° 90° 60° 60° TwowheelerCarBus03_Chapter 3.indd 68 20-12-2025 16:36:34 68 11. A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days? 12. A car takes 2 hours to reach a destination, travelling at a speed of 60 km/h. How long will the car take if it travels at a speed of 80 km/h? y Ratios in the form a : b : c : d : … indicate that for every a units of the first quantity, there are b units of the second quantity, c units of the third quantity, and so on. y If x is divided into many parts in the ratio p : q : r : s : …, then the y Two quantities are directly proportional when they both change by the same factor, and their quotient remains the same. For example, if x and y are two quantities that are directly proportional, and (x1, x2, x3, …) and ( y1, y2, y3, …) are the y Quantities are inversely proportional if, when one quantity changes by a factor n, the other quantity changes by the inverse 1 n. For example, if x and y are two quantities that are inversely proportional, and ( x1, x2, x3, …) and ( y1, y2, y3, …) are the corresponding values of x and y, then x1 y1 = x2 y2 = x3 y3 = … = n, where n is a constant. quantity of the first part is x × p ( p + q + r + s + …) , the quantity of the second part is x × q ( p + q + r + s + …) , and so on. corresponding values of x and y, then x1 y1 = x2 y2 = x3 y3 = … = k, where k is a constant. SUMMARY Proportional Reasoning–2 03_Chapter 3.indd 69 20-12-2025 16:36:34 69" class_8,11,exploring some geometrics themes,ncert_books/class_8/hegp2dd/hegp204.pdf,"4 In this chapter, we will explore two geometric themes. We will study fractals which are self-similar shapes. They exhibit the same or similar pattern over and over again — but at smaller and smaller scales. We will then look at different ways of visualising solids. 4.1 Fractals One of the most beautiful examples of a fractal that also occurs in nature is the fern. The fern is seen to have smaller copies of itself as its leaves, and these in turn have even smaller copies of themselves in their sub-leaves, and so on! Similar phenomena of self-similarity occur in trees (where a trunk has limbs, and a limb has branches, and the branches have branchlets, and so on), clouds, coastlines, mountains, lightning, and many other objects in nature. Other mathematical fractals can also be very beautiful. We will explore some of them here. EXPLORING SOME GEOMETRIC THEMES 04_Chapter 4.indd 70 20-12-2025 16:43:39 Sierpinski Carpet The Polish mathematician Sierpinski discovered a type of fractal known as the Sierpinski Carpet. It is made by taking a square, breaking it into 9 smaller squares, and then removing the central square (see the figure below); the same procedure is then repeated on the remaining 8 squares, and so on. One then sees the same pattern at smaller and smaller scales. Step 0 Step 1 Step 2 . . . Fern Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet. By its construction, each step in the sequence has Do you see any pattern in the number of holes and squares that remain at each step? Let Rn represent the number of remaining squares at the nth step, and Hn represent the number of holes at the nth step. Let us understand how these numbers grow by analysing how the holes and squares that remain are generated from the previous step. Every square that remains at a given step, say Step n, gives rise to 8 squares that remain at the (n + 1)th step. Thus, we have Can this be used to get a formula for Rn? We have R0 = 1 R1 = 8 × 1 = 8 R2 = 8 × 8 = 82 . In general, Rn = 8n . (ii) square holes that are formed by removing square pieces. (i) squares of the same size that remain in the figure, and the size of these squares becomes smaller and smaller as the step number increases, and Sierpinski Carpet Rn + 1 = 8 Rn. Exploring Some Geometric Themes 04_Chapter 4.indd 71 20-12-2025 16:43:40 Similarly, how do we find the number of holes at a given step? Every square that remains at the nth step gives rise to a hole in the (n + 1)th step. All the holes present at the nth step remain in the (n + 1)th step as well. Thus, Hn + 1 = Hn + Rn. So we have R0 = 1 H0 = 0 R1 = 8 H1 = 1 R2 = 82 H2 = 1 + 8 R3 = 83 H3 = 1 + 8 + 82 . . . 71 Ganita Prakash | Grade 8 | Part-II Sierpinski Gasket Sierpinski came up with another fractal made in a similar way. An equilateral triangle is broken up into 4 identical equilateral triangles by joining the midpoints of the bigger triangle, and then the central triangle is removed. This procedure is repeated on the 3 remaining triangles, and so on. Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles. [Hint: Note that the corner triangles are isosceles.] This fractal is called the Sierpinski Triangle/Gasket. Step 0 Step 1 Step 2 Sierpinski Triangle . . . 04_Chapter 4.indd 72 20-12-2025 16:43:41 72 Figure it Out 1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle. 2. Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle. 3. Find the area of the region remaining at the nth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit. Koch Snowflake The Koch Snowflake is another fractal, named after the Swedish mathematician Von Koch, who first described it in 1904. We have already encountered this fractal in Grade 6, Ganita Prakash. To generate it, we start with an equilateral triangle. Each side is Effectively, each side gets replaced by a ‘bump’ - shaped structure . This procedure is repeated on the sides of the new resulting shape, and so on. (ii) an equilateral triangle is raised over the middle part, and then the middle part is removed. (i) divided into 3 equal parts, and Step 0 Step 1 Step 2 Koch Snowflake Exploring Some Geometric Themes . . . 04_Chapter 4.indd 73 20-12-2025 16:43:41 Figure it Out 1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake. 2. Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake. 3. Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit. 73 Ganita Prakash | Grade 8 | Part-II Fractals in Art Fractals have also long been used in human-made art! Perhaps the oldest such fractals appear in the temples of India. An example occurs in the Kandariya Mahadev Temple in Khajuraho, Madhya Pradesh which was completed in around 1025 C.E.; there one sees a tall temple structure, which is made up of smaller copies of the full structure, on which there are even smaller copies of the same structure, and so on. Fractal-like patterns also occur in temples in Madurai, Hampi, Rameswaram, Varanasi, among many others. Fractals are also common in traditional African cultures. For example, patterns on Nigerian Fulani wedding blankets often exhibit fractal structures. Kandariya Mahadev Temple 04_Chapter 4.indd 74 20-12-2025 16:43:44 74 Nigerian Fulani Wedding Blanket In the blanket shown in the figure, there are diamond shaped patterns, inside of which there are smaller diamond shaped patterns, and so on. The modern maestro of fractal art is undoubtedly the Dutch artist M.C. Escher. We have already seen that some of his prints explore the mathematical theme of tiling. One famous example involving fractals is his work ‘Smaller and Smaller’, which exhibits the identical pattern of lizards but at smaller and smaller scales. Here is a print inspired by Escher’s ‘Smaller and Smaller’. 4.2 Visualising Solids Build it in Your Imagination We will start this section by practising visualisation. For each prompt, feel free to talk to your partner, gesture, draw it in the air — but do not actually draw on paper! Exploring Some Geometric Themes 04_Chapter 4.indd 75 20-12-2025 16:43:45 1. Picture your name, then read off the letters backwards. Make sure to do this by sight, not by sound — really see your name! Now try with your friend’s name. My method is different. I do not rush into actual work. When I get an idea I start at once building it up in my imagination. I change the construction, make improvements, and operate the device entirely in my mind. — Nikolas Tesla (1856 – 1943), the great Serbian-American engineer and inventor, who made fundamental contributions to electrical engineering and other fields. 75 Ganita Prakash | Grade 8 | Part-II When we see a solid object, we are really seeing its profile from a specific viewpoint. Depending on our viewpoint, the outline of this profile can vary dramatically! Perhaps you’ve noticed interesting profiles of objects yourself, while taking a photograph or observing the shadow of an object? Or, maybe you’ve seen a cartoon like the following, in which a character bursts through a wall and leaves a hole shaped like their outline! 2. Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square? 3. Mark the sides of an equilateral triangle into thirds. Cut off each corner of the triangle, as far as the marks. What shape do you get? 4. Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left? 04_Chapter 4.indd 76 20-12-2025 16:43:49 76 In previous classes, you’ve seen solids that are much simpler than an elephant or cat, such as cubes, spheres, cylinders, and cones. What would the profiles of these look like, from different viewpoints? Can you describe a solid and a viewpoint that would result in each of the following cases? If it helps, you can imagine the solid passing through a wall like Tom did, and leaving a hole of the appropriate shape. As we saw with the elephant, a given solid might have very different profiles from different viewpoints. Can you visualise solids that have the following contrasting profiles? Spend some time on this, and if you are finding it difficult to visualise, you may look around and use objects that are around you, or that you will make in the next section. Feel free to consider viewpoints from any direction, including directly above the object. 10. A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint 11. A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint 5. A solid whose profile has a square outline 6. A solid whose profile has a circular outline 7. A solid whose profile has a triangular outline 8. A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint 9. A solid with a circular profile from one viewpoint and a triangular one from another viewpoint Exploring Some Geometric Themes 04_Chapter 4.indd 77 20-12-2025 16:43:49 12. A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint Are there unique solids for each of the conditions, or can you come up with multiple possibilities? Making Solids Many basic solid shapes such as a cuboid, parallelopiped, cylinder, cone, prism, and pyramid can be made using foldable flat materials like paper, cardboard, and even metallic sheets! This is one of the methods for manufacturing hollow solids. 77 Ganita Prakash | Grade 8 | Part-II While discussing solids that contain plane surfaces in their boundaries, the notion of faces, edges and vertices serves a useful purpose. Cuboid Parallelopiped Cylinder Cone Triangular prism Triangular pyramid Face 04_Chapter 4.indd 78 20-12-2025 16:43:49 78 Faces are the plane/flat surfaces of a solid that form its boundary. Edges are the line segments that form the sides of the faces, and vertices are the points at which the edges meet. For example, a cuboid or cube has 6 faces, 12 edges and 8 vertices. Vertex Edge Vertex Edge Face There are multiple types of prisms and pyramids, whose names depend on the shapes of their faces. Prism: A prism, in general, has two congruent polygons as opposite faces, with edges connecting the corresponding vertices of these polygons. All the other faces are parallelograms. Based on the shape of the congruent polygons, prisms may be called triangular prisms, pentagonal prisms, and so on. Pyramids: A pyramid, in general, has a polygonal base and a point outside it, and edges connect the point with each of the vertices of the base. Based on the shape of the base, pyramids may be called triangular pyramids, square pyramids, pentagonal pyramids, and so on. A triangular pyramid is also known as a tetrahedron. Triangular Prism Pentagonal Prism Hexagonal Prism Exploring Some Geometric Themes 04_Chapter 4.indd 79 20-12-2025 16:43:49 If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides? If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an n-sided polygon? Now we will see how to make different solids using a foldable flat surface such as paper, cardboard, etc. The basic idea is to create a shape on a flat surface that can be folded into the solid. Such a shape is called a net. In other words, a net is obtained by ‘unfolding’ a solid onto a plane. Triangular Pyramid Pentagonal Pyramid Hexagonal Pyramid 79 Ganita Prakash | Grade 8 | Part-II What is a net of a cube? Fig. 4.1 is a net of a cube. Visualise how it can be folded to form a cube. Practical Aspects of Using a Net To make an actual cube from its net, certain practical aspects have to be kept in mind. One is that the material used must have sufficient sturdiness for the cube to stand. The second aspect is that there should be some mechanism for attaching the faces together. If cardboard or a similar material is used, cello tape is one way to join the adjacent faces. For less sturdy materials, such as chart paper, it is useful to have extra ‘flaps’ on some of the faces that can be stuck to the adjacent faces. You might have observed this strategy in packaging boxes. However, when we talk about the net of a solid, we consider only the shape formed by unfolding the solid, and not the other supporting flaps that we may use to actually make the solid. Apart from using a net, there are other ways to make a cube from paper. (i) (ii) (iii) Fig. 4.1 (i) (ii) (iii) 04_Chapter 4.indd 80 22-12-2025 12:53:20 80 Figure it Out 1. Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try. There are multiple possible nets one can use to build a cube. Let us find out the nets of other solids. A tetrahedron whose faces are equilateral triangles is called a regular tetrahedron. What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron? 2. A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. For example, the following nets are all considered the same — 3. Draw a net of a cuboid having sidelengths: (ii) 6 cm, 3 cm, and 2 cm Find all the 11 nets of a cube. (i) 5 cm, 3 cm, and 1 cm (iv) (v) (vi) Exploring Some Geometric Themes Try This 04_Chapter 4.indd 81 20-12-2025 16:43:51 Are there any other possible nets? A regular tetrahedron has only 2 possible nets. Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout. Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout. Math Talk 81 Ganita Prakash | Grade 8 | Part-II What is the net of a cylinder? If the circular faces of a cylinder are unfolded, and if a cut is made along the height of the cylinder, as shown in the figure below, then we get What are the sidelengths of the rectangle obtained? How will the net of a cone look? height O l 04_Chapter 4.indd 82 20-12-2025 16:43:51 82 If the cone is slit open along the line l and then unrolled, what will we get? Observe that all the points on the boundary of the base circle are at equal distances from ‘O’. So after unrolling the cone, the boundary of the net will be a portion of a circle with centre O. What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question! Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout. O Math Talk Here is a solid called the octahedron, made by joining two square pyramids at their square bases. Taking all the triangles in the net to be equilateral, make a cutout of the net and fold it to form an octahedron. As in the case of cubes, an octahedron too has 11 different nets. So far, we have seen solids whose faces are all equilateral triangles (regular tetrahedrons), and all squares (cubes). Does there exist a solid whose faces are all pentagons? Interestingly, the answer is yes! This solid is called a dodecahedron. This solid too can be made from a net! Mathematicians have been even able to determine exactly how many nets a dodecahedron has — 43,380. Can you visualise its net? This is one of its nets. Octahedron Exploring Some Geometric Themes Math Talk Math Talk 04_Chapter 4.indd 83 20-12-2025 16:43:57 Dodecahedron 83 Ganita Prakash | Grade 8 | Part-II Net of a sphere? Experiment and see if you can make a paper cutout that can perfectly wrap around a ball without leaving any wrinkles, gaps or overlaps. Shortest Paths on a Cube Let us consider an interesting problem related to the discussion so far. We know that on a plane, the shortest path between two points is the straight line between them. Now, what is the shortest path between two points on the surface of a cuboid, if we are allowed to travel only along its surface? Let us imagine a hungry ant living on the surface of a cuboid. To its good fortune, there is a laddu on the surface. What is the shortest path for the ant to reach the laddu? What about in the following case? Laddu (at the centre of the top face) Laddu (at the centre of the edge) Ant (at the centre of the side face) Math Talk 04_Chapter 4.indd 84 20-12-2025 16:44:00 84 If we think that a certain path is the shortest, how can we be sure that it truly is, among all the infinite possibilities? For example, are either of these the shortest path? [Hint: Draw the net and see how these paths appear on the net.] What does this show? A path on the surface of the cuboid can be transformed to a path of the same length on the net. Conversely, every path on the net transforms to a path of the same length on the cuboid. In the first case, the path marked on the cuboid is the shortest since it transforms to a straight line between the ant and laddu on the net, which is the shortest path between them. In the second case, the path marked on the cuboid does not transform to a straight line path on the net, and hence is not the shortest path for the ant to take. Exploring Some Geometric Themes Laddu (at the centre of the edge) 04_Chapter 4.indd 85 20-12-2025 16:44:03 Thus, by using a net, we convert the problem of finding the shortest path on a cuboid to the problem of finding the shortest path on the net. Have we now completely analysed the problem of finding the shortest path between two points on a cuboid? The red path is the shortest 85 Ganita Prakash | Grade 8 | Part-II Find the shortest path between the ant and the laddu in the following case: If we unfold the cuboid as before, we get Notice the line segment between the ant and laddu going outside the net! Obviously, this does not correspond to any path on the cuboid. So what do we do now? If the net is unfolded in the following way, then we get a path on the cuboid. 4 cm 4 cm 4 cm 4 cm Ant (at the centre of the face) 8 cm 4 cm 4 cm 4 cm 4 cm 4 cm 2 cm 8 cm 04_Chapter 4.indd 86 20-12-2025 16:44:05 86 Thus, the way a cuboid is unfolded matters! The shortest path 12 c m 12 c m Now we will look at a trickier case. What is the length of the shortest path between the ant and the laddu? Here are some of the ways of unfolding the cuboid. In the second case, the distance d between the ant and the laddu can be calculated using the Baudhayana Theorem, since we have a right triangle here. d2 = 242 + 322 Representation of Solids on a Plane Surface Drawing is one of the oldest human activities. People have been doing it for thousands of years for both aesthetic as well as practical and engineering purposes. Regarding the latter, making a drawing of an object is a useful way to record or convey information about it. Visual representation also helps us in thinking about the objects being represented. Such needs often arise in engineering — while constructing buildings, designing machines, etc. Further, engineering drawings need Try This 12 cm 6 cm 30 cm 1 cm 42 cm laddu (stuck to the back of the box) Exploring Some Geometric Themes 1 cm 32 cm 6 cm 12 cm 24 cm 04_Chapter 4.indd 87 22-12-2025 12:53:40 We see that in each of these unfoldings, the lengths of the line segments between the ant and the laddu are different! So we have to carefully list all the possible different unfoldings to find the answer! d = 1600 = 40 cm 87 Ganita Prakash | Grade 8 | Part-II to be detailed enough so that one can physically construct the objects they depict. We will now explore some ways of drawing solids on a plane. Projections The net of a solid (when it exists) is one way of representing a solid on the plane. But it can be difficult to visualise the solid from the net, without physically cutting out the net and folding it up. Another way to represent a solid is by looking at the projections of all its points on a plane. This idea is closely related to the profile of a solid from a specific viewpoint, which we discussed earlier. Let us see how this technique works. Consider a point P in space, and a plane M. Imagine a line from P intersecting the plane M at a point O. We say that OP is perpendicular to the plane if for any line OX on the plane, ∠POX = 90°. In this case, we say that point O is the projection of point P on the plane. The projections of all the points of an object together form the projection of the object on the plane. M O is the projection of P O X P 04_Chapter 4.indd 88 20-12-2025 16:44:08 88 Let us visualise the projection of a line. Projection Projection Fig. 4.2 What happens to the length of a line in its projection? Let us consider projections in Fig. 4.3. For example let l be the actual length of the line and p be the length of its projection. Draw AE ⊥ BC. AECD is a rectangle (why?). So, AE = DC = p. Also, ∠AEB = 90°. Can you now compare the lengths p and l? When is the length of the projected line equal to its actual length? What do you think are the different possible projections of a square that we get based on its orientation? What do you think is the projection of a parallelogram under different orientations? Can this ever be a quadrilateral that is not a parallelogram? As a starting point, you could think about the projection of a pair of parallel lines. What can you say about the projection of an n-sided regular polygon? [Hint: Projection of a polygon is composed of the projections of its sides.] Let us consider projections of solids now. How would the projections of a cube and a cone look? Fig. 4.4 Fig. 4.5 D Exploring Some Geometric Themes C p Fig. 4.3 A E p B l Math Talk Math Talk 04_Chapter 4.indd 89 20-12-2025 16:44:08 89 Ganita Prakash | Grade 8 | Part-II If an object were to pass perpendicularly through a plane and form a hole, the shape of the hole would be the same as the shape of the projection of the object. See Figures 4.2 – 4.5. In each case, see if you can visualise another object that gives the same projection. 04_Chapter 4.indd 90 20-12-2025 16:44:08 90 We see that a given projection is not made by a unique object. Different lines and different cuboids giving the same projections Fig. 4.6 Find another object that makes the same projection as that of a given cone. For this reason, we often take three mutually perpendicular projections of an object. As shown in the figure, we consider a plane in front of the object called the vertical plane, a plane below the object called the horizontal plane and a plane to the side of the object called the side plane. Vertical Plane Side Plane Side View Top View Exploring Some Geometric Themes Front View 04_Chapter 4.indd 91 20-12-2025 16:44:09 We give names to the projections on these three planes based on the directions from which the object is viewed. The projection on the vertical plane is called the front view, on the horizontal plane the top view and on the side plane the side view. These projections formalise the notion of profiles that we explored in an earlier section. Let us see what these projections are for the objects shown in Fig. 4.6, when the planes shown are taken to be vertical planes. Projections of the different lines in Fig. 4.6 — Horizontal Plane 91 Ganita Prakash | Grade 8 | Part-II Projections of the different cuboids in Fig. 4.6 — Front View Top View Side View Front View Top View Side View 04_Chapter 4.indd 92 20-12-2025 16:44:09 92 Figure it Out 1. Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths? 2. Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid. If needed, see the next problem for clues. Math Talk 3. Match each of the following objects with its projections. FRONT TOP SIDE Exploring Some Geometric Themes 04_Chapter 4.indd 93 20-12-2025 16:44:10 Ganita Prakash | Grade 8 | Part-II Shadows Place an object in front of a plane, such as a wall of your room. Shine a torch light on the object in a direction perpendicular to the wall. What do you see? We will see that the shape of the shadow on a plane is quite similar to the shape of the projection on that plane! However the shadow may be scaled up, stretched, or even distorted slightly, depending on how the object is held. Observe what happens to the size of the shadow as you vary the distance between your torch and your object. Why does this happen? Now, imagine your torch is extremely powerful and will continue casting a shadow of the object on the wall, no matter how far back you move it. If this imaginary torch always points in a direction perpendicular to the wall, then, as the distance between the torch and the object increases, the shadow of the object will become indistinguishable from the projection of the object to the wall plane! Although this experiment may sound fanciful, you have encountered one such extremely powerful and distant torch before … namely, the Sun! Indeed, when sunlight is perpendicular to a plane, the shadows it casts on that plane are indistinguishable from projections! Earlier we asked what the projection of a parallelogram might look like. You can now physically answer this question by making a cutout of a parallelogram and viewing its shadows under sunlight. No matter how you orient the parallelogram, you will see that the shadow remains a parallelogram! Try This 04_Chapter 4.indd 94 20-12-2025 16:44:11 94 More generally, the projection of a pair of parallel lines will always remain parallel. Figure it Out 1. Draw the top view, front view and the side view of each of the following combinations of identical cubes. This property is useful for drawing projections of objects. Front Front Top Top Top Side Side Side Front Front Exploring Some Geometric Themes Top Top Top Side Side 04_Chapter 4.indd 95 20-12-2025 16:44:12 2. Imagine eight identical cubes, glued together along faces to form the letter ‘ ’. (iii) Now, can you glue even more cubes to make it look like ‘ ’ from the front, ‘ ’ from the top, and ‘ ’ from the side? (iv) Can you think of other letter combinations to make with a single combination of cubes in this manner? (ii) Glue additional cubes to make a shape that looks like ‘ ’ from the front and ‘ ’ from the top. (i) This looks like a ‘ ’ from the front. What does it look like from the side? From the top? Front Front Side Math Talk 95 Ganita Prakash | Grade 8 | Part-II 3. Which solid corresponds to the given top view, front view, and side view? Front 4. Using identical cubes, make a solid that gives the following projections. Top (i) (ii) (iii) (iv) Front View Top View Side View Top Top Top (v) (vi) (vii) Front Front Front Top Side Front Top Side Side Side Front Top Side Front Side Side 04_Chapter 4.indd 96 20-12-2025 16:44:14 96 (vii) (iv) (i) Top View Front View Side View Top View Front View Side View Top View Front View Side View (viii) (ii) (v) (iii) (vi) (ix) Isometric Projections In general we lose information when projecting a solid to a plane. But depending on the orientation of the solid, we can sometimes recover much of the information we’ve lost. Let us consider an orientation of a cube in which the lengths of the projections of all the edges of the cube are equal. Such a projection is called the isometric projection of the cube. ‘Isometric’ means ‘equal measure’ in Greek. Imagine balancing a cube perfectly on one of its corner vertices, and then projecting it down to the floor plane. This projection is isometric and appears as shown. 5. Find the number of cubes in this stack of identical cubes. 6. What are the different shapes the projection of a cube can make under different orientations? Exploring Some Geometric Themes Math Talk 04_Chapter 4.indd 97 20-12-2025 16:44:16 Construct a model of a cube and use your hands to keep it balanced on one corner vertex. Can you try to understand why all the projected edges have equal length? The isometric representation of a cube is a regular hexagon. If the cube is made of glass, then all the edges would be visible. This structure is the basis for isometric drawing! Tile the plane with hexagons and we get an isometric grid. Math Talk 97 Ganita Prakash | Grade 8 | Part-II Isometric grids are widely used in engineering. They make it easy to draw projections of solids and to measure lengths along all the 3 primary directions — length, depth and height, as shown in Fig. 4.7. Drawing on Isometric Grids Vertical Plane Side Plane Horizontal Plane depth length Fig. 4.7 height 04_Chapter 4.indd 98 20-12-2025 16:44:17 98 Have you played Tetris? There are five basic shapes in Tetris, corresponding to the different ways of arranging four squares. (iii) (iv) (v) (i) (ii) Fig. 4.8 Imagine these are cubes, not squares. Draw each of these on your isometric paper (you can find it at the end of the book). As seen in Fig. 4.7, there are three principal axes of the solid shape, which we will call the depth axis, length axis, and height axis. The edges of the grid appear in three different orientations: , and |. We will associate height to | , depth to , and length to . While drawing on the grid, it may be useful to draw edge by edge, counting the number of units you want to go along a given axis. For example, you can draw a 1 × 1 × 1 cube as follows. How would you draw a 2 × 2 × 2 cube? Feel free to add shading, if it helps you visualise the solid. depth length height Exploring Some Geometric Themes 04_Chapter 4.indd 99 20-12-2025 16:44:19 The first tetris shape is a row of four cubes joined face-to-face. In order to draw this solid, you will need to choose an orientation for the row of four cubes. Let’s draw it oriented along the depth axis. You could draw it cube-by-cube, but in that case you may need to erase the lines that get hidden by additional cubes. (If you don’t have an eraser, you can also draw faint lines initially, and then darken the visible ones later.) depth length height 99 Ganita Prakash | Grade 8 | Part-II You could also draw it all at once, by counting edges as you go: Similarly, you could draw the shape oriented along the height axis or the length axis. In your drawing, moving in the vertical direction on the isometric paper corresponds to moving along the height axis of the solid. Similarly, moving along the or direction on the paper corresponds to moving along the length or depth axis respectively of the solid. depth length depth length height height 04_Chapter 4.indd 100 20-12-2025 16:44:22 100 Why is this correspondence between directions on isometric paper and axes of the solid so effective for communicating the shape of the solid? Earlier we observed that parallel lines project to parallel lines. This geometric fact ensures that the three families of parallel edges on the solid (corresponding to height, the length, and depth) project to three families of parallel edges on the isometric paper: |, , and . Moreover, the isometric nature of the projection ensures that the projections of unit distances along the three axes are equal. Can you try drawing the other tetris shapes on isometric paper? Figure it Out 1. In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well? Math Talk 2. Draw the following figures on the isometric grid. 3. Is there anything strange about the path of this ball? Recreate it on the isometric grid. [Hint: It may be useful to determine whether the edge to be currently drawn — say, along the height — goes from down to up or up to down. Accordingly, draw the line segment on the grid either in the direction of the height axis or opposite to it.] [Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.] (i) Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle? Math Talk Exploring Some Geometric Themes 04_Chapter 4.indd 101 20-12-2025 16:44:24 4. Observe this triangle. 101 Ganita Prakash | Grade 8 | Part-II y Fractals are self-similar geometric objects found in nature and in art. y The Sierpinski Carpet, Sierpinski Gasket, and Koch Snowflake are some examples of mathematical fractals. They can be obtained by repeatedly applying certain geometric operations that generate a sequence of shapes approaching the fractal. y Cuboids, tetrahedrons, cylinders, cones, prisms, pyramids, and octahedrons are some of the solids that can be obtained by folding suitable nets. y The shortest path between two points on the surface of a cuboid can be found by using a suitable net of the cuboid. y Any object can be represented on a plane surface by using its projections on plane surfaces. For this purpose, we generally use the front view (projection on the vertical plane), top view (projection on the horizontal plane), and side view (projection on the side plane) of the object. y A cube can be oriented such that the lengths of all its edges in the projection are equal. Such a projection is called the isometric projection. Isometric projections of different solids can be drawn using isometric grid paper. (iii) Why does the illusion work? (ii) Recreate this on an isometric grid. SUMMARY Math Talk 04_Chapter 4.indd 102 20-12-2025 16:44:25 102" class_8,12,tales by dots and lines,ncert_books/class_8/hegp2dd/hegp205.pdf,"5 5.1 The Balancing Act Last year, we learnt about the mean and median. Recall that the mean of some data is the sum of all the values divided by the number of values in the data. The median is the middle value when the data is sorted. We shall try to understand the mean and median from a different perspective and see how the mean behaves with changing data. Consider any 2 numbers. Find their average/arithmetic mean. Repeat this by taking other pairs. What do you observe? For example, let the two numbers be 3 and 7. Their average is 3 + 7 2 = 5. Taking another pair of numbers, say 8 and 9, their average is 8 + 9 2 = 8.5. Visualising these as dot plots we get TALES BY DOTS AND LINES 05_Chapter 5.indd 103 20-12-2025 16:48:56 Notice that the mean is exactly halfway between the two numbers. We have learnt earlier that the arithmetic mean is a measure of central tendency and represents the ‘centre’ of the data. Let us see how the mean represents the ‘centre’ in the case of 3 numbers. Calculate and mark the mean of each collection of data below. Ganita Prakash | Grade 8 | Part-II Can you explain how the mean is the centre of each collection? Mark the mean for the collections below. Can you explain how the mean is the centre of each collection? Is the mean the midpoint of the two endpoints/extremes of the data? It is not always so. Instead, the total distances are equal on both the sides of the mean. This is illustrated through the following dot plots. Math Talk Math Talk 05_Chapter 5.indd 104 20-12-2025 16:48:58 104 Verify that this holds for all the collections of data shown earlier. Can there be more than one such ‘centre’? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal? In the case of the collection 10, 10, 11, and 17 whose mean is 12, suppose there is a different centre larger than 12. Clearly, all the distances on the LHS will increase and the distances on the RHS will decrease. Thus, it is no longer the ‘centre’. Similarly, for any value smaller than 12, the distances on the LHS will decrease while those on the RHS will increase. Both these cases are illustrated in the following diagram. Therefore, there is only one centre. Will including a new value in the data increase or decrease the mean? When a new value greater than the mean is included, the mean increases to maintain the balance between the sum of distances on the LHS and RHS, as illustrated below. Tales by Dots and Lines 05_Chapter 5.indd 105 20-12-2025 16:48:58 Similarly, if a value smaller than the mean is included, then the new mean will be less than before. What happens to the mean when an existing value is removed? When will the mean increase, decrease, or stay the same? What happens to the mean if a value equal to the mean is included or removed? Try to explain this using the fair-share interpretation of mean that we studied last year. Unchanging Mean! We saw earlier how the mean varies when a value is included or removed. Math Talk 105 Ganita Prakash | Grade 8 | Part-II Explore if it is possible to include or remove 2 values such that the mean is unchanged. You may use the following data to experiment with. How about including or removing 3 values without changing the mean? Is it possible? Can we include 2 values less than the mean and 1 value greater than the mean, so that the mean remains the same? One of the possibilities is shown here. Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same. Relatively Unchanged! We saw what happens to the mean when values are included or removed from the collection. What happens to the mean if every value in the collection increases by some fixed number? 05_Chapter 5.indd 106 20-12-2025 16:48:59 106 Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean. Now, consider this data with every value increased by 10: 18, 13, 20, 23, 14, 16, 17, 17, 18, 18, 15. What is its mean? Is there a quicker way to find out? [Hint: Observe the following dot plots corresponding to the two data collections.] The mean of the new collection also increases by 10. Notice that the relative position of the mean stays the same. We get the following dot plot if we reduce every value by 1. This can be explained using algebra — Suppose there are n values in the collection. Let these values be represented by x1, x2, x3, … xn. Their average is given by — x1 + x2 + x3 + … + xn n = a. When a fixed number, for example, 3 is added to every value in the collection, the new average becomes — (x1 + 3) + (x2 + 3) + (x3 + 3) + … + (xn + 3) n = x1 + x2 + x3 + … + xn + 3n n = x1 + x2 + x3 + … + xn n + 3n n = a + 3. Tales by Dots and Lines 05_Chapter 5.indd 107 20-12-2025 16:48:59 That is, the new average is 3 more than the previous average. Try to explain, using algebra, what the average is when a fixed number, e.g., 2 is subtracted from every value in the collection. Try to explain this using the fair-share interpretation of average that you learnt last year. What happens to the average if every value in the collection is doubled? You may have guessed that the average also doubles. The following is an example with the data we saw earlier — Try This 107 Ganita Prakash | Grade 8 | Part-II We can see that the average has doubled. Let us prove it using algebra. Suppose there are n values in the collection. Let these values be represented by x1, x2, x3, … xn. Their average is — x1 + x2 + x3 + … + xn n = a. When a fixed number, for example, 5 is multiplied to every value in the collection, the new average becomes — (5x1) + (5x2) + (5x3) + … + (5xn) n Tinkering with Median We know that the median is the middle value in the sorted data — there are an equal number of values less than it and greater than it. Will including a new value to the data increase or decrease the median? Let us consider the following data. The median of this data is 8. That is, the new average is 5 times more than the previous average. = (x1 + x2 + x3 + … + xn) × 5 n (using the distributive property) = (x1 + x2 + x3 + … + xn) n × 5. = 5a. 05_Chapter 5.indd 108 20-12-2025 16:49:00 108 Median = 9.5 Median = 8 Suppose we include a value 11. The new value included is greater than the (earlier) median — the median can no longer be 8 as there are more values greater than it. Therefore, the median will also increase. The median, now, will be the average of the two middle values 8 and 11, which is 9.5. We can similarly argue that when a value less than the median is included, the median will decrease. Finding the Unknown Coach Balwan noted down the weights of the kushti players (wrestlers) and the mean as shown. But one value that was written down got smudged. Can you find out the missing value? Average weight of the players = Sum of weight of all players Number of players . Let the unknown weight be w kg. = 42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 + w 10 = 39.2. Simplifying this we get, The missing value is 43 kg. Venkayya keeps track of the coconut harvest in his farm. He calculates the average harvest per tree as 25.6. His son verifies the counts and finds that one tree’s harvest count is incorrectly noted as 3 more than the actual number. Can you find the correct average if the number of trees is 15? 349 + w = 392, w = 392 – 349 = 43. Tales by Dots and Lines 05_Chapter 5.indd 109 20-12-2025 16:49:01 The average harvest per tree = Total number of coconuts harvested Number of trees . The data of harvest per tree is not given. Can we still find out the number of coconuts harvested? Let the initial number of coconuts harvested be z. Based on what is given, 25.6 = z 15, Simplifying, The initial count of coconuts harvested is 384. We know that one tree’s count is 3 more than the actual. Therefore, the actual total harvest count is 384 – 3 = 381. The correct average harvest is 381 15 = 25.4. z = 25.6 × 15 = 384. 109 Ganita Prakash | Grade 8 | Part-II Mean and Median with Frequencies What is the average family size of students in your class? How would you find this out? We can collect the data of how many family members each student has, add them up, and divide it by the number of students. The family size data of students in a class is shown in the table here. What is the average family size of this class? 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 8 = 52 8 = 6.5.” Remember that finding the average involves adding all the values in the data. The number 3 occurs three times, the number 4 occurs eleven times, and so on. Do these reflect in your calculation? We know that the average = Sum of all the values in the data Number of value in the data . Accounting for the frequencies of each value, the average will be What is the median family size of this class? We know that there are 36 values in the data. The median would be the average of the 18th and 19th value in the sequence when the data is ordered. Some of you may have thought, “Easy! It will be = (3 × 3) + (4 × 11) + (5 × 9) + (6 × 7) + (7 × 3) + (8 × 1) + (9 × 1) + (10 × 1) 3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 = 9 + 44 + 45 + 42 + 21 + 8 + 9 + 10 36 = 188 36 = 5.22. The average family size of this class is 5.22. Number Frequency 10 1 3 3 4 11 5 9 6 7 7 3 8 1 9 1 05_Chapter 5.indd 110 20-12-2025 16:49:01 110 Do we need to write all the 36 numbers in order? Is there a quicker way to find out? We can make use of the table where the frequencies are listed. We successively add the frequencies starting from the smallest value until we reach 18 and 19. Adding the frequencies of 3 and 4, we get 3 + 11 = 14. This means the value in the 14th position when the data is sorted is 4. Adding the frequencies of 3, 4, and 5, we get 3 + 11 + 9 = 23. This means the value in the 23rd position when the data is sorted is 5. We can see that all the values from positions 15 to 23 are 5. Therefore, the median of this data is 5. Frequency 3 11 9 7 3 1 1 1 Number 3 4 5 6 7 8 9 10 Spreadsheets Sudhakar has collected the mid-term exam marks obtained by his Grade 8 students in the following table: Aishwarya 48 45 46 47 44 43 Shravan 12 17 21 20 14 18 Nagesh 41 43 48 39 40 39 Ashwin 29 31 33 34 30 28 Ganesh 39 37 35 38 36 40 Farooq 47 46 38 42 49 44 Mrinal 33 35 28 32 30 36 Pankaj 16 19 22 17 18 20 Gowri 27 29 34 31 32 30 Trupti 29 36 30 33 27 33 Name Odia Telugu English Maths Social Science Science Ratna 25 39 29 36 34 37 Jaya 31 38 40 50 43 46 Hari 25 28 24 21 23 26 Tales by Dots and Lines 05_Chapter 5.indd 111 20-12-2025 16:49:01 Sanskruti 35 42 41 46 38 40 Vyshnavi 37 32 29 33 31 35 Shanker 42 45 39 36 31 39 Veeresh 23 25 28 31 19 22 Govind 15 18 12 20 20 18 Vidhya 34 36 37 40 32 34 Tarun 41 44 39 43 37 42 Shiva 29 24 32 34 28 30 Jyothi 29 30 33 28 34 29 111 Ganita Prakash | Grade 8 | Part-II Sudhakar has to calculate the total marks scored by each student. He is also interested in knowing the average marks scored in each subject. But there are so many numbers! He will have to spend too much time and effort to complete this task. Of course, he can use a calculator to speed up the task. Is there any way to further quicken this process? One way is to use a computer. Computers have different applications/ tools that can be used to perform tasks. One such application is a spreadsheet — it is like a digital notebook with rows and columns of small boxes called cells. In each cell, we can type text or numbers. The picture below shows a snapshot of a spreadsheet where Sudhakar has entered the marks data of his class. Try to get access to a computer before proceeding. Before learning how to calculate the average marks per subject and the total marks scored by each student, let us first understand the structure of spreadsheets and how to read them. Can you tell which cell has the marks obtained by Farooq in Mathematics? Cells are named and referred to using the column headers labelled A, B, C, … , and row headers labelled 1, 2, 3, … Farooq’s score in Mathematics is in cell E5. Can you tell what data is in column B7? 05_Chapter 5.indd 112 20-12-2025 16:49:02 112 In which subjects has Ashwin scored more than 30 marks? How can we use spreadsheets to quickly calculate totals and averages? In addition to text and numbers, we can also enter formulae in a cell. We can write a formula that computes the sum or average of a row, or column of cells. We can describe a row of cells by an expression of the form Start:End, indicating the first and last cell in the row. For instance, Nagesh’s marks are described by the expression B3:G3, while Gowri’s marks in Odiya, Telugu and English are described by the expression B7:D7. We can use similar expressions to describe columns. For instance, D2:D6 describes the marks in English for the first five students. We can then write a formula to compute the sum or the average of a row or column. For instance =SUM(B3:G3) calculates the total marks for Nagesh across all subjects, while =AVERAGE(B7:D7) calculates Gowri’s average marks across Odia, Telugu and English. What formula would you type to find out the class average marks in Science? Find out if the class average marks in Odia is greater than the class average marks in Telugu. Show the average marks in other subjects after the last row by typing the appropriate formulae. Try these out on a computer. You can download the tabular data from this QR code. Get the total scores of each student by typing the appropriate formulae. Figure it Out Note to the Teacher: You can use any spreadsheet software application such as Microsoft Excel, Google Sheets, LibreOffice Calc, etc. If there are not sufficient computers for each student, students can share a computer in groups. If that’s not possible, the computer screen can be projected for the whole class if there is only one computer available. Tales by Dots and Lines 05_Chapter 5.indd 113 20-12-2025 16:49:03 1. Find the mean of the following data and share your observations: 2. The dot plot below shows a collection of data and its average; but one dot is missing. Mark the missing value so that the mean is 9 (as shown below). (iii) The first 50 multiples of 4. (ii) The first 50 odd numbers. (i) The first 50 natural numbers. Math Talk 113 Ganita Prakash | Grade 8 | Part-II 3. Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students in his class and calculate the average height. Shreyas informs the teacher that the average height is 150.2 cm. Sudhakar discovers that the students were wearing uniform shoes when the measurements were taken and the shoes add 1 cm to the height. 4. The three dot plots below show the lengths, in minutes, of songs of different albums. Which of these has a mean of 5.57 minutes? Explain how you arrived at the answer. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 A 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 B (ii) What is the correct average height of the class? (i) Should the teacher get all the heights measured again without the shoes to find the correct average height? Or is there a simpler way? (a) 174.2 cm (b) 126.2 cm (c) 150.2 cm (d) 149.2 cm (e) 151.2 cm (f) None of the above (g) Insufficient information 05_Chapter 5.indd 114 20-12-2025 16:49:03 114 5. Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (iii) If we remove one value from the data without affecting the median what could the value be? 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 C (ii) If we include two values to the data without affecting the median what could the two values be? (i) If we include one value to the data (in the given list) without affecting the median, what could that value be? Math Talk 6. Examine the statements below and justify if the statement is always true, sometimes true, or never true. 7. The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y. 8. The mean of a set of data with 15 values is 134. Find the sum of the data. 9. Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the following number(s) could be p if the median of this data is 29? 10. The number of times students rode their cycles in a week is shown in the dot plot below. Four students rode their cycles twice in that week. (iii) Including any 4 values will not affect the median. (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30 (iv) Including 4 values less than the median will increase the median. (ii) Including a value less than the mean will decrease the mean. (i) Removing a value less than the median will decrease the median. Tales by Dots and Lines 05_Chapter 5.indd 115 20-12-2025 16:49:04 (iii) Which of the following statements are valid? Why? (ii) Find the median number of times students rode their cycles. (i) Find the average number of times students rode their cycles. (a) Everyone used their cycle at least once. (b) Almost everyone used their cycle a few times. (c) There are some students who cycled more than once on some days. (d) Exactly 5 students have used their cycles more than once on some days. 115 Ganita Prakash | Grade 8 | Part-II 11. A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull’s eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median. 5.2 Visualising and Interpreting Data So far we have learnt how to read and make pictographs, bar graphs, clustered-bar graphs, and dot plots. We now examine some more ways of visualising data. Line Graphs Temperature The following figure is a clustered-column graph that shows the monthly maximum temperature in Kerala and Punjab in 2023. No. of students 1 0 0 1 4 9 12 15 10 10 No. of trials 1 2 3 4 5 6 7 8 9 10 (e) The following week, if all of them cycled 1 more time than they did the previous week, what would be the average and median of the next week’s data? 05_Chapter 5.indd 116 20-12-2025 16:49:05 116 Now, observe the following graph. Do both these graphs represent the same information? We call such a graph made up of lines a line graph. Line graphs are generally used to visualise data across time. How do we get the maximum temperature over a month in a state? There could be a few weather stations across the state that regularly track the local temperature. We can get the monthly maximum temperature by looking at the maximum value among all the values recorded across the state. When we try to understand how the data is collected or produced, we gain a clearer idea of its scope and can interpret it more meaningfully. It also helps us identify any limitations, such as bias or missing information, and decide how confidently we can draw conclusions from the data. Tales by Dots and Lines 05_Chapter 5.indd 117 20-12-2025 16:49:07 To identify and interpret the information presented, let us follow a two-step process. Step 1: Identify what is given Notice how the graph is organised, what scale is used, and what patterns the data shows. • The data for each month for a city is marked and connected by lines to show the change over time. Kerala’s data is shown using blue circle marks connected by blue lines and Punjab’s data is shown in red. The different shape markers help in easily distinguishing if the graph is printed in greyscale/black-and-white, or for people who find it difficult to distinguish colour. 117 Ganita Prakash | Grade 8 | Part-II Step 2: Infer and interpret from what is given Analyse and interpret each of the observations you made. Share appropriate summary/conclusion statements. • In Punjab, the monthly maximum temperature increases from January to June, reaching a high of 38oC. Then, it reduces to just under 35oC in July, stays mostly flat till September, and then falls continuously till December, reaching about 23oC. January has the lowest among the monthly maximum temperature of about 19oC. • The horizontal line shows the months of the year. The vertical line shows the temperature in oC. • Kerala’s trend is different — it stays mostly flat throughout the year. The peak is around 33oC in April and the lowest point is around 29oC in July. Notice that the monthly maximum temperatures in Kerala are similar both in summer and winter! 05_Chapter 5.indd 118 20-12-2025 16:49:09 118 • In short, the temperature in Punjab varies more, reaching colder and warmer temperatures than in Kerala. This can trigger questions in different directions, some of which can be answered using data. Here are some directions to think about — • Why are the trends so distinct in these two states? What factors determine a region’s temperature? • You might be curious to look at the trends of a few other states. Are there other types of trends that states exhibit? • Which states show trends similar to Punjab’s? Is there anything common between these states? • What would a plot of the monthly minimum temperatures for these states look like? What thoughts or questions occur to you? Space Jam: A Traffic Problem in the Future? Take a look at the line graph below showing the annual number of objects launched into space. What could be the possible method used to derive this data? Discuss. Step 1: Identify what is given Tales by Dots and Lines Math Talk Math Talk 05_Chapter 5.indd 119 20-12-2025 16:49:09 Notice how the graph is organised, what scale is used, and what patterns the data shows. Step 2: Infer from and interpret what is given Analyse and interpret each of the observations you made. Once all interpretations are made, summarising/concluding statements can be made. • In 2024, the worldwide count is around 2800, whereas in 2023, it was around 2900. We can say that 2023 saw the highest number of objects launched into space worldwide (assuming the trend before 2012 was decreasing). • The counts of the three countries don’t add up to the worldwide count. Therefore, we can infer that the counts of other countries are not shown in this visualisation. 119 Ganita Prakash | Grade 8 | Part-II Which of the following statements are valid inferences? Identify two consecutive years where the worldwide count increased by 2 times or more. Catch the (Pattern in) Rain The following graphs show monthly average rainfall data of a few cities. Imagine the same data being shown as a clustered column graph. There would be 13 clusters  — one for each year — and within each cluster, 4 columns, making a total of 52 bars! Such a graph would look crowded and difficult to read, making it hard to interpret the trends clearly. A line graph, on the other hand, is better suited for illustrating changes over time. By connecting data points with line segments, it provides a clear visual representation of trends and variations, allowing the reader to easily track how a parameter evolves across different years. • For the USA, the increase from 2022 to 2023 is more than from 2023 to 2024. We can say this by looking at how steep the line segments are — the steeper the line is, the greater the increase. • From 2012 till 2024, the worldwide count of space object launches increased every year. • USA is a major contributor in the years 2022 – 24, launching about 3 4th of the worldwide count. • Nepal did not launch any object in the period 2012 – 24. • The combined count of object launches by China and Russia in 2024 is about 400. 05_Chapter 5.indd 120 20-12-2025 16:49:11 120 What could be the possible method to compile this data? This data shows the monthly average rainfall in 6 cities. This means rainfall data is collected over a few years in every city. The total rainfall in a month, say June, across years is averaged to get the monthly average rainfall in June in a city. Mark these cities on a map of India. What is common to how they are grouped in the graphs? Share your observations and inferences about the graphs. Kovalam, Udupi, and Mumbai are along the west coast. Rameswaram, Chennai, and Puri are along the east coast. It appears that regions along the west coast receive more rain. Source: weather-and-climate.com Tales by Dots and Lines Math Talk 05_Chapter 5.indd 121 20-12-2025 16:49:13 Identify the peak months and low months of rainfall for each city. Udupi, Mumbai, and Kovalam have peak rainfall during June – August. Rameswaram gets most of its rain during October – December. Chennai starts getting rain from June onwards, which peaks in November, and continues till December. Puri, although on the east coast, gets its peak rainfall during July – September. January – March are dry months, with low rainfall, for all the cities. Rameswaram receives very little rain from January – September. Read about the south-west monsoon and north-east monsoon and which regions come under the influence of these and when. 121 Rameswaram 2.6 1.3 1.9 3.4 2.5 0.4 1 1 1.9 8.1 10.4 7.8 Mangaluru 0.1 0 0.1 1.8 6.2 24.1 27.7 24.5 14 8.8 3.9 0.9 New Delhi Port Blair 2.4 1.3 0.9 3.3 15.5 18.7 17.3 18.8 16.8 14.1 11.3 5.4 Ganita Prakash | Grade 8 | Part-II Figure it Out 1. The average number of customers visiting a shop and the average number of customers actually purchasing items over different days of the week is shown in the table below. Visualise this data on a line graph. 2. The average number of days of rainfall in each month for a few cities is shown in the table below: (ii) Mark the data for Mangaluru, Port Blair, and Rameswaram in the line graph shown below. You can round off the values to the nearest integer. (i) What could be the possible method to compile this data? Purchasing 10 8 7 11 12 16 26 Visiting 16 19 10 14 20 22 35 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Mon Tue Wed Thu Fri Sat Sun 05_Chapter 5.indd 122 20-12-2025 16:49:13 122 Source: weather-and-climate.com 2. The following line graph shows the number of births in every month in India over a time period: (iii) Based on the line for New Delhi in the graph fill the data in the table. (iv) Which city among these receives the most number of days of rainfall per year? Which city gets the least number of days of rainfall per year? (v) Looking at the table, when is the rainy season in New Delhi and Rameswaram? (ii) What was the approximate number of births in July 2017? (i) What are your observations? Tales by Dots and Lines 05_Chapter 5.indd 123 20-12-2025 16:49:14 Infographics In Grade 6, we saw an example of how infographics can be used to communicate information and insights more clearly and quickly in a visually appealing way. Take a look at the following infographic. Are you able to understand the information presented here? (iii) What time period does the graph capture? (iv) Compare the number of births in the month of January in the years 2018, 2019, and 2020. (v) Estimate the number of births in the year 2019. 123 Ganita Prakash | Grade 8 | Part-II This infographic compares the preference between rice and wheat in different states. The colour scale at the top right indicates how to interpret the different shades in terms of preferences. The difference between the per capita consumption of rice and wheat is mapped to values between – 100 and + 100. A value of + 100 doesn’t mean that the state doesn’t consume wheat at all. It means that this state prefers mostly rice and the difference between per capita rice and wheat consumption is the highest. Isn’t it interesting how there is a clear geographical split in preferences of rice vs. wheat shown by the red line? Map not to scale 05_Chapter 5.indd 124 20-12-2025 16:49:15 124 Share your observations. Based on this infographic, answer the following: (ii) Which are the top 5 states where rice is the most popular? (iii) Which are the top 5 states where wheat is the most popular? (iv) List a few states where the preference between rice and wheat is more or less balanced. (i) The value of Karnataka is hidden. Can you guess what it could be? What can a Strip Say? Manoj has an interesting hobby. He makes note of what he does throughout the day. He records his activities by colouring a strip of paper with 48 boxes, marking time in 30 minute intervals from midnight to midnight. He has recorded five types of activities for three different days of the week on three coloured strips, shown to the right. (iii) Meeting friends, hobbies, media, time with family (ii) Eating (i) Sleeping I wonder if there is anything common in all the states preferring rice or states preferring wheat. I want to know if the preferences were similar even 100 years ago. Tales by Dots and Lines 05_Chapter 5.indd 125 20-12-2025 16:49:17 (vi) Travelling Look at the three coloured strips carefully. (iii) On one of these days, he went out with friends to watch a long movie. When do you think this happened? (iv) Attending classes, studying and homework (v) Showering and getting dressed, yoga or exercise (iv) At what time does his school break for lunch? (ii) The three strips correspond to the days Friday – Sunday in some order. Which day do you think each strip represents? (v) What more can the strips tell us? (i) What activity does each colour stand for? 125 Ganita Prakash | Grade 8 | Part-II What would your strip for a weekday look like? How similar or different is it to Manoj’s? What would a strip of your typical day during your vacation look like? How similar/different would it look? What would a strip for any of the adults in your family look like? Make a strip of a day for any adult at home. Compare your strip with theirs. What do you find interesting? Data Story: Sleepy-Deepy Do you remember the sleep time pie chart from the Proportionality chapter? Isn’t it amusing that there are animals that sleep as little as 2 hours per day and animals that sleep as much as 20 hours per day? How about insects — have you seen any insect sleep/rest? Humans typically sleep for about 7 –  9 hours a day. The sleep duration can vary greatly among people. The amount of sleep people need depends on age, living conditions, lifestyle (food they consume, the activities they engage in, etc.) among other factors. You may have observed that babies sleep longer than adults. The line graphs below show typical sleep durations of Indians across ages 6 to 75. The second picture is a zoomed-in version of the first picture. Try This Math Talk 05_Chapter 5.indd 126 20-12-2025 16:49:17 126 Share your observations on this graph. What do you find interesting? Unlike the earlier graphs that were made up of a few connected line segments, this line graph looks like a smooth curve because it contains 80 data points that are placed very close to each other. Representing the same data using a column graph would require 70 columns, making the graph look cluttered and heavy. A line graph is a better choice here. It is light and captures the pattern well. From the graph we can see that the average sleep time for 6-year olds is about 9.5 hours per day. The daily sleep time decreases as we grow through teenage years and step into adulthood, touching about 8 hours a day between ages 30 and 50. After 50, the daily sleep time increases, reaching about 8.5 hours. You might wonder — why do newborns and infants sleep for so long? Do people in different countries sleep differently? Do animals also exhibit such patterns in sleep durations across age? Figure it Out 1. Mean Grids: 2. Give two examples of data that satisfy each of the following conditions: (iii) 5 numbers whose mean is 13.6. (ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions? (ii) 4 numbers whose median is 15.5. (i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10. (i) 3 numbers whose mean is 8. Tales by Dots and Lines 05_Chapter 5.indd 127 20-12-2025 16:49:17 3. Fill in the blanks such that the median of the collection is 13: 5, 21, 14, _____, ______, ______. How many possibilities exist if only counting numbers are allowed? 4. Fill in the blanks such that the mean of the collection is 6.5: 3, 11, ____, _____, 15, 6. How many possibilities exist if only counting numbers are allowed? 5. Check whether each of the statements below is true. Justify your reasoning. Use algebra, if necessary, to justify. (ii) The average of any two multiples of 5 will be a multiple of 5. (iii) The average of any 5 multiples of 5 will also be a multiple of 5. (iv) 6 numbers whose mean = median. (v) 6 numbers whose mean > median. (i) The average of two even numbers is even. 127 Ganita Prakash | Grade 8 | Part-II 6. There were 2 new admissions to Sudhakar’s class just a couple of days after the class average height was found to be 150.2 cm. (iii) Which of the following statements about the new class average height are correct? Why? (ii) The heights of the two new joinees are 149 cm and 152 cm. Which of the following statements about the class’ average height are correct? Why? (i) Which of the following statements are correct? Why? (a) The average height of the class will increase as there are 2 new values. (b) The average height of the class will remain the same. (c) The heights of the new students have to be measured to find out the new average height. (d) The heights of everyone in the class has to be measured again to calculate the new average height. (a) The average will remain the same. (b) The average will increase. (c) The average will decrease. (d) The information is not sufficient to make a claim about the average. (a) The median will remain the same. (b) The median will increase. (c) The median will decrease. (d) The information is not sufficient to make a claim about median. 05_Chapter 5.indd 128 20-12-2025 16:49:17 128 7. Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question. 8. The weights of people in a group were measured every month. The average weight for the previous month was 65.3 kg and the median weight was 67 kg. The data for this month showed that 14 15 16 17 18 19 20 21 22 × × × × × × × × × × × × × × × × × × × × × × × 23 × × Math Talk Math Talk 9. The following table shows the retail price (in ₹) of iodised salt in the month of January in a few states over 10 years. For your calculations and plotting you may round off values to the nearest counting number. 2016 16 6 16.5 20 16.15 9.47 2017 12 12 14.75 20 16.97 11.65 2018 12 12 14.75 22 16.18 11.63 2019 12 12 14.75 22 18.24 11.43 2020 13.88 12 13 20 18.96 11.11 2021 18.22 15 14.45 22 20.63 12.79 2022 18.73 14 14.28 25 21.3 16.14 2023 20.63 12.02 14.54 27.65 25.39 18.43 2024 19.73 13.72 14.8 29.03 26.9 21.66 one person has lost 2 kg and two have gained 1 kg. What can we say about the change in mean weight and median weight this month? Andaman and Nicobar Islands Assam Gujarat Mizoram Uttar Pradesh West Bengal Tales by Dots and Lines 05_Chapter 5.indd 129 20-12-2025 16:49:17 2025 20.99 12.35 19.2 29.8 24.81 23.99 (iii) Compare the price variation in Gujarat and Uttar Pradesh. (iv) In which state has the price increased the most from 2016 to 2025? (v) What are you curious to explore further? (ii) What do you find interesting in this data? Share your observations. (i) Choose data from any 3 states you find interesting and present it through a line graph using an appropriate scale. 129 Ganita Prakash | Grade 8 | Part-II 10. Referring to the graph below, which of the following statements are valid? Why? (iii) In the year 2000, 10% of the urban households used electricity as a primary lighting source. (iv) In 2023, there were no power cuts. (ii) The use of kerosene as a primary lighting source has decreased over time in both rural and urban areas. (i) In 1983, the majority in rural areas used kerosene as a primary lighting source while the majority in urban areas used electricity. 05_Chapter 5.indd 130 20-12-2025 16:49:19 130 11. Answer the following questions based on the line graph. (ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 hours? (i) How long do children aged 10 in urban areas spend each day on hobbies and games? (a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years 12. Individual project: Make your own activity strip for different days of the week. 13. Small group project: Make a group of 3 – 4 members. Do at least one of the following: (iii) Are the following statements correct? (iii) Similarly, track the activities of any adult at home. Compare your data with theirs. (ii) Calculate the average time spent per activity. Represent this average day using a strip. (i) Do you eat and sleep at regular times every day? Typically how long do you spend outdoors? (i) Track daily sleep time of all your family members for a week. Daily sleep time includes night sleep, naps, and any sleep during the day. (a) The average time spent daily on hobbies and games by kids aged 15 is twice that of kids aged 10. (b) All rural kids aged 15 spend at least 1 hour on hobbies and games everyday. (a) Represent this on strips. (b) Put together the data of all your group members. Calculate the average and median sleep time of children, adults, elderly. (c) Share your findings and observations. Hey, come play with us — in real life! Tales by Dots and Lines Try This 05_Chapter 5.indd 131 20-12-2025 16:49:20 14. The following graphs show the sunrise and sunset times across the year at 4 locations in India. Observe how the graphs are organised. Are you able to identify which lines indicate the sunrise and which indicate the sunset? (ii) When do schools start and end? On a weekday, Manoj’s school starts at 9:30 am and ends at 4:30 pm, i.e., 7 hours which include class time and breaks. Collect information on the daily timings of different schools for Grade 8, including class time and break time (the schools can be anywhere in the country. You can ask your neighbours, relatives, parents and friends to find out). Analyse and present the data collected. 131 Ganita Prakash | Grade 8 | Part-II Answer the following questions based on the graphs: 15. We all know the typical sunrise and sunset timings. Do you know when the moon rises and sets? Does it follow a regular pattern like the sun? Let’s find out. The following graph shows the moonrise and moonset time over a month: (iii) Share your observations — what do you find interesting? What are you curious to find out? (ii) Which place has the longest day length over the year? (i) At which place does the sun rise the earliest in January? What is the approximate day length at this place in January? (i) Find out on what dates amavasya (new moon) and purnima (full moon) were in this month. Try This 05_Chapter 5.indd 132 20-12-2025 16:49:22 132 (ii) What do you notice? What do you wonder? y Last year we looked at mean as a fair-share. Here, we learnt how the sum of the distances of the values to its left and right are the same. y We saw that when values greater than the mean are inserted, the mean increases. When values less than the mean are inserted, the mean decreases. Similar phenomena can be observed with the median. y Line graphs can be used to visualise change over time. y We saw that examining data can lead to new questions and directions to probe further. SUMMARY Tales by Dots and Lines 05_Chapter 5.indd 133 23-12-2025 17:24:16 133 Hex is a two-player strategy game played on a rhombus-shaped board made of hexagonal cells, usually of size 11 × 11. Each player is assigned a colour and two opposite sides of the board. Players take turns placing a piece of their colour on any empty cell. Once placed, pieces can not be moved or removed. The objective is to create an unbroken chain of one’s own pieces connecting the two assigned sides. The first player to complete such a connection wins the game. The pictures below show two possible gameplays where blue wins in the first and red wins in the second board. Here is an empty board. You can use pencils to play each round and erase the marks to play a fresh round. Game of Hex 05_Chapter 5.indd 134 20-12-2025 16:49:28" class_8,13,algebra play,ncert_books/class_8/hegp2dd/hegp206.pdf,"ALGEBRA PLAY 6 6.1 Algebra Play Over the last two years, we have used algebra to model different situations. We have learned how to solve algebraic equations and find the values of unknown letter-numbers. Let’s now have some fun with algebra. We shall investigate tricks and puzzles, and explain why they work using algebra. We will also see how to invent new tricks and puzzles to entertain others. 6.2 Thinking about ‘Think of a Number’ Tricks In Grade 7, we learned about ‘Think of a Number’ tricks, like this one. 1. Think of a number. 2. Double it. 3. Add four. 06_Chapter 6.indd 135 20-12-2025 16:51:41 I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why? We saw that we can understand such tricks through algebra. Therefore, no matter what the starting number is, the end result will always be 2! 4. Divide by two. 5. Subtract the original number you thought of. 1. Think of a number: x 2. Double it: 2x 3. Add four: 2x + 4 4. Divide by 2: x + 2 5. Subtract the original number you thought of: x + 2 – x = 2 Ganita Prakash | Grade 8 | Part-II How would you change this game to make the final answer 3? What about 5? Can you come up with more complicated steps that always lead to the same final value? Let us now look at a different trick of this type. Multiply the month by 5. Tell me your answer. Think of a date. 26/01 Multiply by 4. Multiply by 5. Add the day. Add 6. Add 9. 11 × 4 = 44 53 × 5 = 265 44 + 9 = 53 5 + 6 = 11 1 × 5 = 5 06_Chapter 6.indd 136 20-12-2025 16:51:43 136 How did Shubham figure out the date chosen by Mukta? • Let the month be M and the day be D. • Multiply M by 5: 5M You thought of our Republic Day, 26/01. 265 + 26 = 291. Mukta’s answer was 291. • 126 = 100M + D Since D is a day within a month, it is atmost 31 and requires only 2 digits. So the last 2 digits are D and what comes before that is M. In this case, M is 1 and D is 26, i.e., the 26th of January. Mukta thinks of another date, follows the same steps, and reports her answer as 1390. What date did Mukta start with this time? Find the dates if the final answers are the following: You can try this trick with your friends. Ask them to choose the starting date as their birthday. Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number. (iii) 296 (ii) 394 (i) 1269 • Add 6: 5M + 6 • Multiply by 4: 20M + 24 • Add 9: 20M + 33 • Multiply by 5: 100M + 165 • Add the day: 100M + 165 + D • 291 = 100M + 165 + D • 291 – 165 = 100M + D • Subtracting 165 from 1390, we get 1225. • This means that the date she thought of was 25th December. Algebra Play 06_Chapter 6.indd 137 20-12-2025 16:51:43 Try to devise your own ‘Think of a Number’ trick. 6.3 Number Pyramids In a number pyramid, each number is the sum of the two numbers directly below it (see the figure). 1 10 23 9 4 13 Math Talk Math Talk 137 Ganita Prakash | Grade 8 | Part-II Use the same rule to fill these pyramids: How do we fill this pyramid? 1 1 4 4 10 10 3 3 And here? 6 6 6 ? 2 What will go here? 3 ? 1 ? 4 3 4 ? 10 ? ? ? 1 1 5 4 5 0 4 4 10 ? 10 4 – 1 = 3 ? ? ? ? ? 10 – 4 = 6 06_Chapter 6.indd 138 20-12-2025 16:51:43 138 What about filling in the numbers in this pyramid? Where do we start? What will go here? 1 12 4 ? 10 3 3 60 ? 8 6 ? 6 – 3 = 3 Let us fill the empty boxes with letter-numbers. From the rules for filling up pyramids, we get the following equations. From this, we see that, Hence, Once we know c, we can find a and b, and complete the pyramid. Fill the following pyramids: • a + b = 60 • 12 + c = a • c + 8 = b • (12 + c) + (c + 8) = a + b = 60 • 20 + 2c = 60 • 2c = 60 – 20 = 40 • c = 20. 12 12 32 a 60 60 20 8 c 8 28 b Algebra Play 06_Chapter 6.indd 139 20-12-2025 16:51:43 What is the relationship between the numbers in the bottom row and the number at the top? Let us start with the simplest pyramid. 4 6 50 22 5 7 9 a b 40 a + b 3 5 35 7 139 Ganita Prakash | Grade 8 | Part-II What about a pyramid with three rows? Using letter numbers for the bottom row, we can write an expression for the top row. Figure it Out 1. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases. 2. Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row. 3. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases. 4. If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci numbers? Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, … where each number is the sum of the two numbers before it. 8 19 21 13 7 18 19 6 9 7 5 11 4 13 8 7 11 3 10 14 25 a b c a + b b + c a + 2b + c 06_Chapter 6.indd 140 20-12-2025 16:51:43 140 5. What can you say about the numbers in the pyramid and the number at the top in the following cases? (ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid. 6. If the bottom row of an n row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top? (i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid. 6.4 Fun with Grids Calendar Magic A page from a calendar is given below. Your friend picks a 2 × 2 grid from this calendar, adds the 4 numbers in this grid and tells you the sum. Can we find the 4 numbers in the grid from just knowing this sum? Let us use algebra. Consider a 2 × 2 grid. Let a represent the top left number. What are the other numbers in terms of a? Adding all four numbers, the sum is a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. Suppose you are told that the sum is 36. Can you find the 4 numbers in the grid? SUN MON TUE WED THU FRI SAT 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3 4 5 6 7 8 9 AUGUST 2025 1 2 a + 7 a + 8 a a + 1 13 14 6 + 7 + 13 + 14 = 40 6 7 Algebra Play 06_Chapter 6.indd 141 20-12-2025 16:51:43 Now that we have found a, we know that the other three numbers are a + 1, a + 7 and a + 8. Therefore, the grid must be the following: • 4a + 16 = 36 • 4a = 20 (subtracting 16 from both sides) • a = 5 (dividing both sides by 4) 12 13 5 6 141 SUN MON TUE WED THU FRI SAT Ganita Prakash | Grade 8 | Part-II 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3 4 5 6 7 8 9 Create your own calendar trick. For instance, choose a grid of a different size and shape. Algebra Grids In the following grid, shapes represent numbers. In each row, the last column is the sum of the values to its left. How do we find the values of the shapes? + + AUGUST 2025 1 2 Now, 27 19 So, 9 12 13 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 32 33 34 35 36 37 38 39 40 42 43 44 45 46 47 48 49 50 2 3 4 5 6 7 8 9 10 = 2 × 9 19 + + = + + + = 27 9 19 5 = = = 19 Math Talk 06_Chapter 6.indd 142 20-12-2025 16:51:44 142 In the following grids, find the values of the shapes and fill in the empty squares: 6.5 The Largest Product Fill the digits 2, 3, and 5 in × , using each digit once. What is the largest product possible? 27 21 18 15 Let us approach this problem systematically. There are six ways to place three digits: • The six options are 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2. How do we find the largest product among these six options? We can group them in pairs where the multiplier is the same. In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions. • 32 × 5 It is clear that 52 × 3 is bigger than 53 × 2, so we only need to compare 52 × 3 and 32 × 5. Let us expand these. The first terms in both expressions are the same. The second term shows that 32 × 5 is larger, and hence the largest of the six possible products we can form with 2, 3 and 5. In this case, we used the largest digit as the multiplier. The other two digits were arranged in decreasing order to form the multiplicand. Will this always be the case? Let us find out using algebra. Suppose p, q, and r are the three digits such that p < q < r. As before, we have six possible products, which we group by the multiplier: • We can fill the first box with 2, 3 or 5. • For each of these choices, we have 2 ways of filling the remaining 2 digits. • 35 × 2 and 53 × 2 • 25 × 3 and 52 × 3 • 23 × 5 and 32 × 5 • 53 × 2 • 52 × 3 • 32 × 5 = (3 × 10 × 5) + (2 × 5) • 52 × 3 = (5 × 10 × 3) + (2 × 3) Algebra Play 06_Chapter 6.indd 143 20-12-2025 16:51:44 In each pair, the multiplicand with the larger tens digit forms the larger product. So we have three products to compare: • qr × p, rq × p • pr × q, rp × q • pq × r, qp × r • rq × p • rp × q • qp × r 143 Ganita Prakash | Grade 8 | Part-II Once again, the first term is the same in both expressions. Since r > q the second term of the first expression is larger, so the largest product is qp × r. This matches our earlier observation that the largest digit should be the multiplier and the other two digits should be arranged in decreasing order to form the multiplicand. Figure it Out 1. Fill the digits 1, 3, and 7 in × to make the largest product possible. 2. Fill the digits 3, 5, and 9 in × to make the largest product possible. 6.6 Decoding Divisibility Tricks It is now Mukta’s turn to show a mathematics trick to Shubham. Since q > p, we can see that rp × q is bigger than rq × p. This leaves us with a comparison between qp × r and rp × q. If we expand these, we get • qp × r = (10 × q × r) + (p × r) • rp × q = (10 × r × q) + (p × q) Choose a 2-digit number of different digits and don’t reveal it! 47 Reverse the digits to get another number. Find their difference. 74 06_Chapter 6.indd 144 20-12-2025 16:51:45 144 Divide the result by 9. There won’t be any remainder. Yes, you are right. But how did you know? 74 – 47 = 27 27 9 = 3 If we choose other 2-digit numbers, and follow the steps, will there always be no remainder? Let’s try to understand how Mukta’s trick works. Suppose the two-digit number is ab. When it is reversed, the new number is ba. The difference is divisible by 9. Can you work out what happens if a > b? Figure it Out 1. In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient? 2. In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example: • If b > a, then ba > ab. So, the difference is Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra? • We start with 31. After reversing we get 13. Adding 31 and 13, we get 44. • We start with 28. After reversing we get 82. Adding 28 and 82, we get 110. • We start with 12. After reversing we get 21. Adding 12 and 21, we get 33. (10b + a) – (10a + b) = 10b – b – 10a + a = 9b – 9a = 9 (b – a). Algebra Play 06_Chapter 6.indd 145 20-12-2025 16:51:45 3. Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.] 4. Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.] 5. There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places Math Talk 145 Ganita Prakash | Grade 8 | Part-II 7. A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now? 8. Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have? 6. A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?] some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine? Math Talk 06_Chapter 6.indd 146 20-12-2025 16:51:46 146 10. Evaluate the following sequence of fractions: 9. I run a small dosa cart and my expenses are as follows: • Rent for the dosa cart is `5000 per day. • The cost of making one dosa (including all the ingredients and fuel) is `10. 1 3 , (1 + 3) (5 + 7) , (1 + 3 + 5) (7 + 9 + 11) What do you observe? Can you explain why this happens? [Hint: Recall what you know about the sum of the first n odd numbers.] (ii) If my customers are willing to pay only `50 for a dosa, how many dosas should I aim to sell in a day to make a profit of `2000? (i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of `2000? 11. Karim and the Genie Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, “I have come to serve you, Oh master”. He woke up and to his surprise, it was a genie! “Do you want to make money?”, asked the genie. Karim nodded dumbly in bewilderment. The genie continued, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”. Karim immediately started towards the tree, only to be stopped by the genie. “One moment!”, said the genie. “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.” Thinking that was a trifling amount, Karim readily agreed. He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie! As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared. (ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has? (i) How many coins did Karim initially have? Algebra Play 06_Chapter 6.indd 147 20-12-2025 16:51:47 (iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins? y Algebra is very useful in modeling and understanding numerical scenarios. Because of this, it occurs in almost all areas of mathematics, science and beyond. y Algebra is an indispensable tool in justifying mathematical statements. y We applied algebra to analyse ‘Think of a Number’ tricks, number pyramids, grids, ways of forming numbers using given digits to maximise certain products, divisibility tricks, and various other problems. SUMMARY Math Talk 147" class_8,14,area,ncert_books/class_8/hegp2dd/hegp207.pdf,"7 7.1 Rectangle and Squares How many different ways can you divide a square into 4 parts of equal area? One can actually think of infinitely many such ways! Consider a division, such as — and alter each part as follows. AREA 07_Chapter 7.indd 148 20-12-2025 16:59:54 In each part, the area is compressed along one edge and expanded along another edge. If both the compression and expansion are of the same magnitude, then all 4 parts still have the same area! Try to think of different creative ways to divide a square into 4 parts of equal area. Math Talk 7 You might have seen the rangoli art form, in which regions of different shapes are beautifully coloured using rangoli powder. Which of these rectangles requires more rangoli powder to be coloured, if the colouring is done evenly? We can answer this by counting the number of non-overlapping unit squares (squares of sidelength 1 cm in this case) that can be packed into each of the rectangles. Clearly, the rectangle having sidelengths 7 cm and 4 cm contains 7 × 4 = 28 unit squares, and the rectangle having sidelengths 8 cm and 3 cm contains 8 × 3 = 24 unit squares. Thus, the rectangle of sidelengths 7 cm and 4 cm requires more powder to be coloured. 7 cm 4 cm 8 cm 3 cm Area 07_Chapter 7.indd 149 20-12-2025 16:59:57 Recall that we measure the area of a region by finding the number of unit squares (which can also be a fraction) whose area equals that of the given region. We have seen that the number of unit squares contained in a rectangle is given by the product of its length and width — The areas of the rectangles as seen in the previous problem are generally written as 28 sq. cm and 24 sq. cm, or as 28 cm2 and 24 cm2 . Area of a rectangle = length × width. 149 Area of a rectangle = length × width. Ganita Prakash | Grade 8 | Part-II What is the area of each triangle in this rectangle? We have seen that the diagonal of a rectangle divides it into two congruent triangles. So, the area of each triangle is half the area of the rectangle. In terms of unit squares, half the area fills exactly half the number of unit squares. So the area of each triangle is 1 2 × 7 × 4 = 14 cm2 . Why Can’t Perimeter be a Measure of Area? Why do we count the number of unit squares to assign measures for area? Couldn’t we have just used the perimeter of a region, i.e., the length of its boundary as a measure of its area? If two regions have the same perimeter, can’t we conclude that they have the same area? Or, if one region has a larger perimeter than another region, can’t we conclude that it also has a larger area? The perimeter of a region is not indicative of its area. The reason is that regions can have the same perimeter but different areas, and vice versa. We can even find two regions, Region 1 and Region 2, such that Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this. Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area! This property should be visually clear in your example. Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < Area of Region 2. 7 cm 4 cm Math Talk 07_Chapter 7.indd 150 20-12-2025 16:59:57 150 Figure it Out 1. Identify the missing sidelengths. (i) 3 in 14 in2 35 in2 28 in2 2 in ? in 4 in 21 in2 7 in Area = 50 m2 (ii) 4 m 29 m2 ? ? ? 11 m2 2. The figure shows a path (the shaded portion) laid around a rectangular park EFGH. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.] D C A B (iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown? H G E F (ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. An example of a formula — Area of a rectangle = length × width. [Hint: Break the path into rectangles.] D C A B H G E F D C A B D C A B H G E F H G E F Area 07_Chapter 7.indd 151 20-12-2025 16:59:57 3. The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose. Math Talk 151 Ganita Prakash | Grade 8 | Part-II 4. Find the area of the spiral tube shown in the figure. The tube has the same width throughout. [Hint: There are different ways of finding the area. Here is one method.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left? 1 5 15 5 5 ? 15 5 10 20 20 1 10 20 07_Chapter 7.indd 152 20-12-2025 16:59:57 152 5. In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons. 6. Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials. 1 2 3 Math Talk Triangles In the given figure, which triangle has a greater area: ∆XDC or ∆YDC, if both the rectangles are identical? A X In the given figure, which triangle has a greater area: ∆XDC or ∆YBC, if both the rectangles are identical? A X In each case, by dropping the altitudes from X and Y, it becomes clear that each triangle has exactly half the area of the rectangle ABCD. Find the area of ∆ XDC. D D 4 A X B C B C Y A D D A Y B B C B C Area 07_Chapter 7.indd 153 20-12-2025 16:59:58 To find the area of a triangle, what measurements do we need? We need the sidelengths of the outer rectangle, as in Fig. 7.1. How do we get the outer rectangle from the given triangle? B C A D B C 5 Fig. 7.1 Y A C B C l E D l ‖ BC A 153 Ganita Prakash | Grade 8 | Part-II BCDE is a rectangle (how?). Let us take its sidelengths to be height and base. Then, Since BXAE is a rectangle (how?), the height of the rectangle is the same as the height of the triangle. Thus, if the height and the base of a triangle are known, we can find its area. Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with BC as the base? Area of a triangle = 1 2 × base × height. height Area ( ∆ABC) = 1 2 × base × height A B C height B X C E D base A base A 07_Chapter 7.indd 154 20-12-2025 16:59:58 154 Here is one way to look at it. The area of ∆ABC is the difference of the areas of ∆ADC and ∆ADB, each of which can be enclosed in a rectangle, as in Fig. 7.1. = 1 2 × h × BC Thus, the area formula holds for all types of triangles. Area (∆ABC) = 1 2 × h × DC – 1 2 × h × DB h D C = 1 2 × h (DC – DB) B Some Applications of the Area Formula Find BY. BY can be found using the formula for the area of a triangle. What is Area (∆ABC)? Area (∆ABC) = 1 2 × BY × AC = 1 2 × 4 × BY = 2 BY. Thus, Are the 4 triangles obtained by drawing the diagonals of a rectangle (regions 1 – 4 in the figure) of equal areas? Clearly, the triangles 1 – 4 are not four congruent triangles. To compare their areas, let us consider any two adjacent triangles, say 1 and 2. Since the area of a triangle depends on its height and the base, let us consider suitable height-base pairs in each of the triangles. If OD and OB are taken as the bases, then we see that both triangles have the same altitude! The area of the triangle can also be written as Area (∆ABC) = 1 2 × AX × BC = 15 2 sq. units. So, BY = 15 4 = 3.75 units. 2 BY = 15 2 . D C A B A B B X 3 4 1 3 A 4 5 O 2 Y Area C 07_Chapter 7.indd 155 20-12-2025 16:59:58 We also have OB = OD, since the diagonals of a rectangle bisect each other. So, the triangles 1 and 2 have equal areas. Arguing this way, we can show that all four triangles have equal areas. From this problem, we can make the following general statement. In a triangle, the line joining a vertex to the midpoint of its opposite side divides the triangle into two triangles of equal areas. Triangles 1 and 2 have equal areas as they have the same measures for height and base D C 1 2 X O 155 Ganita Prakash | Grade 8 | Part-II Triangles between Parallel Lines with a Common Base Line l‖BC. Consider the different triangles that have BC as their base, and with their third vertex lying anywhere on l. Here, we will only show how to find the triangle with the minimum perimeter, and leave the other questions as exercises. Intuition might suggest that the triangle with the minimum perimeter can be obtained by constructing the perpendicular bisector of BC. But how do we justify that this triangle has the least perimeter among all the triangles? (ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter? (i) Which of these triangles has the maximum area, and which has the minimum area? B C B C A l l 07_Chapter 7.indd 156 20-12-2025 16:59:58 156 Firstly, note that in all these triangles, BC is a common side. Therefore, it is enough to consider the sum of the other two sides. Let us imagine the line l as a mirror. Then we get a reflection of all the points and lines below it. While studying the properties of a plane mirror, we experimentally observed that the distance of the image behind the mirror is the same as the distance of the object (that creates the image) in front of the mirror. This law can be used to locate the reflections of the points lying below the line l. What can we say about the lengths of AB and its reflection AB ́? Since ∆AXB ≅ ∆AXB ́, AB = AB ́. Similarly, AC = AC ́. Therefore, the length of the path B A C is the same as the length of the path B A C ́. So, finding point A that gives the shortest possible path B A C is the same as finding a point A that gives the shortest possible path B A C ́. But the solution to the latter problem is clear — choose A on the straight line BC ́, since BC ́ is the shortest possible path from B to C ́. Therefore, this triangle ∆ABC has the minimum perimeter. Analyse whether A lies on the perpendicular bisector of BC. B B´ C´ C A l This is true no matter where point A is on line l. X Y A B C B´ C´ B C B´ C´ A Area l l 07_Chapter 7.indd 157 20-12-2025 16:59:58 Figure it Out 1. Find the areas of the following triangles: B E C A 3 cm (i) 4 cm (iii) D F E 3.2 cm N (ii) 5cm 4 cm N A 3 cm T Math Talk 157 Ganita Prakash | Grade 8 | Part-II 2. Find the length of the altitude BY. 3. Find the area of ∆SUB, given that it is isosceles, SE is perpendicular to UB, and the area of ∆SEB is 24 sq. units. In the Śulba-Sūtras, which are ancient Indian geometric texts that deal with the construction of altars, we can find many interesting problems on the topic of areas. When altars are built, they must have the exact prescribed shape and area. This gives rise to problems of the kind where one has to transform a given shape into another of the same area. The Śulba-Sūtras give solutions to many such problems. Such problems are also posed and solved in Euclid’s Elements. Here are two problems of this kind. 4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area. 5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area. 4 units A X S B C 6 units Y 8 units U E B 07_Chapter 7.indd 158 20-12-2025 16:59:58 158 6. ABCD, BCEF, and BFGH are identical squares. (ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square? (i) If the area of the red region is 49 sq. units, then what is the area of the blue region? D E A H G C B F Math Talk 7. If M and N are the midpoints of XY and XZ, what fraction of the area of ∆XYZ is the area of ∆XMN? [Hint: Join NY] X 8. Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path. Area of any Polygon How do we find the area of this quadrilateral? What measurements do we need for this? If we join BD, the quadrilateral ABCD gets divided into two triangles. By finding their areas, we can find the area of ABCD. Y Z House M N River Water tank A B Area Try This Math Talk 07_Chapter 7.indd 159 20-12-2025 16:59:59 How do we find the area of this pentagon? Can any polygon be divided into triangles? It can be seen that any polygon can be divided into triangles. Thus, by knowing how to compute the area of a triangle, we can find the area of any polygon. D C 159 Ganita Prakash | Grade 8 | Part-II Figure it Out 1. Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC. 2. Find the area of the shaded region given that ABCD is a rectangle. 3. What measurements would you need to find the area of a regular hexagon? 4. What fraction of the total area of the rectangle is the area of the blue region? 6 cm 4 cm A D F C N 22 cm M 3 cm 3 cm D C A B 10 cm 18 cm E 8 cm 10 cm B Math Talk Math Talk 07_Chapter 7.indd 160 20-12-2025 16:59:59 160 5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral. One can derive special formulae to find the areas of a parallelogram, rhombus and trapezium. Math Talk Parallelogram We can derive a special formula for the area of a parallelogram by converting it into a rectangle of equal area. Give a method to convert a parallelogram into a rectangle of equal area. You can try this using a cut-out of a parallelogram. Construct AX perpendicular to CD — represented in short as AX ^ CD . We call this a height of the parallelogram. Cut the parallelogram into ∆AXD and trapezium ABCX. A B Can ∆AXD and ABCX fit together, as shown in the figure, to get a rectangle? One simple way to check this is to identify the triangle that can complete ABCX to a rectangle, and then check if this triangle is congruent to ∆AXD. Observe that ∠X = 90°, and so ∠A = 90°, since AB‖XC. We need another right angle to get a rectangle (what about the fourth angle?). To get the third right angle, extend XC to the right and then construct a line perpendicular to XC that passes through B. The ∆BYC completes ABCX to a rectangle. Is ∆AXD congruent to it? D X C D X A A X D X C A B A X C Y B Area 07_Chapter 7.indd 161 20-12-2025 17:00:00 So, by the RHS congruency criterion, ∆BYC ≅ ∆AXD. Thus, ∆AXD will fit exactly over the region occupied by ∆BYC, and convert the parallelogram into a rectangle. The process of cutting a figure into pieces and rearranging them to get a different figure of equal area is called dissection. BY = AX (since ABYX is a rectangle) ∠BYC = ∠AXD = 90° BC = AD (since ABCD is a parallelogram) 161 Ganita Prakash | Grade 8 | Part-II How do we find the area of a parallelogram by dissecting it into a rectangle? We have Area of the parallelogram ABCD = Area of the rectangle ABYX. Area of the rectangle ABYX = AX × XY. AX is the height of the parallelogram. Is there a relation between XY and DC? Since DX = CY, we get DC = XY by adding the common part XC to DX and CY. Since DC is the base of the parallelogram, we get Can the area of the parallelogram be determined by taking another side as the base and its corresponding height? Can the parallelogram be cut along CZ and rearranged to form a rectangle? It can be seen that this is indeed possible, and so any side and its corresponding height can be used to find the area of a parallelogram. Figure it Out Area of the parallelogram = base × height. Z height base D C A B D X C Y A B heig ht base 07_Chapter 7.indd 162 20-12-2025 17:00:00 162 1. Observe the parallelograms in the figure below. (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter? (i) What can we say about the areas of all these parallelograms? (a) (b) (c) (d) (e) (f) (g) 2. Find the areas of the following parallelograms: 3. Find QN. 4 cm 7 cm (i) 6 cm 7.6 cm P Q (iii) 4.8 cm S M R N 5 cm 12 cm 3 cm (iv) 4.4 cm 2 cm 5 cm (ii) Area 07_Chapter 7.indd 163 20-12-2025 17:00:00 4. Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.] 5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of? 4 cm 4 cm 5 cm 163 Ganita Prakash | Grade 8 | Part-II 6. [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle. 7. [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it? [Hint: Show that triangles ∆ADB and ∆ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.] 8. [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection. 9. Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons. Rhombus Since a rhombus is a parallelogram, the area formula for a parallelogram holds for a rhombus as well. However, the additional properties of a rhombus give us another method to transform a rhombus into a rectangle of the same area by dissection. This method occurs in one of the Śulba-Sūtras. B D C A 07_Chapter 7.indd 164 20-12-2025 17:00:00 164 Try working this out! B D O A C B O B C O O A A O D C D Since ABCD is a rhombus, all its sides have equal length, and the diagonals are perpendicular bisectors of each other. Therefore, ∆ABD and ∆CBD are isosceles triangles. Each of them can be transformed into a rectangle of equal area, and the two rectangles can then be joined to form a single rectangle. This rectangle, say WXYZ, has the same area as the rhombus ABCD. What are the sidelengths of the rectangle WXYZ? From the dissection, we can see that Thus, we have O D B O A O D C B O = AC × BD 2 Area of rhombus ABCD = Area of rectangle WXYZ = XW × WZ XW = length of the diagonal AC, and WZ = half the length of the other diagonal BD. C A X Y W Z Area 07_Chapter 7.indd 165 20-12-2025 17:00:00 = 1 2 × AC × BD Therefore, Area of rhombus ABCD can also be determined by finding the areas of ∆ADB and ∆CDB. What formula does this give us? Since the diagonals are perpendicular to each other, we have Area of rhombus ABCD = Area ( ∆ADB) + Area ( ∆CDB). Area ( ∆ADB) = 1 2 × AO × BD, and Area ( ∆CDB) = 1 2 × CO × BD, and Area of a rhombus = 1 2 × product of diagonals. D C A B O 165 Ganita Prakash | Grade 8 | Part-II Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals. Trapezium Find the areas of the following trapeziums by breaking them into figures whose areas can be computed. One way of finding the area of a trapezium is by breaking it into a rectangle and triangles. Consider a trapezium WXYZ with WX ‖ ZY. Find its area. Construct WM ^ ZY , and XN ^ ZY (WM and XN perpendicular to ZY). Is WXNM a rectangle? ∠MWX = ∠NXW = 90°, since WX‖ZY and the interior angles on the same side of a transversal (WM and XN) add up to 180°. Therefore, WXNM is a rectangle. We have D M C A B = h( 1 2 x + a + 1 2 y) = h( x + y + 2a 2 ) = 1 2 h(x + y + 2a). W X Z M N Y S T U R P Q Z M N Y W X 07_Chapter 7.indd 166 20-12-2025 17:00:01 166 = 1 2 × MZ × WM + WX × WM + 1 2 × NY × XN Let us assign letter numbers to the lengths that we need to find the area of the trapezium. Let MZ = x, WM = XN = h, WX = a, NY = y. So we get Area WXYZ = 1 2 hx + ha + 1 2 hy Area WXYZ = Area (∆WMZ) + Area WXNM + Area (∆XNY) Z x y M N Y W X h h a b We have taken the length of one of the parallel sides as a. Let b be the length of the other parallel side. Can the area of the trapezium be expressed in terms of a, b and h? To replace x + y in the area expression with a and b, observe that b = x + y + a. Subtracting a from both sides, we get x + y = b – a. Using this, we get = 1 2 h(a + b). Therefore, Will this formula hold for a trapezium that looks like this? There are different ways of approaching this, which are sketched below. Complete the arguments. Approach 1: Rectangle and Triangles Area of a trapezium = 1 2 × height × sum of the parallel sides. h A h Area WXYZ = 1 2 h(b – a + 2a) a B a b Area 07_Chapter 7.indd 167 20-12-2025 17:00:01 Approach 2: Parallelogram and Triangle Will Approach 2 work for any type of trapezium? Math Talk Area ABCD = Area ABED + Area ∆BEC Area ABED = Area ABEF – Area ∆AFD h F E F G b Draw BG ‖ AD A D C D C a B b 167 D Ganita Prakash | Grade 8 | Part-II Finding the Area Using Two Copies of the Trapezium There is another interesting way of finding the area of a trapezium. Consider two copies of the given trapezium in which AB‖CD. Rotate the second copy as shown. What figure will we get when the two trapeziums are joined along BC? The possibilities are either a 6-sided figure or a 4-sided figure (quadrilateral). Possibilities We can rule out the first possibility by looking at the sum of angles x and y. Since AB ‖ CD, we have x + y = 180, since they are internal angles along the same side of the transversal (BC). So ABD́ and A ́CD are straight lines. Therefore, the resulting figure is a quadrilateral. What type of a quadrilateral is this? Let us look at the other two angles of the trapezium. A 6-sided figure D B C y x A x y A´ x B C y D´ D C´ D´ A y B´ A´ x B x Quadrilateral C y x y A´ D´ 07_Chapter 7.indd 168 20-12-2025 17:00:01 168 We have u + v = 180. Therefore, AD‖D́ A ́, since the sum of the internal angles along the same side of the transversal A ́D is 180°. Since we already have AD́  ‖ A ́D, the quadrilateral AD́ A ́D is actually a parallelogram. a h b D A v u B h a C b u v A´ D´ a b Figure it Out 1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm. 2. Give a method to convert a rectangle into a rhombus of equal area using dissection. 3. Find the areas of the following figures: 10 in 6 in 14 in 10 ft Area of the trapezium = 1 2 × Area of the parallelogram 16 ft (i) 7 ft = 1 2 h(a + b). 8 ft 24 m 36 m 14 m 12 ft (ii) 18 ft Area 07_Chapter 7.indd 169 20-12-2025 17:00:01 4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection. 5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area — A B Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ, then the trapezium and rectangle have equal areas.] (iii) D C H E I J G F (iv) Math Talk 169 Ganita Prakash | Grade 8 | Part-II 6. Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm2 . 7. A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas. 8. ZYXW is a trapezium with ZY‖WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ∆ZWB. Areas in Real Life What do you think is the area of an A4 sheet? Its sidelengths are 21 cm and 29.7 cm. Now find its area. What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise how many A4 sheets can fit on your table. The dimensions of furniture like tables and chairs are sometimes measured in inches (in) and feet (ft). 1 in = 2.54 cm W X Z Y A B Math Talk 07_Chapter 7.indd 170 20-12-2025 17:00:02 170 Express the following lengths in centimeters: (i) 5 in (ii) 7.4 in Express the following lengths in inches: (i) 5.08 cm (ii) 11.43 cm How many cm2 is 1 in2 ? So, 1 in2 = 2.542 cm2 = 6.4516 cm2 . 1 in 2.54 cm 1 in 2.54 cm 1 ft = 12 in = How many cm2 is 10 in2 ? Convert 161.29 cm2 to in2 . Every 6.4516 cm2 gives an in2 . Hence, 161.29 cm2 = 161.29 6.4516 in2 . Evaluate the quotient. What do you think is the area of your classroom? Areas of classroom, house, etc., are generally measured in ft2 or m2 . How many in2 is 1 ft2 ? What do you think is the area of your school? Make an estimate and compare it with the actual data. Larger areas of land are also measured in acres. 1 acre = 43,560 ft2 . Besides these units, different parts of India use different local units for measuring area, such as bigha, gaj, katha, dhur, cent, ankanam, etc. Find out the local unit of area measurement in your region. What do you think is the area of your village/town/city? Make an estimate and compare it with the actual data. Larger areas are measured in km2 . How many m2 is a km2  ? 10 in2 = 10 × 6.4516 cm2 = 64.516 cm2 . Area 07_Chapter 7.indd 171 20-12-2025 17:00:02 How many times is your village/town/city bigger than your school? Find the city with the largest area in (i) India, and (ii) the world. Find the city with the smallest area in (i) India, and (ii) the world. y Area of a triangle = 1 2 × base × height. y The area of any polygon can be evaluated by breaking it into triangles. y Area of a parallelogram = base × height. y Area of a rhombus = 1 2 × product of its diagonals. y Area of a trapezium = 1 2 × height × sum of parallel sides. SUMMARY 171 LEARNING MATERIAL SHEETS 07_Chapter 7.indd 172 20-12-2025 17:00:02 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 07_Chapter 7.indd 173 20-12-2025 17:00:02 17 18 19 20 21 22 23 24 17 18 19 20 21 22 23 24 17 18 19 20 21 22 23 24 07_Chapter 7.indd 174 20-12-2025 17:00:03 Isometric Grid 07_Chapter 7.indd 175 20-12-2025 17:00:03 07_Chapter 7.indd 176 20-12-2025 17:00:03" class_9,1,Number Systems,ncert_books/class_9/iemh1dd/iemh101.pdf,"NUMBER SYSTEMS 1 NUMBER SYSTEMS 1.1 Introduction In your earlier classes, you have learnt about the number line and how to represent various types of numbers on it (see Fig. 1.1). positive direction. As far as your eyes can see, there are numbers, numbers and numbers! Just imagine you start from zero and go on walking along this number line in the Fig. 1.1 : The number line CHAPTER 1 numbers. Get a bag ready to store them! Now suppose you start walking along the number line, and collecting some of the Reprint 2025-26 Fig. 1.2 2 MATHEMATICS N You might begin with picking up only natural numbers like 1, 2, 3, and so on. You know that this list goes on for ever. (Why is this true?) So, now your bag contains infinitely many natural numbers! Recall that we denote this collection by the symbol N. Now turn and walk all the way back, pick up zero and put it into the bag. You now have the collection of whole numbers which is denoted by the symbol W. negative integers into your bag. What is your new collection? Recall that it is the collection of all integers, and it is denoted by the symbol Z. Now, stretching in front of you are many, many negative integers. Put all the 74 5 2 601 422 58 0 -3 16 31 71 -757 -66-21 -40 0 Z 3 22 -75 2 1 9 3 -40 166 Why Z ? “zahlen”, which means Z comes from the German word “to count”. 74 5 2 601 4652 58 0 31 1 71 10 9 40 16 74 5 2 601 29 58 3 40 16 0 W Are there some numbers still left on the line? Of course! There are numbers like 1 3 , 2 4 , or even 2005 2006 − . If you put all such numbers also into the bag, it will now be the 40 27 71 89 3 16 2005 2006 60 999 4 –8 –6625 – –6 7 2 3 9 14 0 –12 13 58 –5 17 981 Reprint 2025-26 2005 2006 –12 13 9 14 –6625 –6721 12 1 3 –1 9 81 16 1 4 999 0 –6 7 27 58 -65 60 19 19 Q NUMBER SYSTEMS 3 collection of rational numbers. The collection of rational numbers is denoted by Q. ‘Rational’ comes from the word ‘ratio’, and Q comes from the word ‘quotient’. where p and q are integers and q ≠ 0. (Why do we insist that q ≠ 0?) and q are integers and q ≠ 0. For example, –25 can be written as 25 ; 1 − here p = –25 and q = 1. Therefore, the rational numbers also include the natural numbers, whole numbers and integers. the form p q , where p and q are integers and q ≠ 0. For example, 1 2 = 2 4 = 10 20 = 25 50 = 47 94 , and so on. These are equivalent rational numbers (or fractions). However, when we say that p q is a rational number, or when we represent p q on the number line, we assume that q ≠ 0 and that p and q have no common factors other than 1 (that is, p and q are co-prime). So, on the number line, among the infinitely many You may recall the definition of rational numbers: A number ‘r’ is called a rational number, if it can be written in the form p q , Notice that all the numbers now in the bag can be written in the form p q , where p You also know that the rational numbers do not have a unique representation in fractions equivalent to 1 2 , we will choose 1 2 to represent all of them. have studied in earlier classes. Example 1 : Are the following statements true or false? Give reasons for your answers. (i) Every whole number is a natural number. (ii) Every integer is a rational number. (iii) Every rational number is an integer. Solution : (i) False, because zero is a whole number but not a natural number. (ii) True, because every integer m can be expressed in the form 1 m , and so it is a rational number. Now, let us solve some examples about the different types of numbers, which you Reprint 2025-26 4 MATHEMATICS (iii) False, because 3 5 is not an integer. Example 2 : Find five rational numbers between 1 and 2. We can approach this problem in at least two ways. Solution 1 : Recall that to find a rational number between r and s, you can add r and s and divide the sum by 2, that is 2 r s + lies between r and s. So, 3 2 is a number between 1 and 2. You can proceed in this manner to find four more rational numbers between 1 and 2. These four numbers are 5 11 13 7 , , and . 4 8 8 4 Solution 2 : The other option is to find all the five rational numbers in one step. Since we want five numbers, we write 1 and 2 as rational numbers with denominator 5 + 1, i.e., 1 = 6 6 and 2 = 12 6 . Then you can check that 7 6 , 8 6 , 9 6 , 10 6 and 11 6 are all rational numbers between 1 and 2. So, the five numbers are 7 4 3 5 11 , , , and 6 3 2 3 6 . Remark : Notice that in Example 2, you were asked to find five rational numbers between 1 and 2. But, you must have realised that in fact there are infinitely many rational numbers between 1 and 2. In general, there are infinitely many rational numbers between any two given rational numbers. Let us take a look at the number line again. Have you picked up all the numbers? Not, yet. The fact is that there are infinitely many more numbers left on the number line! There are gaps in between the places of the numbers you picked up, and not just one or two but infinitely many. The amazing thing is that there are infinitely many numbers lying between any two of these gaps too! So we are left with the following questions: 1. What are the numbers, that are left on the number line, called? 2. How do we recognise them? That is, how do we distinguish them from the rationals (rational numbers)? These questions will be answered in the next section. Reprint 2025-26 NUMBER SYSTEMS 5 1.2 Irrational Numbers We saw, in the previous section, that there may be numbers on the number line that are not rationals. In this section, we are going to investigate these numbers. So far, all the numbers you have come across, are of the form p q , where p and q are integers and q ≠ 0. So, you may ask: are there numbers which are not of this form? There are indeed such numbers. The Pythagoreans in Greece, followers of the famous mathematician and philosopher Pythagoras, were the first to discover the numbers which were not rationals, around 400 BC. These numbers are called irrational numbers (irrationals), because they cannot be written in the form of a ratio of integers. There are many myths surrounding the discovery of irrational numbers by the Pythagorean, Hippacus of Croton. In all the myths, Hippacus has an 1. Is zero a rational number? Can you write it in the form p q , where p and q are integers 2. Find six rational numbers between 3 and 4. 3. Find five rational numbers between 3 5 and 4 5 . 4. State whether the following statements are true or false. Give reasons for your answers. and q ≠ 0? (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number. EXERCISE 1.1 and q are integers and q ≠ 0. unfortunate end, either for discovering that 2 is irrational or for disclosing the secret about 2 to people outside the secret Pythagorean sect! Let us formally define these numbers. A number ‘s’ is called irrational, if it cannot be written in the form p q , where p Reprint 2025-26 Pythagoras (569 BCE – 479 BCE) Fig. 1.3 6 MATHEMATICS are infinitely many irrational numbers too. Some examples are: Remark : Recall that when we use the symbol , we assume that it is the positive square root of the number. So 4 = 2, though both 2 and –2 are square roots of 4. have already come across many of the square roots listed above and the number π. The Pythagoreans proved that 2 is irrational. Later in approximately 425 BC, Theodorus of Cyrene showed that 3, 5, 6, 7, 10, 11, 12, 13, 14, 15 and 17 are also irrationals. Proofs of irrationality of 2 , 3 , 5 , etc., shall be discussed in Class X. As to π, it was known to various cultures for thousands of years, it was proved to be irrational by Lambert and Legendre only in the late 1700s. In the next section, we will discuss why 0.10110111011110... and π are irrational. Let us return to the questions raised at the end of the previous section. Remember the bag of rational numbers. If we now put all irrational numbers into the bag, will there be any number left on the number line? The answer is no! It turns out that the collection of all rational numbers and irrational numbers together make up what we call the collection of real numbers, which is denoted by R. Therefore, a real number is either rational or irrational. So, we can say that every real number is represented by a unique point on the number line. Also, every point on the number line represents a unique real number. This is why we call the number line, the real number line. You already know that there are infinitely many rationals. It turns out that there Some of the irrational numbers listed above are familiar to you. For example, you 2, 3, 15,, π, 0.10110111011110... 16 2005 2006 3 19 -65 –66 26 -45 71 17 981 –12 13 89 –6 7 27 9 14 60 999 4 –8 –6625 36 0 2 3 0 58 –5 R G. Cantor (1845-1918) Fig. 1.5 R. Dedekind (1831-1916) Fig. 1.4 In the 1870s two German mathematicians, Cantor and Dedekind, showed that : Corresponding to every real number, there is a point on the real number line, and corresponding to every point on the number line, there exists a unique real number. Reprint 2025-26 NUMBER SYSTEMS 7 Example 3 : Locate 2 on the number line. Solution : It is easy to see how the Greeks might have discovered 2 . Consider a square OABC, with each side 1 unit in length (see Fig. 1.6). Then you can see by the Pythagoras theorem that OB = 2 2 1 1 2 + = . How do we represent 2 on the number line? This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex O coincides with zero (see Fig. 1.7). We have just seen that OB = 2 . Using a compass with centre O and radius OB, draw an arc intersecting the number line at the point P. Then P corresponds to 2 on the number line. Example 4 : Locate 3 on the number line. Solution : Let us return to Fig. 1.7. Let us see how we can locate some of the irrational numbers on the number line. Fig. 1.7 Fig. 1.6 Construct BD of unit length perpendicular to OB (as in Fig. 1.8). Then using the Pythagoras theorem, we see that OD = ( )2 2 2 1 3 + = . Using a compass, with centre O and radius OD, draw an arc which intersects the number line at the point Q. Then Q corresponds to 3 . Reprint 2025-26 Fig. 1.8 8 MATHEMATICS In the same way, you can locate n for any positive integer n, after n − 1 has been located. 1. State whether the following statements are true or false. Justify your answers. 2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number. 3. Show how 5 can be represented on the number line. 4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1 P2 perpendicular to OP1 of unit length (see Fig. 1.9). Now draw a line segment P2 P3 perpendicular to OP2 . Then draw a line segment P3 P4 perpendicular to OP3 . Continuing in this manner, you can get the line segment Pn–1Pn by drawing a line segment of unit length perpendicular to OPn–1. In this manner, you will have created the points P2 , P3 ,...., Pn ,... ., and joined them to create a beautiful spiral (i) Every irrational number is a real number. (ii) Every point on the number line is of the form m , where m is a natural number. (iii) Every real number is an irrational number. EXERCISE 1.2 Fig. 1.9 : Constructing square root spiral 1.3 Real Numbers and their Decimal Expansions In this section, we are going to study rational and irrational numbers from a different point of view. We will look at the decimal expansions of real numbers and see if we can use the expansions to distinguish between rationals and irrationals. We will also explain how to visualise the representation of real numbers on the number line using their decimal expansions. Since rationals are more familiar to us, let us start with them. Let us take three examples : 10 7 1 , , 3 8 7 . Pay special attention to the remainders and see if you can find any pattern. depicting 2, 3, 4, ... Reprint 2025-26 NUMBER SYSTEMS 9 Example 5 : Find the decimal expansions of 10 3 , 7 8 and 1 7 . Solution : Remainders : 1, 1, 1, 1, 1... Remainders : 6, 4, 0 Remainders : 3, 2, 6, 4, 5, 1, Divisor : 3 Divisor : 8 3, 2, 6, 4, 5, 1,... 3.333... 0.875 0.142857... 3 10 8 7.0 7 1.0 9 64 7 10 60 30 9 56 28 10 40 20 9 40 14 10 0 60 9 56 1 40 35 50 49 1 What have you noticed? You should have noticed at least three things: (i) The remainders either become 0 after a certain stage, or start repeating themselves. (ii) The number of entries in the repeating string of remainders is less than the divisor (iii) If the remainders repeat, then we get a repeating block of digits in the quotient (in 10 3 one number repeats itself and the divisor is 3, in 1 7 there are six entries 326451 in the repeating string of remainders and 7 is the divisor). (for 10 3 , 3 repeats in the quotient and for 1 7 , we get the repeating block 142857 in the quotient). Reprint 2025-26 Divisor : 7 10 MATHEMATICS Although we have noticed this pattern using only the examples above, it is true for all rationals of the form p q (q ≠ 0). On division of p by q, two main things happen – either the remainder becomes zero or never becomes zero and we get a repeating string of remainders. Let us look at each case separately. Case (i) : The remainder becomes zero In the example of 7 8 , we found that the remainder becomes zero after some steps and the decimal expansion of 7 8 = 0.875. Other examples are 1 2 = 0.5, 639 250 = 2.556. In all these cases, the decimal expansion terminates or ends after a finite number of steps. We call the decimal expansion of such numbers terminating. Case (ii) : The remainder never becomes zero In the examples of 10 3 and 1 7 , we notice that the remainders repeat after a certain stage forcing the decimal expansion to go on for ever. In other words, we have a repeating block of digits in the quotient. We say that this expansion is non-terminating recurring. For example, 10 3 = 3.3333... and 1 7 = 0.142857142857142857... The usual way of showing that 3 repeats in the quotient of 10 3 is to write it as 3.3 . Similarly, since the block of digits 142857 repeats in the quotient of 1 7 , we write 1 7 as 0.142857 , where the bar above the digits indicates the block of digits that repeats. Also 3.57272... can be written as 3.572 . So, all these examples give us non-terminating recurring (repeating) decimal expansions. Thus, we see that the decimal expansion of rational numbers have only two choices: either they are terminating or non-terminating recurring. Now suppose, on the other hand, on your walk on the number line, you come across a number like 3.142678 whose decimal expansion is terminating or a number like 1.272727... that is, 1.27 , whose decimal expansion is non-terminating recurring, can you conclude that it is a rational number? The answer is yes! Reprint 2025-26 NUMBER SYSTEMS 11 We will not prove it but illustrate this fact with a few examples. The terminating cases are easy. Example 6 : Show that 3.142678 is a rational number. In other words, express 3.142678 in the form p q , where p and q are integers and q ≠ 0. Solution : We have 3.142678 = 3142678 1000000 , and hence is a rational number. Now, let us consider the case when the decimal expansion is non-terminating recurring. Example 7 : Show that 0.3333... = 0 3. can be expressed in the form p q , where p and q are integers and q ≠ 0. Solution : Since we do not know what 0 3. is , let us call it ‘x’ and so Now here is where the trick comes in. Look at Now, 3.3333... = 3 + x, since x = 0.3333... Therefore, 10 x = 3 + x Solving for x, we get 10 x = 10 × (0.333...) = 3.333... x = 0.3333... Example 8 : Show that 1.272727... = 1 27 . can be expressed in the form p q , where p and q are integers and q ≠ 0. Solution : Let x = 1.272727... Since two digits are repeating, we multiply x by 100 to get So, 100 x = 126 + 1.272727... = 126 + x Therefore, 100 x – x = 126, i.e., 99 x = 126 100 x = 127.2727... Reprint 2025-26 9x = 3, i.e., x = 1 3 12 MATHEMATICS i.e., x = 126 14 99 11 = You can check the reverse that 14 11 = 1 27 . . Example 9 : Show that 0.2353535... = 0 235 . can be expressed in the form p q , where p and q are integers and q ≠ 0. Solution : Let x = 0 235 . . Over here, note that 2 does not repeat, but the block 35 repeats. Since two digits are repeating, we multiply x by 100 to get So, 100 x = 23.3 + 0.23535... = 23.3 + x Therefore, 99 x = 23.3 i.e., 99 x = 233 10 , which gives x = 233 990 You can also check the reverse that 233 990 = 0 235 . . So, every number with a non-terminating recurring decimal expansion can be expressed in the form p q (q ≠ 0), where p and q are integers. Let us summarise our results in the 100 x = 23.53535... following form : The decimal expansion of a rational number is either terminating or nonterminating recurring. Moreover, a number whose decimal expansion is terminating or non-terminating recurring is rational. So, now we know what the decimal expansion of a rational number can be. What about the decimal expansion of irrational numbers? Because of the property above, we can conclude that their decimal expansions are non-terminating non-recurring. So, the property for irrational numbers, similar to the property stated above for rational numbers, is The decimal expansion of an irrational number is non-terminating non-recurring. Moreover, a number whose decimal expansion is non-terminating non-recurring is irrational. Reprint 2025-26 NUMBER SYSTEMS 13 Recall s = 0.10110111011110... from the previous section. Notice that it is nonterminating and non-recurring. Therefore, from the property above, it is irrational. Moreover, notice that you can generate infinitely many irrationals similar to s. What about the famous irrationals 2 and π? Here are their decimal expansions up to a certain stage. (Note that, we often take 22 7 as an approximate value for π, but π ≠ 22 7 .) Over the years, mathematicians have developed various techniques to produce more and more digits in the decimal expansions of irrational numbers. For example, you might have learnt to find digits in the decimal expansion of 2 by the division method. Interestingly, in the Sulbasutras (rules of chord), a mathematical treatise of the Vedic period (800 BC - 500 BC), you find an approximation of 2 as follows: 2 = 1 1 1 1 1 1 1 1 4142156 3 4 3 34 4 3 . + + × − × × = Notice that it is the same as the one given above for the first five decimal places. The history of the hunt for digits in the decimal expansion of π is very interesting. The Greek genius Archimedes was the first to compute digits in the decimal expansion of π. He showed 3.140845 < π < 3.142857. Aryabhatta (476 – 550 C.E.), the great Indian mathematician and astronomer, found the value of π correct to four decimal places (3.1416). Using high speed computers and advanced algorithms, π has been computed to over 1.24 trillion decimal places! π = 3.14159265358979323846264338327950... 2 = 1.4142135623730950488016887242096... Now, let us see how to obtain irrational numbers. Example 10 : Find an irrational number between 1 7 and 2 7 . Solution : We saw that 1 7 = 0142857 . . So, you can easily calculate 2 0 285714 7 = . . To find an irrational number between 1 7 and 2 7 , we find a number which is Reprint 2025-26 Archimedes (287 BCE – 212 BCE) Fig. 1.10 14 MATHEMATICS non-terminating non-recurring lying between them. Of course, you can find infinitely many such numbers. An example of such a number is 0.150150015000150000... 1. Write the following in decimal form and say what kind of decimal expansion each has : 2. You know that 1 7 = 0142857 . . Can you predict what the decimal expansions of 2 7 , 3 7 , 3. Express the following in the form p q , where p and q are integers and q ≠ 0. 4. Express 0.99999 .... in the form p q . Are you surprised by your answer? With your 5. What can the maximum number of digits be in the repeating block of digits in the (i) 36 100 (ii) 1 11 (iii) 1 4 8 (iv) 3 13 (v) 2 11 (vi) 329 400 [Hint : Study the remainders while finding the value of 1 7 carefully.] (i) 0 6. (ii) 0 47 . (iii) 0 001 . teacher and classmates discuss why the answer makes sense. 4 7 , 5 7 , 6 7 are, without actually doing the long division? If so, how? EXERCISE 1.3 6. Look at several examples of rational numbers in the form p q (q ≠ 0), where p and q are 7. Write three numbers whose decimal expansions are non-terminating non-recurring. 8. Find three different irrational numbers between the rational numbers 5 7 and 9 11 . 9. Classify the following numbers as rational or irrational : decimal expansion of 1 17 ? Perform the division to check your answer. integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy? (i) 23 (ii) 225 (iii) 0.3796 (iv) 7.478478... (v) 1.101001000100001... Reprint 2025-26 NUMBER SYSTEMS 15 1.4 Operations on Real Numbers You have learnt, in earlier classes, that rational numbers satisfy the commutative, associative and distributive laws for addition and multiplication. Moreover, if we add, subtract, multiply or divide (except by zero) two rational numbers, we still get a rational number (that is, rational numbers are ‘closed’ with respect to addition, subtraction, multiplication and division). It turns out that irrational numbers also satisfy the commutative, associative and distributive laws for addition and multiplication. However, the sum, difference, quotients and products of irrational numbers are not always irrational. For example, ( 6 6 ) + −( ) ,( 2 2 3 3 ) − ⋅ ( ),( ) ( ) and 17 17 are rationals. irrational number. For example, 3 is irrational. What about 2 3 + and 2 3 ? Since Example 11 : Check whether 7 5 , 7 2 21 2 5 , + π − , are irrational numbers or not. Solution : 5 = 2.236... , 2 = 1.4142..., π = 3.1415... Then 7 5 = 15.652..., 7 2 3 + and 2 3 . Therefore, both 2 3 + and 2 3 are also irrational numbers. 3 has a non-terminating non-recurring decimal expansion, the same is true for Let us look at what happens when we add and multiply a rational number with an 5 = 7 5 7 5 5 5 5 = = 3.1304... All these are non-terminating non-recurring decimals. So, all these are irrational numbers. Now, let us see what generally happens if we add, subtract, multiply, divide, take square roots and even nth roots of these irrational numbers, where n is any natural number. Let us look at some examples. Example 12 : Add 2 2 5 3 + and 2 3 3 – . Solution : (2 2 5 3 2 3 3 + +) ( – ) = (2 2 2 5 3 3 3 + +) ( – ) = (2 + 1) 2 (5 3) 3 3 2 2 3 + − = + 2 + 21 = 22.4142..., π – 2 = 1.1415... Reprint 2025-26 16 MATHEMATICS Example 13 : Multiply 6 5 by 2 5 . Solution : 6 5 × 2 5 = 6 × 2 × 5 × 5 = 12 × 5 = 60 Example 14 : Divide 8 15 by 2 3 . Solution : 8 3 5 8 15 2 3 4 5 2 3 × ÷ = = These examples may lead you to expect the following facts, which are true: (i) The sum or difference of a rational number and an irrational number is irrational. (ii) The product or quotient of a non-zero rational number with an irrational number is irrational. (iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational. We now turn our attention to the operation of taking square roots of real numbers. Recall that, if a is a natural number, then a b = means b 2 = a and b > 0. The same definition can be extended for positive real numbers. Let a > 0 be a real number. Then a = b means b 2 = a and b > 0. In Section 1.2, we saw how to represent n for any positive integer n on the number line. We now show how to find x for any given positive real number x geometrically. For example, let us find it for x = 3.5, i.e., we find 3 5. geometrically. Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such that AB = 3.5 units (see Fig. 1.11). From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = 3.5 . Reprint 2025-26 Fig. 1.11 NUMBER SYSTEMS 17 More generally, to find x , for any positive real number x, we mark B so that AB = x units, and, as in Fig. 1.12, mark C so that BC = 1 unit. Then, as we have done for the case x = 3.5, we find BD = x (see Fig. 1.12). We can prove this result using the Pythagoras Theorem. Notice that, in Fig. 1.12, ∆ OBD is a right-angled triangle. Also, the radius of the circle is 1 2 x + units. Therefore, OC = OD = OA = 1 2 x + units. Now, OB = 1 1 2 2 x x x + − − = ⋅ So, by the Pythagoras Theorem, we have This shows that BD = x . This construction gives us a visual, and geometric way of showing that x exists for all real numbers x > 0. If you want to know the position of x on the number line, then let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with centre B and radius BD, which intersects the number line in E BD2 = OD2 – OB2 = 2 2 1 1 4 2 2 4 x x x x + − − = = . Fig. 1.12 (see Fig. 1.13). Then, E represents x . Reprint 2025-26 Fig. 1.13 18 MATHEMATICS and in general nth roots, where n is a positive integer. Recall your understanding of square roots and cube roots from earlier classes. What is 3 8 ? Well, we know it has to be some positive number whose cube is 8, and you must have guessed 3 8 = 2. Let us try 5 243 . Do you know some number b such that b 5 = 243? The answer is 3. Therefore, 5 243 = 3. integer n? Let a > 0 be a real number and n be a positive integer. Then n a = b, if bn = a and b > 0. Note that the symbol ‘ ’ used in 3 2, 8, n a , etc. is called the radical sign. ways. You are already familiar with some of these from your earlier classes. The remaining ones follow from the distributive law of multiplication over addition of real numbers, and from the identity (x + y) (x – y) = x 2 – y 2 , for any real numbers x and y. Let a and b be positive real numbers. Then (i) ab a b = (ii) a a b b = (iii) ( a b a b a b + − = − ) ( ) (iv) ( ) ( ) 2 a b a b a b + − = − We would like to now extend the idea of square roots to cube roots, fourth roots, From these examples, can you define n a for a real number a > 0 and a positive We now list some identities relating to square roots, which are useful in various (v) ( a b c d ac ad bc bd + + = + + + ) ( ) (vi) ( ) 2 a b a ab b + = + + 2 Let us look at some particular cases of these identities. Example 15 : Simplify the following expressions: (i) (5 7 2 5 + + ) ( ) (ii) (5 5 5 5 + − ) ( ) (iii) ( ) 2 3 7 + (iv) ( 11 7 11 7 − + ) ( ) Reprint 2025-26 NUMBER SYSTEMS 19 Solution : (i) (5 7 2 5 10 5 5 2 7 35 + + = + + + ) ( ) (ii) ( ) ( ) ( ) 2 2 5 5 5 5 5 5 25 5 20 + − = − = = – (iii) ( ) ( ) ( ) 2 2 2 3 7 3 2 3 7 7 3 2 21 7 10 2 21 + = + + = + + = + (iv) ( ) ( ) ( ) ( ) 2 2 11 7 11 7 11 7 11 7 4 − + = − = − = Remark : Note that ‘simplify’ in the example above has been used to mean that the expression should be written as the sum of a rational and an irrational number. We end this section by considering the following problem. Look at 1 2 ⋅ Can you tell where it shows up on the number line? You know that it is irrational. May be it is easier to handle if the denominator is a rational number. Let us see, if we can ‘rationalise’ the denominator, that is, to make the denominator into a rational number. To do so, we need the identities involving square roots. Let us see how. Example 16 : Rationalise the denominator of 1 2 ⋅ Solution : We want to write 1 is a rational number. We know that 2 . 2 is rational. We also know that multiplying 2 as an equivalent expression in which the denominator facts together to get In this form, it is easy to locate 1 and 2 . 1 2 by 2 2 will give us an equivalent expression, since 2 2 = 1. So, we put these two 1 1 2 2 2 2 2 2 = × = ⋅ 2 on the number line. It is half way between 0 Reprint 2025-26 20 MATHEMATICS Example 17 : Rationalise the denominator of 1 2 3 ⋅ + Solution : We use the Identity (iv) given earlier. Multiply and divide 1 2 3 + by Example 18 : Rationalise the denominator of 5 3 5 ⋅ − Solution : Here we use the Identity (iii) given earlier. So, 5 3 5 − = ( ) ( ) 5 3 5 5 3 5 5 3 5 3 5 3 5 3 5 2 + + − × = = + − + − Example 19 : Rationalise the denominator of 1 Solution : 1 1 7 3 2 7 3 2 7 3 2 7 3 2 7 3 2 7 3 2 49 18 31 − − − = × = = + + − − So, when the denominator of an expression contains a term with a square root (or a number under a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator. 2 3 − to get 1 2 3 2 3 2 3 2 3 2 3 4 3 − − × = = − + − − . 7 3 2 ⋅ + 1. Classify the following numbers as rational or irrational: (i) 2 5 − (ii) (3 23 23 + −) (iii) 2 7 (iv) 1 2 (v) 2π EXERCISE 1.4 Reprint 2025-26 7 7 NUMBER SYSTEMS 21 1.5 Laws of Exponents for Real Numbers Do you remember how to simplify the following? 2. Simplify each of the following expressions: 3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter 4. Represent 9 3. on the number line. 5. Rationalise the denominators of the following: (i) 172 . 175 = (ii) (52 ) 7 = (iii) 10 7 23 23 = (iv) 7 3 . 93 = (i) (3 3 2 2 + + ) ( ) (ii) (3 3 3 3 + − ) ( ) (iii) ( ) 2 5 2 + (iv) ( 5 2 5 2 − + ) ( ) (say d). That is, π = c d ⋅ This seems to contradict the fact that π is irrational. How will you resolve this contradiction? (i) 1 7 (ii) 1 7 6 − (iii) 1 5 2 + (iv) 1 7 2 − Did you get these answers? They are as follows: which you have learnt in your earlier classes. (Here a, n and m are natural numbers. Remember, a is called the base and m and n are the exponents.) (i) 172 . 175 = 177 (ii) (52 ) 7 = 514 (iii) 10 3 7 23 23 23 = (iv) 7 3 . 93 = 633 To get these answers, you would have used the following laws of exponents, (i) a m . an = am + n (ii) (a m ) n = a mn (iii) m m n n a a , m n a − = > (iv) a mb m = (ab) m Reprint 2025-26 22 MATHEMATICS get 1 . n n a a − = We can now extend the laws to negative exponents too. that we have studied earlier, even when the base is a positive real number and the exponents are rational numbers. (Later you will study that it can further to be extended when the exponents are real numbers.) But before we state these laws, and to even make sense of these laws, we need to first understand what, for example 3 2 4 is. So, we have some work to do! What is (a) 0 ? Yes, it is 1! So you have learnt that (a) 0 = 1. So, using (iii), we can So, for example : (i) 2 –5 –3 3 1 17 17 17 17 ⋅ = = (ii) 2 –7 –14 (5 ) 5 = (iii) –10 –17 7 23 23 23 = (iv) –3 –3 –3 (7) (9) (63) ⋅ = Suppose we want to do the following computations: (i) 2 1 3 3 2 2⋅ (ii) 4 1 5 3 (iii) How would we go about it? It turns out that we can extend the laws of exponents We define n a for a real number a > 0 as follows: 1 5 1 3 7 7 (iv) 1 1 5 5 13 17 ⋅ b > 0. There are now two ways to look at 3 2 4 . Let a > 0 be a real number and n a positive integer. Then n a = b, if b n = a and In the language of exponents, we define n a = 1 n a . So, in particular, 1 3 3 2 2 = . Reprint 2025-26 3 2 4 = 3 2 4 = ( ) ( ) 1 1 3 4 64 8 2 = = 2 3 1 2 3 4 2 8 = = NUMBER SYSTEMS 23 common factors other than 1, and n > 0. Then, Example 20 : Simplify (i) 2 1 3 3 2 2⋅ (ii) 4 1 5 3 Solution : (i) 2 1 3 2 1 3 3 3 3 3 1 2 2 2 2 2 2 + ⋅ = = = = (ii) 4 1 4 5 5 3 3 = (iii) Therefore, we have the following definition: Let a > 0 be a real number. Let m and n be integers such that m and n have no We now have the following extended laws of exponents: Let a > 0 be a real number and p and q be rational numbers. Then, we have You can now use these laws to answer the questions asked earlier. 1 1 1 3 5 2 5 5 3 15 15 1 3 7 7 7 7 7 (i) a p . a q = a p+q (ii) (a p ) q = a pq (iii) p p q q a a a − = (iv) a pb p = (ab) p − − − = = = (iv) 1 1 1 1 5 5 5 5 13 17 (13 17) 221 ⋅ = × = (iii) 1 5 1 3 7 7 m n a = ( ) m n n m a a = (iv) 1 1 5 5 13 17 ⋅ 1. Find : (i) 1 2 64 (ii) 1 5 32 (iii) 1 3 125 2. Find : (i) 3 2 9 (ii) 2 5 32 (iii) 3 4 16 (iv) 1 3 125 − 3. Simplify : (i) 2 1 3 5 2 2⋅ (ii) 7 3 1 3 (iii) EXERCISE 1.5 Reprint 2025-26 1 2 1 4 11 11 (iv) 1 1 2 2 7 8⋅ 24 MATHEMATICS 1.6 Summary In this chapter, you have studied the following points: 1. A number r is called a rational number, if it can be written in the form p q , where p and q are 2. A number s is called a irrational number, if it cannot be written in the form p q , where p and 3. The decimal expansion of a rational number is either terminating or non-terminating recurring. Moreover, a number whose decimal expansion is terminating or non-terminating recurring is rational. 4. The decimal expansion of an irrational number is non-terminating non-recurring. Moreover, a number whose decimal expansion is non-terminating non-recurring is irrational. 5. All the rational and irrational numbers make up the collection of real numbers. 6. If r is rational and s is irrational, then r + s and r – s are irrational numbers, and rs and r s are 7. For positive real numbers a and b, the following identities hold: integers and q ≠ 0. q are integers and q ≠ 0. irrational numbers, r ≠ 0. (i) ab a b = (ii) a a b b = (iii) ( a b a b a b + − = − ) ( ) (iv) ( ) ( ) 2 a b a b a b + − = − (v) ( ) 2 a b a ab b + = + + 2 8. To rationalise the denominator of 1 , a b + we multiply this by , a b a b − 9. Let a > 0 be a real number and p and q be rational numbers. Then integers. (i) a p . a q = a p + q (ii) (a p ) q = a pq (iii) p p q q a a a − = (iv) a pb p = (ab) p Reprint 2025-26 − where a and b are" class_9,2,Polynomials,ncert_books/class_9/iemh1dd/iemh102.pdf,"POLYNOMIALS 25 POLYNOMIALS 2.1 Introduction You have studied algebraic expressions, their addition, subtraction, multiplication and division in earlier classes. You also have studied how to factorise some algebraic expressions. You may recall the algebraic identities : and x 2 – y 2 = (x + y) (x – y) and their use in factorisation. In this chapter, we shall start our study with a particular type of algebraic expression, called polynomial, and the terminology related to it. We shall also study the Remainder Theorem and Factor Theorem and their use in the factorisation of polynomials. In addition to the above, we shall study some more algebraic identities and their use in factorisation and in evaluating some given expressions. (x + y) 2 = x 2 + 2xy + y 2 (x – y) 2 = x 2 – 2xy + y 2 CHAPTER 2 2.2 Polynomials in One Variable Let us begin by recalling that a variable is denoted by a symbol that can take any real value. We use the letters x, y, z, etc. to denote variables. Notice that 2x, 3x, – x, – 1 2 x are algebraic expressions. All these expressions are of the form (a constant) × x. Now suppose we want to write an expression which is (a constant) × (a variable) and we do not know what the constant is. In such cases, we write the constant as a, b, c, etc. So the expression will be ax, say. denoting a variable. The values of the constants remain the same throughout a particular situation, that is, the values of the constants do not change in a given problem, but the value of a variable can keep changing. However, there is a difference between a letter denoting a constant and a letter Reprint 2025-26 26 MATHEMATICS Now, consider a square of side 3 units (see Fig. 2.1). What is its perimeter? You know that the perimeter of a square is the sum of the lengths of its four sides. Here, each side is 3 units. So, its perimeter is 4 × 3, i.e., 12 units. What will be the perimeter if each side of the square is 10 units? The perimeter is 4 × 10, i.e., 40 units. In case the length of each side is x units (see Fig. 2.2), the perimeter is given by 4x units. So, as the length of the side varies, the perimeter varies. Can you find the area of the square PQRS? It is x × x = x 2 square units. x 2 is an algebraic expression. You are also familiar with other algebraic expressions like 2x, x 2 + 2x, x 3 – x 2 + 4x + 7. Note that, all the algebraic expressions we have considered so far have only whole numbers as the exponents of the variable. Expressions of this form are called polynomials in one variable. In the examples above, the variable is x. For instance, x 3 – x 2 + 4x + 7 is a polynomial in x. Similarly, 3y 2 + 5y is a polynomial in the variable y and t 2 + 4 is a polynomial in the variable t. polynomial. Similarly, the polynomial 3y 2 + 5y + 7 has three terms, namely, 3y 2 , 5y and 7. Can you write the terms of the polynomial –x 3 + 4x 2 + 7x – 2 ? This polynomial has 4 terms, namely, –x 3 , 4x 2 , 7x and –2. coefficient of x 3 is –1, the coefficient of x 2 is 4, the coefficient of x is 7 and –2 is the coefficient of x 0 (Remember, x 0 = 1). Do you know the coefficient of x in x 2 – x + 7? In the polynomial x 2 + 2x, the expressions x 2 and 2x are called the terms of the Each term of a polynomial has a coefficient. So, in –x 3 + 4x 2 + 7x – 2, the x x 3 3 x S R P Q Fig. 2.1 Fig. 2.2 3 3 x It is –1. The constant polynomial 0 is called the zero polynomial. This plays a very important role in the collection of all polynomials, as you will see in the higher classes. know that you can write x + 1 x = x + x –1? Here, the exponent of the second term, i.e., x –1 is –1, which is not a whole number. So, this algebraic expression is not a polynomial. not a whole number. So, is x + 3 a polynomial? No, it is not. What about 3 y + y 2? It is also not a polynomial (Why?). 2 is also a polynomial. In fact, 2, –5, 7, etc. are examples of constant polynomials. Now, consider algebraic expressions such as x + 2 3 1 , x y y + + 3 and . x Do you Again, x + 3 can be written as 1 2 x + 3 . Here the exponent of x is 1 2 , which is Reprint 2025-26 POLYNOMIALS 27 or r(x), etc. So, for example, we may write : + x 2 + x + 1 is a polynomial with 151 terms. polynomials has only one term? Polynomials having only one term are called monomials (‘mono’ means ‘one’). two terms. Polynomials having only two terms are called binomials (‘bi’ means ‘two’). (‘tri’ means ‘three’). Some examples of trinomials are highest power of x ? It is 3x 7 . The exponent of x in this term is 7. Similarly, in the polynomial q(y) = 5y 6 – 4y 2 – 6, the term with the highest power of y is 5y 6 and the If the variable in a polynomial is x, we may denote the polynomial by p(x), or q(x), A polynomial can have any (finite) number of terms. For instance, x 150 + x 149 + ... Consider the polynomials 2x, 2, 5x 3 , –5x 2 , y and u 4 . Do you see that each of these Now observe each of the following polynomials: p(x) = x + 1, q(x) = x 2 – x, r(y) = y 9 + 1, t(u) = u 15 – u 2 How many terms are there in each of these? Each of these polynomials has only Similarly, polynomials having only three terms are called trinomials p(x) = x + x 2 + π, q(x) = 2 + x – x 2 , r(u) = u + u 2 – 2, t(y) = y 4 + y + 5. Now, look at the polynomial p(x) = 3x 7 – 4x 6 + x + 9. What is the term with the p(x) = 2x 2 + 5x – 3 q(x) = x 3 –1 r(y) = y 3 + y + 1 s(u) = 2 – u – u 2 + 6u 5 exponent of y in this term is 6. We call the highest power of the variable in a polynomial as the degree of the polynomial. So, the degree of the polynomial 3x 7 – 4x 6 + x + 9 is 7 and the degree of the polynomial 5y 6 – 4y 2 – 6 is 6. The degree of a non-zero constant polynomial is zero. Example 1 : Find the degree of each of the polynomials given below: Solution : (i) The highest power of the variable is 5. So, the degree of the polynomial is 5. (ii) The highest power of the variable is 8. So, the degree of the polynomial is 8. (iii)The only term here is 2 which can be written as 2x 0 . So the exponent of x is 0. Therefore, the degree of the polynomial is 0. (i) x 5 – x 4 + 3 (ii) 2 – y 2 – y 3 + 2y 8 (iii) 2 Reprint 2025-26 28 MATHEMATICS s(u) = 3 – u. Do you see anything common among all of them? The degree of each of these polynomials is one. A polynomial of degree one is called a linear polynomial. Some more linear polynomials in one variable are 2x – 1, 2 y + 1, 2 – u. Now, try and find a linear polynomial in x with 3 terms? You would not be able to find it because a linear polynomial in x can have at most two terms. So, any linear polynomial in x will be of the form ax + b, where a and b are constants and a ≠ 0 (why?). Similarly, ay + b is a linear polynomial in y. a quadratic polynomial. Some examples of a quadratic polynomial are 5 – y 2 , 4y + 5y 2 and 6 – y – y 2 . Can you write a quadratic polynomial in one variable with four different terms? You will find that a quadratic polynomial in one variable will have at most 3 terms. If you list a few more quadratic polynomials, you will find that any quadratic polynomial in x is of the form ax2 + bx + c, where a ≠ 0 and a, b, c are constants. Similarly, quadratic polynomial in y will be of the form ay2 + by + c, provided a ≠ 0 and a, b, c are constants. cubic polynomial in x are 4x 3 , 2x 3 + 1, 5x 3 + x 2 , 6x 3 – x, 6 – x 3 , 2x 3 + 4x 2 + 6x + 7. How many terms do you think a cubic polynomial in one variable can have? It can have at most 4 terms. These may be written in the form ax3 + bx2 + cx + d, where a ≠ 0 and a, b, c and d are constants. Now observe the polynomials p(x) = 4x + 5, q(y) = 2y, r(t) = t + 2 and Now consider the polynomials : Do you agree that they are all of degree two? A polynomial of degree two is called We call a polynomial of degree three a cubic polynomial. Some examples of a 2x 2 + 5, 5x 2 + 3x + π, x 2 and x 2 + 2 5 x looks like, can you write down a polynomial in one variable of degree n for any natural number n? A polynomial in one variable x of degree n is an expression of the form an x n + an–1x n–1 + . . . + a1 x + a0 where a0 , a1 , a2 , . . ., an are constants and an ≠ 0. the zero polynomial, which is denoted by 0. What is the degree of the zero polynomial? The degree of the zero polynomial is not defined. polynomials in more than one variable. For example, x 2 + y 2 + xyz (where variables are x, y and z) is a polynomial in three variables. Similarly p 2 + q 10 + r (where the variables are p, q and r), u 3 + v 2 (where the variables are u and v) are polynomials in three and two variables, respectively. You will be studying such polynomials in detail later. Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3 In particular, if a0 = a1 = a2 = a3 = . . . = an = 0 (all the constants are zero), we get So far we have dealt with polynomials in one variable only. We can also have Reprint 2025-26 POLYNOMIALS 29 2.3 Zeroes of a Polynomial Consider the polynomial p(x) = 5x 3 – 2x 2 + 3x – 2. If we replace x by 1 everywhere in p(x), we get 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. 2. Write the coefficients of x 2 in each of the following: 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100. 4. Write the degree of each of the following polynomials: 5. Classify the following as linear, quadratic and cubic polynomials: (i) 4x 2 – 3x + 7 (ii) y 2 + 2 (iii) 3 2 t t + (iv) y + 2 y (v) x 10 + y 3 + t 50 (i) 2 + x 2 + x (ii) 2 – x 2 + x 3 (iii) 2 2 x x π + (iv) 2 1 x − (i) 5x 3 + 4x 2 + 7x (ii) 4 – y 2 (iii) 5t – 7 (iv) 3 (i) x 2 + x (ii) x – x 3 (iii) y + y 2 + 4 (iv) 1 + x (v) 3t (vi) r 2 (vii) 7x 3 p(1) = 5 × (1)3 – 2 × (1)2 + 3 × (1) – 2 EXERCISE 2.1 So, we say that the value of p(x) at x = 1 is 4. Similarly, p(0) = 5(0)3 – 2(0)2 + 3(0) –2 Can you find p(–1)? Example 2 : Find the value of each of the following polynomials at the indicated value of variables: (i) p(x) = 5x 2 – 3x + 7 at x = 1. (ii) q(y) = 3y 3 – 4y + 11 at y = 2. (iii) p(t) = 4t 4 + 5t 3 – t 2 + 6 at t = a. = –2 = 5 – 2 + 3 –2 = 4 Reprint 2025-26 30 MATHEMATICS Solution : (i) p(x) = 5x 2 – 3x + 7 The value of the polynomial p(x) at x = 1 is given by (ii) q(y) = 3y 3 – 4y + 11 The value of the polynomial q(y) at y = 2 is given by (iii) p(t) = 4t 4 + 5t 3 – t 2 + 6 The value of the polynomial p(t) at t = a is given by Now, consider the polynomial p(x) = x – 1. What is p(1)? Note that : p(1) = 1 – 1 = 0. As p(1) = 0, we say that 1 is a zero of the polynomial p(x). Similarly, you can check that 2 is a zero of q(x), where q(x) = x – 2. In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0. equating it to 0, i.e., x – 1 = 0, which gives x = 1. We say p(x) = 0 is a polynomial equation and 1 is the root of the polynomial equation p(x) = 0. So we say 1 is the zero of the polynomial x – 1, or a root of the polynomial equation x – 1 = 0. You must have observed that the zero of the polynomial x – 1 is obtained by Now, consider the constant polynomial 5. Can you tell what its zero is? It has no p(1) = 5(1)2 – 3(1) + 7 q(2) = 3(2)3 – 4(2) + 11 = 24 – 8 + 11 = 16 + 11 = 5 – 3 + 7 = 9 p(a) = 4a 4 + 5a 3 – a 2 + 6 zero because replacing x by any number in 5x 0 still gives us 5. In fact, a non-zero constant polynomial has no zero. What about the zeroes of the zero polynomial? By convention, every real number is a zero of the zero polynomial. Example 3 : Check whether –2 and 2 are zeroes of the polynomial x + 2. Solution : Let p(x) = x + 2. Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0 Therefore, –2 is a zero of the polynomial x + 2, but 2 is not. Example 4 : Find a zero of the polynomial p(x) = 2x + 1. Solution : Finding a zero of p(x), is the same as solving the equation p(x) = 0 Reprint 2025-26 POLYNOMIALS 31 Now, 2x + 1 = 0 gives us x = 1 – 2 So, 1 – 2 is a zero of the polynomial 2x + 1. p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x), amounts to solving the polynomial equation p(x) = 0. Now, p(x) = 0 means ax + b = 0, a ≠ 0 So, ax = –b i.e., x = – b a . So, x = b a − is the only zero of p(x), i.e., a linear polynomial has one and only one zero. Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2. Example 5 : Verify whether 2 and 0 are zeroes of the polynomial x 2 – 2x. Solution : Let p(x) = x 2 – 2x Then p(2) = 2 2 – 4 = 4 – 4 = 0 and p(0) = 0 – 0 = 0 Hence, 2 and 0 are both zeroes of the polynomial x 2 – 2x. Let us now list our observations: (i) A zero of a polynomial need not be 0. Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of (iv) A polynomial can have more than one zero. (iii) Every linear polynomial has one and only one zero. (ii) 0 may be a zero of a polynomial. 1. Find the value of the polynomial 5x – 4x 2 + 3 at 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) x = 0 (ii) x = –1 (iii) x = 2 (i) p(y) = y 2 – y + 1 (ii) p(t) = 2 + t + 2t 2 – t 3 (iii) p(x) = x 3 (iv) p(x) = (x – 1) (x + 1) EXERCISE 2.2 Reprint 2025-26 32 MATHEMATICS 2.4 Factorisation of Polynomials Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, 1 2 q − = 0, (2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t) for some polynomial g(t). This is a particular case of the following theorem. Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x). 3. Verify whether the following are zeroes of the polynomial, indicated against them. (vii) p(x) = 3x 2 – 1, x = 1 2 , 3 3 − (viii) p(x) = 2x + 1, x = 1 2 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = 3x + 1, x = 1 – 3 (ii) p(x) = 5x – π, x = 4 5 (iii) p(x) = x 2 – 1, x = 1, –1 (iv) p(x) = (x + 1) (x – 2), x = – 1, 2 (v) p(x) = x 2 , x = 0 (vi) p(x) = lx + m, x = – m l (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Proof: By the Remainder Theorem, p(x)=(x – a) q(x) + p(a). Example 6 : Examine whether x + 2 is a factor of x 3 + 3x 2 + 5x + 6 and of 2x + 4. Solution : The zero of x + 2 is –2. Let p(x) = x 3 + 3x 2 + 5x + 6 and s(x) = 2x + 4 Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 (i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x). (ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x). In this case, p(a) = (a – a) g(a) = 0. = –8 + 12 – 10 + 6 = 0 Reprint 2025-26 POLYNOMIALS 33 So, by the Factor Theorem, x + 2 is a factor of x 3 + 3x 2 + 5x + 6. Again, s(–2) = 2(–2) + 4 = 0 So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor Theorem, since 2x + 4 = 2(x + 2). Example 7 : Find the value of k, if x – 1 is a factor of 4x 3 + 3x 2 – 4x + k. Solution : As x – 1 is a factor of p(x) = 4x 3 + 3x 2 – 4x + k, p(1) = 0 Now, p(1) = 4(1)3 + 3(1)2 – 4(1) + k So, 4 + 3 – 4 + k = 0 i.e., k = –3 We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3. You are already familiar with the factorisation of a quadratic polynomial like x 2 + lx + m. You had factorised it by splitting the middle term lx as ax + bx so that ab = m. Then x 2 + lx + m = (x + a) (x + b). We shall now try to factorise quadratic polynomials of the type ax2 + bx + c, where a ≠ 0 and a, b, c are constants. Factorisation of the polynomial ax2 + bx + c by splitting the middle term is as follows: Let its factors be (px + q) and (rx + s). Then Comparing the coefficients of x 2 , we get a = pr. Similarly, comparing the coefficients of x, we get b = ps + qr. And, on comparing the constant terms, we get c = qs. ax2 + bx + c = (px + q) (rx + s) = pr x2 + (ps + qr) x + qs 2 3x x = 3x = first term of quotient (ps)(qr) = (pr)(qs) = ac. numbers whose product is ac. This will be clear from Example 13. Example 8 : Factorise 6x 2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem. Solution 1 : (By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors. So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17. This shows us that b is the sum of two numbers ps and qr, whose product is Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two Reprint 2025-26 34 MATHEMATICS So, 6x 2 + 17x + 5 = 6x 2 + (2 + 15)x + 5 Solution 2 : (Using the Factor Theorem) 6x 2 + 17x + 5 = 2 17 5 6 6 6 x x + + = 6 p(x), say. If a and b are the zeroes of p(x), then 6x 2 + 17x + 5 = 6(x – a) (x – b). So, ab = 5 . 6 Let us look at some possibilities for a and b. They could be 1 1 5 5 , , , , 1 2 3 3 2 ± ± ± ± ± . Now, 1 1 17 1 5 2 4 6 2 6 p = + + ≠ 0. But Therefore, 6x 2 + 17x + 5 = 6 1 5 3 2 x x + + 5 2 x + is a factor of p(x). 1 3 p − = 0. So, 1 3 x + is a factor of p(x). Similarly, by trial, you can find that = 6x 2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5) = 3 1 2 5 6 3 2 x x + + = (3x + 1) (2x + 5) For the example above, the use of the splitting method appears more efficient. However, let us consider another example. Example 9 : Factorise y 2 – 5y + 6 by using the Factor Theorem. Solution : Let p(y) = y 2 – 5y + 6. Now, if p(y) = (y – a) (y – b), you know that the constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the factors of 6. The factors of 6 are 1, 2 and 3. Now, p(2) = 22 – (5 × 2) + 6 = 0 So, y – 2 is a factor of p(y). Reprint 2025-26 POLYNOMIALS 35 Also, p(3) = 32 – (5 × 3) + 6 = 0 So, y – 3 is also a factor of y 2 – 5y + 6. Therefore, y 2 – 5y + 6 = (y – 2)(y – 3) Note that y 2 – 5y + 6 can also be factorised by splitting the middle term –5y. Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example. Example 10 : Factorise x 3 – 23x 2 + 142x – 120. Solution : Let p(x) = x 3 – 23x 2 + 142x – 120 We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60. By trial, we find that p(1) = 0. So x – 1 is a factor of p(x). Now we see that x 3 – 23x 2 + 142x – 120 = x 3 – x 2 – 22x 2 + 22x + 120x – 120 = x 2 (x –1) – 22x(x – 1) + 120(x – 1) (Why?) = (x – 1) (x 2 – 22x + 120) [Taking (x – 1) common] We could have also got this by dividing p(x) by x – 1. Now x 2 – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have: x 2 – 22x + 120 = x 2 – 12x – 10x + 120 So, x 3 – 23x 2 – 142x – 120 = (x – 1)(x – 10)(x – 12) 1. Determine which of the following polynomials has (x + 1) a factor : 2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) x 3 + x 2 + x + 1 (ii) x 4 + x 3 + x 2 + x + 1 (iii) x 4 + 3x 3 + 3x 2 + x + 1 (iv) x 3 – x 2 – ( 2 2 2 + + ) x (i) p(x) = 2x 3 + x 2 – 2x – 1, g(x) = x + 1 EXERCISE 2.3 Reprint 2025-26 = x(x – 12) – 10(x – 12) = (x – 12) (x – 10) 36 MATHEMATICS 2.5 Algebraic Identities From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes: 3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: (i) p(x) = x 2 + x + k (ii) p(x) = 2x 2 + kx + 2 (iii) p(x) = kx2 – 2 x + 1 (iv) p(x) = kx2 – 3x + k 4. Factorise : 5. Factorise : Identity I : (x + y) 2 = x 2 + 2xy + y 2 Identity II : (x – y) 2 = x 2 – 2xy + y 2 Identity III : x 2 – y 2 = (x + y) (x – y) Identity IV : (x + a) (x + b) = x 2 + (a + b)x + ab You must have also used some of these algebraic identities to factorise the algebraic (ii) p(x) = x 3 + 3x 2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x 3 – 4x 2 + x + 6, g(x) = x – 3 (i) 12x 2 – 7x + 1 (ii) 2x 2 + 7x + 3 (iii) 6x 2 + 5x – 6 (iv) 3x 2 – x – 4 (i) x 3 – 2x 2 – x + 2 (ii) x 3 – 3x 2 – 9x – 5 (iii) x 3 + 13x 2 + 32x + 20 (iv) 2y 3 + y 2 – 2y – 1 expressions. You can also see their utility in computations. Example 11 : Find the following products using appropriate identities: Solution : (i) Here we can use Identity I : (x + y) 2 = x 2 + 2xy + y 2 . Putting y = 3 in it, we get (ii) Using Identity IV above, i.e., (x + a) (x + b) = x 2 + (a + b)x + ab, we have (i) (x + 3) (x + 3) (ii) (x – 3) (x + 5) (x + 3) (x + 3) = (x + 3)2 = x 2 + 2(x)(3) + (3)2 (x – 3) (x + 5) = x 2 + (–3 + 5)x + (–3)(5) Reprint 2025-26 = x 2 + 6x + 9 = x 2 + 2x – 15 POLYNOMIALS 37 Example 12 : Evaluate 105 × 106 without multiplying directly. Solution : 105 × 106 = (100 + 5) × (100 + 6) You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples. Example 13 : Factorise: Solution : (i) Here you can see that Comparing the given expression with x 2 + 2xy + y 2 , we observe that x = 7a and y = 5b. Using Identity I, we get (ii) We have 2 2 2 25 5 2 – – 4 9 2 3 y y x x = Now comparing it with Identity III, we get (i) 49a 2 + 70ab + 25b 2 (ii) 2 25 2 4 9 y x − 49a 2 + 70ab + 25b 2 = (7a + 5b) 2 = (7a + 5b) (7a + 5b) = (100)2 + (5 + 6) (100) + (5 × 6), using Identity IV = 10000 + 1100 + 30 = 11130 49a 2 = (7a) 2 , 25b 2 = (5b) 2 , 70ab = 2(7a) (5b) So far, all our identities involved products of binomials. Let us now extend the Identity I to a trinomial x + y + z. We shall compute (x + y + z) 2 by using Identity I. Let x + y = t. Then, (x + y + z) 2 = (t + z) 2 = t 2 + 2tz + t 2 (Using Identity I) = (x + y) 2 + 2(x + y)z + z 2 (Substituting the value of t) 2 25 2 – 4 9 y x = 2 2 5 – 2 3 y x Reprint 2025-26 = 5 5 2 3 2 3 y y x x + − 38 MATHEMATICS So, we get the following identity: Identity V : (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx Remark : We call the right hand side expression the expanded form of the left hand side expression. Note that the expansion of (x + y + z) 2 consists of three square terms and three product terms. Example 14 : Write (3a + 4b + 5c) 2 in expanded form. Solution : Comparing the given expression with (x + y + z) 2 , we find that Therefore, using Identity V, we have Example 15 : Expand (4a – 2b – 3c) 2 . Solution : Using Identity V, we have Example 16 : Factorise 4x 2 + y 2 + z 2 – 4xy – 2yz + 4xz. Solution : We have 4x 2 + y 2 + z2 – 4xy – 2yz + 4xz = (2x) 2 + (–y) 2 + (z) 2 + 2(2x)(–y) (3a + 4b + 5c) 2 = (3a) 2 + (4b) 2 + (5c) 2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) (4a – 2b – 3c) 2 = [4a + (–2b) + (–3c)]2 = 9a 2 + 16b 2 + 25c 2 + 24ab + 40bc + 30ac = (4a) 2 + (–2b) 2 + (–3c) 2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a) = 16a 2 + 4b 2 + 9c 2 – 16ab + 12bc – 24ac x = 3a, y = 4b and z = 5c. = x 2 + 2xy + y 2 + 2xz + 2yz + z 2 (Using Identity I) = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx (Rearranging the terms) extend Identity I to compute (x + y) 3 . We have: So far, we have dealt with identities involving second degree terms. Now let us (x + y) 3 = (x + y) (x + y) 2 Reprint 2025-26 = [2x + (–y) + z] 2 (Using Identity V) = (2x – y + z) 2 = (2x – y + z)(2x – y + z) = (x + y)(x 2 + 2xy + y 2 ) = x(x 2 + 2xy + y 2 ) + y(x 2 + 2xy + y 2 ) = x 3 + 2x 2 y + xy2 + x 2 y + 2xy2 + y 3 = x 3 + 3x 2 y + 3xy2 + y 3 = x 3 + y 3 + 3xy(x + y) + 2(–y)(z) + 2(2x)(z) POLYNOMIALS 39 So, we get the following identity: Identity VI : (x + y) 3 = x 3 + y 3 + 3xy (x + y) Also, by replacing y by –y in the Identity VI, we get Identity VII : (x – y) 3 = x 3 – y 3 – 3xy(x – y) Example 17 : Write the following cubes in the expanded form: Solution : (i) Comparing the given expression with (x + y) 3 , we find that So, using Identity VI, we have: (ii) Comparing the given expression with (x – y) 3 , we find that So, using Identity VII, we have: Example 18 : Evaluate each of the following using suitable identities: Solution : (i) We have (i) (3a + 4b) 3 (ii) (5p – 3q) 3 (i) (104)3 (ii) (999)3 x = 3a and y = 4b. = x 3 – 3x 2y + 3xy2 – y 3 (3a + 4b) 3 = (3a) 3 + (4b) 3 + 3(3a)(4b)(3a + 4b) (5p – 3q) 3 = (5p) 3 – (3q) 3 – 3(5p)(3q)(5p – 3q) x = 5p, y = 3q. = 27a 3 + 64b 3 + 108a 2b + 144ab2 = 125p 3 – 27q 3 – 225p 2q + 135pq2 (ii) We have (104)3 = (100 + 4)3 (999)3 = (1000 – 1)3 Reprint 2025-26 = (100)3 + (4)3 + 3(100)(4)(100 + 4) = 1000000 + 64 + 124800 = 1124864 = (1000)3 – (1)3 – 3(1000)(1)(1000 – 1) = 1000000000 – 1 – 2997000 = 997002999 (Using Identity VII) (Using Identity VI) 40 MATHEMATICS Example 19 : Factorise 8x 3 + 27y 3 + 36x 2 y + 54xy2 Solution : The given expression can be written as Now consider (x + y + z)(x 2 + y 2 + z 2 – xy – yz – zx) On expanding, we get the product as x(x 2 + y 2 + z 2 – xy – yz – zx) + y(x 2 + y 2 + z 2 – xy – yz – zx) + z(x 2 + y 2 + z 2 – xy – yz – zx) = x 3 + xy2 + xz 2 – x 2y – xyz – zx2 + x 2y + y 3 + yz2 – xy2 – y 2 z – xyz + x 2 z + y 2 z + z 3 – xyz – yz2 – xz2 So, we obtain the following identity: Identity VIII : x 3 + y 3 + z 3 – 3xyz = (x + y + z)(x 2 + y 2 + z 2 – xy – yz – zx) Example 20 : Factorise : 8x 3 + y 3 + 27z 3 – 18xyz Solution : Here, we have 8x 3 + y 3 + 27z 3 – 18xyz = (2x) 3 + y 3 + (3z) 3 – 3(2x)(y)(3z) = (2x + y + 3z)[(2x) 2 + y 2 + (3z) 2 – (2x)(y) – (y)(3z) – (2x)(3z)] = (2x + y + 3z) (4x 2 + y 2 + 9z 2 – 2xy – 3yz – 6xz) = x 3 + y 3 + z 3 – 3xyz (On simplification) (2x) 3 + (3y) 3 + 3(4x 2 )(3y) + 3(2x)(9y 2 ) = (2x) 3 + (3y) 3 + 3(2x) 2 (3y) + 3(2x)(3y) 2 = (2x + 3y) 3 (Using Identity VI) = (2x + 3y)(2x + 3y)(2x + 3y) 1. Use suitable identities to find the following products: 2. Evaluate the following products without multiplying directly: 3. Factorise the following using appropriate identities: (i) (x + 4) (x + 10) (ii) (x + 8) (x – 10) (iii) (3x + 4) (3x – 5) (iv) (y 2 + 3 2 ) (y 2 – 3 2 ) (v) (3 – 2x) (3 + 2x) (i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96 (i) 9x 2 + 6xy + y 2 (ii) 4y 2 – 4y + 1 (iii) x 2 – 2 EXERCISE 2.4 Reprint 2025-26 100 y POLYNOMIALS 41 4. Expand each of the following, using suitable identities: (iv) (3a – 7b – c) 2 (v) (–2x + 5y – 3z) 2 (vi) 2 1 1 1 4 2 a b − + 5. Factorise: 6. Write the following cubes in expanded form: 3 2 3 x y − 7. Evaluate the following using suitable identities: 8. Factorise each of the following: 9. Verify : (i) x 3 + y 3 = (x + y) (x 2 – xy + y 2 ) (ii) x 3 – y 3 = (x – y) (x 2 + xy + y 2 ) 10. Factorise each of the following: (i) 27y 3 + 125z 3 (ii) 64m3 – 343n 3 (i) (x + 2y + 4z) 2 (ii) (2x – y + z) 2 (iii) (–2x + 3y + 2z) 2 (i) 4x 2 + 9y 2 + 16z 2 + 12xy – 24yz – 16xz (ii) 2x 2 + y 2 + 8z 2 – 2 2 xy + 4 2 yz – 8xz (i) (2x + 1)3 (ii) (2a – 3b) 3 (iii) 3 3 1 2 x + (iv) (i) (99)3 (ii) (102)3 (iii) (998)3 (i) 8a 3 + b 3 + 12a 2b + 6ab2 (ii) 8a 3 – b 3 – 12a 2b + 6ab2 (iii) 27 – 125a 3 – 135a + 225a 2 (iv) 64a 3 – 27b 3 – 144a 2b + 108ab2 (v) 27p 3 – 1 216 – 9 1 2 2 4 p p + 11. Factorise : 27x 3 + y 3 + z 3 – 9xyz 12. Verify that x 3 + y 3 + z 3 – 3xyz = 1 2 2 2 ( ) ( ) ( ) ( ) 2 x y z x y y z z x + + − + − + − 13. If x + y + z = 0, show that x 3 + y 3 + z 3 = 3xyz. 14. Without actually calculating the cubes, find the value of each of the following: 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: [Hint : See Question 9.] (i) (–12)3 + (7)3 + (5)3 (ii) (28)3 + (–15)3 + (–13)3 (i) (ii) Area : 25a 2 – 35a + 12 Area : 35y 2 + 13y –12 Reprint 2025-26 42 MATHEMATICS 2.6 Summary In this chapter, you have studied the following points: 1. A polynomial p(x) in one variable x is an algebraic expression in x of the form 2. A polynomial of one term is called a monomial. 3. A polynomial of two terms is called a binomial. 4. A polynomial of three terms is called a trinomial. 5. A polynomial of degree one is called a linear polynomial. 6. A polynomial of degree two is called a quadratic polynomial. 7. A polynomial of degree three is called a cubic polynomial. 8. A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a root of the equation p(x) = 0. 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? p(x) = an x n + an–1x n – 1 + . . . + a2 x 2 + a1 x + a0 , where a0 , a1 , a2 , . . ., an are constants and an ≠ 0. a0 , a1 , a2 , . . ., an are respectively the coefficients of x 0 , x, x 2 , . . ., x n , and n is called the degree of the polynomial. Each of an x n , an–1 x n–1 , ..., a0 , with an ≠ 0, is called a term of the polynomial p(x). (i) (ii) Volume : 3x 2 – 12x Volume : 12ky2 + 8ky – 20k 9. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial. 10. Factor Theorem : x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor of p(x), then p(a) = 0. 11. (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx 12. (x + y) 3 = x 3 + y 3 + 3xy(x + y) 13. (x – y) 3 = x 3 – y 3 – 3xy(x – y) 14. x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx) Reprint 2025-26" class_9,3,Coordinate Geometry,ncert_books/class_9/iemh1dd/iemh103.pdf,"COORDINATE GEOMETRY LEWIS CARROLL, The Hunting of the Snark 3.1 Introduction You have already studied how to locate a point on a number line. You also know how to describe the position of a point on the line. There are many other situations, in which to find a point we are required to describe its position with reference to more than one line. For example, consider the following situations: I. In Fig. 3.1, there is a main road running in the East-West direction and streets with numbering from West to East. Also, on each street, house numbers are marked. To look for a friend’s house here, is it enough to know only one reference point? For instance, if we only know that she lives on Street 2, will we be able to find her house easily? Not as easily as when we know two pieces of information about it, namely, the number of the street on which it is situated, and the house number. If we want to reach the house which is situated in the 2nd street and has the number 5, first of all we would identify the 2nd street and then the house numbered 5 on it. In Fig. 3.1, H shows the location of the house. Similarly, P shows the location of the house corresponding to Street number 7 and House number 4. What’s the good of Mercator’s North Poles and Equators, Tropics, Zones and Meridian Lines?’ So the Bellman would cry; and crew would reply ‘ They are merely conventional signs!’ CHAPTER 3 Reprint 2025-26 Fig. 3.1 44 MATHEMATICS the position of the dot on the paper, how will you do this? Perhaps you will try in some such manner: “The dot is in the upper half of the paper”, or “It is near the left edge of the paper”, or “It is very near the left hand upper corner of the sheet”. Do any of these statements fix the position of the dot precisely? No! But, if you say “ The dot is nearly 5 cm away from the left edge of the paper”, it helps to give some idea but still does not fix the position of the dot. A little thought might enable you to say that the dot is also at a distance of 9 cm above the bottom line. We now know exactly where the dot is! For this purpose, we fixed the position of the dot by specifying its distances from two fixed lines, the left edge of the paper and the bottom line of the paper [Fig.3.2 (b)]. In other words, we need two independent informations for finding the position of the dot. II. Suppose you put a dot on a sheet of paper [Fig.3.2 (a)]. If we ask you to tell us Now, perform the following classroom activity known as ‘Seating Plan’. Fig. 3.2 Activity 1 (Seating Plan) : Draw a plan of the seating in your classroom, pushing all the desks together. Represent each desk by a square. In each square, write the name of the student occupying the desk, which the square represents. Position of each student in the classroom is described precisely by using two independent informations: the shaded square in Fig. 3.3), your position could be written as (5, 3), first writing the column number, and then the row number. Is this the same as (3, 5)? Write down the names and positions of other students in your class. For example, if Sonia is sitting in the 4th column and 1st row, write S(4,1). The teacher’s desk is not part of your seating plan. We are treating the teacher just as an observer. (i) the column in which she or he sits, (ii) the row in which she or he sits. If you are sitting on the desk lying in the 5th column and 3rd row (represented by Reprint 2025-26 COORDINATE GEOMETRY 45 can be represented with the help of two perpendicular lines. In case of ‘dot’, we require distance of the dot from bottom line as well as from left edge of the paper. In case of seating plan, we require the number of the column and that of the row. This simple idea has far reaching consequences, and has given rise to a very important branch of Mathematics known as Coordinate Geometry. In this chapter, we aim to introduce some basic concepts of coordinate geometry. You will study more about these in your higher classes. This study was initially developed by the French philosopher and mathematician René Déscartes. René Déscartes, the great French mathematician of the seventeenth century, liked to lie in bed and think! One day, when resting in bed, he solved the problem of describing the position of a point in a plane. His method was a development of the older idea of latitude and longitude. In honour of Déscartes, the system used for describing the position of a point in a plane is also known as the Cartesian system. In the discussion above, you observe that position of any object lying in a plane Fig. 3.3 1. How will you describe the position of a table lamp on your study table to another person? 2. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. EXERCISE 3.1 Reprint 2025-26 René Déscartes (1596 -1650) Fig. 3.4 46 MATHEMATICS 3.2 Cartesian System You have studied the number line in the chapter on ‘Number System’. On the number line, distances from a fixed point are marked in equal units positively in one direction and negatively in the other. The point from which the distances are marked is called the origin. We use the number line to represent the numbers by marking points on a line at equal distances. If one unit distance represents the number ‘1’, then 3 units distance represents the number ‘3’, ‘0’ being at the origin. The point in the positive direction at a distance r from the origin represents the number r. The point in the negative direction at a distance r from the origin represents the number −r. Locations of different numbers on the number line are shown in Fig. 3.5. All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find: (i) how many cross - streets can be referred to as (4, 3). (ii) how many cross - streets can be referred to as (3, 4). on a plane, and locating points on the plane by referring them to these lines. The perpendicular lines may be in any direction such as in Fig.3.6. But, when we choose Descartes invented the idea of placing two such lines perpendicular to each other Reprint 2025-26 Fig. 3.5 Fig. 3.6 COORDINATE GEOMETRY 47 these two lines to locate a point in a plane in this chapter, one line will be horizontal and the other will be vertical, as in Fig. 3.6(c). These lines are actually obtained as follows : Take two number lines, calling them X′X and Y′Y. Place X′X horizontal [as in Fig. 3.7(a)] and write the numbers on it just as written on the number line. We do the same thing with Y′Y except that Y′Y is vertical, not horizontal [Fig. 3.7(b)]. Combine both the lines in such a way that the two lines cross each other at their zeroes, or origins (Fig. 3.8). The horizontal line X′X is called the x - axis and the vertical line YY′ is called the y - axis. The point where X′X and Y′Y cross is called the origin, and is denoted by O. Since the positive numbers lie on the directions OX and OY, OX and OY are called the positive directions of the x - axis and the y - axis, respectively. Similarly, OX′ and OY′ are called the negative directions of the x - axis and the y - axis, respectively. Fig. 3.8 Fig. 3.7 Reprint 2025-26 48 MATHEMATICS You observe that the axes (plural of the word ‘axis’) divide the plane into four parts. These four parts are called the quadrants (one fourth part), numbered I, II, III and IV anticlockwise from OX (see Fig.3.9). So, the plane consists of the axes and these quadrants. We call the plane, the Cartesian plane, or the coordinate plane, or the xy-plane. The axes are called the coordinate axes. Consider the following diagram where the axes are drawn on graph paper. Let us see the distances of the points P and Q from the axes. For this, we draw perpendiculars PM on the x - axis and PN on the y - axis. Similarly, we draw perpendiculars QR and QS as shown in Fig. 3.10. Now, let us see why this system is so basic to mathematics, and how it is useful. (ii) The perpendicular distance of the point P from the x - axis measured along the positive direction of the y - axis is PM = ON = 3 units. Fig. 3.9 You find that (i) The perpendicular distance of the point P from the y - axis measured along the positive direction of the x - axis is PN = OM = 4 units. Reprint 2025-26 Fig.3.10 COORDINATE GEOMETRY 49 same as (4, 3). Example 1 : See Fig. 3.11 and complete the following statements: (i) The abscissa and the ordinate of the point B are _ _ _ and _ _ _, respectively. Hence, the coordinates of B are (_ _, _ _). (iii) The perpendicular distance of the point Q from the y - axis measured along the negative direction of the x - axis is OR = SQ = 6 units. (iv) The perpendicular distance of the point Q from the x - axis measured along the negative direction of the y - axis is OS = RQ = 2 units. Now, using these distances, how can we describe the points so that there is no confusion? We write the coordinates of a point, using the following conventions: (i) The x - coordinate of a point is its perpendicular distance from the y - axis measured along the x -axis (positive along the positive direction of the x - axis and negative along the negative direction of the x - axis). For the point P, it is + 4 and for Q, it is – 6. The x - coordinate is also called the abscissa. (ii) The y - coordinate of a point is its perpendicular distance from the x - axis measured along the y - axis (positive along the positive direction of the y - axis and negative along the negative direction of the y - axis). For the point P, it is + 3 and for Q, it is –2. The y - coordinate is also called the ordinate. (iii) In stating the coordinates of a point in the coordinate plane, the x - coordinate comes first, and then the y - coordinate. We place the coordinates in brackets. Hence, the coordinates of P are (4, 3) and the coordinates of Q are (– 6, – 2). Note that the coordinates describe a point in the plane uniquely. (3, 4) is not the (ii) The x-coordinate and the y-coordinate of the point M are _ _ _ and _ _ _, respectively. Hence, the coordinates of M are (_ _, _ _). (iii) The x-coordinate and the y-coordinate of the point L are _ _ _ and _ _ _, respectively. Hence, the coordinates of L are (_ _, _ _). (iv) The x-coordinate and the y-coordinate of the point S are _ _ _ and _ _ _, respectively. Hence, the coordinates of S are (_ _, _ _). Reprint 2025-26 50 MATHEMATICS Solution: (i) Since the distance of the point B from the y - axis is 4 units, the x - coordinate or abscissa of the point B is 4. The distance of the point B from the x - axis is 3 units; therefore, the y - coordinate, i.e., the ordinate, of the point B is 3. Hence, the coordinates of the point B are (4, 3). Fig. 3.11 As in (i) above : (ii) The x - coordinate and the y - coordinate of the point M are –3 and 4, respectively. Hence, the coordinates of the point M are (–3, 4). (iii) The x - coordinate and the y - coordinate of the point L are –5 and – 4, respectively. Hence, the coordinates of the point L are (–5, – 4). (iv) The x - coordinate and the y- coordinate of the point S are 3 and – 4, respectively. Hence, the coordinates of the point S are (3, – 4). Reprint 2025-26 COORDINATE GEOMETRY 51 Example 2 : Write the coordinates of the points marked on the axes in Fig. 3.12. Solution : You can see that : (i) The point A is at a distance of + 4 units from the y - axis and at a distance zero from the x - axis. Therefore, the x - coordinate of A is 4 and the y - coordinate is 0. Hence, the coordinates of A are (4, 0). (ii) The coordinates of B are (0, 3). Why? (iii) The coordinates of C are (– 5, 0). Why? (iv) The coordinates of D are (0, – 4). Why? (v) The coordinates of E are 2 , 0 3 . Why? therefore, the y - coordinate of every point lying on the x - axis is always zero. Thus, the coordinates of any point on the x - axis are of the form (x, 0), where x is the distance of the point from the y - axis. Similarly, the coordinates of any point on the y - axis are of the form (0, y), where y is the distance of the point from the x - axis. Why? axes so that its abscissa and ordinate are both zero. Therefore, the coordinates of the origin are (0, 0). Since every point on the x - axis has no distance (zero distance) from the x - axis, What are the coordinates of the origin O? It has zero distance from both the Fig. 3.12 the signs of the coordinates of a point and the quadrant of a point in which it lies. (i) If a point is in the 1st quadrant, then the point will be in the form (+, +), since the 1st quadrant is enclosed by the positive x - axis and the positive y - axis. (ii) If a point is in the 2nd quadrant, then the point will be in the form (–, +), since the 2nd quadrant is enclosed by the negative x - axis and the positive y - axis. (iii) If a point is in the 3rd quadrant, then the point will be in the form (–, –), since the 3rd quadrant is enclosed by the negative x - axis and the negative y - axis. (iv) If a point is in the 4th quadrant, then the point will be in the form (+, –), since the 4th quadrant is enclosed by the positive x - axis and the negative y - axis (see Fig. 3.13). In the examples above, you may have observed the following relationship between Reprint 2025-26 52 MATHEMATICS Remark : The system we have discussed above for describing a point in a plane is only a convention, which is accepted all over the world. The system could also have been, for example, the ordinate first, and the abscissa second. However, the whole world sticks to the system we have described to avoid any confusion. Fig. 3.13 1. Write the answer of each of the following questions: 2. See Fig.3.14, and write the following: (i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? (ii) What is the name of each part of the plane formed by these two lines? (iii) Write the name of the point where these two lines intersect. (i) The coordinates of B. (ii) The coordinates of C. (iii) The point identified by the coordinates (–3, –5). EXERCISE 3.2 Reprint 2025-26 COORDINATE GEOMETRY 53 (iv) The point identified by the coordinates (2, – 4). (v) The abscissa of the point D. (vi) The ordinate of the point H. (vii) The coordinates of the point L. (viii) The coordinates of the point M. 3.3 Summary In this chapter, you have studied the following points : 1. To locate the position of an object or a point in a plane, we require two perpendicular lines. One of them is horizontal, and the other is vertical. 2. The plane is called the Cartesian, or coordinate plane and the lines are called the coordinate axes. 3. The horizontal line is called the x -axis, and the vertical line is called the y - axis. Reprint 2025-26 Fig. 3.14 54 MATHEMATICS 4. The coordinate axes divide the plane into four parts called quadrants. 5. The point of intersection of the axes is called the origin. 6. The distance of a point from the y - axis is called its x-coordinate, or abscissa, and the distance of the point from the x-axis is called its y-coordinate, or ordinate. 7. If the abscissa of a point is x and the ordinate is y, then (x, y) are called the coordinates of the point. 8. The coordinates of a point on the x-axis are of the form (x, 0) and that of the point on the y-axis are (0, y). 9. The coordinates of the origin are (0, 0). 10. The coordinates of a point are of the form (+ , +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant, where + denotes a positive real number and – denotes a negative real number. 11. If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y. Reprint 2025-26" class_9,4,Linear Equations in Two Variables,ncert_books/class_9/iemh1dd/iemh104.pdf,"LINEAR EQUATIONS IN TWO VARIABLES 55 LINEAR EQUATIONS IN TWO VARIABLES 4.1 Introduction In earlier classes, you have studied linear equations in one variable. Can you write down a linear equation in one variable? You may say that x + 1 = 0, x + 2 = 0 and 2 y + 3 = 0 are examples of linear equations in one variable. You also know that such equations have a unique (i.e., one and only one) solution. You may also remember how to represent the solution on a number line. In this chapter, the knowledge of linear equations in one variable shall be recalled and extended to that of two variables. You will be considering questions like: Does a linear equation in two variables have a solution? If yes, is it unique? What does the solution look like on the Cartesian plane? You shall also use the concepts you studied in Chapter 3 to answer these questions. The principal use of the Analytic Art is to bring Mathematical Problems to Equations and to exhibit those Equations in the most simple terms that can be. —Edmund Halley CHAPTER 4 4.2 Linear Equations Let us first recall what you have studied so far. Consider the following equation: Its solution, i.e., the root of the equation, is 5 2 − . This can be represented on the number line as shown below: 2x + 5 = 0 Reprint 2025-26 Fig. 4.1 56 MATHEMATICS Nagpur, two Indian batsmen together scored 176 runs. Express this information in the form of an equation. unknown quantities. Let us use x and y to denote them. So, the number of runs scored by one of the batsmen is x, and the number of runs scored by the other is y. We know that which is the required equation. the variables in such equations by x and y, but other letters may also be used. Some examples of linear equations in two variables are: Note that you can put these equations in the form 1.2s + 3t – 5 = 0, p + 4q – 7 = 0, πu + 5v – 9 = 0 and 2 x – 7y – 3 = 0, respectively. While solving an equation, you must always keep the following points in mind: The solution of a linear equation is not affected when: (i) the same number is added to (or subtracted from) both the sides of the equation. (ii) you multiply or divide both the sides of the equation by the same non-zero number. Let us now consider the following situation: In a One-day International Cricket match between India and Sri Lanka played in Here, you can see that the score of neither of them is known, i.e., there are two This is an example of a linear equation in two variables. It is customary to denote So, any equation which can be put in the form ax + by + c = 0, where a, b and c 1.2s + 3t = 5, p + 4q = 7, πu + 5v = 9 and 3 = 2 x – 7y. x + y = 176, are real numbers, and a and b are not both zero, is called a linear equation in two variables. This means that you can think of many many such equations. Example 1 : Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: Solution : (i) 2x + 3y = 4.37 can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 and c = – 4.37. (ii) The equation x – 4 = 3 y can be written as x – 3 y – 4 = 0. Here a = 1, b = – 3 and c = – 4. (iii) The equation 4 = 5x – 3y can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 and c = – 4. Do you agree that it can also be written as –5x + 3y + 4 = 0 ? In this case a = –5, b = 3 and c = 4. (i) 2x + 3y = 4.37 (ii) x – 4 = 3 y (iii) 4 = 5x – 3y (iv) 2x = y Reprint 2025-26 LINEAR EQUATIONS IN TWO VARIABLES 57 (iv) The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0. Equations of the type ax + b = 0 are also examples of linear equations in two variables because they can be expressed as For example, 4 – 3x = 0 can be written as –3x + 0.y + 4 = 0. Example 2 : Write each of the following as an equation in two variables: Solution : (i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0. (ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0. (iii) 2x = 3 can be written as 2x + 0.y – 3 = 0. (iv) 5y = 2 can be written as 0.x + 5y – 2 = 0. 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ` x and that of a pen to be ` y). 2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (i) x = –5 (ii) y = 2 (iii) 2x = 3 (iv) 5y = 2 (i) 2x + 3y = 9.35 (ii) x – 5 y – 10 = 0 (iii) –2x + 3y = 6 (iv) x = 3y (v) 2x = –5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x ax + 0.y + b = 0 EXERCISE 4.1 4.3 Solution of a Linear Equation You have seen that every linear equation in one variable has a unique solution. What can you say about the solution of a linear equation involving two variables? As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation. Let us consider the equation 2x + 3y = 12. Here, x = 3 and y = 2 is a solution because when you substitute x = 3 and y = 2 in the equation above, you find that 2x + 3y = (2 × 3) + (3 × 2) = 12 then the value for y. Similarly, (0, 4) is also a solution for the equation above. This solution is written as an ordered pair (3, 2), first writing the value for x and Reprint 2025-26 58 MATHEMATICS x = 1 and y = 4 we get 2x + 3y = 14, which is not 12. Note that (0, 4) is a solution but not (4, 0). you find any other solution? Do you agree that (6, 0) is another solution? Verify the same. In fact, we can get many many solutions in the following way. Pick a value of your choice for x (say x = 2) in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12, which is a linear equation in one variable. On solving this, you get y = 8 3 . So 8 2, 3 is another solution of 2x + 3y = 12. Similarly, choosing x = – 5, you find that the equation becomes –10 + 3y = 12. This gives y = 22 3 . So, 22 5, 3 − is another solution of 2x + 3y = 12. So there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions. Example 3 : Find four different solutions of the equation x + 2y = 6. Solution : By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2 Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are: On the other hand, (1, 4) is not a solution of 2x + 3y = 12, because on putting You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4). Can x + 2y = 2 + 4 = 6 Remark : Note that an easy way of getting a solution is to take x = 0 and get the corresponding value of y. Similarly, we can put y = 0 and obtain the corresponding value of x. Example 4 : Find two solutions for each of the following equations: Solution : (i) Taking x = 0, we get 3y = 12, i.e., y = 4. So, (0, 4) is a solution of the given equation. Similarly, by taking y = 0, we get x = 3. Thus, (3, 0) is also a solution. (i) 4x + 3y = 12 (ii) 2x + 5y = 0 (iii) 3y + 4 = 0 (2, 2), (0, 3), (6, 0) and (4, 1). Reprint 2025-26 LINEAR EQUATIONS IN TWO VARIABLES 59 (ii) Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation. Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the earlier one. To get another solution, take x = 1, say. Then you can check that the corresponding value of y is 2 . 5 − So 2 1, 5 − is another solution of 2x + 5y = 0. (iii) Writing the equation 3y + 4 = 0 as 0.x + 3y + 4 = 0, you will find that y = 4 – 3 for any value of x. Thus, two solutions can be given as 4 4 0, – and 1, – 3 3 . 1. Which one of the following options is true, and why? 2. Write four solutions for each of the following equations: 3. Check which of the following are solutions of the equation x – 2y = 4 and which are not: (i) (0, 2) (ii) (2, 0) (iii) (4, 0) 4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. y = 3x + 5 has (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions (i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y (iv) ( 2 , 4 2) (v) (1, 1) EXERCISE 4.2 4.4 Summary In this chapter, you have studied the following points: 1. An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. 2. A linear equation in two variables has infinitely many solutions. 3. Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation. Reprint 2025-26" class_9,5,Introduction to Euclid's Geometry,ncert_books/class_9/iemh1dd/iemh105.pdf,"60 MATHEMATICS INTRODUCTION TO EUCLID’S GEOMETRY 5.1 Introduction The word ‘geometry’ comes form the Greek words ‘geo’, meaning the ‘earth’, and ‘metrein’, meaning ‘to measure’. Geometry appears to have originated from the need for measuring land. This branch of mathematics was studied in various forms in every ancient civilisation, be it in Egypt, Babylonia, China, India, Greece, the Incas, etc. The people of these civilisations faced several practical problems which required the development of geometry in various ways. For example, whenever the river Nile overflowed, it wiped out the boundaries between the adjoining fields of different land owners. After such flooding, these boundaries had to be redrawn. For this purpose, the Egyptians developed a number of geometric techniques and rules for calculating simple areas and also for doing simple constructions. The knowledge of geometry was also used by them for computing volumes of granaries, and for constructing canals and pyramids. They also knew the correct formula to find the volume of a truncated pyramid (see Fig. 5.1).You know that a pyramid is a solid figure, the base of which is a triangle, or square, or some other polygon, and its side faces are triangles converging to a point at the top. Fig. 5.1 : A Truncated Pyramid CHAPTER 5 Reprint 2025-26 INTRODUCTION TO EUCLID’S GEOMETRY 61 In the Indian subcontinent, the excavations at Harappa and Mohenjo-Daro, etc. show that the Indus Valley Civilisation (about 3000 BCE) made extensive use of geometry. It was a highly organised society. The cities were highly developed and very well planned. For example, the roads were parallel to each other and there was an underground drainage system. The houses had many rooms of different types. This shows that the town dwellers were skilled in mensuration and practical arithmetic. The bricks used for constructions were kiln fired and the ratio length : breadth : thickness, of the bricks was found to be 4 : 2 : 1. In ancient India, the Sulbasutras (800 BCE to 500 BCE) were the manuals of geometrical constructions. The geometry of the Vedic period originated with the construction of altars (or vedis) and fireplaces for performing Vedic rites. The location of the sacred fires had to be in accordance to the clearly laid down instructions about their shapes and areas, if they were to be effective instruments. Square and circular altars were used for household rituals, while altars whose shapes were combinations of rectangles, triangles and trapeziums were required for public worship. The sriyantra (given in the Atharvaveda) consists of nine interwoven isosceles triangles. These triangles are arranged in such a way that they produce 43 subsidiary triangles. Though accurate geometric methods were used for the constructions of altars, the principles behind them were not discussed. These examples show that geometry was being developed and applied everywhere in the world. But this was happening in an unsystematic manner. What is interesting about these developments of geometry in the ancient world is that they were passed on from one generation to the next, either orally or through palm leaf messages, or by other ways. Also, we find that in some civilisations like Babylonia, geometry remained a very practical oriented discipline, as was the case in India and Rome. The geometry developed by Egyptians mainly consisted of the statements of results. There were no general rules of the procedure. In fact, Babylonians and Egyptians used geometry mostly for practical purposes and did very little to develop it as a systematic science. But in civilisations like Greece, the emphasis was on the reasoning behind why certain constructions work. The Greeks were interested in establishing the truth of the statements they discovered using deductive reasoning (see Appendix 1). A Greek mathematician, Thales is credited with giving the first known proof. This proof was of the statement that a circle is bisected (i.e., cut into two equal parts) by its diameter. One of Thales’ most famous pupils was Pythagoras (572 BCE), whom you have heard about. Pythagoras and his group discovered many geometric properties and developed the theory of geometry to a great extent. This process continued till 300 BCE. At that time Euclid, a teacher of mathematics at Alexandria in Egypt, collected all the known work and arranged it in his famous treatise, Thales (640 BCE – 546 BCE) Fig. 5.2 Reprint 2025-26 62 MATHEMATICS called ‘Elements’. He divided the ‘Elements’ into thirteen chapters, each called a book. These books influenced the whole world’s understanding of geometry for generations to come. In this chapter, we shall discuss Euclid’s approach to geometry and shall try to link it with the present day geometry. 5.2 Euclid’s Definitions, Axioms and Postulates The Greek mathematicians of Euclid’s time thought of geometry as an abstract model of the world in which they lived. The notions of point, line, plane (or surface) and so on were derived from what was seen around them. From studies of the space and solids in the space around them, an abstract geometrical notion of a solid object was developed. A solid has shape, size, position, and can be moved from one place to another. Its boundaries are called surfaces. They separate one part of the space from another, and are said to have no thickness. The boundaries of the surfaces are curves or straight lines. These lines end in points. each step we lose one extension, also called a dimension. So, a solid has three dimensions, a surface has two, a line has one and a point has none. Euclid summarised these statements as definitions. He began his exposition by listing 23 definitions in Book 1 of the ‘Elements’. A few of them are given below : Consider the three steps from solids to points (solids-surfaces-lines-points). In 1. A point is that which has no part. Euclid (325 BCE – 265 BCE) Fig. 5.3 breadth, length, evenly, etc. need to be further explained clearly. For example, consider his definition of a point. In this definition, ‘a part’ needs to be defined. Suppose if you define ‘a part’ to be that which occupies ‘area’, again ‘an area’ needs to be defined. So, to define one thing, you need to define many other things, and you may get a long chain of definitions without an end. For such reasons, mathematicians agree to leave 2. A line is breadthless length. 3. The ends of a line are points. 4. A straight line is a line which lies evenly with the points on itself. 5. A surface is that which has length and breadth only. 6. The edges of a surface are lines. 7. A plane surface is a surface which lies evenly with the straight lines on itself. If you carefully study these definitions, you find that some of the terms like part, Reprint 2025-26 INTRODUCTION TO EUCLID’S GEOMETRY 63 some geometric terms undefined. However, we do have a intuitive feeling for the geometric concept of a point than what the ‘definition’ above gives us. So, we represent a point as a dot, even though a dot has some dimension. neither of which has been defined. Because of this, a few terms are kept undefined while developing any course of study. So, in geometry, we take a point, a line and a plane (in Euclid‘s words a plane surface) as undefined terms. The only thing is that we can represent them intuitively, or explain them with the help of ‘physical models’. be proved. These assumptions are actually ‘obvious universal truths’. He divided them into two types: axioms and postulates. He used the term ‘postulate’ for the assumptions that were specific to geometry. Common notions (often called axioms), on the other hand, were assumptions used throughout mathematics and not specifically linked to geometry. For details about axioms and postulates, refer to Appendix 1. Some of Euclid’s axioms, not in his order, are given below : A similar problem arises in Definition 2 above, since it refers to breadth and length, Starting with his definitions, Euclid assumed certain properties, which were not to (1) Things which are equal to the same thing are equal to one another. (2) If equals are added to equals, the wholes are equal. (3) If equals are subtracted from equals, the remainders are equal. (4) Things which coincide with one another are equal to one another. (5) The whole is greater than the part. (6) Things which are double of the same things are equal to one another. (7) Things which are halves of the same things are equal to one another. notion could be applied to plane figures. For example, if an area of a triangle equals the area of a rectangle and the area of the rectangle equals that of a square, then the area of the triangle also equals the area of the square. different kinds cannot be compared. For example, a line cannot be compared to a rectangle, nor can an angle be compared to a pentagon. they are the same), then they are equal. In other words, everything equals itself. It is the justification of the principle of superposition. Axiom (5) gives us the definition of ‘greater than’. For example, if a quantity B is a part of another quantity A, then A can be written as the sum of B and some third quantity C. Symbolically, A > B means that there is some C such that A = B + C. These ‘common notions’ refer to magnitudes of some kind. The first common Magnitudes of the same kind can be compared and added, but magnitudes of The 4th axiom given above seems to say that if two things are identical (that is, Reprint 2025-26 64 MATHEMATICS Postulate 1 : A straight line may be drawn from any one point to any other point. distinct points, but it does not say that there cannot be more than one such line. However, in his work, Euclid has frequently assumed, without mentioning, that there is a unique line joining two distinct points. We state this result in the form of an axiom as follows: Axiom 5.1 : Given two distinct points, there is a unique line that passes through them. that is, the line PQ. How many lines passing through Q also pass through P? Only one, that is, the line PQ. Thus, the statement above is self-evident, and so is taken as an axiom. Postulate 2 : A terminated line can be produced indefinitely. Now let us discuss Euclid’s five postulates. They are : Note that this postulate tells us that at least one straight line passes through two How many lines passing through P also pass through Q (see Fig. 5.4)? Only one, Note that what we call a line segment now-a-days is what Euclid called a terminated Fig. 5.4 line. So, according to the present day terms, the second postulate says that a line segment can be extended on either side to form a line (see Fig. 5.5). Postulate 3 : A circle can be drawn with any centre and any radius. Postulate 4 : All right angles are equal to one another. Postulate 5 : If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. Reprint 2025-26 Fig. 5.5 INTRODUCTION TO EUCLID’S GEOMETRY 65 For example, the line PQ in Fig. 5.6 falls on lines AB and CD such that the sum of the interior angles 1 and 2 is less than 180° on the left side of PQ. Therefore, the lines AB and CD will eventually intersect on the left side of PQ. complex than any other postulate. On the other hand, Postulates 1 through 4 are so simple and obvious that these are taken as ‘self-evident truths’. However, it is not possible to prove them. So, these statements are accepted without any proof (see Appendix 1). Because of its complexity, the fifth postulate will be given more attention in the next section. and in the same sense. ‘Postulate’ is actually a verb. When we say “let us postulate”, we mean, “let us make some statement based on the observed phenomenon in the Universe”. Its truth/validity is checked afterwards. If it is true, then it is accepted as a ‘Postulate’. deduce from these axioms a statement that contradicts any axiom or previously proved statement. So, when any system of axioms is given, it needs to be ensured that the system is consistent. Then using these results, he proved some more results by applying deductive reasoning. The statements that were proved are called propositions or theorems. Euclid deduced 465 propositions in a logical chain using his axioms, postulates, definitions and theorems proved earlier in the chain. In the next few chapters on geometry, you will be using these axioms to prove some theorems. A brief look at the five postulates brings to your notice that Postulate 5 is far more Now-a-days, ‘postulates’ and ‘axioms’ are terms that are used interchangeably A system of axioms is called consistent (see Appendix 1), if it is impossible to After Euclid stated his postulates and axioms, he used them to prove other results. Fig. 5.6 for proving some of the results: Example 1 : If A, B and C are three points on a line, and B lies between A and C (see Fig. 5.7), then prove that AB + BC = AC. Now, let us see in the following examples how Euclid used his axioms and postulates Reprint 2025-26 Fig. 5.7 66 MATHEMATICS Solution : In the figure given above, AC coincides with AB + BC. Also, Euclid’s Axiom (4) says that things which coincide with one another are equal to one another. So, it can be deduced that Note that in this solution, it has been assumed that there is a unique line passing through two points. Example 2 : Prove that an equilateral triangle can be constructed on any given line segment. Solution : In the statement above, a line segment of any length is given, say AB [see Fig. 5.8(i)]. Here, you need to do some construction. Using Euclid’s Postulate 3, you can draw a circle with point A as the centre and AB as the radius [see Fig. 5.8(ii)]. Similarly, draw another circle with point B as the centre and BA as the radius. The two circles meet at a point, say C. Now, draw the line segments AC and BC to form ∆ ABC [see Fig. 5.8 (iii)]. AB + BC = AC Fig. 5.8 So, you have to prove that this triangle is equilateral, i.e., AB = AC = BC. Now, AB = AC, since they are the radii of the same circle (1) Similarly, AB = BC (Radii of the same circle) (2) From these two facts, and Euclid’s axiom that things which are equal to the same thing are equal to one another, you can conclude that AB = BC = AC. So, ∆ ABC is an equilateral triangle. Note that here Euclid has assumed, without mentioning anywhere, that the two circles drawn with centres A and B will meet each other at a point. Now we prove a theorem, which is frequently used in different results: Reprint 2025-26 INTRODUCTION TO EUCLID’S GEOMETRY 67 Theorem 5.1 : Two distinct lines cannot have more than one point in common. Proof : Here we are given two lines l and m. We need to prove that they have only one point in common. say P and Q. So, you have two lines passing through two distinct points P and Q. But this assumption clashes with the axiom that only one line can pass through two distinct points. So, the assumption that we started with, that two lines can pass through two distinct points is wrong. lines cannot have more than one point in common. For the time being, let us suppose that the two lines intersect in two distinct points, From this, what can we conclude? We are forced to conclude that two distinct 1. Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. (ii) There are an infinite number of lines which pass through two distinct points. (iii) A terminated line can be produced indefinitely on both the sides. (iv) If two circles are equal, then their radii are equal. (v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY. EXERCISE 5.1 2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them? 3. Consider two ‘postulates’ given below: (i) parallel lines (ii) perpendicular lines (iii) line segment (iv) radius of a circle (v) square (i) Given any two distinct points A and B, there exists a third point C which is in between A and B. (ii) There exist at least three points that are not on the same line. Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain. Reprint 2025-26 Fig. 5.9 68 MATHEMATICS 5.3 Summary In this chapter, you have studied the following points: 1. Though Euclid defined a point, a line, and a plane, the definitions are not accepted by mathematicians. Therefore, these terms are now taken as undefined. 2. Axioms or postulates are the assumptions which are obvious universal truths. They are not proved. 3. Theorems are statements which are proved, using definitions, axioms, previously proved statements and deductive reasoning. 4. Some of Euclid’s axioms were : 4. If a point C lies between two points A and B such that AC = BC, then prove that 5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point. 6. In Fig. 5.10, if AC = BD, then prove that AB = CD. 7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.) AC = 1 2 AB. Explain by drawing the figure. Fig. 5.10 5. Euclid’s postulates were : (1) Things which are equal to the same thing are equal to one another. (2) If equals are added to equals, the wholes are equal. (3) If equals are subtracted from equals, the remainders are equal. (4) Things which coincide with one another are equal to one another. (5) The whole is greater than the part. (6) Things which are double of the same things are equal to one another. (7) Things which are halves of the same things are equal to one another. Postulate 1 : A straight line may be drawn from any one point to any other point. Postulate 2 : A terminated line can be produced indefinitely. Postulate 3 : A circle can be drawn with any centre and any radius. Postulate 4 : All right angles are equal to one another. Reprint 2025-26" class_9,6,Lines and Angles,ncert_books/class_9/iemh1dd/iemh106.pdf,"LINES AND ANGLES 6.1 Introduction In Chapter 5, you have studied that a minimum of two points are required to draw a line. You have also studied some axioms and, with the help of these axioms, you proved some other statements. In this chapter, you will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Further you will use these properties to prove some statements using deductive reasoning (see Appendix 1). You have already verified these statements through some activities in the earlier classes. plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles. For instance, suppose you want to make a model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you would make it? You would keep some of the sticks parallel to each other, and some sticks would be kept slanted. Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles. Without the knowledge of the properties of these lines and angles, do you think she can draw the layout of the building? In your daily life, you see different types of angles formed between the edges of CHAPTER 6 For example, to study the refraction property of light when it enters from one medium to the other medium, you use the properties of intersecting lines and parallel lines. When two or more forces act on a body, you draw the diagram in which forces are represented by directed line segments to study the net effect of the forces on the body. At that time, you need to know the relation between the angles when the rays (or line segments) are parallel to or intersect each other. To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle In science, you study the properties of light by drawing the ray diagrams. Reprint 2025-26 70 MATHEMATICS formed between the horizontal and the line of sight. Plenty of other examples can be given where lines and angles are used. In the subsequent chapters of geometry, you will be using these properties of lines and angles to deduce more and more useful properties. earlier classes. 6.2 Basic Terms and Definitions Recall that a part (or portion) of a line with two end points is called a line-segment and a part of a line with one end point is called a ray. Note that the line segment AB is denoted by AB , and its length is denoted by AB. The ray AB is denoted by AB , and a line is denoted by AB . However, we will not use these symbols, and will denote the line segment AB, ray AB, length AB and line AB by the same symbol, AB. The meaning will be clear from the context. Sometimes small letters l, m, n, etc. will be used to denote lines. otherwise they are called non-collinear points. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle. You have studied different types of angles, such as acute angle, right angle, obtuse angle, straight angle and reflex angle in earlier classes (see Fig. 6.1). Let us first revise the terms and definitions related to lines and angles learnt in If three or more points lie on the same line, they are called collinear points; Recall that an angle is formed when two rays originate from the same end point. (i) acute angle : 0° < x < 90° (ii) right angle : y = 90° (iii) obtuse angle : 90° < z < 180° (iv) straight angle : s = 180° (v) reflex angle : 180° < t < 360° Fig. 6.1 : Types of Angles Reprint 2025-26 LINES AND ANGLES 71 equal to 90°. An angle greater than 90° but less than 180° is called an obtuse angle. Also, recall that a straight angle is equal to 180°. An angle which is greater than 180° but less than 360° is called a reflex angle. Further, two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles. You have also studied about adjacent angles in the earlier classes (see Fig. 6.2). Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm. In Fig. 6.2, ∠ ABD and ∠ DBC are adjacent angles. Ray BD is their common arm and point B is their common vertex. Ray BA and ray BC are non common arms. Moreover, when two angles are adjacent, then their sum is always equal to the angle formed by the two noncommon arms. So, we can write ∠ ABC = ∠ ABD + ∠ DBC. Note that ∠ ABC and ∠ ABD are not adjacent angles. Why? Because their noncommon arms BD and BC lie on the same side of the common arm BA. If the non-common arms BA and BC in Fig. 6.2, form a line then it will look like Fig. 6.3. In this case, ∠ ABD and ∠ DBC are called linear pair of angles. An acute angle measures between 0° and 90°, whereas a right angle is exactly Fig. 6.2 : Adjacent angles You may also recall the vertically opposite angles formed when two lines, say AB and CD, intersect each other, say at the point O (see Fig. 6.4). There are two pairs of vertically opposite angles. One pair is ∠AOD and ∠BOC. Can you find the other pair? Reprint 2025-26 Fig. 6.3 : Linear pair of angles Fig. 6.4 : Vertically opposite angles 72 MATHEMATICS 6.3 Intersecting Lines and Non-intersecting Lines Draw two different lines PQ and RS on a paper. You will see that you can draw them in two different ways as shown in Fig. 6.5 (i) and Fig. 6.5 (ii). and RS in Fig. 6.5 (i) are intersecting lines and in Fig. 6.5 (ii) are parallel lines. Note that the lengths of the common perpendiculars at different points on these parallel lines is the same. This equal length is called the distance between two parallel lines. 6.4 Pairs of Angles In Section 6.2, you have learnt the definitions of some of the pairs of angles such as complementary angles, supplementary angles, adjacent angles, linear pair of angles, etc. Can you think of some relations between these angles? Now, let us find out the relation between the angles formed when a ray stands on a line. Draw a figure in which a ray stands on a line as shown in Fig. 6.6. Name the line as AB and the ray as OC. What are the angles formed at the point O? They are ∠ AOC, ∠ BOC and ∠ AOB. Recall the notion of a line, that it extends indefinitely in both directions. Lines PQ (i) Intersecting lines (ii) Non-intersecting (parallel) lines Fig. 6.5 : Different ways of drawing two lines Can we write ∠ AOC + ∠ BOC = ∠ AOB? (1) Yes! (Why? Refer to adjacent angles in Section 6.2) What is the measure of ∠ AOB? It is 180°. (Why?) (2) From (1) and (2), can you say that ∠ AOC + ∠ BOC = 180°? Yes! (Why?) From the above discussion, we can state the following Axiom: Reprint 2025-26 Fig. 6.6 : Linear pair of angles LINES AND ANGLES 73 Axiom 6.1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. linear pair of angles. concluded that ‘the sum of two adjacent angles so formed is 180°’. Can we write Axiom 6.1 the other way? That is, take the ‘conclusion’ of Axiom 6.1 as ‘given’ and the ‘given’ as the ‘conclusion’. So it becomes: the non-common arms form a line). each others. We call each as converse of the other. We do not know whether the statement (A) is true or not. Let us check. Draw adjacent angles of different measures as shown in Fig. 6.7. Keep the ruler along one of the non-common arms in each case. Does the other non-common arm also lie along the ruler? Recall that when the sum of two adjacent angles is 180°, then they are called a In Axiom 6.1, it is given that ‘a ray stands on a line’. From this ‘given’, we have (A) If the sum of two adjacent angles is 180°, then a ray stands on a line (that is, Now you see that the Axiom 6.1 and statement (A) are in a sense the reverse of Fig. 6.7 : Adjacent angles with different measures Reprint 2025-26 74 MATHEMATICS ruler, that is, points A, O and B lie on the same line and ray OC stands on it. Also see that ∠ AOC + ∠ COB = 125° + 55° = 180°. From this, you may conclude that statement (A) is true. So, you can state in the form of an axiom as follows: Axiom 6.2 : If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line. Axiom. angles are equal. Let us prove this result now. See Appendix 1 for the ingredients of a proof, and keep those in mind while studying the proof given below. Theorem 6.1 : If two lines intersect each other, then the vertically opposite angles are equal. Proof : In the statement above, it is given that ‘two lines intersect each other’. So, let AB and CD be two lines intersecting at O as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely, (i) ∠ AOC and ∠ BOD (ii) ∠ AOD and ∠ BOC. We need to prove that ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC. You will find that only in Fig. 6.7 (iii), both the non-common arms lie along the For obvious reasons, the two axioms above together is called the Linear Pair Let us now examine the case when two lines intersect each other. Recall, from earlier classes, that when two lines intersect, the vertically opposite Fig. 6.8 : Vertically opposite angles Now, ray OA stands on line CD. Therefore, ∠ AOC + ∠ AOD = 180° (Linear pair axiom) (1) Can we write ∠ AOD + ∠ BOD = 180°? Yes! (Why?) (2) From (1) and (2), we can write ∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD This implies that ∠ AOC = ∠ BOD (Refer Section 5.2, Axiom 3) Similarly, it can be proved that ∠AOD = ∠BOC Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1. Reprint 2025-26 LINES AND ANGLES 75 Example 1 : In Fig. 6.9, lines PQ and RS intersect each other at point O. If ∠ POR : ∠ ROQ = 5 : 7, find all the angles. Solution : ∠ POR +∠ ROQ = 180° (Linear pair of angles) But ∠ POR : ∠ ROQ = 5 : 7 (Given) Therefore, ∠ POR = 5 12 × 180° = 75° Similarly, ∠ ROQ = 7 12 × 180° = 105° Now, ∠ POS = ∠ROQ = 105° (Vertically opposite angles) and ∠ SOQ = ∠POR = 75° (Vertically opposite angles) Example 2 : In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠ POS and ∠ SOQ, respectively. If ∠ POS = x, find ∠ ROT. Solution : Ray OS stands on the line POQ. Therefore, ∠ POS + ∠ SOQ = 180° But, ∠ POS = x Therefore, x + ∠ SOQ = 180° So, ∠ SOQ = 180° – x Now, ray OR bisects ∠ POS, therefore, Fig. 6.10 Fig. 6.9 Similarly, ∠ SOT = 1 2 × ∠ SOQ ∠ ROS = 1 2 × ∠ POS Reprint 2025-26 = 1 2 × x = 2 x = 1 2 × (180° – x) = 90 2 x ° − 76 MATHEMATICS Now, ∠ ROT = ∠ ROS + ∠ SOT Example 3 : In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°. Solution : In Fig. 6.11, you need to produce any of the rays OP, OQ, OR or OS backwards to a point. Let us produce ray OQ backwards to a point T so that TOQ is a line (see Fig. 6.12). Now, ray OP stands on line TOQ. Therefore, ∠ TOP + ∠ POQ = 180° (1) Similarly, ray OS stands on line TOQ. Therefore, ∠ TOS + ∠ SOQ = 180° (2) But ∠ SOQ = ∠ SOR + ∠ QOR So, (2) becomes ∠ TOS + ∠ SOR + ∠ QOR = 180° (3) Now, adding (1) and (3), you get ∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = 360° (4) (Linear pair axiom) = 90 – 2 2 x x + ° = 90° Fig. 6.11 Fig. 6.12 But ∠ TOP + ∠ TOS = ∠ POS Therefore, (4) becomes ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360° 1. In Fig. 6.13, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE. EXERCISE 6.1 Reprint 2025-26 Fig. 6.13 LINES AND ANGLES 77 2. In Fig. 6.14, lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c. 3. In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT. 4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line. Fig. 6.14 Fig. 6.15 5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that 6. It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP. ∠ ROS = 1 2 (∠ QOS – ∠ POS). Reprint 2025-26 Fig. 6.16 Fig. 6.17 78 MATHEMATICS 6.5 Lines Parallel to the Same Line If two lines are parallel to the same line, will they be parallel to each other? Let us check it. See Fig. 6.18 in which line m || line l and line n || line l. line m || line l and line n || line l. Therefore, ∠ 1 = ∠ 2 and ∠ 1 = ∠ 3 So, ∠ 2 = ∠ 3 (Why?) But ∠ 2 and ∠ 3 are corresponding angles and they are equal. Therefore, you can say that This result can be stated in the form of the following theorem: Theorem 6.6 : Lines which are parallel to the same line are parallel to each other. Note : The property above can be extended to more than two lines also. Now, let us solve some examples related to parallel lines. Example 4 : In Fig. 6.19, if PQ || RS, ∠ MXQ = 135° and ∠ MYR = 40°, find ∠ XMY. Let us draw a line t transversal for the lines, l, m and n. It is given that (Converse of corresponding angles axiom) (Corresponding angles axiom) Line m || Line n Fig. 6.18 Solution : Here, we need to draw a line AB parallel to line PQ, through point M as shown in Fig. 6.20. Now, AB || PQ and PQ || RS. Fig. 6.19 Fig. 6.20 Reprint 2025-26 LINES AND ANGLES 79 Therefore, AB || RS (Why?) Now, ∠ QXM + ∠ XMB = 180° But ∠ QXM = 135° So, 135° + ∠ XMB = 180° Therefore, ∠ XMB = 45° (1) Now, ∠ BMY = ∠ MYR (AB || RS, Alternate angles) Therefore, ∠ BMY = 40° (2) Adding (1) and (2), you get ∠ XMB + ∠ BMY = 45° + 40° That is, ∠ XMY = 85° Example 5 : If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel. Solution : In Fig. 6.21, a transversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of ∠ ABQ and ray CG is the bisector of ∠ BCS; and BE || CG. We are to prove that PQ || RS. It is given that ray BE is the bisector of ∠ ABQ. Therefore, ∠ ABE = 1 2 ∠ ABQ (1) (AB || PQ, Interior angles on the same side of the transversal XM) Similarly, ray CG is the bisector of ∠ BCS. Therefore, ∠ BCG = 1 2 ∠ BCS (2) But BE || CG and AD is the transversal. Therefore, ∠ ABE = ∠ BCG (Corresponding angles axiom) (3) Substituting (1) and (2) in (3), you get That is, ∠ ABQ = ∠ BCS 1 2 ∠ ABQ = 1 2 ∠ BCS Reprint 2025-26 Fig. 6.21 80 MATHEMATICS But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal. Therefore, PQ || RS Example 6 : In Fig. 6.22, AB || CD and CD || EF. Also EA ⊥ AB. If ∠ BEF = 55°, find the values of x, y and z. Solution : y + 55° = 180° Therefore, y = 180º – 55º = 125º Again x = y Therefore x = 125º Now, since AB || CD and CD || EF, therefore, AB || EF. So, ∠ EAB + ∠ FEA = 180° (Interior angles on the same Therefore, 90° + z + 55° = 180° Which gives z = 35° (Interior angles on the same side of the transversal ED) (AB || CD, Corresponding angles axiom) EXERCISE 6.2 (Converse of corresponding angles axiom) side of the transversal EA) Fig. 6.22 1. In Fig. 6.23, if AB || CD, CD || EF and y : z = 3 : 7, find x. Reprint 2025-26 Fig. 6.23 LINES AND ANGLES 81 2. In Fig. 6.24, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE. 3. In Fig. 6.25, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. 4. In Fig. 6.26, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y. [Hint : Draw a line parallel to ST through point R.] Fig. 6.24 Fig. 6.25 5. In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD. Reprint 2025-26 Fig. 6.26 Fig. 6.27 82 MATHEMATICS 6.6 Summary In this chapter, you have studied the following points: 1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and viceversa. This property is called as the Linear pair axiom. 2. If two lines intersect each other, then the vertically opposite angles are equal. 3. Lines which are parallel to a given line are parallel to each other. Reprint 2025-26" class_9,7,Triangles,ncert_books/class_9/iemh1dd/iemh107.pdf,"TRIANGLES 7.1 Introduction You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a triangle. (‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For example, in triangle ABC, denoted as ∆ ABC (see Fig. 7.1); AB, BC, CA are the three sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices. In Chapter 6, you have also studied some properties of triangles. In this chapter, you will study in details about the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle. You have already verified most of these properties in earlier classes. We will now prove some of them. CHAPTER 7 7.2 Congruence of Triangles You must have observed that two copies of your photographs of the same size are identical. Similarly, two bangles of the same size, two ATM cards issued by the same bank are identical. You may recall that on placing a one rupee coin on another minted in the same year, they cover each other completely. figures (‘congruent’ means equal in all respects or figures whose shapes and sizes are both the same). you observe? They cover each other completely and we call them as congruent circles. Do you remember what such figures are called? Indeed they are called congruent Now, draw two circles of the same radius and place one on the other. What do Reprint 2025-26 Fig. 7.1 84 MATHEMATICS Repeat this activity by placing one square on the other with sides of the same measure (see Fig. 7.2) or by placing two equilateral triangles of equal sides on each other. You will observe that the squares are congruent to each other and so are the equilateral triangles. tray in your refrigerator. Observe that the moulds for making ice are all congruent. The cast used for moulding in the tray also has congruent depressions (may be all are rectangular or all circular or all triangular). So, whenever identical objects have to be produced, the concept of congruence is used in making the cast. and this is so when the new refill is not of the same size as the one you want to remove. Obviously, if the two refills are identical or congruent, the new refill fits. daily life situations. Fig 7.3 (i) : You may wonder why we are studying congruence. You all must have seen the ice Sometimes, you may find it difficult to replace the refill in your pen by a new one So, you can find numerous examples where congruence of objects is applied in Can you think of some more examples of congruent figures? Now, which of the following figures are not congruent to the square in Fig. 7.2 Fig 7.3 (i), but the square in Fig 7.3 (iv) is congruent to the one given in Fig 7.3 (i). triangle are equal to the corresponding sides and angles of the other triangle. The large squares in Fig. 7.3 (ii) and (iii) are obviously not congruent to the one in Let us now discuss the congruence of two triangles. You already know that two triangles are congruent if the sides and angles of one Reprint 2025-26 Fig. 7.3 TRIANGLES 85 Fig. 7.4 (i)? try to cover ∆ ABC. Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent to ∆ ABC while ∆ TSU of Fig 7.4 (v) is not congruent to ∆ ABC. Now, which of the triangles given below are congruent to triangle ABC in Cut out each of these triangles from Fig. 7.4 (ii) to (v) and turn them around and Fig. 7.4 equal sides of ∆ ABC and so is the case for the angles. That is, PQ covers AB, QR covers BC and RP covers CA; ∠ P covers ∠ A, ∠ Q covers ∠ B and ∠ R covers ∠ C. Also, there is a one-one correspondence between the vertices. That is, P corresponds to A, Q to B, R to C and so on which is written as write ∆QRP ≅ ∆ ABC. If ∆ PQR is congruent to ∆ ABC, we write ∆ PQR ≅ ∆ ABC. Notice that when ∆ PQR ≅ ∆ ABC, then sides of ∆ PQR fall on corresponding Note that under this correspondence, ∆ PQR ≅ ∆ ABC; but it will not be correct to Similarly, for Fig. 7.4 (iii), P ↔ A, Q ↔ B, R ↔ C Reprint 2025-26 86 MATHEMATICS congruence of triangles in symbolic form. in short ‘CPCT’ for corresponding parts of congruent triangles. 7.3 Criteria for Congruence of Triangles In earlier classes, you have learnt four criteria for congruence of triangles. Let us recall them. that they are not congruent (see Fig. 7.5). So, ∆ FDE ≅ ∆ ABC but writing ∆ DEF ≅ ∆ ABC is not correct. Give the correspondence between the triangle in Fig. 7.4 (iv) and ∆ ABC. So, it is necessary to write the correspondence of vertices correctly for writing of Note that in congruent triangles corresponding parts are equal and we write Draw two triangles with one side 3 cm. Are these triangles congruent? Observe FD ↔ AB, DE ↔ BC and EF ↔ CA and F ↔ A, D ↔ B and E ↔ C they congruent? Now, draw two triangles with one side 4 cm and one angle 50° (see Fig. 7.6). Are Reprint 2025-26 Fig. 7.5 Fig. 7.6 TRIANGLES 87 sufficient to give us congruent triangles. equal? In Fig 7.7, BC = QR, ∠ B = ∠ Q and also, AB = PQ. Now, what can you say about congruence of ∆ ABC and ∆ PQR? Verify this for ∆ ABC and ∆ PQR in Fig. 7.7. of two sides and the included angle is enough for the congruence of triangles? Yes, it is enough. Axiom 7.1 (SAS congruence rule) : Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle. See that these two triangles are not congruent. Repeat this activity with some more pairs of triangles. So, equality of one pair of sides or one pair of sides and one pair of angles is not What would happen if the other pair of arms (sides) of the equal angles are also Recall from your earlier classes that, in this case, the two triangles are congruent. Repeat this activity with other pairs of triangles. Do you observe that the equality This is the first criterion for congruence of triangles. Fig. 7.7 accepted true as an axiom (see Appendix 1). Example 1 : In Fig. 7.8, OA = OB and OD = OC. Show that Solution : (i) You may observe that in ∆ AOD and ∆ BOC, This result cannot be proved with the help of previously known results and so it is Let us now take some examples. (i) ∆ AOD ≅ ∆ BOC and (ii) AD || BC. OA = OB (Given) OD = OC Fig. 7.8 Reprint 2025-26 88 MATHEMATICS Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite angles, we have ∠ AOD = ∠ BOC. So, ∆ AOD ≅ ∆ BOC (by the SAS congruence rule) (ii) In congruent triangles AOD and BOC, the other corresponding parts are also equal. So, ∠ OAD = ∠ OBC and these form a pair of alternate angles for line segments AD and BC. Example 2 : AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B. Solution : Line l ⊥ AB and passes through C which is the mid-point of AB (see Fig. 7.9). You have to show that PA = PB. Consider ∆ PCA and ∆ PCB. We have AC = BC (C is the mid-point of AB) ∠ PCA = ∠ PCB = 90° (Given) So, ∆ PCA ≅ ∆ PCB (SAS rule) and so, PA = PB, as they are corresponding sides of congruent triangles. Therefore, AD || BC. Now, let us construct two triangles, whose sides are 4 cm and 5 cm and one of the PC = PC (Common) Fig. 7.9 angles is 50° and this angle is not included in between the equal sides (see Fig. 7.10). Are the two triangles congruent? Reprint 2025-26 Fig. 7.10 TRIANGLES 89 to be congruent, it is very important that the equal angles are included between the pairs of equal sides. the side included between these angles is 4 cm (see Fig. 7.11). See that one triangle covers the other completely; that is, the two triangles are congruent. Repeat this activity with more pairs of triangles. You will observe that equality of two angles and the included side is sufficient for congruence of triangles. ASA criterion. You have verified this criterion in earlier classes, but let us state and prove this result. Notice that the two triangles are not congruent. Repeat this activity with more pairs of triangles. You will observe that for triangles So, SAS congruence rule holds but not ASS or SSA rule. Next, try to construct the two triangles in which two angles are 60° and 45° and Cut out these triangles and place one triangle on the other. What do you observe? This result is the Angle-Side-Angle criterion for congruence and is written as Fig. 7.11 SAS axiom for congruence. Theorem 7.1 (ASA congruence rule) : Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. Proof : We are given two triangles ABC and DEF in which: and BC = EF We need to prove that ∆ ABC ≅ ∆ DEF For proving the congruence of the two triangles see that three cases arise. Since this result can be proved, it is called a theorem and to prove it, we use the ∠ B = ∠ E, ∠ C = ∠ F Reprint 2025-26 90 MATHEMATICS Case (i) : Let AB = DE (see Fig. 7.12). Now what do you observe? You may observe that So, ∆ ABC ≅ ∆ DEF (By SAS rule) Case (ii) : Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider ∆ PBC and ∆ DEF (see Fig. 7.13). AB = DE (Assumed) ∠ B = ∠ E (Given) BC = EF (Given) Fig. 7.12 Observe that in ∆ PBC and ∆ DEF, So, we can conclude that: ∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence. PB = DE (By construction) ∠ B = ∠ E (Given) Reprint 2025-26 BC = EF (Given) Fig. 7.13 TRIANGLES 91 Since the triangles are congruent, their corresponding parts will be equal. So, ∠ PCB = ∠ DFE But, we are given that ∠ ACB = ∠ DFE So, ∠ ACB = ∠ PCB Is this possible? This is possible only if P coincides with A. or, BA = ED So, ∆ ABC ≅ ∆ DEF (by SAS axiom) Case (iii) : If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (ii), we can conclude that AB = DE and so, ∆ ABC ≅ ∆ DEF. sides are equal but the side is not included between the corresponding equal pairs of angles. Are the triangles still congruent? You will observe that they are congruent. Can you reason out why? angles are equal, the third pair is also equal (180° – sum of equal angles). corresponding sides are equal. We may call it as the AAS Congruence Rule. Suppose, now in two triangles two pairs of angles and one pair of corresponding You know that the sum of the three angles of a triangle is 180°. So if two pairs of So, two triangles are congruent if any two pairs of angles and one pair of Now let us perform the following activity : draw? sides (see Fig. 7.14). Draw triangles with angles 40°, 50° and 90°. How many such triangles can you In fact, you can draw as many triangles as you want with different lengths of Reprint 2025-26 Fig. 7.14 92 MATHEMATICS for congruence of triangles out of three equal parts, one has to be a side. Example 3 : Line-segment AB is parallel to another line-segment CD. O is the mid-point of AD (see Fig. 7.15). Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the mid-point of BC. Solution : (i) Consider ∆ AOB and ∆ DOC. Therefore, ∆AOB ≅ ∆DOC (AAS rule) (ii) OB = OC (CPCT) So, O is the mid-point of BC. Observe that the triangles may or may not be congruent to each other. So, equality of three angles is not sufficient for congruence of triangles. Therefore, Let us now take some more examples. ∠ ABO = ∠ DCO and BC is the transversal) ∠ AOB = ∠ DOC (Alternate angles as AB || CD OA = OD (Given) (Vertically opposite angles) EXERCISE 7.1 Fig. 7.15 1. In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD? Reprint 2025-26 Fig. 7.16 TRIANGLES 93 2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that 3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB. 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ∆ ABC ≅ ∆ CDA. (i) ∆ ABD ≅ ∆ BAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC. Fig. 7.17 Fig. 7.18 5. Line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20). Show that: (i) ∆ APB ≅ ∆ AQB (ii) BP = BQ or B is equidistant from the arms of ∠ A. Reprint 2025-26 Fig. 7.19 Fig. 7.20 94 MATHEMATICS 6. In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE. 7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig. 7.22). Show that 8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that: (i) ∆ DAP ≅ ∆ EBP (ii) AD = BE (i) ∆ AMC ≅ ∆ BMD Fig. 7.22 Fig. 7.21 7.4 Some Properties of a Triangle In the above section you have studied two criteria for congruence of triangles. Let us now apply these results to study some properties related to a triangle whose two sides are equal. (ii) ∠ DBC is a right angle. (iii) ∆ DBC ≅ ∆ ACB (iv) CM = 1 2 AB Reprint 2025-26 Fig. 7.23 TRIANGLES 95 Construct a triangle in which two sides are equal, say each equal to 3.5 cm and the third side equal to 5 cm (see Fig. 7.24). You have done such constructions in earlier classes. Do you remember what is such a triangle called? A triangle in which two sides are equal is called an isosceles triangle. So, ∆ ABC of Fig. 7.24 is an isosceles triangle with AB = AC. are equal. be proved as shown below. Theorem 7.2 : Angles opposite to equal sides of an isosceles triangle are equal. This result can be proved in many ways. One of the proofs is given here. Proof : We are given an isosceles triangle ABC in which AB = AC. We need to prove that ∠ B = ∠ C. Perform the activity given below: Now, measure ∠ B and ∠ C. What do you observe? Repeat this activity with other isosceles triangles with different sides. You may observe that in each such triangle, the angles opposite to the equal sides This is a very important result and is indeed true for any isosceles triangle. It can Fig. 7.24 Let us draw the bisector of ∠ A and let D be the point of intersection of this bisector of ∠ A and BC (see Fig. 7.25). In ∆ BAD and ∆ CAD, So, ∆ BAD ≅ ∆ CAD (By SAS rule) So, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles. So, ∠ B = ∠ C AB = AC (Given) ∠ BAD = ∠ CAD (By construction) AD = AD (Common) Reprint 2025-26 Fig. 7.25 96 MATHEMATICS them are also equal? bisector of ∠ A and let it intersect BC at D (see Fig. 7.26). falls on vertex B. Repeat this activity with some more triangles. Each time you will observe that the sides opposite to equal angles are equal. So we have the following: Theorem 7.3 : The sides opposite to equal angles of a triangle are equal. Example 4 : In ∆ ABC, the bisector AD of ∠ A is perpendicular to side BC (see Fig. 7.27). Show that AB = AC and ∆ ABC is isosceles. Is the converse also true? That is: If two angles of any triangle are equal, can we conclude that the sides opposite to Perform the following activity. Construct a triangle ABC with BC of any length and ∠ B = ∠ C = 50°. Draw the Cut out the triangle from the sheet of paper and fold it along AD so that vertex C What can you say about sides AC and AB? Observe that AC covers AB completely So, AC = AB This is the converse of Theorem 7.2. You can prove this theorem by ASA congruence rule. Let us take some examples to apply these results. Fig. 7.26 Solution : In ∆ABD and ∆ACD, So, ∆ ABD ≅ ∆ ACD (ASA rule) So, AB = AC (CPCT) or, ∆ ABC is an isosceles triangle. ∠ BAD = ∠ CAD (Given) AD = AD (Common) ∠ ADB = ∠ ADC = 90° (Given) Reprint 2025-26 Fig. 7.27 TRIANGLES 97 Example 5 : E and F are respectively the mid-points of equal sides AB and AC of ∆ ABC (see Fig. 7.28). Show that BF = CE. Solution : In ∆ ABF and ∆ ACE, So, ∆ ABF ≅ ∆ ACE (SAS rule) Therefore, BF = CE (CPCT) Example 6 : In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see Fig. 7.29). Show that AD = AE. Solution : In ∆ ABD and ∆ ACE, Also, BE = CD So, BE – DE = CD – DE That is, BD = CE (3) So, ∆ ABD ≅ ∆ ACE AB = AC (Given) ∠ A = ∠ A (Common) AB = AC (Given) (1) ∠ B = ∠ C AF = AE (Halves of equal sides) (Angles opposite to equal sides) (2) Fig. 7.28 Fig. 7.29 This gives AD = AE (CPCT) 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that : 2. In ∆ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ∆ ABC is an isosceles triangle in which AB = AC. (i) OB = OC (ii) AO bisects ∠ A (Using (1), (2), (3) and SAS rule). EXERCISE 7.2 Reprint 2025-26 Fig. 7.30 98 MATHEMATICS 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal. 4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that 5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD = ∠ ACD. (i) ∆ ABE ≅ ∆ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle. Fig. 7.32 Fig. 7.31 6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle. 7. ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C. 8. Show that the angles of an equilateral triangle are 60° each. Reprint 2025-26 Fig. 7.34 Fig. 7.33 TRIANGLES 99 7.5 Some More Criteria for Congruence of Triangles You have seen earlier in this chapter that equality of three angles of one triangle to three angles of the other is not sufficient for the congruence of the two triangles. You may wonder whether equality of three sides of one triangle to three sides of another triangle is enough for congruence of the two triangles. You have already verified in earlier classes that this is indeed true. (see Fig. 7.35). Cut them out and place them on each other. What do you observe? They cover each other completely, if the equal sides are placed on each other. So, the triangles are congruent. congruence. To be sure, construct two triangles with sides 4 cm, 3.5 cm and 4.5 cm Repeat this activity with some more triangles. We arrive at another rule for Fig. 7.35 Theorem 7.4 (SSS congruence rule) : If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent. has to be the included angle between the pairs of corresponding pair of equal sides and if this is not so, the two triangles may not be congruent. equal to 4 cm each (see Fig. 7.36). This theorem can be proved using a suitable construction. You have already seen that in the SAS congruence rule, the pair of equal angles Perform this activity: Construct two right angled triangles with hypotenuse equal to 5 cm and one side Reprint 2025-26 100 MATHEMATICS other. Turn the triangles, if necessary. What do you observe? this activity with other pairs of right triangles. What do you observe? hypotenuse are equal. You have verified this in earlier classes. Theorem 7.5 (RHS congruence rule) : If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent. Cut them out and place one triangle over the other with equal side placed on each The two triangles cover each other completely and so they are congruent. Repeat You will find that two right triangles are congruent if one pair of sides and the Note that, the right angle is not the included angle in this case. So, you arrive at the following congruence rule: Note that RHS stands for Right angle - Hypotenuse - Side. Let us now take some examples. Fig. 7.36 Example 7 : AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular bisector of AB. Solution : You are given that PA = PB and QA = QB and you are to show that PQ ⊥ AB and PQ bisects AB. Let PQ intersect AB at C. Can you think of two congruent triangles in this figure? Let us take ∆ PAQ and ∆ PBQ. In these triangles, Reprint 2025-26 Fig. 7.37 TRIANGLES 101 So, ∆ PAQ ≅ ∆ PBQ (SSS rule) Therefore, ∠ APQ = ∠ BPQ (CPCT). Now let us consider ∆ PAC and ∆ PBC. You have : AP = BP (Given) ∠ APC = ∠ BPC (∠ APQ = ∠ BPQ proved above) So, ∆ PAC ≅ ∆ PBC (SAS rule) Therefore, AC = BC (CPCT) (1) and ∠ ACP = ∠ BCP (CPCT) Also, ∠ ACP + ∠ BCP = 180° (Linear pair) So, 2∠ ACP = 180° or, ∠ ACP = 90° (2) From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB. [Note that, without showing the congruence of ∆ PAQ and ∆ PBQ, you cannot show that ∆ PAC ≅ ∆ PBC even though AP = BP (Given) AQ = BQ (Given) AP = BP (Given) PQ = PQ (Common) PC = PC (Common) PC = PC (Common) and ∠ PAC = ∠ PBC (Angles opposite to equal sides in ∆APB) It is because these results give us SSA rule which is not always valid or true for congruence of triangles. Also the angle is not included between the equal pairs of sides.] Let us take some more examples. Example 8 : P is a point equidistant from two lines l and m intersecting at point A (see Fig. 7.38). Show that the line AP bisects the angle between them. Solution : You are given that lines l and m intersect each other at A. Let PB ⊥ l, PC ⊥ m. It is given that PB = PC. You are to show that ∠ PAB = ∠ PAC. Reprint 2025-26 102 MATHEMATICS Let us consider ∆ PAB and ∆ PAC. In these two triangles, So, ∆ PAB ≅ ∆ PAC (RHS rule) So, ∠ PAB = ∠ PAC (CPCT) Note that this result is the converse of the result proved in Q.5 of Exercise 7.1. 1. ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see Fig. 7.40). Show that: (i) ∆ ABD ≅ ∆ ACD (ii) ∆ ABP≅ ∆ ACP (iii) AP bisects ∠ A as well as ∠ D. (iv) AP is the perpendicular bisector of BC. (i) AD bisects BC (ii) AD bisects ∠ A. PB = PC (Given) ∠ PBA = ∠ PCA = 90° (Given) PA = PA (Common) EXERCISE 7.3 Fig. 7.38 Fig. 7.39 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C. (i) ∆ ABM ≅ ∆ PQN (ii) ∆ ABC ≅ ∆ PQR Reprint 2025-26 Fig. 7.40 TRIANGLES 103 7.6 Summary In this chapter, you have studied the following points : 1. Two figures are congruent, if they are of the same shape and of the same size. 2. Two circles of the same radii are congruent. 3. Two squares of the same sides are congruent. 4. If two triangles ABC and PQR are congruent under the correspondence A ↔ P, B ↔ Q and C ↔ R, then symbolically, it is expressed as ∆ ABC ≅ ∆ PQR. 5. If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent (SAS Congruence Rule). 6. If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA Congruence Rule). 7. If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent (AAS Congruence Rule). 8. Angles opposite to equal sides of a triangle are equal. 9. Sides opposite to equal angles of a triangle are equal. 10. Each angle of an equilateral triangle is of 60°. 11. If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent (SSS Congruence Rule). 12. If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS Congruence Rule). Reprint 2025-26" class_9,8,Quadrilaterals,ncert_books/class_9/iemh1dd/iemh108.pdf,"104 MATHEMATICS QUADRILATERALS 8.1 Properties of a Parallelogram You have already studied quadrilaterals and their types in Class VIII. A quadrilateral has four sides, four angles and four vertices. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Let us perform an activity. Cut out a parallelogram from a sheet of paper and cut it along a diagonal (see Fig. 8.1). You obtain two triangles. What can you say about these triangles? Place one triangle over the other. Turn one around, if necessary. What do you observe? CHAPTER 8 Observe that the two triangles are congruent to each other. that each diagonal divides the parallelogram into two congruent triangles. Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent triangles. Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.2). Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC and ∆ CDA. We need to prove that these triangles are congruent. Fig. 8.1 Repeat this activity with some more parallelograms. Each time you will observe Let us now prove this result. Reprint 2025-26 QUADRILATERALS 105 So, ∠ BCA = ∠ DAC (Pair of alternate angles) Also, AB || DC and AC is a transversal. So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) So, ∆ ABC ≅ ∆ CDA (ASA rule) or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA. Now, measure the opposite sides of parallelogram ABCD. What do you observe? Theorem 8.2 : In a parallelogram, opposite sides are equal. triangles; so what can you say about the corresponding parts say, the corresponding sides? They are equal. So, AB = DC and AD = BC in a theorem, the same is to be proved in the converse and whatever is proved in the theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below : In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal. You will find that AB = DC and AD = BC. This is another property of a parallelogram stated below: You have already proved that a diagonal divides the parallelogram into two congruent Now what is the converse of this result? You already know that whatever is given If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So Fig. 8.2 its converse is : Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Let sides AB and CD of the quadrilateral ABCD be equal and also AD = BC (see Fig. 8.3). Draw diagonal AC. Can you reason out why? Clearly, ∆ ABC ≅ ∆ CDA (Why?) So, ∠ BAC = ∠ DCA and ∠ BCA = ∠ DAC (Why?) Can you now say that ABCD is a parallelogram? Why? Reprint 2025-26 Fig. 8.3 106 MATHEMATICS conversely if each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Can we conclude the same result for the pairs of opposite angles? given below. Theorem 8.4 : In a parallelogram, opposite angles are equal. a quadrilateral and the results of parallel lines intersected by a transversal, we can see that the converse is also true. So, we have the following theorem : Theorem 8.5 : If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. parallelogram ABCD and draw both its diagonals intersecting at the point O (see Fig. 8.4). Measure the lengths of OA, OB, OC and OD. What do you observe? You will observe that or, O is the mid-point of both the diagonals. Repeat this activity with some more parallelograms. You have just seen that in a parallelogram each pair of opposite sides is equal and Draw a parallelogram and measure its angles. What do you observe? Each pair of opposite angles is equal. Repeat this with some more parallelograms. We arrive at yet another result as Now, is the converse of this result also true? Yes. Using the angle sum property of There is yet another property of a parallelogram. Let us study the same. Draw a OA = OC and OB = OD. Each time you will find that O is the mid-point of both the diagonals. So, we have the following theorem : Theorem 8.6 : The diagonals of a parallelogram bisect each other. Will it be aparallelogram? Indeed this is true. Theorem 8.7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Now, what would happen, if in a quadrilateral the diagonals bisect each other? This result is the converse of the result of Theorem 8.6. It is given below: You can reason out this result as follows: Reprint 2025-26 Fig. 8.4 QUADRILATERALS 107 Note that in Fig. 8.5, it is given that OA = OC and OB = OD. So, ∆ AOB ≅ ∆ COD (Why?) Therefore, ∠ ABO = ∠ CDO (Why?) From this, we get AB || CD Similarly, BC || AD Therefore ABCD is a parallelogram. Let us now take some examples. Example 1 : Show that each angle of a rectangle is a right angle. Solution : Let us recall what a rectangle is. A rectangle is a parallelogram in which one angle is a right angle. Let ABCD be a rectangle in which ∠ A = 90°. We have to show that ∠ B = ∠ C = ∠ D = 90° We have, AD || BC and AB is a transversal (see Fig. 8.6). So, ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal) But, ∠ A = 90° So, ∠ B = 180° – ∠ A = 180° – 90° = 90° Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram) Fig. 8.5 Fig. 8.6 So, ∠ C = 90° and ∠ D = 90°. Therefore, each of the angles of a rectangle is a right angle. Example 2 : Show that the diagonals of a rhombus are perpendicular to each other. Solution : Consider the rhombus ABCD (see Fig. 8.7). You know that AB = BC = CD = DA (Why?) Now, in ∆ AOD and ∆ COD, OA = OC (Diagonals of a parallelogram bisect each other) OD = OD (Common) AD = CD Reprint 2025-26 Fig. 8.7 108 MATHEMATICS Therefore, ∆ AOD ≅ ∆ COD (SSS congruence rule) This gives, ∠ AOD = ∠ COD (CPCT) But, ∠ AOD + ∠ COD = 180° (Linear pair) So, 2∠ AOD = 180° or, ∠ AOD = 90° So, the diagonals of a rhombus are perpendicular to each other. Example 3 : ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB (see Fig. 8.8). Show that Solution : (i) ∆ ABC is isosceles in which AB = AC (Given) So, ∠ ABC = ∠ ACB (Angles opposite to equal sides) Also, ∠ PAC = ∠ ABC + ∠ ACB (Exterior angle of a triangle) or, ∠ PAC = 2∠ ACB (1) Now, AD bisects ∠ PAC. So, ∠ PAC = 2∠ DAC (2) Therefore, (i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram. 2∠ DAC = 2∠ ACB [From (1) and (2)] Fig. 8.8 or, ∠ DAC = ∠ ACB (ii) Now, these equal angles form a pair of alternate angles when line segments BC and AD are intersected by a transversal AC. So, BC || AD Also, BA || CD (Given) Now, both pairs of opposite sides of quadrilateral ABCD are parallel. So, ABCD is a parallelogram. Example 4 : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.9). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. Reprint 2025-26 QUADRILATERALS 109 Solution : It is given that PS || QR and transversal p intersects them at points A and C respectively. The bisectors of ∠ PAC and ∠ ACQ intersect at B and bisectors of ∠ ACR and ∠ SAC intersect at D. We are to show that quadrilateral ABCD is a rectangle. Now, ∠ PAC = ∠ ACR (Alternate angles as l || m and p is a transversal) So, 1 2 ∠ PAC = 1 2 ∠ ACR i.e., ∠ BAC = ∠ ACD These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also. So, AB || DC Similarly, BC || AD (Considering ∠ ACB and ∠ CAD) Therefore, quadrilateral ABCD is a parallelogram. Also, ∠ PAC + ∠ CAS = 180° (Linear pair) So, 1 2 ∠ PAC + 1 2 ∠ CAS = 1 2 × 180° = 90° or, ∠ BAC + ∠ CAD = 90° Fig. 8.9 or, ∠ BAD = 90° So, ABCD is a parallelogram in which one angle is 90°. Therefore, ABCD is a rectangle. Example 5 : Show that the bisectors of angles of a parallelogram form a rectangle. Solution : Let P, Q, R and S be the points of intersection of the bisectors of ∠ A and ∠ B, ∠ B and ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectively of parallelogram ABCD (see Fig. 8.10). In ∆ ASD, what do you observe? Since DS bisects ∠ D and AS bisects ∠ A, therefore, Reprint 2025-26 Fig. 8.10 110 MATHEMATICS Also, ∠ DAS + ∠ ADS + ∠ DSA = 180° (Angle sum property of a triangle) or, 90° + ∠ DSA = 180° or, ∠ DSA = 90° So, ∠ PSR = 90° (Being vertically opposite to ∠ DSA) Similarly, it can be shown that ∠ APB = 90° or ∠ SPQ = 90° (as it was shown for ∠ DSA). Similarly, ∠ PQR = 90° and ∠ SRQ = 90°. So, PQRS is a quadrilateral in which all angles are right angles. Can we conclude that it is a rectangle? Let us examine. We have shown that ∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°. So both pairs of opposite angles are equal. Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and so, PQRS is a rectangle. ∠ DAS + ∠ ADS = 1 2 ∠ A + 1 2 ∠ D = 1 2 (∠ A + ∠ D) = 1 2 × 180° (∠ A and ∠ D are interior angles on the same side of the transversal) = 90° 1. If the diagonals of a parallelogram are equal, then show that it is a rectangle. 2. Show that the diagonals of a square are equal and bisect each other at right angles. 3. Diagonal AC of a parallelogram ABCD bisects ∠ A (see Fig. 8.11). Show that 4. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that: (i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D. Fig. 8.11 (i) it bisects ∠ C also, (ii) ABCD is a rhombus. EXERCISE 8.1 Reprint 2025-26 QUADRILATERALS 111 5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that: 6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.13). Show that 7. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that (i) ∆ APD ≅ ∆ CQB (ii) AP = CQ (iii) ∆ AQB ≅∆ CPD (iv) AQ = CP (v) APCQ is a parallelogram (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ (i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) ∆ ABC ≅ ∆ BAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] Fig. 8.12 Fig. 8.14 Fig. 8.13 8.2The Mid-point Theorem You have studied many properties of a triangle as well as a quadrilateral. Now let us study yet another result which is related to the mid-point of sides of a triangle. Perform the following activity. the points E and F (see Fig. 8.15). Measure EF and BC. Measure ∠ AEF and ∠ ABC. Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join What do you observe? You will find that : so, EF || BC Repeat this activity with some more triangles. So, you arrive at the following theorem: EF = 1 2 BC and ∠ AEF = ∠ ABC Reprint 2025-26 Fig. 8.15 112 MATHEMATICS Theorem 8.8 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side. You can prove this theorem using the following clue: Observe Fig 8.16 in which E and F are mid-points of AB and AC respectively and CD || BA. ∆ AEF ≅ ∆ CDF (ASA Rule) Can you state the converse of Theorem 8.8? Is the converse true? You will see that converse of the above theorem is also true which is stated as below: Theorem 8.9 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. In Fig 8.17, observe that E is the mid-point of AB, line l is passsing through E and is parallel to BC and CM || BA. Prove that AF = CF by using the congruence of ∆ AEF and ∆ CDF. So, EF = DF and BE = AE = DC (Why?) Therefore, BCDE is a parallelogram. (Why?) This gives EF || BC. In this case, also note that EF = 1 2 ED = 1 2 BC. Fig. 8.16 Example 6 : In ∆ ABC, D, E and F are respectively the mid-points of sides AB, BC and CA (see Fig. 8.18). Show that ∆ ABC is divided into four congruent triangles by joining D, E and F. Solution : As D and E are mid-points of sides AB and BC of the triangle ABC, by Theorem 8.8, Similarly, DF || BC and EF || AB DE || AC Reprint 2025-26 Fig. 8.18 Fig. 8.17 QUADRILATERALS 113 Therefore ADEF, BDFE and DFCE are all parallelograms. Now DE is a diagonal of the parallelogram BDFE, therefore, ∆ BDE ≅ ∆ FED Similarly ∆ DAF ≅ ∆ FED and ∆ EFC ≅ ∆ FED So, all the four triangles are congruent. Example 7 : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see Fig. 8.19). Show that l, m and n cut off equal intercepts DE and EF on q also. Solution : We are given that AB = BC and have to prove that DE = EF. Let us join A to F intersecting m at G.. The trapezium ACFD is divided into two triangles; namely ∆ ACF and ∆ AFD. In ∆ ACF, it is given that B is the mid-point of AC (AB = BC) and BG || CF (since m || n). So, G is the mid-point of AF (by using Theorem 8.9) Now, in ∆ AFD, we can apply the same argument as G is the mid-point of AF, GE || AD and so by Theorem 8.9, E is the mid-point of DF, Fig. 8.19 i.e., DE = EF. In other words, l, m and n cut off equal intercepts on q also. 1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.20). AC is a diagonal. Show that : (i) SR || AC and SR = 1 2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. EXERCISE 8.2 Reprint 2025-26 Fig. 8.20 114 MATHEMATICS 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC. 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD. Fig. 8.21 Fig. 8.22 6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = 1 2 AB Reprint 2025-26 QUADRILATERALS 115 8.3 Summary In this chapter, you have studied the following points : 1. A diagonal of a parallelogram divides it into two congruent triangles. 2. In a parallelogram, 3. Diagonals of a rectangle bisect each other and are equal and vice-versa. 4. Diagonals of a rhombus bisect each other at right angles and vice-versa. 5. Diagonals of a square bisect each other at right angles and are equal, and vice-versa. 6. The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it. 7. A line through the mid-point of a side of a triangle parallel to another side bisects the third side. (i) opposite sides are equal (ii) opposite angles are equal (iii) diagonals bisect each other Reprint 2025-26" class_9,9,Circles,ncert_books/class_9/iemh1dd/iemh109.pdf,"116 MATHEMATICS CIRCLES 9.1 Angle Subtended by a Chord at a Point You have already studied about circles and its parts in Class VI. Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR (see Fig. 9.1). Then ∠ PRQ is called the angle subtended by the line segment PQ at the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ∠ POQ is the angle subtended by the chord PQ at the centre O, ∠ PRQ and ∠ PSQ are respectively the angles subtended by PQ at points R and S on the major and minor arcs PQ. CHAPTER 9 subtended by it at the centre. You may see by drawing different chords of a circle and angles subtended by them at the centre that the longer is the chord, the bigger will be the angle subtended by it at the centre. What will happen if you take two equal chords of a circle? Will the angles subtended at the centre be the same or not? Let us examine the relationship between the size of the chord and the angle Fig. 9.1 Fig. 9.2 Reprint 2025-26 CIRCLES 117 Draw two or more equal chords of a circle and measure the angles subtended by them at the centre (see Fig.9.3). You will find that the angles subtended by them at the centre are equal. Let us give a proof of this fact. Theorem 9.1 : Equal chords of a circle subtend equal angles at the centre. Proof : You are given two equal chords AB and CD of a circle with centre O (see Fig.9.4). You want to prove that ∠ AOB = ∠ COD. In triangles AOB and COD, Therefore, ∆ AOB ≅ ∆ COD (SSS rule) This gives ∠ AOB = ∠ COD (Corresponding parts of congruent triangles) Remark : For convenience, the abbreviation CPCT will be used in place of ‘Corresponding parts of congruent triangles’, because we use this very frequently as you will see. say about the chords? Are they equal or not? Let us examine this by the following activity: Now if two chords of a circle subtend equal angles at the centre, what can you OA = OC (Radii of a circle) OB = OD (Radii of a circle) AB = CD (Given) Fig. 9.3 Fig. 9.4 Take a tracing paper and trace a circle on it. Cut it along the circle to get a disc. At its centre O, draw an angle AOB where A, B are points on the circle. Make another angle POQ at the centre equal to ∠AOB. Cut the disc along AB and PQ (see Fig. 9.5). You will get two segments ACB and PRQ of the circle. If you put one on the other, what do you observe? They cover each other, i.e., they are congruent. So AB = PQ. Reprint 2025-26 Fig. 9.5 118 MATHEMATICS too. The chords will all turn out to be equal because of the following theorem: Theorem 9.2 : If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. you take ∠ AOB = ∠ COD, then ∆ AOB ≅ ∆ COD (Why?) 9.2 Perpendicular from the Centre to a Chord Activity : Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. Does the point B coincide with A (see Fig.9.6)? Though you have seen it for this particular case, try it out for other equal angles The above theorem is the converse of the Theorem 9.1. Note that in Fig. 9.4, if Can you now see that AB = CD? 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. EXERCISE 9.1 Fig. 9.6 Yes it will. So MA = MB. and OMB to be congruent. This example is a particular instance of the following result: Theorem 9.3 : The perpendicular from the centre of a circle to a chord bisects the chord. assumed in Theorem 9.3 and what is proved. Given that the perpendicular from the centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the converse, what the hypothesis is ‘if a line from the centre bisects a chord of a circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the converse is: Give a proof yourself by joining OA and OB and proving the right triangles OMA What is the converse of this theorem? To write this, first let us be clear what is Reprint 2025-26 CIRCLES 119 Theorem 9.4 : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Is this true? Try it for few cases and see. You will see that it is true for these cases. See if it is true, in general, by doing the following exercise. We will write the stages and you give the reasons. Let AB be a chord of a circle with centre O and O is joined to the mid-point M of AB. You have to prove that OM ⊥ AB. Join OA and OB (see Fig. 9.7). In triangles OAM and OBM, 9.3 Equal Chords and their Distances from the Centre Let AB be a line and P be a point. Since there are infinite numbers of points on a line, if you join these points to P, you will get infinitely many line segments PL1 , PL2 , PM, PL3 , PL4 , etc. Which of these is the distance of AB from P? You may think a while and get the answer. Out of these line segments, the perpendicular from P to AB, namely PM in Fig. 9.8, will be the least. In Mathematics, we define this least length PM to be the distance of AB from P. So you may say that: Therefore, ∆OAM ≅ ∆OBM (How ?) This gives ∠OMA = ∠OMB = 90° (Why ?) AM = BM (Why ?) OM = OM (Common) OA = OB (Why ?) Fig. 9.7 line from the point. a circle that longer chord is nearer to the centre than the smaller chord. You may observe it by drawing several chords of a circle of different lengths and measuring their distances from the centre. What is the distance of the diameter, which is the The length of the perpendicular from a point to a line is the distance of the Note that if the point lies on the line, the distance of the line from the point is zero. A circle can have infinitely many chords. You may observe by drawing chords of Reprint 2025-26 Fig. 9.8 120 MATHEMATICS longest chord from the centre? Since the centre lies on it, the distance is zero. Do you think that there is some relationship between the length of chords and their distances from the centre? Let us see if this is so. Activity : Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O. Fold the figure so that D falls on B and C falls on A [see Fig.9.9 (i)]. You may observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat the activity by drawing congruent circles with centres O and O′ and taking equal chords AB and CD one on each. Draw perpendiculars OM and O′N on them [see Fig. 9.9(ii)]. Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that O coincides with O′ and M coincides with N. In this way you verified the following: Fig. 9.9 Theorem 9.5 : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). this, draw a circle with centre O. From the centre O, draw two line segments OL and OM of equal length and lying inside the circle [see Fig. 9.10(i)]. Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig 9.10(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are equal. Repeat the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 9.5 which is stated as follows: Next, it will be seen whether the converse of this theorem is true or not. For Reprint 2025-26 CIRCLES 121 Theorem 9.6 : Chords equidistant from the centre of a circle are equal in length. We now take an example to illustrate the use of the above results: Example 1 : If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal. Solution : Given that AB and CD are two chords of a circle, with centre O intersecting at a point E. PQ is a diameter through E, such that ∠ AEQ = ∠ DEQ (see Fig.9.11). You have to prove that AB = CD. Draw perpendiculars OL and OM on chords AB and CD, respectively. Now ∠ LOE = 180° – 90° – ∠ LEO = 90° – ∠ LEO (Angle sum property of a triangle) Fig. 9.10 In triangles OLE and OME, ∠ LEO = ∠ MEO (Why ?) ∠ LOE = ∠ MOE (Proved above) Therefore, ∆ OLE ≅ ∆ OME (Why ?) This gives OL = OM (CPCT) So, AB = CD (Why ?) = 90° – ∠ AEQ = 90° – ∠ DEQ = 90° – ∠ MEO = ∠ MOE EO = EO (Common) Reprint 2025-26 Fig. 9.11 122 MATHEMATICS 9.4Angle Subtended by an Arc of a Circle You have seen that the end points of a chord other than diameter of a circle cuts it into two arcs – one major and other minor. If you take two equal chords, what can you say about the size of arcs? Is one arc made by first chord equal to the corresponding arc made by another chord? In fact, they are more than just equal in length. They are congruent in the sense that if one arc is put on the other, without bending or twisting, one superimposes the other completely. 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 9.12). 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip? 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. EXERCISE 9.2 Fig. 9.12 You can verify this fact by cutting the arc, corresponding to the chord CD from the circle along CD and put it on the corresponding arc made by equal chord AB. You will find that the arc CD superimpose the arc AB completely (see Fig. 9.13). This shows that equal chords make congruent arcs and conversely congruent arcs make equal chords of a circle. You can state it as follows: and conversely, if two arcs are congruent, then their corresponding chords are equal. If two chords of a circle are equal, then their corresponding arcs are congruent Reprint 2025-26 Fig. 9.13 CIRCLES 123 Also the angle subtended by an arc at the centre is defined to be angle subtended by the corresponding chord at the centre in the sense that the minor arc subtends the angle and the major arc subtends the reflex angle. Therefore, in Fig 9.14, the angle subtended by the minor arc PQ at O is ∠POQ and the angle subtended by the major arc PQ at O is reflex angle POQ. In view of the property above and Theorem 9.1, the following result is true: angle subtended by the corresponding (minor) arc at the centre. The following theorem gives the relationship between the angles subtended by an arc at the centre and at a point on the circle. Theorem 9.7 : The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Proof : Given an arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle. We need to prove that ∠ POQ = 2 ∠ PAQ. Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre. Therefore, the angle subtended by a chord of a circle at its centre is equal to the Fig. 9.14 Consider the three different cases as given in Fig. 9.15. In (i), arc PQ is minor; in (ii), arc PQ is a semicircle and in (iii), arc PQ is major. Let us begin by joining AO and extending it to a point B. In all the cases, ∠ BOQ = ∠ OAQ + ∠ AQO because an exterior angle of a triangle is equal to the sum of the two interior opposite angles. Reprint 2025-26 Fig. 9.15 124 MATHEMATICS Also in ∆ OAQ, Therefore, ∠ OAQ = ∠ OQA (Theorem 7.5) This gives ∠ BOQ = 2 ∠ OAQ (1) Similarly, ∠ BOP = 2 ∠ OAP (2) From (1) and (2), ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ) This is the same as ∠ POQ = 2 ∠ PAQ (3) For the case (iii), where PQ is the major arc, (3) is replaced by Remark : Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP. In Theorem 9.7, A can be any point on the remaining part of the circle. So if you take any other point C on the remaining part of the circle (see Fig. 9.16), you have ∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ Therefore, ∠ PCQ = ∠ PAQ. This proves the following: Theorem 9.8 : Angles in the same segment of a circle are equal. reflex angle POQ = 2 ∠ PAQ OA = OQ (Radii of a circle) Fig. 9.16 Again let us discuss the case (ii) of Theorem 10.8 separately. Here ∠PAQ is an angle in the segment, which is a semicircle. Also, ∠ PAQ = 1 2 ∠ POQ = 1 2 × 180° = 90°. If you take any other point C on the semicircle, again you get that ∠ PCQ = 90° Therefore, you find another property of the circle as: Angle in a semicircle is a right angle. The converse of Theorem 9.8 is also true. It can be stated as: Theorem 9.9 : If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic). Reprint 2025-26 CIRCLES 125 You can see the truth of this result as follows: In Fig. 9.17, AB is a line segment, which subtends equal angles at two points C and D. That is ∠ ACB = ∠ ADB To show that the points A, B, C and D lie on a circle let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say E (or E′). If points A, C, E and B lie on a circle, ∠ ACB = ∠ AEB (Why?) But it is given that ∠ ACB = ∠ ADB. Therefore, ∠ AEB = ∠ ADB. This is not possible unless E coincides with D. (Why?) Similarly, E′ should also coincide with D. 9.5 Cyclic Quadrilaterals A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle (see Fig 9.18). You will find a peculiar property in such quadrilaterals. Draw several cyclic quadrilaterals of different sides and name each of these as ABCD. (This can be done by drawing several circles of different radii and taking four points on each of them.) Measure the opposite angles and write your observations in the following table. Fig. 9.17 S.No. of Quadrilateral ∠ A ∠ B ∠ C ∠ D ∠ A +∠ C ∠ B +∠ D What do you infer from the table? 1. 2. 3. 4. 5. 6. Reprint 2025-26 Fig. 9.18 126 MATHEMATICS measurements. This verifies the following: Theorem 9.10 : The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. Theorem 9.11 : If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic. adopted for Theorem 9.9. Example 2 : In Fig. 9.19, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that ∠ AEB = 60°. Solution : Join OC, OD and BC. Triangle ODC is equilateral (Why?) Therefore, ∠ COD = 60° Now, ∠ CBD = 1 2 ∠ COD (Theorem 9.7) This gives ∠ CBD = 30° Again, ∠ ACB = 90° (Why ?) So, ∠ BCE = 180° – ∠ ACB = 90° You find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, neglecting the error in In fact, the converse of this theorem, which is stated below is also true. You can see the truth of this theorem by following a method similar to the method Which gives ∠ CEB = 90° – 30° = 60°, i.e., ∠ AEB = 60° Example 3 : In Fig 9.20, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠ DBC = 55° and ∠ BAC = 45°, find ∠ BCD. Solution : ∠ CAD = ∠ DBC = 55° (Angles in the same segment) Therefore, ∠ DAB = ∠ CAD + ∠ BAC But ∠ DAB + ∠ BCD = 180° So, ∠ BCD = 180° – 100° = 80° (Opposite angles of a cyclic quadrilateral) = 55° + 45° = 100° Reprint 2025-26 Fig. 9.20 Fig. 9.19 CIRCLES 127 Example 4 : Two circles intersect at two points A and B. AD and AC are diameters to the two circles (see Fig. 9.21). Prove that B lies on the line segment DC. Solution : Join AB. ∠ ABD = 90° (Angle in a semicircle) ∠ ABC = 90° (Angle in a semicircle) So, ∠ ABD + ∠ ABC = 90° + 90° = 180° Therefore, DBC is a line. That is B lies on the line segment DC. Example 5 : Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic. Solution : In Fig. 9.22, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH. Now, ∠ FEH = ∠ AEB = 180° – ∠ EAB – ∠ EBA (Why ?) and ∠ FGH = ∠ CGD = 180° – ∠ GCD – ∠ GDC (Why ?) Therefore, ∠ FEH + ∠ FGH = 180° – 1 2 (∠ A + ∠ B) + 180° – 1 2 (∠ C + ∠ D) = 180° – 1 2 (∠ C + ∠ D) = 180° – 1 2 (∠ A + ∠ B) Fig. 9.21 Fig. 9.22 Therefore, by Theorem 9.11, the quadrilateral EFGH is cyclic. 1. In Fig. 9.23, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. = 360° – 1 2 (∠ A+ ∠ B +∠ C +∠ D) = 360° – 1 2 × 360° = 360° – 180° = 180° EXERCISE 9.3 Reprint 2025-26 Fig. 9.23 128 MATHEMATICS 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. 3. In Fig. 9.24, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR. 4. In Fig. 9.25, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC. 5. In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC. Fig. 9.24 Fig. 9.25 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD. 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Reprint 2025-26 Fig. 9.26 CIRCLES 129 9.6 Summary In this chapter, you have studied the following points: 1. A circle is the collection of all points in a plane, which are equidistant from a fixed point in the plane. 2. Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. 3. If the angles subtended by two chords of a circle (or of congruent circles) at the centre (corresponding centres) are equal, the chords are equal. 4. The perpendicular from the centre of a circle to a chord bisects the chord. 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD. 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD. 12. Prove that a cyclic parallelogram is a rectangle. Fig. 9.27 5. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. 6. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres). 7. Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are equal. 8. If two arcs of a circle are congruent, then their corresponding chords are equal and conversely if two chords of a circle are equal, then their corresponding arcs (minor, major) are congruent. 9. Congruent arcs of a circle subtend equal angles at the centre. 10. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. 11. Angles in the same segment of a circle are equal. Reprint 2025-26 130 MATHEMATICS 12. Angle in a semicircle is a right angle. 13. If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle. 14. The sum of either pair of opposite angles of a cyclic quadrilateral is 1800 . 15. If sum of a pair of opposite angles of a quadrilateral is 1800 , the quadrilateral is cyclic. Reprint 2025-26" class_9,10,Heron's Formula,ncert_books/class_9/iemh1dd/iemh110.pdf,"HERON’S FORMULA 10.1Area of a Triangle — by Heron’s Formula We know that the area of triangle when its height is given, is 1 2 × base × height. Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? For instance, you have a triangular park whose sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height. Try doing so. If you are not able to get it, then go to the next section. Heron was born in about 10AD possibly in Alexandria in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields. His geometrical works deal largely with problems on mensuration written in three books. Book I deals with the area of squares, rectangles, triangles, trapezoids (trapezia), various other specialised quadrilaterals, the regular polygons, circles, surfaces of cylinders, cones, spheres etc. In this book, Heron has derived the famous formula for the area of a triangle in terms of its three sides. CHAPTER 10 formula. It is stated as: The formula given by Heron about the area of a triangle, is also known as Hero’s Area of a triangle = s s a s b s c ( ) ( ) ( ) − − − (I) Reprint 2025-26 Heron (10 C.E. – 75 C.E.) Fig. 10.1 132 MATHEMATICS where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = 2 a b c + + , easily. Let us apply it to calculate the area of the triangular park ABC, mentioned above (see Fig. 10.2). Let us take a = 40 m, b = 24 m, c = 32 m, so that we have s = 40 24 32 2 + + m = 48 m. make a right triangle. The largest side, i.e., BC which is 40 m will be the hypotenuse and the angle between the sides AB and AC will be 90°. This formula is helpful where it is not possible to find the height of the triangle s – a = (48 – 40) m = 8 m, s – b = (48 – 24) m = 24 m, s – c = (48 – 32) m = 16 m. Therefore, area of the park ABC = s s a s b s c ( ) ( ) ( ) − − − = 2 2 48 8 24 16 m 384m × × × = We see that 322 + 242 = 1024 + 576 = 1600 = 402 . Therefore, the sides of the park We can check that the area of the park is 1 2 × 32 × 24 m2 = 384 m2 . Fig. 10.2 formula. triangles discussed earlier viz., We find that the area we have got is the same as we found by using Heron’s Now using Heron’s formula, you verify this fact by finding the areas of other (i) equilateral triangle with side 10 cm. (ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm. You will see that For (i), we have s = 10 10 10 2 + + cm = 15 cm. Reprint 2025-26 HERON’S FORMULA 133 Example 1 : Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm (see Fig. 10.3). Solution : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm. Therefore, area of the triangle = s s a s b s c ( ) ( ) ( ) − − − Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of `20 per metre leaving a space 3m wide for a gate on one side. Area of triangle = 15(15 10) (15 10) (15 10) − − − cm2 = 2 2 15 5 5 5 cm 25 3 cm × × × = For (ii), we have s = 8 5 5 cm 9 cm 2 + + = Area of triangle = 9(9 8) (9 5) (9 5) − − − cm2 = 2 2 9 1 4 4 cm 12 cm . × × × = Let us now solve some more examples: Third side c = 32 cm – (8 + 11) cm = 13 cm So, 2s = 32, i.e., s = 16 cm, s – a = (16 – 8) cm = 8 cm, s – b = (16 – 11) cm = 5 cm, s – c = (16 – 13) cm = 3 cm. = 2 2 16 8 5 3 cm 8 30 cm × × × = Fig. 10.3 Solution : For finding area of the park, we have i.e., s = 125 m Now, s – a = (125 – 120) m = 5 m, s – b = (125 – 80) m = 45 m, s – c = (125 – 50) m = 75 m. 2s = 50 m + 80 m + 120 m = 250 m. Reprint 2025-26 Fig. 10.4 134 MATHEMATICS Therefore, area of the park = s s a s b s c ( ) ( ) ( ) − − − Also, perimeter of the park = AB + BC + CA = 250 m Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate) And so the cost of fencing = `20 × 247 = `4940 Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area. Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 10.5). Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle) Therefore, 15x = 300, which gives x = 20. So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m i.e., 60 m, 100 m and 140 m. Can you now find the area [Using Heron’s formula]? We have s = 60 100 140 2 + + m = 150 m, and area will be 150(150 60) (150 100) (150 140) − − − m2 = 125 5 45 75 × × × m2 = 2 375 15 m = 247 m Fig. 10.5 1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board? = 150 90 50 10 × × × m2 = 2 1500 3 m EXERCISE 10.1 Reprint 2025-26 HERON’S FORMULA 135 2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 10.6). The advertisements yield an earning of ` 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay? 3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEPTHE PARK GREEN AND CLEAN” (see Fig. 10.7 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. Fig. 10.6 4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm. 5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area. 6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle. Reprint 2025-26 Fig. 10.7 136 MATHEMATICS 10.2 Summary In this chapter, you have studied the following points : 1. Area of a triangle with its sides as a, b and c is calculated by using Heron’s formula, stated as where s = 2 a b c + + Area of triangle = s s a s b s c ( ) ( ) ( ) − − − Reprint 2025-26" class_9,11,Surface Areas and Volumes,ncert_books/class_9/iemh1dd/iemh111.pdf,"SURFACE AREAS AND VOLUMES 137 SURFACE AREAS AND VOLUMES 11.1 Surface Area of a Right Circular Cone We have already studied the surface areas of cube, cuboid and cylinder. We will now study the surface area of cone. So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called prisms. Now let us look at another kind of solid which is not a prism (These kinds of solids are called pyramids.). Let us see how we can generate them. Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)]. Hold the string with your hands on either sides of the triangle and rotate the triangle about the string a number of times. What happens? Do you recognize the shape that the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind you of the time you had eaten an ice-cream heaped into a container of that shape [see Fig. 11.1 (c) and (d)]? CHAPTER 11 Reprint 2025-26 Fig. 11.1 138 MATHEMATICS This is called a right circular cone. In Fig. 11.1(c) of the right circular cone, the point A is called the vertex, AB is called the height, BC is called the radius and AC is called the slant height of the cone. Here B will be the centre of circular base of the cone. The height, radius and slant height of the cone are usually denoted by h, r and l respectively. Once again, let us see what kind of cone we can not call a right circular cone. Here, you are (see Fig. 11.2)! What you see in these figures are not right circular cones; because in (a), the line joining its vertex to the centre of its base is not at right angle to the base, and in (b) the base is not circular. As in the case of cylinder, since we will be studying only about right circular cones, remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’ Activity : (i) Cut out a neatly made paper cone that does not have any overlapped paper, straight along its side, and opening it out, to see the shape of paper that forms the surface of the cone. (The line along which you cut the cone is the slant height of the cone which is represented by l). It looks like a part of a round cake. (ii) If you now bring the sides marked A and B at the tips together, you can see that the curved portion of Fig. 11.3 (c) will form the circular base of the cone. Fig. 11.2 (iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces, along the lines drawn from the point O, each cut portion is almost a small triangle, whose height is the slant height l of the cone. (iv) Now the area of each triangle = 1 So, area of the entire piece of paper Reprint 2025-26 Fig. 11.3 2 × base of each triangle × l. SURFACE AREAS AND VOLUMES 139 But the curved portion of the figure makes up the perimeter of the base of the cone and the circumference of the base of the cone = 2πr, where r is the base radius of the cone. So, Curved Surface Area of a Cone = 1 2 × l × 2πr = πrl where r is its base radius and l its slant height. Note that l 2 = r 2 + h 2 (as can be seen from Fig. 11.4), by applying Pythagoras Theorem. Here h is the height of the cone. Therefore, l = 2 2 r h + Now if the base of the cone is to be closed, then a circular piece of paper of radius r is also required whose area is πr 2 . So, Total Surface Area of a Cone = πrl + πr 2 = πr(l + r) (as b1 + b2 + b3 + . . . makes up the curved portion of the figure) = sum of the areas of all the triangles = 1 2 3 1 1 1 2 2 2 b l b l b l + + + ⋯ = ( 1 2 3 ) 1 2 l b b b + + + ⋯ = 1 2 × l × length of entire curved boundary of Fig. 11.3(c) Fig. 11.4 Example 1 : Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm. Solution : Curved surface area = πrl Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14). Solution : Here, h = 16 cm and r = 12 cm. So, from l 2 = h 2 + r 2 , we have l = 2 2 16 12 + cm = 20 cm = 22 7 × 7 × 10 cm2 = 220 cm2 Reprint 2025-26 140 MATHEMATICS So, curved surface area = πrl Further, total surface area = πrl + πr 2 Example 3 : A corn cob (see Fig. 11.5), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob. Solution : Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height. Here, l = 2 2 r h + = 2 2 (2.1) 20 + cm Therefore, the curved surface area of the corn cob = πrl = 404.41 cm = 20.11 cm = 22 7 × 2.1 × 20.11 cm2 = 132.726 cm2 = 132.73 cm2 (approx.) = 3.14 × 12 × 20 cm2 = 753.6 cm2 = (753.6 + 3.14 × 12 × 12) cm2 = (753.6 + 452.16) cm2 = 1205.76 cm2 Fig. 11.5 Number of grains of corn on 1 cm2 of the surface of the corn cob = 4 Therefore, number of grains on the entire curved surface of the cob So, there would be approximately 531 grains of corn on the cob. 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. 2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. = 132.73 × 4 = 530.92 = 531 (approx.) Assume π = 22 7 , unless stated otherwise. EXERCISE 11.1 Reprint 2025-26 SURFACE AREAS AND VOLUMES 141 11.2 Surface Area of a Sphere What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes, you can, because a circle is a plane closed figure whose every point lies at a constant distance (called radius) from a fixed point, which is called the centre of the circle. Now if you paste a string along a diameter of a circular disc and rotate it as you had rotated the triangle in the previous section, you see a new solid (see Fig 11.6). What does it resemble? A ball? Yes. It is called a sphere. 3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. 4. A conical tent is 10 m high and the radius of its base is 24 m. Find 5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14). 6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ` 210 per 100 m2 . 7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. 8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ` 12 per m2 , what will be the cost of painting all these cones? (Use π = 3.14 and take (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ` 70. 1.04 = 1.02) Reprint 2025-26 Fig. 11.6 142 MATHEMATICS rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three dimensional figure (solid figure), which is made up of all points in the space, which lie at a constant distance called the radius, from a fixed point called the centre of the sphere. solid whose surface is a sphere. Activity : Have you ever played with a top or have you at least watched someone play with one? You must be aware of how a string is wound around it. Now, let us take a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep the string in place, and continue to wind the string around the remaining part of the ball, till you have completely covered the ball [see Fig. 11.7(a)]. Mark the starting and finishing points on the string, and slowly unwind the string from the surface of the ball. Now, ask your teacher to help you in measuring the diameter of the ball, from which you easily get its radius. Then on a sheet of paper, draw four circles with radius equal to the radius of the ball. Start filling the circles one by one, with the string you had wound around the ball [see Fig. 11.7(b)]. Can you guess what happens to the centre of the circle, when it forms a sphere on Note : A sphere is like the surface of a ball. The word solid sphere is used for the What have you achieved in all this? The string, which had completely covered the surface area of the sphere, has been used to completely fill the regions of four circles, all of the same radius as of the sphere. So, what does that mean? This suggests that the surface area of a sphere of radius r So, Surface Area of a Sphere = 4 π r 2 = 4 times the area of a circle of radius r = 4 × (π r 2 ) Reprint 2025-26 Fig. 11.7 SURFACE AREAS AND VOLUMES 143 where r is the radius of the sphere. How many faces do you see in the surface of a sphere? There is only one, which is curved. Now, let us take a solid sphere, and slice it exactly ‘through the middle’ with a plane that passes through its centre. What happens to the sphere? Yes, it gets divided into two equal parts (see Fig. 11.8)! What will each half be called? It is called a hemisphere. (Because ‘hemi’ also means ‘half’) And what about the surface of a hemisphere? How many faces does it have? Two! There is a curved face and a flat face (base). The curved surface area of a hemisphere is half the surface area of the sphere, which is 1 2 of 4πr 2 . Therefore, Curved Surface Area of a Hemisphere = 2πr 2 where r is the radius of the sphere of which the hemisphere is a part. Now taking the two faces of a hemisphere, its surface area 2πr 2 + πr 2 Fig. 11.8 So, Total Surface Area of a Hemisphere = 3πr 2 Example 4 : Find the surface area of a sphere of radius 7 cm. Solution : The surface area of a sphere of radius 7 cm would be Example 5 : Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm. Solution : The curved surface area of a hemisphere of radius 21 cm would be = 2πr 2 = 2 × 22 7 × 21 × 21 cm2 = 2772 cm2 4πr 2 = 4 × 22 7 × 7 × 7 cm2 = 616 cm2 Reprint 2025-26 144 MATHEMATICS (ii) the total surface area of the hemisphere would be Example 6 : The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding. Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding space available for the motorcyclist is the surface area of the ‘sphere’ which is given by Example 7 : A hemispherical dome of a building needs to be painted (see Fig. 11.9). If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ` 5 per 100 cm2 . Solution : Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr. So, the radius of the dome = 17.6 × 7 2 22 × m = 2.8 m The curved surface area of the dome = 2πr 2 = 2 × 22 7 × 2.8 × 2.8 m2 3πr 2 = 3 × 22 7 × 21 × 21 cm2 = 4158 cm2 4πr 2 = 4 × 22 7 × 3.5 × 3.5 m2 = 154 m2 Now, cost of painting 100 cm2 is ` 5. So, cost of painting 1 m2 = ` 500 Therefore, cost of painting the whole dome 1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm Assume π = 22 7 , unless stated otherwise. = ` 500 × 49.28 = ` 24640 = 49.28 m2 EXERCISE 11.2 Reprint 2025-26 Fig. 11.9 SURFACE AREAS AND VOLUMES 145 11.3 Volume of a Right Circular Cone In earlier classes we have studied the volumes of cube, cuboid and cylinder In Fig 11.11, can you see that there is a right circular cylinder and a right circular cone of the same base radius and the same height? 2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m 3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14) 4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. 5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ` 16 per 100 cm2 . 6. Find the radius of a sphere whose surface area is 154 cm2 . 7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. 8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. 9. A right circular cylinder just encloses a sphere of radius r (see Fig. 11.10). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii). Fig. 11.10 Activity : Try to make a hollow cylinder and a hollow cone like this with the same base radius and the same height (see Fig. 11.11). Then, we can try out an experiment that will help us, to see practically what the volume of a right circular cone would be! Reprint 2025-26 Fig. 11.12 Fig. 11.11 146 MATHEMATICS that it fills up only a part of the cylinder [see Fig. 11.12(a)]. that the cylinder is still not full [see Fig. 11.12(b)]. seen that the cylinder is also full to the brim [see Fig. 11.12(c)]. cone, makes up the volume of a cylinder, which has the same base radius and the same height as the cone, which means that the volume of the cone is one-third the volume of the cylinder. So, Volume of a Cone = 1 3 πr 2h where r is the base radius and h is the height of the cone. Example 8 : The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone. Solution : From l 2 = r 2 + h 2 , we have So, volume of the cone = 1 3 πr 2h = 1 3 × 22 7 7 7 7 21 7 × × × cm3 So, let us start like this. Fill the cone up to the brim with sand once, and empty it into the cylinder. We find When we fill up the cone again to the brim, and empty it into the cylinder, we see When the cone is filled up for the third time, and emptied into the cylinder, it can be With this, we can safely come to the conclusion that three times the volume of a r = 2 2 2 2 l h − = − = 28 21 cm 7 7 cm = 7546 cm3 Example 9 : Monica has a piece of canvas whose area is 551 m2 . She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2 , find the volume of the tent that can be made with it. Solution : Since the area of the canvas = 551 m2 and area of the canvas lost in wastage is 1 m2 , therefore the area of canvas available for making the tent is (551 – 1) m2 = 550 m2 . Now, the surface area of the tent = 550 m2 and the required base radius of the conical tent = 7 m Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!). Reprint 2025-26 SURFACE AREAS AND VOLUMES 147 Therefore, curved surface area of tent = 550 m2 . That is, πrl = 550 or, 22 7 × 7 × l = 550 or, l = 3 550 22 m = 25 m Now, l 2 = r 2 + h 2 Therefore, h = 2 2 l r − = 2 2 25 7 m 625 49 m 576 m − = − = = 24 m So, the volume of the conical tent = 1 1 22 2 3 7 7 24 m 3 3 7 π = × × × × r h = 1232 m3 . 1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm 2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm 3. The height of a cone is 15 cm. If its volume is 1570 cm3 , find the radius of the base. (Use π = 3.14) 4. If the volume of a right circular cone of height 9 cm is 48 π cm3 , find the diameter of its base. Assume π = 22 7 , unless stated otherwise. EXERCISE 11.3 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? 6. The volume of a right circular cone is 9856 cm3 . If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone 7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. 8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8. 9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. Reprint 2025-26 148 MATHEMATICS 11.4 Volume of a Sphere Now, let us see how to go about measuring the volume of a sphere. First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container. Then, fill the container up to the brim with water [see Fig. 11.13(a)]. the container will over flow into the trough in which it is kept [see Fig. 11.13(b)]. Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water over flowed [see Fig. 11.13(c)]. Suppose the radius of the immersed sphere is r (you can find the radius by measuring the diameter of the sphere). Then evaluate 4 3 πr 3 . Do you find this value almost equal to the measure of the volume over flowed? Find the radius R of this sphere and then calculate the value of 4 3 R . 3 π Once again this Now, carefully place one of the spheres in the container. Some of the water from Once again repeat the procedure done just now, with a different size of sphere. Fig. 11.13 value is nearly equal to the measure of the volume of the water displaced (over flowed) by the sphere. What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it. By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to 4 3 π times the cube of its radius. This gives us the idea that where r is the radius of the sphere. as true. Later, in higher classes it can be proved also. But at this stage, we will just take it Volume of a Sphere = 4 3 3 π r Reprint 2025-26 SURFACE AREAS AND VOLUMES 149 hemisphere will be? Yes, it is 1 4 3 of 2 3 πr = 2 3 πr 3 . where r is the radius of the hemisphere. Example 10 : Find the volume of a sphere of radius 11.2 cm. Solution : Required volume = 4 3 πr 3 Example 11 : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3 , find the mass of the shot-putt. Solution : Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere. Now, volume of the sphere = 4 3 3 πr Since a hemisphere is half of a sphere, can you guess what the volume of a So, Volume of a Hemisphere = 2 3 3 πr Let us take some examples to illustrate the use of these formulae. = 4 22 3 4.9 4.9 4.9 cm 3 7 × × × × = 4 22 11.2 11.2 11.2 3 7 × × × × cm3 = 5887.32 cm3 Further, mass of 1 cm3 of metal is 7.8 g. Therefore, mass of the shot-putt = 7.8 × 493 g Example 12 : A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain? Solution : The volume of water the bowl can contain = 493 cm3 (nearly) Reprint 2025-26 = 3845.44 g = 3.85 kg (nearly) = 2 3 3 πr = 2 22 3.5 3.5 3.5 3 7 × × × × cm3 = 89.8 cm3 150 MATHEMATICS 11.5 Summary 1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m 2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m 3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3 ? 4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? 5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? 6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. 7. Find the volume of a sphere whose surface area is 154 cm2 . 8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ` 4989.60. If the cost of white-washing is ` 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. 9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the (i) radius r′ of the new sphere, (ii) ratio of S and S′. 10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3 ) is needed to fill this capsule? Assume π = 22 7 , unless stated otherwise. EXERCISE 11.4 In this chapter, you have studied the following points: 1. Curved surface area of a cone = πrl 2. Total surface area of a right circular cone = πrl + πr 2 , i.e., πr (l + r) 3. Surface area of a sphere of radius r = 4 π r 2 4. Curved surface area of a hemisphere = 2πr 2 5. Total surface area of a hemisphere = 3πr 2 6. Volume of a cone = 1 3 πr 2h 7. Volume of a sphere of radius r = 4 3 3 π r 8. Volume of a hemisphere = 2 3 3 π r [Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the context.] Reprint 2025-26" class_9,12,Statistics,ncert_books/class_9/iemh1dd/iemh112.pdf,"STATISTICS 151 STATISTICS 12.1 Graphical Representation of Data The representation of data by tables has already been discussed. Now let us turn our attention to another representation of data, i.e., the graphical representation. It is well said that one picture is better than a thousand words. Usually comparisons among the individual items are best shown by means of graphs. The representation then becomes easier to understand than the actual data. We shall study the following graphical representations in this section. (A) Bar graphs (A) Bar Graphs (B) Histograms of uniform width, and of varying widths (C) Frequency polygons In earlier classes, you have already studied and constructed bar graphs. Here we CHAPTER 12 shall discuss them through a more formal approach. Recall that a bar graph is a pictorial representation of data in which usually bars of uniform width are drawn with equal spacing between them on one axis (say, the x-axis), depicting the variable. The values of the variable are shown on the other axis (say, the y-axis) and the heights of the bars depend on the values of the variable. Example 1 : In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: Reprint 2025-26 152 MATHEMATICS Observe the bar graph given above and answer the following questions: (i) How many students were born in the month of November? (ii) In which month were the maximum number of students born? Solution : Note that the variable here is the ‘month of birth’, and the value of the variable is the ‘Number of students born’. (i) 4 students were born in the month of November. (ii) The Maximum number of students were born in the month of August. Let us now recall how a bar graph is constructed by considering the following example. Example 2 : A family with a monthly income of ` 20,000 had planned the following expenditures per month under various heads: Fig. 12.1 Entertainment 1 Miscellaneous 1 Draw a bar graph for the data above. Heads Expenditure (in thousand rupees) Grocery 4 Rent 5 Education of children 5 Medicine 2 Fuel 2 Reprint 2025-26 Table 12.1 STATISTICS 153 Solution : We draw the bar graph of this data in the following steps. Note that the unit in the second column is thousand rupees. So, ‘4’ against ‘grocery’ means `4000. 1. We represent the Heads (variable) on the horizontal axis choosing any scale, since the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one Head be represented by one unit. 2. We represent the expenditure (value) on the vertical axis. Since the maximum expenditure is `5000, we can choose the scale as 1 unit = `1000. 3. To represent our first Head, i.e., grocery, we draw a rectangular bar with width 1 unit and height 4 units. 4. Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars. The bar graph is drawn in Fig. 12.2. Here, you can easily visualise the relative characteristics of the data at a glance, e.g., the expenditure on education is more than double that of medical expenses. Therefore, in some ways it serves as a better representation of data than the tabular form. Activity 1 : Continuing with the same four groups of Activity 1, represent the data by suitable bar graphs. can be represented graphically. Let us now see how a frequency distribution table for continuous class intervals Reprint 2025-26 Fig. 12.2 154 MATHEMATICS (B) Histogram This is a form of representation like the bar graph, but it is used for continuous class intervals. For instance, consider the frequency distribution Table 12.2, representing the weights of 36 students of a class: Let us represent the data given above graphically as follows: (i) We represent the weights on the horizontal axis on a suitable scale. We can choose the scale as 1 cm = 5 kg. Also, since the first class interval is starting from 30.5 and not zero, we show it on the graph by marking a kink or a break on the axis. (ii) We represent the number of students (frequency) on the vertical axis on a suitable scale. Since the maximum frequency is 15, we need to choose the scale to accomodate this maximum frequency. Weights (in kg) Number of students 30.5 - 35.5 9 35.5 - 40.5 6 40.5 - 45.5 15 45.5 - 50.5 3 50.5 - 55.5 1 55.5 - 60.5 2 Total 36 Table 12.2 (iii) We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals. For example, the rectangle for the class interval 30.5 - 35.5 will be of width 1 cm and length 4.5 cm. (iv) In this way, we obtain the graph as shown in Fig. 12.3: Reprint 2025-26 STATISTICS 155 Observe that since there are no gaps in between consecutive rectangles, the resultant graph appears like a solid figure. This is called a histogram, which is a graphical representation of a grouped frequency distribution with continuous classes. Also, unlike a bar graph, the width of the bar plays a significant role in its construction. frequencies. However, since the widths of the rectangles are all equal, the lengths of the rectangles are proportional to the frequencies. That is why, we draw the lengths according to (iii) above. Here, in fact, areas of the rectangles erected are proportional to the corresponding Fig. 12.3 Example 3 : A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, 70 - 100. Then she formed the following table: Now, consider a situation different from the one above. Reprint 2025-26 156 MATHEMATICS A histogram for this table was prepared by a student as shown in Fig. 12.4. 0 - 20 7 20 - 30 10 30 - 40 10 40 - 50 20 50 - 60 20 60 - 70 15 70 - above 8 Marks Number of students Total 90 Table 12.3 Carefully examine this graphical representation. Do you think that it correctly represents the data? No, the graph is giving us a misleading picture. As we have mentioned earlier, the areas of the rectangles are proportional to the frequencies in a histogram. Earlier this problem did not arise, because the widths of all the rectangles were equal. But here, since the widths of the rectangles are varying, the histogram above does not Reprint 2025-26 Fig. 12.4 STATISTICS 157 give a correct picture. For example, it shows a greater frequency in the interval 70 - 100, than in 60 - 70, which is not the case. the areas are again proportional to the frequencies. the class-size is 10, the length of the rectangle will be 7 20 × 10 = 3.5. Similarly, proceeding in this manner, we get the following table: So, we need to make certain modifications in the lengths of the rectangles so that The steps to be followed are as given below: 1. Select a class interval with the minimum class size. In the example above, the minimum class-size is 10. 2. The lengths of the rectangles are then modified to be proportionate to the class-size 10. For instance, when the class-size is 20, the length of the rectangle is 7. So when Marks Frequency Width of Length of the rectangle the class 0 - 20 7 20 7 20 × 10 = 3.5 20 - 30 10 10 10 10 × 10 = 10 Table 12.4 30 - 40 10 10 10 10 × 10 = 10 40 - 50 20 10 20 10 × 10 = 20 50 - 60 20 10 20 10 × 10 = 20 60 - 70 15 10 15 10 × 10 = 15 70 - 100 8 30 8 30 × 10 = 2.67 Reprint 2025-26 158 MATHEMATICS we may call these lengths as “proportion of students per 10 marks interval”. Since we have calculated these lengths for an interval of 10 marks in each case, So, the correct histogram with varying width is given in Fig. 12.5. (C) Frequency Polygon There is yet another visual way of representing quantitative data and its frequencies. This is a polygon. To see what we mean, consider the histogram represented by Fig. 12.3. Let us join the mid-points of the upper sides of the adjacent rectangles of this histogram by means of line segments. Let us call these mid-points B, C, D, E, F and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 12.6). To complete the polygon, we assume that there is a class interval with frequency zero Reprint 2025-26 Fig. 12.5 STATISTICS 159 before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H, respectively. ABCDEFGH is the frequency polygon corresponding to the data shown in Fig. 12.3. We have shown this in Fig. 12.6. the highest class, addition of the two class intervals with zero frequency enables us to make the area of the frequency polygon the same as the area of the histogram. Why is this so? (Hint : Use the properties of congruent triangles.) preceding the first class? Let us consider such a situation. Although, there exists no class preceding the lowest class and no class succeeding Now, the question arises: how do we complete the polygon when there is no class Reprint 2025-26 Fig. 12.6 160 MATHEMATICS Example 4 : Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table 12.5. Draw a frequency polygon corresponding to this frequency distribution table. Solution : Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is 0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end point, i.e., B is joined to this mid-point with zero frequency on the negative direction of the horizontal axis. The point where this line segment meets the vertical axis is marked as A. Let L be the mid-point of the class succeeding the last class of the given data. Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 12.7. 0 - 10 5 10 - 20 10 20 - 30 4 30 - 40 6 40 - 50 7 50 - 60 3 60 - 70 2 70 - 80 2 80 - 90 3 90 - 100 9 Total 51 Marks Number of students Table 12.5 Reprint 2025-26 Fig. 12.7 STATISTICS 161 histograms. For this, we require the mid-points of the class-intervals used in the data. These mid-points of the class-intervals are called class-marks. lower limit of a class and divide it by 2. Thus, Example 5 : In a city, the weekly observations made in a study on the cost of living index are given in the following table: Frequency polygons can also be drawn independently without drawing To find the class-mark of a class interval, we find the sum of the upper limit and Let us consider an example. Cost of living index Number of weeks 140 - 150 5 150 - 160 10 160 - 170 20 170 - 180 9 180 - 190 6 190 - 200 2 Class-mark = Upper limit + Lower limit 2 Total 52 Table 12.6 Solution : Since we want to draw a frequency polygon without a histogram, let us find the class-marks of the classes given above, that is of 140 - 150, 150 - 160,.... For 140 - 150, the upper limit = 150, and the lower limit = 140 So, the class-mark = 150 + 140 2 = 290 2 = 145. Continuing in the same manner, we find the class-marks of the other classes as well. So, the new table obtained is as shown in the following table: Draw a frequency polygon for the data above (without constructing a histogram). Reprint 2025-26 162 MATHEMATICS We can now draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis, and then plotting and joining the points B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments. We should not forget to plot the point corresponding to the class-mark of the class 130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is, A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant frequency polygon will be ABCDEFGH (see Fig. 12.8). 140 - 150 145 5 150 - 160 155 10 160 - 170 165 20 170 - 180 175 9 180 - 190 185 6 190 - 200 195 2 Classes Class-marks Frequency Total 52 Table 12.7 Reprint 2025-26 Fig. 12.8 STATISTICS 163 very useful for comparing two different sets of data of the same nature, for example, comparing the performance of two different sections of the same class. Frequency polygons are used when the data is continuous and very large. It is 1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %): 2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below. (i) Represent the information given above graphically. (ii) Which condition is the major cause of women’s ill health and death worldwide? (iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause. S.No. Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0 EXERCISE 12.1 Section Number of girls per thousand boys Scheduled Caste (SC) 940 Scheduled Tribe (ST) 970 Non SC/ST 920 Backward districts 950 Non-backward districts 920 Rural 930 Urban 910 Reprint 2025-26 164 MATHEMATICS 3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections: 4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table: (i) Represent the information above by a bar graph. (ii) In the classroom discuss what conclusions can be arrived at from the graph. (i) Draw a bar graph to represent the polling results. (ii) Which political party won the maximum number of seats? (i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous] (ii) Is there any other suitable graphical representation for the same data? (iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why? Political Party A B C D E F Seats Won 75 55 37 29 10 37 Length (in mm) Number of leaves 118 - 126 3 127 - 135 5 136 - 144 9 145 - 153 12 154 - 162 5 163 - 171 4 172 - 180 2 5. The following table gives the life times of 400 neon lamps: Life time (in hours) Number of lamps 900 - 1000 48 300 - 400 14 400 - 500 56 500 - 600 60 600 - 700 86 700 - 800 74 800 - 900 62 Reprint 2025-26 STATISTICS 165 6. The following table gives the distribution of students of two sections according to the marks obtained by them: 7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below: (i) Represent the given information with the help of a histogram. (ii) How many lamps have a life time of more than 700 hours? Section A Section B Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections. Marks Frequency Marks Frequency 0 - 10 3 0 - 10 5 10 - 20 9 10 - 20 19 20 - 30 17 20 - 30 15 30 - 40 12 30 - 40 10 40 - 50 9 40 - 50 1 Number of balls Team A Team B 1 - 6 2 5 7 - 12 1 6 13 - 18 8 2 19 - 24 9 10 25 - 30 4 5 31 - 36 5 6 37 - 42 6 3 43 - 48 10 4 49 - 54 6 8 55 - 60 2 10 Represent the data of both the teams on the same graph by frequency polygons. [Hint : First make the class intervals continuous.] Reprint 2025-26 166 MATHEMATICS 8. A random survey of the number of children of various age groups playing in a park was found as follows: 9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows: Draw a histogram to represent the data above. (i) Draw a histogram to depict the given information. (ii) Write the class interval in which the maximum number of surnames lie. Number of letters Number of surnames Age (in years) Number of children 1 - 4 6 4 - 6 30 6 - 8 44 8 - 12 16 12 - 20 4 1 - 2 5 2 - 3 3 3 - 5 6 5 - 7 12 7 - 10 9 10 - 15 10 15 - 17 4 12.2 Summary In this chapter, you have studied the following points: 1. How data can be presented graphically in the form of bar graphs, histograms and frequency polygons. Reprint 2025-26" class_10,1,Real Numbers,ncert_books/class_10/jemh1dd/jemh101.pdf,"1.1 Introduction In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers. We continue our discussion on real numbers in this chapter. We begin with very important properties of positive integers in Sections 1.2, namely the Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. Euclid’s division algorithm, as the name suggests, has to do with divisibility of integers. Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b. Many of you probably recognise this as the usual long division process. Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers. We touch upon a few of them, and use it mainly to compute the HCF of two positive integers. The Fundamental Theorem of Arithmetic, on the other hand, has to do something with multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way—this important fact is the Fundamental Theorem of Arithmetic. Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics. We use the Fundamental Theorem of Arithmetic for two main applications. First, we use it to prove the irrationality of many of the numbers you studied in Class IX, such as 2 , 3 and 5 . Second, we apply this theorem to explore when exactly the decimal REAL NUMBERS 1 REAL NUMBERS 1 expansion of a rational number, say ( 0) p q q , is terminating and when it is nonterminating repeating. We do so by looking at the prime factorisation of the denominator q of p q . You will see that the prime factorisation of q will completely reveal the nature of the decimal expansion of p q . So let us begin our exploration. Reprint 2025-26 2 MATHEMATICS 1.2 The Fundamental Theorem of Arithmetic In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on. Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers? Let us see. some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many). Let us list a few : 2 2 × 3 × 7 × 11 × 23 = 21252 and so on. Now, let us suppose your collection of primes includes all the possible primes. What is your guess about the size of this collection? Does it contain only a finite number of integers, or infinitely many? Infact, there are infinitely many primes. So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes. The question is – can we produce all the composite numbers this way? What do you think? Do you think that there may be a composite number which is not the product of powers of primes? Before we answer this, let us factorise positive integers, that is, do the opposite of what we have done so far. Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply 7 × 11 × 23 = 1771 3 × 7 × 11 × 23 = 5313 2 × 3 × 7 × 11 × 23 = 10626 2 3 × 3 × 73 = 8232 We are going to use the factor tree with which you are all familiar. Let us take some large number, say, 32760, and factorise it as shown. Reprint 2025-26 REAL NUMBERS 3 primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try another number, say, 123456789. This can be written as 32 × 3803 × 3607. Of course, you have to check that 3803 and 3607 are primes! (Try it out for several other natural numbers yourself.) This leads us to a conjecture that every composite number can be written as the product of powers of primes. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem. number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. An equivalent version of Theorem 1.2 was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic. However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae. Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science. So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of Theorem 1.1 (Fundamental Theorem of Arithmetic) : Every composite Carl Friedrich Gauss (1777 – 1855) be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written. This fact is also stated in the following form: of its factors. The Fundamental Theorem of Arithmetic says that every composite number can The prime factorisation of a natural number is unique, except for the order Reprint 2025-26 4 MATHEMATICS p1 , p2 ,..., pn are primes and written in ascending order, i.e., p1 p2 . . . pn . If we combine the same primes, we will get powers of primes. For example, is factorised, is unique. mathematics and in other fields. Let us look at some examples. Example 1 : Consider the numbers 4n , where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Solution : If the number 4n , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n ; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n . So, there is no natural number n for which 4n ends with the digit zero. using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! This method is also called the prime factorisation method. Let us recall this method through an example. Example 2 : Find the LCM and HCF of 6 and 20 by the prime factorisation method. Solution : We have : 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51 . In general, given a composite number x, we factorise it as x = p1 p2 ... pn , where 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 Once we have decided that the order will be ascending, then the way the number The Fundamental Theorem of Arithmetic has many applications, both within You have already learnt how to find the HCF and LCM of two positive integers You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes. Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers. LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers. = 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers. Example 3: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) Reprint 2025-26 REAL NUMBERS 5 Solution : The prime factorisation of 96 and 404 gives : Therefore, the HCF of these two integers is 22 = 4. Also, LCM (96, 404) = 96 404 96 404 9696 HCF(96, 404) 4 Example 4 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method. Solution : We have : Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively. So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 23 , 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM (6, 72, 120) = 23 × 32 × 51 = 360 Remark : Notice, 6 × 72 × 120 HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM. 1. Express each number as a product of its prime factors: 96 = 25 × 3, 404 = 22 × 101 6 = 2 × 3, 72 = 23 × 32 , 120 = 23 × 3 × 5 EXERCISE 1.1 2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 4. Given that HCF (306, 657) = 9, find LCM (306, 657). 5. Check whether 6n can end with the digit 0 for any natural number n. 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. 7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Reprint 2025-26 6 MATHEMATICS 1.3 Revisiting Irrational Numbers In Class IX, you were introduced to irrational numbers and many of their properties. You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that 2 , 3 , 5 and, in general, p is irrational, where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic. Recall, a number ‘s’ is called irrational if it cannot be written in the form , p q where p and q are integers and q ¹ 0. Some examples of irrational numbers, with which you are already familiar, are : proof is based on the Fundamental Theorem of Arithmetic. Theorem 1.2 : Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer. *Proof : Let the prime factorisation of a be as follows : Before we prove that 2 is irrational, we need the following theorem, whose a = p1 p2 . . . pn , where p1 ,p2 , . . ., pn are primes, not necessarily distinct. same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? 2 2, 3, 15, , 0.10110111011110 . . . , 3 , etc. * Not from the examination point of view. Therefore, a 2 = ( p1 p2 . . . pn )( p1 p2 . . . pn ) = p 2 1 p 2 2 . . . p 2 n . Now, we are given that p divides a 2 . Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a 2 . However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a 2 are p1 , p2 , . . ., pn . So p is one of p1 , p2 , . . ., pn . Now, since a = p1 p2 . . . pn , p divides a. We are now ready to give a proof that 2 is irrational. The proof is based on a technique called ‘proof by contradiction’. (This technique is discussed in some detail in Appendix 1). Theorem 1.3 : 2 is irrational. Proof : Let us assume, to the contrary, that 2 is rational. Reprint 2025-26 REAL NUMBERS 7 So, we can find integers r and s (¹ 0) such that 2 = r s . Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get 2 , a b where a and b are coprime. So, b 2 = a. Squaring on both sides and rearranging, we get 2b 2 = a 2 . Therefore, 2 divides a 2 . Now, by Theorem 1.2, it follows that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b 2 = 4c 2 , that is, b 2 = 2c 2 . This means that 2 divides b 2 , and so 2 divides b (again using Theorem 1.2 with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that 2 is rational. So, we conclude that 2 is irrational. Example 5 : Prove that 3 is irrational. Solution : Let us assume, to the contrary, that 3 is rational. That is, we can find integers a and b (¹ 0) such that 3 = a b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b a 3 Squaring on both sides, and rearranging, we get 3b 2 = a 2 . Therefore, a 2 is divisible by 3, and by Theorem 1.2, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b 2 = 9c 2 , that is, b 2 = 3c 2 . This means that b 2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.2 with p = 3). Reprint 2025-26 8 MATHEMATICS Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that 3 is rational. So, we conclude that 3 is irrational. Example 6 : Show that 5– 3 is irrational. Solution : Let us assume, to the contrary, that 5– 3 is rational. That is, we can find coprime a and b (b 0) such that 5 3 a b Therefore, 5 3 a b Rearranging this equation, we get 5 3 5– a ba b b Since a and b are integers, we get 5 – a b is rational, and so 3 is rational. But this contradicts the fact that 3 is irrational. In Class IX, we mentioned that : We prove some particular cases here. the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational. This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational. So, we conclude that 5 3 is irrational. Example 7 : Show that 3 2 is irrational. Solution : Let us assume, to the contrary, that 3 2 is rational. That is, we can find coprime a and b (b 0) such that 3 2 a b Rearranging, we get 2 3 a b Since 3, a and b are integers, 3 a b is rational, and so 2 is rational. Reprint 2025-26 REAL NUMBERS 9 But this contradicts the fact that 2 is irrational. So, we conclude that 3 2 is irrational. 1.4 Summary In this chapter, you have studied the following points: 1. Prove that 5 is irrational. 2. Prove that 3 25 is irrational. 3. Prove that the following are irrationals : 1. The Fundamental Theorem of Arithmetic : 2. If p is a prime and p divides a2 , then p divides a, where a is a positive integer. 3. To prove that 2, 3 are irrationals. Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. (i) 1 2 (ii) 7 5 (iii) 6 2 EXERCISE 1.2 You have seen that : HCF ( p, q, r) × LCM (p, q, r) p × q × r, where p, q, r are positive integers (see Example 8). However, the following results hold good for three numbers p, q and r : LCM (p, q, r) = HCF( , , ) HCF( , ) HCF( , ) HCF( , ) pqr p q r p q qr pr HCF (p, q, r) = LCM( , , ) LCM( , ) LCM( , ) LCM( , ) pqr p q r pq q r pr A NOTE TO THE READER Reprint 2025-26" class_10,2,Polynomials,ncert_books/class_10/jemh1dd/jemh102.pdf,"10 MATHEMATICS 2.1 Introduction In Class IX, you have studied polynomials in one variable and their degrees. Recall that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x). For example, 4x + 2 is a polynomial in the variable x of degree 1, 2y2 – 3y + 4 is a polynomial in the variable y of degree 2, 5x3 – 4x2 + x – 2 is a polynomial in the variable x of degree 3 and 7u6 – 3 4 2 4 8 2 u uu is a polynomial in the variable u of degree 6. Expressions like 1 x 1 , x 2 , 2 1 x x 2 3 etc., are not polynomials. A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3, POLYNOMIALS 2 such as 2x + 5 – x2 , x3 + 1, etc., are not linear polynomials. has been derived from the word ‘quadrate’, which means ‘square’. 2 2 2 3 , 5 x x y2 – 2, 2 2 3, x x 22 2 2 1 2 5, 5 , 4 3 37 u u v vz are some examples of quadratic polynomials (whose coefficients are real numbers). More generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a 0. A polynomial of degree 3 is called a cubic polynomial. Some examples of 3 5, x y 2 , 2 11 x , 3z + 4, 2 1 3 u , etc., are all linear polynomials. Polynomials A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ Reprint 2025-26 POLYNOMIALS 11 a cubic polynomial are 2 – x3 , x3 , 3 2 , x 3 – x2 + x3 , 3x3 – 2x2 + x – 1. In fact, the most general form of a cubic polynomial is where, a, b, c, d are real numbers and a 0. polynomial, we get p(2) = 22 – 3 × 2 – 4 = – 6. The value ‘– 6’, obtained by replacing x by 2 in x2 – 3x – 4, is the value of x2 – 3x – 4 at x = 2. Similarly, p(0) is the value of p(x) at x = 0, which is – 4. replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). Also, note that p(4) = 42 – (3 4) – 4 = 0. polynomial x2 – 3x – 4. More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0. polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us 2k + 3 = 0, i.e., k = 3 2 Now consider the polynomial p(x) = x2 – 3x – 4. Then, putting x = 2 in the If p(x) is a polynomial in x, and if k is any real number, then the value obtained by What is the value of p(x) = x2 –3x – 4 at x = –1? We have : As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic You have already studied in Class IX, how to find the zeroes of a linear In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e., b k a p(–1) = (–1)2 –{3 × (–1)} – 4 = 0 ax3 + bx2 + cx + d, So, the zero of the linear polynomial ax + b is (Constant term) Coefficient of b a x . happen in the case of other polynomials too? For example, are the zeroes of a quadratic polynomial also related to its coefficients? division algorithm for polynomials. 2.2 Geometrical Meaning of the Zeroes of a Polynomial You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes. Thus, the zero of a linear polynomial is related to its coefficients. Does this In this chapter, we will try to answer these questions. We will also study the Reprint 2025-26 12 MATHEMATICS Consider first a linear polynomial ax + b, a 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7). that the graph of y = 2x + 3 intersects the x -axis mid-way between x = –1 and x = – 2, that is, at the point 3 , 0 2 . You also know that the zero of 2x + 3 is 3 2 . Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis. x –2 2 y = 2x + 3 –1 7 From Fig. 2.1, you can see In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a Fig. 2.1 straight line which intersects the x-axis at exactly one point, namely, , 0 b a . Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis. Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph* of y = x2 – 3x – 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. * Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated. Reprint 2025-26 POLYNOMIALS 13 If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2. In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. (These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis. Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis. x – 2 –1 0 1 2 3 45 y = x2 – 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6 Table 2.1 polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis. following three cases can happen: This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic From our observation earlier about the shape of the graph of y = ax2 + bx + c, the Reprint 2025-26 Fig. 2.2 14 MATHEMATICS Case (i) : Here, the graph cuts x-axis at two distinct points A and A. ax2 + bx + c in this case (see Fig. 2.3). Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A of Case (i) coincide here to become one point A (see Fig. 2.4). The x-coordinates of A and A are the two zeroes of the quadratic polynomial Fig. 2.3 in this case. The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c Reprint 2025-26 Fig. 2.4 POLYNOMIALS 15 Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point (see Fig. 2.5). So, the quadratic polynomial ax2 + bx + c has no zero in this case. distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree 2 has atmost two zeroes. So, you can see geometrically that a quadratic polynomial can have either two Fig. 2.5 polynomial to be? Let us find out. Consider the cubic polynomial x3 – 4x. To see what the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2. that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. Now, what do you expect the geometrical meaning of the zeroes of a cubic Locating the points of the table on a graph paper and drawing the graph, we see x –2 –1 0 1 2 y = x3 – 4x 0 3 0 –3 0 Reprint 2025-26 Table 2.2 16 MATHEMATICS We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x. Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis. Since the curve meets the x-axis in only these 3 points, their x-coordinates are the only zeroes of the polynomial. Let us take a few more examples. Consider the cubic polynomials x3 and x3 – x2 . We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively. Fig. 2.6 Fig. 2.7 Fig. 2.8 Reprint 2025-26 POLYNOMIALS 17 that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis. Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2 . Also, from Fig. 2.8, these values are the x-coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis. polynomial. In other words, any polynomial of degree 3 can have at most three zeroes. Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points. Therefore, a polynomial p(x) of degree n has at most n zeroes. Example 1 : Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x). Note that 0 is the only zero of the polynomial x3 . Also, from Fig. 2.7, you can see From the examples above, we see that there are at most 3 zeroes for any cubic Solution : (iii) The number of zeroes is 3. (Why?) (ii) The number of zeroes is 2 as the graph intersects the x-axis at two points. (i) The number of zeroes is 1 as the graph intersects the x-axis at one point only. Reprint 2025-26 Fig. 2.9 18 MATHEMATICS 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. (iv) The number of zeroes is 1. (Why?) (vi) The number of zeroes is 4. (Why?) (v) The number of zeroes is 1. (Why?) EXERCISE 2.1 2.3 Relationship between Zeroes and Coefficients of a Polynomial You have already seen that zero of a linear polynomial ax + b is b a . We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, say p(x) = 2x2 – 8x + 6. In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term. So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is 6 × 2x2 = 12x2 . So, we write 2x2 – 8x + 6 = 2x2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3) = (2x – 2)(x – 3) = 2(x – 1)(x – 3) Reprint 2025-26 Fig. 2.10 POLYNOMIALS 19 x = 1 or x = 3. So, the zeroes of 2x2 – 8x + 6 are 1 and 3. Observe that : method of splitting the middle term, when x = 1 3 or x = –2. So, the zeroes of 3x2 + 5x – 2 are 1 3 and – 2. Observe that : a 0, then you know that x – and x – are the factors of p(x). Therefore, So, the value of p(x) = 2x2 – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when Let us take one more quadratic polynomial, say, p(x) = 3x2 + 5x – 2. By the Hence, the value of 3x2 + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e., In general, if * and * are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, Sum of its zeroes = 2 ( 8) (Coefficient of ) 134 2 Coefficient of x x Product of its zeroes = 2 6 Constant term 13 3 2 Coefficient of x Sum of its zeroes = 2 1 5 (Coefficient of ) ( 2) 3 3 Coefficient of Product of its zeroes = 2 1 2 Constant term ( 2) 3 3 Coefficient of x ax2 + bx + c = k(x – ) (x – ), where k is a constant 3x2 + 5x – 2 = 3x2 + 6x – x – 2 = 3x(x + 2) –1(x + 2) = (3x – 1)(x + 2) x x * , are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘’ pronounced as ‘gamma’. = kx2 – k( + )x + k Comparing the coefficients of x2 , x and constant terms on both the sides, we get This gives + = –b a , a = k, b = – k( + ) and c = k = c a Reprint 2025-26 = k[x2 – ( + )x + ] 20 MATHEMATICS i.e., sum of zeroes = + = 2 (Coefficient of ) Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients. Solution : We have So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x2 + 7x + 10 are – 2 and – 5. Now, Example 3 : Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients. Solution : Recall the identity a2 – b2 = (a – b)(a + b). Using it, we can write: Let us consider some examples. product of zeroes = 2 10 Constant term ( 2) ( 5) 10 1 Coefficient of x sum of zeroes = 2 (7) – (Coefficient of ) – 2 (–5) – (7) , 1 Coefficient of product of zeroes = = 2 Constant term Coefficient of c a x . x2 + 7x + 10 = (x + 2)(x + 5) x x Coefficient of b x a x , So, the value of x2 – 3 is zero when x = 3 or x = – 3 Therefore, the zeroes of x2 – 3 are 3 and 3 Now, product of zeroes = 2 3 Constant term 3 3 –3 1 Coefficient of x sum of zeroes = 2 (Coefficient of ) 3 30 , Coefficient of x2 – 3 = x x 3 3 x x Reprint 2025-26 POLYNOMIALS 21 Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively. Solution : Let the quadratic polynomial be ax2 + bx + c, and its zeroes be and . We have and = 2 = c a . If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2. be of the form k(x2 + 3x + 2), where k is real. between the zeroes of a cubic polynomial and its coefficients? zeroes, these are the zeores of 2x3 – 5x2 – 14x + 8. Now, You can check that any other quadratic polynomial that fits these conditions will Let us now look at cubic polynomials. Do you think a similar relation holds Let us consider p(x) = 2x3 – 5x2 – 14x + 8. You can check that p(x) = 0 for x = 4, – 2, 1 2 Since p(x) can have atmost three sum of the zeroes = 2 + = – 3 = b a , 3 1 5 ( 5) (Coefficient of ) 4 ( 2) 22 2 Coefficient of x x , of the zeroes taken two at a time. We have ax3 + bx2 + cx + d, then However, there is one more relationship here. Consider the sum of the products In general, it can be proved that if , , are the zeroes of the cubic polynomial 1 1 4 ( 2) ( 2) 4 2 2 product of the zeroes = 3 1 8 – Constant term 4 ( 2) 4 2 2 Coefficient of x . = 14 –8 1 2 7 2 = 3 Coefficient of Coefficient of Reprint 2025-26 x x . 22 MATHEMATICS = – d a . Let us consider an example. Example 5* : Verify that 3, –1, 1 3 are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients. Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get Therefore, 3, –1 and 1 3 are the zeroes of 3x3 – 5x2 – 11x – 3. a = 3, b = – 5, c = –11, d = – 3. Further p(3) = 3 × 33 – (5 × 32 ) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0, p(–1) = 3 × (–1)3 – 5 × (–1)2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0, = 1 5 11 2 2 – 3– 0 99 3 33 3 2 11 1 1 3 5 11 3 33 3 3 p , + + = c a , + + = –b a , So, we take = 3, = –1 and = 1 3 Now, 1 1 5 ( 5) 3 ( 1) 2 3 33 3 * Not from the examination point of view. b a , c a , d a . 1 ( 3) 3 ( 1) 1 3 3 1 1 1 11 3 ( 1) ( 1) 3 3 1 33 33 Reprint 2025-26 POLYNOMIALS 23 2.4 Summary In this chapter, you have studied the following points: 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials respectively. 2. A quadratic polynomial in x with real coefficients is of the form ax2 + bx + c, where a, b, c are real numbers with a 0. 3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y = p(x) intersects the x -axis. 4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes. 5. If and are the zeroes of the quadratic polynomial ax2 + bx + c, then (iv) 4u2 + 8u (v) t 2 – 15 (vi) 3x2 – x – 4 (iv) 1, 1 (v) 1 1 , 4 4 (vi) 4, 1 (i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (i) 1 , 1 4 (ii) 1 2 , 3 (iii) 0, 5 EXERCISE 2.2 6. If , , are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then and d a . b a , c a . a , c a , b Reprint 2025-26" class_10,3,Pair of Linear Equations in Two Variables,ncert_books/class_10/jemh1dd/jemh103.pdf,"24 MATHEMATICS 3.1 Introduction You must have come across situations like the one given below : and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs ` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent ` 20. possible? Is it possible to have two rides? And so on. Or you may use the knowledge of Class IX, to represent such situations as linear equations in two variables. PAIR OF LINEAR EQUATIONS Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel May be you will try it by considering different cases. If she has one ride, is it IN TWO VARIABLES 3 Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25 Let us try this approach. played Hoopla by y. Now the situation can be represented by the two equations: finding these, which we will study in this chapter. 3.2 Graphical Method of Solution of a Pair of Linear Equations A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent. in two variables and the existence of solutions as follows: Denote the number of rides that Akhila had by x, and the number of times she Can we find the solutions of this pair of equations? There are several ways of We can now summarise the behaviour of lines representing a pair of linear equations (ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations). (i) the lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations). 3x + 4y = 20 (2) y = 1 2 x (1) three examples. Here, a1 , b1 , c1 and a2 , b2 , c2 denote the coefficents of equations given in the general form in Section 3.2. Consider the following three pairs of equations. (i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect) (ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide) (iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel) Let us now write down, and compare, the values of 11 1 2 2 2 , and ab c a b c in all the (iii) the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations]. Reprint 2025-26 26 MATHEMATICS Sl Pair of lines 1 2 a a 1 2 b b 1 2 c c Compare the Graphical Algebraic No. ratios representation interpretation 1. x – 2y = 0 1 3 2 4 − 0 −20 1 1 2 2 a b a b ≠ Intersecting Exactly one 3x + 4y – 20 = 0 lines solution (unique) 2. 2x + 3y – 9 = 0 2 4 3. x + 2y – 4 = 0 1 2 From the table above, you can observe that if the lines represented by the equation and a2 x + b2 y + c 2 = 0 are (i) intersecting, then 1 1 2 2 a b a b ≠ ⋅ 4x + 6y – 18 = 0 lines many solutions 2x + 4y – 12 = 0 (ii) coincident, then 1 1 1 2 2 2 a b c a b c = = ⋅ a1 x + b1 y + c1 = 0 2 4 3 6 9 18 − − 1 1 1 2 2 2 a b c a b c = = Coincident Infinitely 4 12 − − Table 3.1 2 2 2 a b c a b c = ≠ Parallel lines No solution 1 1 1 considering some more examples by yourself. Example 1 : Check graphically whether the pair of equations and 2x – 3y = 12 (2) is consistent. If so, solve them graphically. Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two solutions of each of the equations, which are given in Table 3.2 In fact, the converse is also true for any pair of lines. You can verify them by Let us now consider some more examples to illustrate it. (iii) parallel, then 1 1 1 2 2 2 a b c a b c = ≠ ⋅ x + 3y = 6 (1) Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 27 Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig. 3.1. We observe that there is a point B (6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, i.e., the given pair of equations is consistent. Example 2 : Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions: y = 6 3 − x 2 0 y = 2 12 3 x − – 4 –2 x 0 6 x 0 3 5x – 8y + 1 = 0 (1) Table 3.2 Fig. 3.1 Solution : Multiplying Equation (2) by 5 , 3 we get But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions. Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought. Plot few points on the graph and verify it yourself. 3x – 24 5 y + 3 5 = 0 (2) 5x – 8y + 1 = 0 Reprint 2025-26 28 MATHEMATICS Solution : Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are : and y = 4x – 4 (2) Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations. They are given in Table 3.3. as shown in Fig. 3.2. Plot the points and draw the lines passing through them to represent the equations, The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution y = 2x – 2 2 – 2 y = 4x – 4 – 4 0 x 2 0 x 0 1 Table 3.3 y = 2x – 2 (1) Fig. 3.2 of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt. Verify the answer by checking whether it satisfies the conditions of the given problem. 1. Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. EXERCISE 3.1 Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 29 2. On comparing the ratios 11 1 22 2 , and ab c ab c , find out whether the lines representing the 3. On comparing the ratios 1 1 2 2 , a b a b and 1 2 c c , find out whether the following pair of linear 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 following pairs of linear equations intersect at a point, are parallel or coincident: equations are consistent, or inconsistent. (iii) 6x – 3y + 10 = 0 (iii) 3 5 7 2 3 x y ; 9x – 10y = 14 (iv) 5x – 3y = 11 ; – 10x + 6y = –22 (v) 4 2 8 3 x y ;2x + 3y = 12 (ii) 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46. Find the cost of one pencil and that of one pen. (i) 5x – 4y + 8 = 0 (ii) 9x + 3y + 12 = 0 (i) 3x + 2y = 5 ; 2x – 3y = 7 (ii) 2x – 3y = 8 ; 4x – 6y = 9 7x + 6y – 9 = 0 18x + 6y + 24 = 0 2x – y + 9 = 0 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. 6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines 7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. Reprint 2025-26 30 MATHEMATICS 3.3 Algebraic Methods of Solving a Pair of Linear Equations In the previous section, we discussed how to solve a pair of linear equations graphically. The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ( 3, 2 7 ) , (–1.75, 3.3), 4 1 , 13 19 , etc. There is every possibility of making mistakes while reading such coordinates. Is there any alternative method of finding the solution? There are several algebraic methods, which we shall now discuss. 3.3.1 Substitution Method : We shall explain the method of substitution by taking some examples. Example 4 : Solve the following pair of equations by substitution method: Solution : Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) : and write it as x = 3 – 2y (3) Step 2 : Substitute the value of x in Equation (1). We get 7x – 15y = 2 (1) x + 2y = 3 (2) x + 2y = 3 i.e., 21 – 14y – 15y = 2 i.e., – 29y = –19 Therefore, y = 19 29 Step 3 : Substituting this value of y in Equation (3), we get Therefore, the solution is x = 49 29 , y = 19 29 . 7(3 – 2y) – 15y = 2 Reprint 2025-26 x = 3 – 19 2 29 = 49 29 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 31 Verification : Substituting x = 49 29 and y = 19 29 , you can verify that both the Equations (1) and (2) are satisfied. Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient. Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent. Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable. Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method. Example 5 : Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution. Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is To understand the substitution method more clearly, let us consider it stepwise: and s + 3 = 3 (t + 3), i.e., s – 3t = 6 (2) Using Equation (2), we get s = 3t + 6. Putting this value of s in Equation (1), we get i.e., 4t = 48, which gives t = 12. Putting this value of t in Equation (2), we get s – 7 = 7 (t – 7), i.e., s – 7t + 42 = 0 (1) (3t + 6) – 7t + 42 = 0, Reprint 2025-26 s = 3 (12) + 6 = 42 32 MATHEMATICS So, Aftab and his daughter are 42 and 12 years old, respectively. Example 6 : In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 pencils and 6 erasers is `18. Find the cost of each pencil and each eraser. Solution : The pair of linear equations formed were: We first express the value of x in terms of y from the equation 2x + 3y = 9, to get Now we substitute this value of x in Equation (2), to get i.e., 18 – 6y + 6y = 18 i.e., 18 = 18 This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation. Verify this answer by checking if it satisfies the conditions of the given problems. 4(9 3 ) 2 − y + 6y = 18 2x + 3y = 9 (1) 4x + 6y = 18 (2) x = 9 3 2 − y (3) Example 7 : Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other? Solution : The pair of linear equations formed were: x + 2y – 4 = 0 (1) 2x + 4y – 12 = 0 (2) We express x in terms of y from Equation (1) to get x = 4 – 2y Now, we substitute this value of x in Equation (2) to get 2(4 – 2y) + 4y – 12 = 0 Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 33 i.e., 8 – 12 = 0 i.e., – 4 = 0 which is a false statement. Therefore, the equations do not have a common solution. So, the two rails will not cross each other. 1. Solve the following pair of linear equations by the substitution method. 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. 3. Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) 3x – y = 3 (iv) 0.2x + 0.3y = 1.3 9x – 3y = 9 0.4x + 0.5y = 2.3 (v) 2 30 x y (vi) 3 5 2 2 3 x y (i) x + y = 14 (ii) s – t = 3 (i) The difference between two numbers is 26 and one number is three times the other. Find them. x – y = 4 6 3 2 s t 3 80 x y 13 32 6 x y EXERCISE 3.2 (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes 9 11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5 6 . Find the fraction. (iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Reprint 2025-26 34 MATHEMATICS 3.3.2 Elimination Method Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works. Example 8 : The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ` 2000 per month, find their monthly incomes. Solution : Let us denote the incomes of the two person by ` 9x and ` 7x and their expenditures by ` 4y and ` 3y respectively. Then the equations formed in the situation is given by : and 7x – 3y = 2000 (2) Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations: Step 2 : Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? (28x – 27x) – (12y – 12y) = 8000 – 6000 27x – 12y = 6000 (3) 28x – 12y = 8000 (4) 9x – 4y = 2000 (1) i.e., x = 2000 Step 3 : Substituting this value of x in (1), we get i.e., y = 4000 So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are ` 18,000 and ` 14,000, respectively. Verification : 18000 : 14000 = 9 : 7. Also, the ratio of their expenditures = 18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3 Remarks : 1. The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable. 9(2000) – 4y = 2000 Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 35 Step 1 : First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal. Step 2 : Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent. Step 3 : Solve the equation in one variable (x or y) so obtained to get its value. Step 4 : Substitute this value of x (or y) in either of the original equations to get the value of the other variable. Now to illustrate it, we shall solve few more examples. Example 9 : Use elimination method to find all possible solutions of the following pair of linear equations : 2x + 3y = 8 (1) 4x + 6y = 7 (2) Solution : In the example above, we eliminated y. We could also have eliminated x. Try doing it that way. 2. You could also have used the substitution, or graphical method, to solve this problem. Try doing so, and see which method is more convenient. Let us now note down these steps in the elimination method : Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as : 4x + 6y = 16 (3) 4x + 6y = 7 (4) Step 2 : Subtracting Equation (4) from Equation (3), (4x – 4x) + (6y – 6y) = 16 – 7 i.e., 0 = 9, which is a false statement. Therefore, the pair of equations has no solution. Example 10 : The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? Reprint 2025-26 36 MATHEMATICS Solution : Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6). digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5). According to the given condition. (10x + y) + (10y + x) = 66 i.e., 11(x + y) = 66 i.e., x + y = 6 (1) We are also given that the digits differ by 2, therefore, either x – y = 2 (2) or y – x = 2 (3) If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24. Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2. 1. Solve the following pair of linear equations by the elimination method and the substitution method : When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s EXERCISE 3.3 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) 2 1 and 3 23 3 xy y x (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1 2 if we only add 1 to the denominator. What is the fraction? Reprint 2025-26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 37 3.4 Summary In this chapter, you have studied the following points: 1. A pair of linear equations in two variables can be represented, and solved, by the: 2. Graphical Method : 3. Algebraic Methods : We have discussed the following methods for finding the solution(s) of a pair of linear equations : (i) Substitution Method (ii) Elimination Method 4. If a pair of linear equations is given by a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0, then the following situations can arise : The graph of a pair of linear equations in two variables is represented by two lines. (iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of ` 50 and ` 100 she received. (iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. (ii) algebraic method (ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent). (i) graphical method (i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent. 5. There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations. (iii) 111 222 abc abc : In this case, the pair of linear equations is dependent and consistent. (ii) 111 222 abc abc : In this case, the pair of linear equations is inconsistent. (i) 1 1 2 1 a b a b : In this case, the pair of linear equations is consistent. Reprint 2025-26" class_10,4,Quadratic Equations,ncert_books/class_10/jemh1dd/jemh104.pdf,"38 MATHEMATICS 4.1 Introduction In Chapter 2, you have studied different types of polynomials. One type was the quadratic polynomial of the form ax2 + bx + c, a 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. For instance, suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres. We can depict this information pictorially as shown in Fig. 4.1. QUADRATIC EQUATIONS Fig. 4.1 4 Now, area of the hall = (2x + 1). x m2 = (2x2 + x) m2 So, 2x2 + x = 300 (Given) Therefore, 2x2 + x – 300 = 0 quadratic equation. For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x2 – px + q = 0. Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations. Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians. In fact, Brahmagupta (C.E.598–665) gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, So, the breadth of the hall should satisfy the equation 2x2 + x – 300 = 0 which is a Many people believe that Babylonians were the first to solve quadratic equations. Reprint 2025-26 QUADRATIC EQUATIONS 39 Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book ‘Liber embadorum’ published in Europe in C.E. 1145 gave complete solutions of different quadratic equations. their roots. You will also see some applications of quadratic equations in daily life situations. 4.2 Quadratic Equations A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a 0. For example, 2x2 + x – 300 = 0 is a quadratic equation. Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic equations. 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax2 + bx + c = 0, a 0 is called the standard form of a quadratic equation. different fields of mathematics. Let us consider a few examples. Example 1 : Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. In this chapter, you will study quadratic equations, and various ways of finding In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree Quadratic equations arise in several situations in the world around us and in (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day. Solution : (i) Let the number of marbles John had be x. Then the number of marbles Jivanti had = 45 – x (Why?). The number of marbles left with John, when he lost 5 marbles = x – 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 Reprint 2025-26 = 40 – x 40 MATHEMATICS (ii) Let the number of toys produced on that day be x. Therefore, their product = (x – 5) (40 – x) So, – x2 + 45x – 200 = 124 (Given that product = 124) i.e., – x2 + 45x – 324 = 0 i.e., x2 – 45x + 324 = 0 Therefore, the number of marbles John had, satisfies the quadratic equation Therefore, the cost of production (in rupees) of each toy that day = 55 – x So, the total cost of production (in rupees) that day = x (55 – x) Therefore, x (55 – x) = 750 i.e., 55x – x2 = 750 i.e., – x2 + 55x – 750 = 0 i.e., x2 – 55x + 750 = 0 Therefore, the number of toys produced that day satisfies the quadratic equation which is the required representation of the problem mathematically. which is the required representation of the problem mathematically. x2 – 45x + 324 = 0 x2 – 55x + 750 = 0 = 40x – x2 – 200 + 5x = – x2 + 45x – 200 Example 2 : Check whether the following are quadratic equations: Solution : (i) LHS = (x – 2)2 + 1 = x2 – 4x + 4 + 1 = x2 – 4x + 5 Therefore, (x – 2)2 + 1 = 2x – 3 can be rewritten as i.e., x2 – 6x + 8 = 0 It is of the form ax2 + bx + c = 0. Therefore, the given equation is a quadratic equation. (iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4 (i) (x – 2)2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2) x2 – 4x + 5 = 2x – 3 Reprint 2025-26 QUADRATIC EQUATIONS 41 (ii) Since x(x + 1) + 8 = x2 + x + 8 and (x + 2)(x – 2) = x2 – 4 Therefore, x2 + x + 8 = x2 – 4 i.e., x + 12 = 0 It is not of the form ax2 + bx + c = 0. Therefore, the given equation is not a quadratic equation. (iii) Here, LHS = x (2x + 3) = 2x2 + 3x So, x (2x + 3) = x2 + 1 can be rewritten as 2x2 + 3x = x2 + 1 Therefore, we get x2 + 3x – 1 = 0 It is of the form ax2 + bx + c = 0. So, the given equation is a quadratic equation. (iv) Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8 Therefore, (x + 2)3 = x3 – 4 can be rewritten as x3 + 6x2 + 12x + 8 = x3 – 4 i.e., 6x2 + 12x + 12 = 0 or, x2 + 2x + 2 = 0 It is of the form ax2 + bx + c = 0. So, the given equation is a quadratic equation. Remark : Be careful! In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation. In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not. 1. Check whether the following are quadratic equations : 2. Represent the following situations in the form of quadratic equations : (vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3 (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5) (v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x) (i) The area of a rectangular plot is 528 m2 . The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. EXERCISE 4.1 Reprint 2025-26 42 MATHEMATICS 4.3 Solution of a Quadratic Equation by Factorisation Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get (2 × 12 ) – (3 × 1) + 1 = 0 = RHS of the equation. We say that 1 is a root of the quadratic equation 2x2 – 3x + 1 = 0. This also means that 1 is a zero of the quadratic polynomial 2x2 – 3x + 1. In general, a real number is called a root of the quadratic equation ax2 + bx + c = 0, a 0 if a 2 + b + c = 0. We also say that x = is a solution of the quadratic equation, or that satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. two zeroes. So, any quadratic equation can have atmost two roots. their middle terms. We shall use this knowledge for finding the roots of a quadratic equation. Let us see how. Example 3 : Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation. You have observed, in Chapter 2, that a quadratic polynomial can have at most You have learnt in Class IX, how to factorise quadratic polynomials by splitting (iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. (iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. (ii) The product of two consecutive positive integers is 306. We need to find the integers. Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x2 = (2x2 ) × 3]. So, 2x2 – 5x + 3 = 2x2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0. So, the values of x for which 2x2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0, i.e., either 2x – 3 = 0 or x – 1 = 0. Now, 2x – 3 = 0 gives 3 2 x and x – 1 = 0 gives x = 1. So, 3 2 x and x = 1 are the solutions of the equation. In other words, 1 and 3 2 are the roots of the equation 2x2 – 5x + 3 = 0. Verify that these are the roots of the given equation. Reprint 2025-26 QUADRATIC EQUATIONS 43 2x2 – 5x + 3 into two linear factors and equating each factor to zero. Example 4 : Find the roots of the quadratic equation 6x2 – x – 2 = 0. Solution : We have The roots of 6x2 – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 Therefore, 3x – 2 = 0 or 2x + 1 = 0, i.e., x = 2 3 or x = 1 2 Therefore, the roots of 6x2 – x – 2 = 0 are 2 1 and – . 3 2 Example 5 : Find the roots of the quadratic equation 2 3 26 2 0 x x . Solution : 2 3 26 2 x x = 2 3 6 62 x xx Note that we have found the roots of 2x2 – 5x + 3 = 0 by factorising We verify the roots, by checking that 2 1 and 3 2 satisfy 6x2 – x – 2 = 0. 6x2 – x – 2 = 6x2 + 3x – 4x – 2 = 3 3 2 23 2 xx x = 3x (2x + 1) – 2 (2x + 1) = (3x – 2)(2x + 1) So, the roots of the equation are the values of x for which Now, 3 20 x for 2 3 x . So, this root is repeated twice, one for each repeated factor 3 2 x . Therefore, the roots of 2 3 26 2 0 x x are 2 3 , 2 3 . 3 23 20 x x = 3 23 2 x x Reprint 2025-26 44 MATHEMATICS Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1. Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x satisfies the equation 2x2 + x – 300 = 0. Applying the factorisation method, we write this equation as i.e., (x – 12)(2x + 25) = 0 of the hall, it cannot be negative. Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m. 1. Find the roots of the following quadratic equations by factorisation: 2. Solve the problems given in Example 1. 3. Find two numbers whose sum is 27 and product is 182. 4. Find two consecutive positive integers, sum of whose squares is 365. 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article. So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth (iii) 2 2 7 52 0 x x (iv) 2x2 – x + 1 8 = 0 (v) 100 x2 – 20x + 1 = 0 (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 2x (x – 12) + 25 (x – 12) = 0 2x2 – 24x + 25x – 300 = 0 EXERCISE 4.2 4.4 Nature of Roots The equation ax2 + bx + c = 0 are given by 2 4 – 2 2 b b ac a a . If b2 – 4ac > 0, we get two distinct real roots 2 4 2 2 b b ac a a and x = 2 – 4 2 b b ac a Reprint 2025-26 QUADRATIC EQUATIONS 45 So, the roots of the equation ax2 + bx + c = 0 are both 2 b a real roots in this case. there are no real roots for the given quadratic equation in this case. real roots or not, b2 – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax2 + bx + c = 0 has Example 7: Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots. Solution : The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and c = 3. Therefore, the discriminant (iii) no real roots, if b2 – 4ac < 0. (ii) two equal real roots, if b2 – 4ac = 0, (i) two distinct real roots, if b2 – 4ac > 0, If b2 – 4ac = 0, then x = 0 2 b a , i.e., or – 2 2 b b x a a Therefore, we say that the quadratic equation ax2 + bx + c = 0 has two equal If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has Let us consider some examples. b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0 So, the given equation has no real roots. Example 8 : A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution : Let us first draw the diagram (see Fig. 4.2). Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP – AP) = 7 m. Therefore, AP = (x + 7) m. Reprint 2025-26 Fig. 4.2 46 MATHEMATICS Therefore, AP2 + PB2 = AB2 (By Pythagoras theorem) i.e., (x + 7)2 + x2 = 132 i.e., x2 + 14x + 49 + x2 = 169 i.e., 2x2 + 14x – 120 = 0 So, the distance ‘x’ of the pole from gate B satisfies the equation So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is pole on the boundary of the park. Therefore, x = 5 or – 12. Therefore, x = – 12 will have to be ignored. So, x = 5. from the gate B and 12m from the gate A. Now, AB = 13m, and since AB is a diameter, So, the given quadratic equation has two real roots, and it is possible to erect the Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get Since x is the distance between the pole and the gate B, it must be positive. Thus, the pole has to be erected on the boundary of the park at a distance of 5m b2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0. x2 + 7x – 60 = 0 APB = 90° (Why?) x = 7 289 2 = 7 17 2 Example 9 : Find the discriminant of the equation 3x2 – 2x + 1 3 = 0 and hence find the nature of its roots. Find them, if they are real. Solution : Here a = 3, b = – 2 and 1 3 c . Therefore, discriminant b2 – 4ac = (– 2)2 – 4 × 3 × 1 3 = 4 – 4 = 0. Hence, the given quadratic equation has two equal real roots. The roots are 22 11 , , , , i.e., , i.e., . 2 2 66 33 b b a a Reprint 2025-26 QUADRATIC EQUATIONS 47 4.5 Summary In this chapter, you have studied the following points: 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2 ? If so, find its length and breadth. 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth. 1. A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a 0. 2. A real number is said to be a root of the quadratic equation ax2 + bx + c = 0, if a2 + b + c = 0. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. (iii) 2x2 – 6x + 3 = 0 (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4 3 x + 4 = 0 (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0 EXERCISE 4.3 3. If we can factorise ax2 + bx + c, a 0, into a product of two linear factors, then the roots of the quadratic equation ax2 + bx + c = 0 can be found by equating each factor to zero. 4. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 are given by 5. A quadratic equation ax2 + bx + c = 0 has 2 4 , 2 b b ac a provided b2 – 4ac 0. (iii) no real roots, if b2 – 4ac < 0. (ii) two equal roots (i.e., coincident roots), if b2 – 4ac = 0, and (i) two distinct real roots, if b2 – 4ac > 0, Reprint 2025-26 48 M ATHEMATICS NOTE Reprint 2025-26" class_10,5,Arithmetic Progressions,ncert_books/class_10/jemh1dd/jemh105.pdf,"5.1 Introduction You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone, etc. examples are : ARITHMETIC PROGRESSIONS 49 (i) Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of ` 8000, with an annual increment of ` 500 in her salary. Her salary (in `) for the 1st, 2nd, 3rd, . . . years will be, respectively ARITHMETIC PROGRESSIONS We now look for some patterns which occur in our day-to-day life. Some such 8000, 8500, 9000, . . . . 5 (iii) In a savings scheme, the amount becomes 5 4 times of itself after every 3 years. (ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1). The bottom rung is 45 cm in length. The lengths (in cm) of the 1st, 2nd, 3rd, . . ., 8th rung from the bottom to the top are, respectively The maturity amount (in `) of an investment of ` 8000 after 3, 6, 9 and 12 years will be, respectively : 45, 43, 41, 39, 37, 35, 33, 31 10000, 12500, 15625, 19531.25 Reprint 2025-26 Fig. 5.1 50 MATHEMATICS (iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2) are, respectively (vi) A pair of rabbits are too young to produce in their first month. In the second, and every subsequent month, they produce a new pair. Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, . . ., 6th month, respectively are : (v) Shakila puts ` 100 into her daughter’s money box when she was one year old and increased the amount by ` 50 every year. The amounts of money (in `) in the box on the 1st, 2nd, 3rd, 4th, . . . birthday were 12 , 22 , 32 , . . . . 100, 150, 200, 250, . . ., respectively. 1, 1, 2, 3, 5, 8 Fig. 5.2 Reprint 2025-26 Fig. 5.3 ARITHMETIC PROGRESSIONS 51 succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on. are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems. 5.2 Arithmetic Progressions Consider the following lists of numbers : will you write it? Perhaps by following a pattern or rule. Let us observe and write the rule. In (i), each term is 1 more than the term preceding it. (iv) 3, 3, 3, 3, . . . (iii) – 3, –2, –1, 0, . . . (v) –1.0, –1.5, –2.0, –2.5, . . . (ii) 100, 70, 40, 10, . . . (i) 1, 2, 3, 4, . . . In the examples above, we observe some patterns. In some, we find that the In this chapter, we shall discuss one of these patterns in which succeeding terms Each of the numbers in the list is called a term. Given a term, can you write the next term in each of the lists above? If so, how In (ii), each term is 30 less than the term preceding it. In (iii), each term is obtained by adding 1 to the term preceding it. (or subtracting) 0 to the term preceding it. In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it. In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression ( AP ). So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. it can be positive, negative or zero. In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding This fixed number is called the common difference of the AP. Remember that Reprint 2025-26 52 MATHEMATICS an and the common difference by d. Then the AP becomes a1 , a2 , a3 , . . ., an . So, a2 – a1 = a3 – a2 = . . . = an – an – 1 = d. Some more examples of AP are: represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP. Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions (APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term. (a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149, . . ., 157. (b) The minimum temperatures ( in degree celsius ) recorded for a week in the month of January in a city, arranged in ascending order are (c) The balance money ( in ` ) after paying 5 % of the total loan of ` 1000 every month is 950, 900, 850, 800, . . ., 50. (d) The cash prizes ( in ` ) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, . . ., 750. (e) The total savings (in `) after every month for 10 months when ` 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500. Let us denote the first term of an AP by a1 , second term by a2 , . . ., nth term by It is left as an exercise for you to explain why each of the lists above is an AP. You can see that a, a + d, a + 2d, a + 3d, . . . Note that in examples (a) to (e) above, there are only a finite number of terms. – 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5 enough to know the first term? Or, is it enough to know only the common difference? You will find that you will need to know both – the first term a and the common difference d. the AP is 6, 9,12, 15, . . . and if a is 6 and d is – 3, then the AP is Now, to know about an AP, what is the minimum information that you need? Is it For instance if the first term a is 6 and the common difference d is 3, then 6, 3, 0, –3, . . . Reprint 2025-26 ARITHMETIC PROGRESSIONS 53 round? That is, if you are given a list of numbers can you say that it is an AP and then find a and d? Since a is the first term, it can easily be written. We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, d found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP. We have a2 – a1 = 9 – 6 = 3, a3 – a2 = 12 – 9 = 3, a4 – a3 = 15 – 12 = 3 given list is an AP whose first term a is 6 and common difference d is 3. For the list of numbers : 6, 3, 0, – 3, . . ., a2 – a1 = 3 – 6 = – 3 a3 – a2 = 0 – 3 = – 3 a4 – a3 = –3 – 0 = –3 Similarly this is also an AP whose first term is 6 and the common difference is –3. Similarly, when So, if you know what a and d are, you can list the AP. What about the other way For example, for the list of numbers : Here the difference of any two consecutive terms in each case is 3. So, the a = – 7, d = – 2, the AP is – 7, – 9, – 11, – 13, . . . a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, . . . a = 0, d = 1 1 2 , the AP is 0, 1 1 2 , 3, 4 1 2 , 6, . . . a = 2, d = 0, the AP is 2, 2, 2, 2, . . . 6, 9, 12, 15, . . . , d = ak + 1 – ak where ak + 1 and ak are the ( k + 1)th and the kth terms respectively. To obtain d in a given AP, we need not find all of a2 – a1 , a3 – a2 , a4 – a3 , . . . . It is enough to find only one of them. difference between any two consecutive terms is not the same. So, this is not an AP. In general, for an AP a1 , a2 , . . ., an , we have Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the Reprint 2025-26 54 MATHEMATICS and not 3 from 6, i.e., we should subtract the kth term from the (k + 1) th term even if the (k + 1) th term is smaller. Example 1 : For the AP : 3 2 , 1 2 , – 1 2 , – 3 2 , . . ., write the first term a and the common difference d. Solution : Here, a = 3 2 , d = 1 2 – 3 2 = – 1. Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP. Example 2 : Which of the following list of numbers form an AP? If they form an AP, write the next two terms : Solution : (i) We have a2 – a1 = 10 – 4 = 6 a3 – a2 = 16 – 10 = 6 a4 – a3 = 22 – 16 = 6 i.e., ak + 1 – ak is the same every time. So, the given list of numbers forms an AP with the common difference d = 6. (iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . (i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . . Note that to find d in the AP : 6, 3, 0, – 3, . . ., we have subtracted 6 from 3 Let us make the concept more clear through some examples. The next two terms are: 22 + 6 = 28 and 28 + 6 = 34. (ii) a2 – a1 = – 1 – 1 = – 2 a3 – a2 = – 3 – ( –1 ) = – 3 + 1 = – 2 a4 – a3 = – 5 – ( –3 ) = – 5 + 3 = – 2 i.e., ak + 1 – ak is the same every time. So, the given list of numbers forms an AP with the common difference d = – 2. The next two terms are: – 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9 (iii) a2 – a1 = 2 – (– 2) = 2 + 2 = 4 a3 – a2 = – 2 – 2 = – 4 As a2 – a1 a3 – a2 , the given list of numbers does not form an AP. Reprint 2025-26 ARITHMETIC PROGRESSIONS 55 (iv) a2 – a1 = 1 – 1 = 0 a3 – a2 = 1 – 1 = 0 a4 – a3 = 2 – 1 = 1 Here, a2 – a1 = a3 – a2 a4 – a3 . So, the given list of numbers does not form an AP. 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ` 15 for the first km and ` 8 for each additional km. 2. Write first four terms of the AP, when the first term a and the common difference d are given as follows: 3. For the following APs, write the first term and the common difference: (iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum. (iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre. (iii) a = 4, d = – 3 (iv) a = – 1, d = 1 2 (v) a = – 1.25, d = – 0.25 (ii) The amount of air present in a cylinder when a vacuum pump removes 1 4 of the (i) a = 10, d = 10 (ii) a = –2, d = 0 air remaining in the cylinder at a time. EXERCISE 5.1 4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (vii) 0, – 4, – 8, –12, . . . (viii) – 1 2 , – 1 2 , – 1 2 , – 1 2 , . . . (iii) 1 5 9 13 , ,, , 333 3 . . . (iv) 0.6, 1.7, 2.8, 3.9, . . . (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . (iv) – 10, – 6, – 2, 2, . . . (v) 3, 3 2 , 3 22 , 3 3 2, . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . . (i) 3, 1, – 1, – 3, . . . (ii) – 5, – 1, 3, 7, . . . (i) 2, 4, 8, 16, . . . (ii) 5 7 2, , 3, , 2 2 . . . Reprint 2025-26 56 MATHEMATICS 5.3 nth Term of an AP Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of ` 8000, with an annual increment of ` 500. What would be her monthly salary for the fifth year? would be. salary for the 3rd, 4th and 5th year by adding ` 500 to the salary of the previous year. So, the salary for the 3rd year = ` (8500 + 500) = ` (8000 + 500 + 500) = ` (8000 + 2 × 500) To answer this, let us first see what her monthly salary for the second year It would be ` (8000 + 500) = ` 8500. In the same way, we can find the monthly = ` [8000 + (3 – 1) × 500] (for the 3rd year) = ` 9000 Salary for the 4th year = ` (9000 + 500) = ` (8000 + 500 + 500 + 500) = ` (8000 + 3 × 500) (xiii) 3, 6, 9 , 12 , . . . (xiv) 12 , 32 , 52 , 72 , . . . (xv) 12 , 52 , 72 , 73, . . . (ix) 1, 3, 9, 27, . . . (x) a, 2a, 3a, 4a, . . . (xi) a, a2 , a3 , a4 , . . . (xii) 2, 8, 18 , 32, . . . Salary for the 5th year = ` (9500 + 500) Observe that we are getting a list of numbers These numbers are in AP. (Why?) 8000, 8500, 9000, 9500, 10000, . . . = ` [8000 + (4 – 1) × 500] (for the 4th year) = ` 9500 = ` (8000+500+500+500 + 500) = ` (8000 + 4 × 500) = ` [8000 + (5 – 1) × 500] (for the 5th year) = ` 10000 Reprint 2025-26 ARITHMETIC PROGRESSIONS 57 Now, looking at the pattern formed above, can you find her monthly salary for the 6th year? The 15th year? And, assuming that she will still be working in the job, what about the monthly salary for the 25th year? You would calculate this by adding ` 500 each time to the salary of the previous year to give the answer. Can we make this process shorter? Let us see. You may have already got some idea from the way we have obtained the salaries above. i.e., First salary + (15 – 1) × Annual increment. This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP. Let a1 , a2 , a3 , . . . be an AP whose first term a1 is a and the common difference is d. Salary for the 15th year In the same way, her monthly salary for the 25th year would be = First salary + (25 – 1) × Annual increment = Salary for the 14th year + ` 500 = = ` [8000 + 14 × 500] = ` [8000 + (15 – 1) × 500] = ` 15000 ` [8000 + (25 – 1) × 500] = ` 20000 Then, given by an = a + (n – 1) d. Looking at the pattern, we can say that the nth term an = a + (n – 1) d. So, the nth term an of the AP with first term a and common difference d is the second term a2 = a + d = a + (2 – 1) d the third term a3 = a2 + d = (a + d) + d = a + 2d = a + (3 – 1) d the fourth term a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d . . . . . . . . . . . . . . . . Reprint 2025-26 58 MATHEMATICS am represents the last term which is sometimes also denoted by l. Let us consider some examples. Example 3 : Find the 10th term of the AP : 2, 7, 12, . . . Solution : Here, a = 2, d = 7 – 2 = 5 and n = 10. We have an = a + (n – 1) d So, a10 = 2 + (10 – 1) × 5 = 2 + 45 = 47 Therefore, the 10th term of the given AP is 47. Example 4 : Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer. Solution : Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n. As an = a + ( n – 1) d, we have – 81 = 21 + (n – 1)(– 3) So, n = 35 Therefore, the 35th term of the given AP is – 81. Next, we want to know if there is any n for which an = 0. If such an n is there, then an is also called the general term of the AP. If there are m terms in the AP, then – 105 = – 3n – 81 = 24 – 3n 21 + (n – 1) (–3) = 0, i.e., 3(n – 1) = 21 i.e., n =8 So, the eighth term is 0. Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9. Solution : We have and a7 = a + (7 – 1) d = a + 6d = 9 (2) Solving the pair of linear equations (1) and (2), we get Hence, the required AP is 3, 4, 5, 6, 7, . . . a3 = a + (3 – 1) d = a + 2d = 5 (1) a = 3, d = 1 Reprint 2025-26 ARITHMETIC PROGRESSIONS 59 Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . . Solution : We have : As ak + 1 – ak is the same for k = 1, 2, 3, etc., the given list of numbers is an AP. Now, a = 5 and d = 6. Let 301 be a term, say, the nth term of this AP. We know that So, 301 = 5 + (n – 1) × 6 i.e., 301 = 6n – 1 So, n = 302 151 6 3 But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers. Example 7 : How many two-digit numbers are divisible by 3? Solution : The list of two-digit numbers divisible by 3 is : Is this an AP? Yes it is. Here, a = 12, d = 3, an = 99. As an = a + (n – 1) d, a2 – a1 = 11 – 5 = 6, a3 – a2 = 17 – 11 = 6, a4 – a3 = 23 – 17 = 6 12, 15, 18, . . . , 99 an = a + (n – 1) d we have 99 = 12 + (n – 1) × 3 i.e., 87 = (n – 1) × 3 i.e., n – 1 = 87 3 = 29 i.e., n = 29 + 1 = 30 So, there are 30 two-digit numbers divisible by 3. Example 8 : Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62. Solution : Here, a = 10, d = 7 – 10 = – 3, l = – 62, where l = a + (n – 1) d Reprint 2025-26 60 MATHEMATICS the AP. So, – 62 = 10 + (n – 1)(–3) i.e., – 72 = (n – 1)(–3) i.e., n – 1 = 24 or n = 25 So, there are 25 terms in the given AP. the 14th term. Why?) So, a15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32 i.e., the 11th term from the last term is – 32. Alternative Solution : So, the question now becomes finding the 11th term with these a and d. So, a11 = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32 So, the 11th term, which is now the required term, is – 32. Example 9 : A sum of ` 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact. Solution : We know that the formula to calculate simple interest is given by To find the 11th term from the last term, we will find the total number of terms in The 11th term from the last term will be the 15th term. (Note that it will not be If we write the given AP in the reverse order, then a = – 62 and d = 3 (Why?) So, the interest at the end of the 1st year = 1000 × 8×1 100 ` = ` 80 The interest at the end of the 2nd year = 1000 × 8× 2 100 ` = ` 160 The interest at the end of the 3rd year = 1000 × 8× 3 100 ` = ` 240 Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on. So, the interest (in `) at the end of the 1st, 2nd, 3rd, . . . years, respectively are Simple Interest = P×R×T 100 80, 160, 240, . . . Reprint 2025-26 ARITHMETIC PROGRESSIONS 61 d = 80. Also, a = 80. So, to find the interest at the end of 30 years, we shall find a30. Now, a30 = a + (30 – 1) d = 80 + 29 × 80 = 2400 So, the interest at the end of 30 years will be ` 2400. Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed? Solution : The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are : It forms an AP (Why?). Let the number of rows in the flower bed be n. Then a = 23, d = 21 – 23 = – 2, an = 5 As, an = a + (n – 1) d We have, 5 = 23 + (n – 1)(– 2) i.e., – 18 = (n – 1)(– 2) i.e., n = 10 So, there are 10 rows in the flower bed. It is an AP as the difference between the consecutive terms in the list is 80, i.e., EXERCISE 5.2 23, 21, 19, . . ., 5 1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP: (iv) – 18.9 2.5 . . . 3.6 (iii) . . . – 3 18 – 5 (v) 3.5 0 105 . . . (ii) – 18 . . . 10 0 (i) 7 3 8 . . . adnan Reprint 2025-26 62 MATHEMATICS 2. Choose the correct choice in the following and justify : (i) 30th term of the AP: 10, 7, 4, . . . , is (A) 28 (B) 22 (C) –38 (D) – 48 1 2 3. In the following APs, find the missing terms in the boxes : 4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78? 5. Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 (ii) 18, 1 15 2 , 13, . . . , – 47 6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . . 7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. (iv) – 4, , , , ,6 (iii) 5, , , 1 9 2 (v) , 38, , , , – 22 (ii) 11th term of the AP: – 3, 1 2 , 2, . . ., is (ii) , 13, ,3 (i) 2, , 26 (A) 97 (B) 77 (C) –77 (D) – 87 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. 12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms? 13. How many three-digit numbers are divisible by 7? 14. How many multiples of 4 lie between 10 and 250? 15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal? 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term? 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? Reprint 2025-26 ARITHMETIC PROGRESSIONS 63 5.4 Sum of First n Terms of an AP Let us consider the situation again given in Section 5.1 in which Shakila put ` 100 into her daughter’s money box when she was one year old, ` 150 on her second birthday, ` 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old? fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see. 17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253. 18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP. 19. Subba Rao started work in 1995 at an annual salary of ` 5000 and received an increment of ` 200 each year. In which year did his income reach ` 7000? 20. Ramkali saved ` 5 in the first week of a year and then increased her weekly savings by ` 1.75. If in the nth week, her weekly savings become ` 20.75, find n. Here, the amount of money (in `) put in the money box on her first, second, third, Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote : S = 1 + 2 + 3 + . . . + 99 + 100 And then, reversed the numbers to write S = 100 + 99 + . . . + 3 + 2 + 1 Adding these two, he got 2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100) So, S = 100 101 5050 2 , i.e., the sum = 5050. We consider the problem given to Gauss (about whom you read in = 101 + 101 + . . . + 101 + 101 (100 times) Reprint 2025-26 64 MATHEMATICS of the AP. We have Rewriting the terms in reverse order, we have On adding (1) and (2), term-wise. we get or, 2S = n [2a + (n – 1) d ] (Since, there are n terms) or, S = 2 n [2a + (n – 1) d ] So, the sum of the first n terms of an AP is given by We can also write this as S = 2 n [a + a + (n – 1) d ] i.e., S = 2 n (a + an ) (3) We will now use the same technique to find the sum of the first n terms of an AP : The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms S = a + (a + d ) + (a + 2d) + . . . + [a + (n – 1) d ] (1) S = [a + (n – 1) d] + [a + (n – 2) d ] + . . . + (a + d) + a (2) 2S = [2 ( 1) ] [2 ( 1) ] ... [2 ( 1) ] [2 ( 1) ] times an d a n d a n d an d n a, a + d, a + 2d, . . . S = 2 n [2a + (n – 1) d ] From (3), we see that given and the common difference is not given. of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday, . . ., were 100, 150, 200, 250, . . ., respectively. the sum of the first 21 terms of this AP. Now, if there are only n terms in an AP, then an = l, the last term. This form of the result is useful when the first and the last terms of an AP are Now we return to the question that was posed to us in the beginning. The amount This is an AP. We have to find the total money collected on her 21st birthday, i.e., S = 2 n (a + l ) (4) Reprint 2025-26 ARITHMETIC PROGRESSIONS 65 Here, a = 100, d = 50 and n = 21. Using the formula : we have S = 21 2 100 (21 1) 50 2 = 21 200 1000 2 So, the amount of money collected on her 21st birthday is ` 12600. write S20 to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth. Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., an = Sn – Sn – 1. Let us consider some examples. Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2, . . . Solution : Here, a = 8, d = 3 – 8 = –5, n = 22. We know that Hasn’t the use of the formula made it much easier to solve the problem? We also use Sn in place of S to denote the sum of first n terms of the AP. We S = 2 ( 1) 2 n an d , = 21 1200 2 = 12600 S = 2 ( 1) 2 n an d Therefore, S = 22 16 21( 5) 2 = 11(16 – 105) = 11(–89) = – 979 So, the sum of the first 22 terms of the AP is – 979. Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. Solution : Here, S14 = 1050, n = 14, a = 10. As Sn = 2 ( 1) 2 n an d , so, 1050 = 14 20 13 2 d = 140 + 91d Reprint 2025-26 66 MATHEMATICS i.e., 910 = 91d or, d = 10 Therefore, a20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200. Example 13 : How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78? Solution : Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n. We know that Sn = 2 ( 1) 2 n an d So, 78 = 48 ( 1)( 3) 2 n n = 51 3 2 n n or 3n2 – 51n + 156 = 0 or n2 – 17n + 52 = 0 or (n – 4)(n – 13) = 0 or n = 4 or 13 Both values of n are admissible. So, the number of terms is either 4 or 13. Remarks: 1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78. 2. Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other. Example 14 : Find the sum of : Solution : (ii) Let Sn = 1 + 2 + 3 + . . . + n (i) Let S = 1 + 2 + 3 + . . . + 1000 Using the formula Sn = ( ) 2 n a l for the sum of the first n terms of an AP, we have So, the sum of the first 1000 positive integers is 500500. Here a = 1 and the last term l is n. (i) the first 1000 positive integers (ii) the first n positive integers S1000 = 1000 (1 1000) 2 = 500 × 1001 = 500500 Reprint 2025-26 ARITHMETIC PROGRESSIONS 67 Therefore, Sn = (1 ) 2 n n or Sn = ( 1) 2 n n So, the sum of first n positive integers is given by Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by Solution : As an = 3 + 2n, so, a1 = 3 + 2 = 5 List of numbers becomes 5, 7, 9, 11, . . . Here, 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on. So, it forms an AP with common difference d = 2. To find S24, we have n = 24, a = 5, d = 2. Therefore, S24 = 24 2 5 (24 1) 2 2 = 12 10 46 = 672 a2 = 3 + 2 × 2 = 7 a3 = 3 + 2 × 3 = 9 an = 3 + 2n Sn = ( + 1) 2 n n So, sum of first 24 terms of the list of numbers is 672. Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find : Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. Let us denote the number of TV sets manufactured in the nth year by an . Then, a3 = 600 and a7 = 700 (iii) the total production in first 7 years (i) the production in the 1st year (ii) the production in the 10th year Reprint 2025-26 68 MATHEMATICS or, a + 2d = 600 and a + 6d = 700 Solving these equations, we get d = 25 and a = 550. Therefore, production of TV sets in the first year is 550. (ii) Now a10 = a + 9d = 550 + 9 × 25 = 775 So, production of TV sets in the 10th year is 775. (iii) Also, S7 = 7 2 550 (7 1) 25 2 Thus, the total production of TV sets in first 7 years is 4375. 1. Find the sum of the following APs: (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 111 , , 15 12 10 , . . ., to 11 terms. 2. Find the sums given below : (i) 2, 7, 12, . . ., to 10 terms. (ii) –37, –33, –29, . . ., to 12 terms. (i) 7 + 1 10 2 + 14 + . . . + 84 (ii) 34 + 32 + 30 + . . . + 10 EXERCISE 5.3 = 7 1100 150 2 = 4375 3. In an AP: (viii) given an = 4, d = 2, Sn = –14, find n and a. (ix) given a = 3, n = 8, S = 192, find d. (x) given l = 28, S = 144, and there are total 9 terms. Find a. (vi) given a = 2, d = 8, Sn = 90, find n and an . (vii) given a = 8, an = 62, Sn = 210, find n and d. (i) given a = 5, d = 3, an = 50, find n and Sn . (ii) given a = 7, a13 = 35, find d and S13. (iii) given a12 = 37, d = 3, find a and S12. (iv) given a3 = 15, S10 = 125, find d and a10. (v) given d = 5, S9 = 75, find a and a9 . (iii) –5 + (–8) + (–11) + . . . + (–230) Reprint 2025-26 ARITHMETIC PROGRESSIONS 69 10. Show that a1 , a2 , . . ., an , . . . form an AP where an is defined as below : (i) an = 3 + 4n (ii) an = 9 – 5n 12. Find the sum of the first 40 positive integers divisible by 6. 13. Find the sum of the first 15 multiples of 8. 14. Find the sum of the odd numbers between 0 and 50. 15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ` 200 for the first day, ` 250 for the second day, ` 300 for the third day, etc., the penalty for each succeeding day being ` 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? 11. If the sum of the first n terms of an AP is 4n – n2 , what is the first term (that is S1 )? What 4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636? 5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. 6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. 9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. Also find the sum of the first 15 terms in each case. is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms. 16. A sum of ` 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ` 20 less than its preceding prize, find the value of each of the prizes. 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take = 22 7 ) Reprint 2025-26 70 MATHEMATICS 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row? 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6). [Hint : Length of successive semicircles is l 1 , l 2 , l 3 , l 4 , . . . with centres at A, B, A, B, . . ., respectively.] Fig. 5.4 Fig. 5.5 A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)] Reprint 2025-26 Fig. 5.6 ARITHMETIC PROGRESSIONS 71 1. Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0] 2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. 3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. 5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. the bottom rungs are 1 2 2 m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = 250 1 25 ] [Hint : Sx – 1 = S49 – Sx ] Each step has a rise of 1 4 m and a tread of 1 2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. EXERCISE 5.4 (Optional)* Fig. 5.7 * These exercises are not from the examination point of view. [Hint : Volume of concrete required to build the first step = 1 1 3 50 m 4 2 ] Reprint 2025-26 Fig. 5.8 72 MATHEMATICS 5.5 Summary In this chapter, you have studied the following points : 1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference. 2. A given list of numbers a1 , a2 , a3 , . . . is an AP, if the differences a2 – a1 , a3 – a2 , 3. In an AP with first term a and common difference d, the nth term (or the general term) is given by an = a + (n – 1) d. 4. The sum of the first n terms of an AP is given by : 5. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by : If a, b, c are in AP, then b = 2 a c and b is called the arithmetic The general form of an AP is a, a + d, a + 2d, a + 3d, . . . a4 – a3 , . . ., give the same value, i.e., if ak + 1 – ak is the same for different values of k. A NOTE TO THE READER S = 2 ( 1) 2 n an d S = ( ) 2 n a l mean of a and c. Reprint 2025-26" class_10,6,Triangles,ncert_books/class_10/jemh1dd/jemh106.pdf,"6.1 Introduction You are familiar with triangles and many of their properties from your earlier classes. In Class IX, you have studied congruence of triangles in detail. Recall that two figures are said to be congruent, if they have the same shape and the same size. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier. some long distant objects (say moon) have been found out? Do you think these have TRIANGLES 73 Can you guess how heights of mountains (say Mount Everest) or distances of TRIANGLES 6 Reprint 2025-26 74 MATHEMATICS been measured directly with the help of a measuring tape? In fact, all these heights and distances have been found out using the idea of indirect measurements, which is based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3 and also Chapters 8 and 9 of this book). 6.2 Similar Figures In Class IX, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. Now consider any two (or more) circles [see Fig. 6.1 (i)]. Are they congruent? Since all of them do not have the same radius, they are not congruent to each other. Note that some are congruent and some are not, but all of them have the same shape. So they all are, what we call, similar. Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar. What about two (or more) squares or two (or more) equilateral triangles [see Fig. 6.1 (ii) and (iii)]? As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar. From the above, we can say that all congruent figures are similar but the similar figures need not be congruent. Can a circle and a square be similar? Can a triangle and a square be similar? These questions can be answered by just looking at the figures (see Fig. 6.1). Evidently these figures are not similar. (Why?) Reprint 2025-26 Fig. 6.1 Fig. 6.2 TRIANGLES 75 (see Fig 6.2)?Are they similar? These figures appear to be similar but we cannot be certain about it.Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not. For this, let us look at the photographs given in Fig. 6.3: (Taj Mahal) but are in different sizes. Would you say that the three photographs are similar? Yes,they are. person one at the age of 10 years and the other at the age of 40 years? Are these photographs similar? These photographs are of the same size but certainly they are not of the same shape. So, they are not similar. from the same negative? You must have heard about the stamp size, passport size and postcard size photographs. She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line What can you say about the two quadrilaterals ABCD and PQRS You will at once say that they are the photographs of the same monument What can you say about the two photographs of the same size of the same What does the photographer do when she prints photographs of different sizes Fig. 6.3 segment in the bigger photograph (figure) will be 45 35 55 or 35 of that of the line segment. This really means that every line segment of the smaller photograph is enlarged (increased) in the ratio 35:45 (or 35:55). It can also be said that every line segment of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations(or angles) are always equal. This is the essence of the similarity of two figures and in particular of two polygons. We say that: corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Two polygons of the same number of sides are similar, if (i) their Reprint 2025-26 76 MATHEMATICS factor (or the Representative Fraction) for the polygons. You must have heard that world maps (i.e., global maps) and blue prints for the construction of a building are prepared using a suitable scale factor and observing certain conventions. activity: Activity 1 : Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom. Let us cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table. Then a shadow of ABCD is cast on the table. Mark the outline of this shadow as ABCD (see Fig.6.4). Note that the quadrilateral ABCD is an enlargement (or magnification) of the quadrilateral ABCD. This is because of the property of light that light propogates in a straight line. You may also note that A lies on ray OA, B lies on ray OB, C lies on OC and D lies on OD. Thus, quadrilaterals ABCD and ABCD are of the same shape but of different sizes. Note that the same ratio of the corresponding sides is referred to as the scale In order to understand similarity of figures more clearly, let us perform the following Fig. 6.4 that quadrilateral ABCD is similar to the quadrilateral ABCD. corresponds to vertex B, vertex C corresponds to vertex C and vertex D corresponds to vertex D. Symbolically, these correspondences are represented as A A, B B, C C and D D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that similar, if (i) all the corresponding angles are equal and (ii) all the corresponding sides are in the same ratio (or proportion). So, quadrilateral ABCD is similiar to quadrilateral ABCD. We can also say Here, you can also note that vertex A corresponds to vertex A, vertex B (i) A = A, B = B, C = C, D = D and (ii) AB BC CD DA AB BC CD DA . This again emphasises that two polygons of the same number of sides are Reprint 2025-26 TRIANGLES 77 Fig. 6.5 are similar. Remark : You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon. Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the same ratio. From the above, you can easily say that quadrilaterals ABCD and PQRS of You may note that in the two quadrilaterals (a square and a rectangle) of Fig. 6.5 quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the same ratio, but their corresponding angles are not equal. Again, the two polygons (quadrilaterals) are not similar. So, the two quadrilaterals are not similar. Similarly, you may note that in the two Reprint 2025-26 Fig. 6.6 78 MATHEMATICS polygons is not sufficient for them to be similar. 1. Fill in the blanks using the correct word given in brackets : 2. Give two different examples of pair of Thus, either of the above two conditions (i) and (ii) of similarity of two (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are .(equal, proportional) (iii) All triangles are similar. (isosceles, equilateral) (ii) All squares are . (similar, congruent) (i) All circles are . (congruent, similar) EXERCISE 6.1 Fig. 6.7 3. State whether the following quadrilaterals are similar or not: (i) similar figures. (ii) non-similar figures. Reprint 2025-26 Fig. 6.8 TRIANGLES 79 6.3 Similarity of Triangles What can you say about the similarity of two triangles? for the similarity of two triangles. That is: Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same. To understand the Basic Proportionality Theorem, let us perform the following activity: Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB. You may recall that triangle is also a polygon. So, we can state the same conditions Two triangles are similiar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Thales (640 – 546 B.C.) Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9). Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from your constructions that AD 3 DB 2 ? Measure AE and EC. What about AE EC ? Observe that AE EC is also equal to 3 2 . Thus, you can see that in ABC, DE || BC and AD AE DB EC . Is it a coincidence? No, it is due to the following theorem (known as the Basic Proportionality Theorem): Reprint 2025-26 Fig. 6.9 80 MATHEMATICS Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10). We need to prove that AD AE DB EC . Let us join BE and CD and then draw DM AC and EN AB. Now, area of ADE (= 1 2 base × height) = 1 2 AD × EN. Recall from Class IX, that area of ADE is denoted as ar(ADE). So, ar(ADE) = 1 2 AD × EN Similarly, ar(BDE) = 1 2 DB × EN, ar(ADE) = 1 2 AE × DM and ar(DEC) = 1 2 EC × DM. Fig. 6.10 Therefore, ar(ADE) ar(BDE) = and ar(ADE) ar(DEC) = Note that BDE and DEC are on the same base DE and between the same parallels BC and DE. So, ar(BDE) = ar(DEC) (3) 1 AD × EN 2 AD 1 DB DB × EN 2 1 AE × DM 2 AE 1 EC EC × DM 2 Reprint 2025-26 (1) (2) TRIANGLES 81 Therefore, from (1), (2) and (3), we have : Appendix 1)? To examine this, let us perform the following activity: Activity 3 : Draw an angle XAY on your notebook and on ray AX, mark points B1 , B2 , B3 , B4 and B such that AB1 = B1 B2 = B2 B3 = B3 B4 = B4 B. Similarly, on ray AY, mark points C1 , C2 , C3 , C4 and C such that AC1 = C1 C2 = C2 C3 = C3 C4 = C4 C. Then join B1 C1 and BC (see Fig. 6.11). Similarly, by joining B2 C2 , B3 C3 and B4 C4 , you can see that: Is the converse of this theorem also true (For the meaning of converse, see Note that 1 You can also see that lines B1 C1 and BC are parallel to each other, i.e., 2 AB B B = 2 2 AC C C 1 AB B B = 1 AD DB = AE EC 2 1 AC C C (Each equal to 1 4 ) B1 C1 || BC (1) 2 3 and B2 C2 || BC (2) Fig. 6.11 triangle in the same ratio, then the line is parallel to the third side. taking any number of equal parts on arms AX and AY . Each time, you will arrive at the same result. Thus, we obtain the following theorem, which is the converse of Theorem 6.1: From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a You can repeat this activity by drawing any angle XAY of different measure and 4 AB B B = 4 4 AC C C 3 AB B B = 3 4 3 3 AC C C Reprint 2025-26 3 2 and B3 C3 || BC (3) 4 1 and B4 C4 || BC (4) 82 MATHEMATICS Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. This theorem can be proved by taking a line DE such that AD AE DB EC and assuming that DE is not parallel to BC (see Fig. 6.12). If DE is not parallel to BC, draw a line DE parallel to BC. So, AD DB = AE E C (Why ?) Therefore, AE EC = AE E C (Why ?) (Why ?) Let us take some examples to illustrate the use of the above theorems. Example 1 : If a line intersects sides AB and AC of a ABC at D and E respectively and is parallel to BC, prove that AD AB = AE AC (see Fig. 6.13). Solution : DE || BC (Given) Adding 1 to both sides of above, you can see that E and E must coincide. Fig. 6.12 So, AD DB = AE EC (Theorem 6.1) or, DB AD = EC AE or, DB 1 AD = EC 1 AE or, AB AD = AC AE So, AD AB = AE AC Reprint 2025-26 Fig. 6.13 TRIANGLES 83 Example 2 : ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig. 6.14). Show that AE BF ED FC . Solution : Let us join AC to intersect EF at G (see Fig. 6.15). So, EF || DC (Lines parallel to the same line are parallel to each other) Now, in ADC, So, AE ED = AG GC (Theorem 6.1) (1) Similarly, from CAB, i.e., AG GC = BF FC (2) AB || DC and EF || AB (Given) EG || DC (As EF || DC) CG AG = CF BF Fig. 6.15 Fig. 6.14 Therefore, from (1) and (2), Example 3 : In Fig. 6.16, PS SQ = PT TR and PST = PRQ. Prove that PQR is an isosceles triangle. Solution : It is given that PS PT SQ TR So, ST || QR (Theorem 6.2) Therefore, PST = PQR (Corresponding angles) (1) AE ED = BF FC Reprint 2025-26 Fig. 6.16 84 MATHEMATICS Also, it is given that So, PRQ = PQR [From (1) and (2)] Therefore, PQ = PR (Sides opposite the equal angles) i.e., PQR is an isosceles triangle. 1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). 2. E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR : PST = PRQ (2) EXERCISE 6.2 Fig. 6.17 3. In Fig. 6.18, if LM || CB and LN || CD, prove that 4. In Fig. 6.19, DE || AC and DF || AE. Prove that (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm AM AN BF BE FE EC AB AD (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm Reprint 2025-26 Fig. 6.19 Fig. 6.18 TRIANGLES 85 6.4 Criteria for Similarity of Triangles In the previous section, we stated that two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). 10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that 5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR. 6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. 7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). 8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). 9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO CO BO DO AO CO BO DO Show that ABCD is a trapezium. Fig. 6.20 Fig. 6.21 That is, in ABC and DEF, if (i) A = D, B = E, C = F and (ii) AB BC CA , DE EF FD then the two triangles are similar (see Fig. 6.22). Reprint 2025-26 Fig. 6.22 86 MATHEMATICS corresponds to F. Symbolically, we write the similarity of these two triangles as ‘ ABC ~ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’. The symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘’ for ‘is congruent to’ in Class IX. ABC and DEF, should we always look for all the equality relations of their corresponding angles ( A = D, B = E, C = F) and all the equality relations of the ratios of their corresponding sides AB BC CA DE EF FD ? Let us examine. You may recall that in Class IX, you have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts (or elements) of the two triangles. Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts. For this, let us perform the following activity: Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23). It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices. For example, for the triangles ABC and DEF of Fig. 6.22, we cannot write ABC ~ EDF or ABC ~ FED. However, we can write BAC ~ EDF. Here, you can see that A corresponds to D, B corresponds to E and C Now a natural question arises : For checking the similarity of two triangles, say Reprint 2025-26 Fig. 6.23 TRIANGLES 87 other at D. In the two triangles ABC and DEF, you can see that B = E, C = F and A = D. That is, corresponding angles of these two triangles are equal. What can you say about their corresponding sides ? Note that BC 3 0.6. EF 5 What about AB DE and CA FD ? On measuring AB, DE, CA and FD, you will find that AB DE and CA FD are also equal to 0.6 (or nearly equal to 0.6, if there is some error in the measurement). Thus, AB BC CA DE EF FD You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two triangles. Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles. This theorem can be proved by taking two triangles ABC and DEF such that A = D, B = E and C = F (see Fig. 6.24) Cut DP = AB and DQ = AC and join PQ. Let rays BP and CQ intersect each other at A and rays ER and FS intersect each Fig. 6.24 So, ABC DPQ (Why ?) This gives B = P = E and PQ || EF (How?) Therefore, DP PE = DQ QF (Why?) i.e., AB DE = AC DF (Why?) Similarly, AB DE = BC EF and so AB BC AC DE EF DF . Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal. Therefore, AAA similarity criterion can also be stated as follows: Reprint 2025-26 88 MATHEMATICS triangle, then the two triangles are similar. equal to the three angles of another triangle, then their corresponding sides are proportional (i.e., in the same ratio). What about the converse of this statement? Is the converse true? In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal? Let us examine it through an activity : Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25). So, you have : AB BC CA DE EF FD (each equal to 2 3 ) If two angles of one triangle are respectively equal to two angles of another This may be referred to as the AA similarity criterion for two triangles. You have seen above that if the three angles of one triangle are respectively Now measure A, B, C, D, E and F. You will observe that Fig. 6.25 A = D, B = E and C = F, i.e., the corresponding angles of the two triangles are equal. in the same ratio). Everytime you shall see that their corresponding angles are equal. It is due to the following criterion of similarity of two triangles: Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar. two triangles. This theorem can be proved by taking two triangles ABC and DEF such that AB BC CA DE EF FD (< 1) (see Fig. 6.26): You can repeat this activity by drawing several such triangles (having their sides This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for Reprint 2025-26 TRIANGLES 89 Cut DP = AB and DQ = AC and join PQ. It can be seen that DP PE = DQ QF and PQ || EF (How?) So, P = E and Q = F. Therefore, DP DE = DQ DF = PQ EF So, DP DE = DQ DF = BC EF (Why?) So, BC = PQ (Why?) Thus, ABC DPQ (Why ?) Fig. 6.26 So, A = D, B = E and C = F (How ?) Remark : You may recall that either of the two conditions namely, (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other. Class IX. You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles. For this, let us perform an activity. Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm, A = 50°, AC = 4 cm, DE = 3 cm, D = 50° and DF = 6 cm (see Fig.6.27). Let us now recall the various criteria for congruency of two triangles learnt in Reprint 2025-26 90 MATHEMATICS between the sides AB and AC) = D (included between the sides DE and DF). That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion). Now let us measure B, C, E and F. C = F. So, by AAA similarity criterion, ABC ~ DEF. You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional. Everytime, you will find that the triangles are similar. It is due to the following criterion of similarity of triangles: Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. Here, you may observe that AB DE = AC DF (each equal to 2 3 ) and A (included You will find that B = E and C = F. That is, A = D, B = E and Fig. 6.27 This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles. As before, this theorem can be proved by taking two triangles ABC and DEF such that (see Fig. 6.28). Cut DP = AB, DQ = AC and join PQ. Fig. 6.28 AB AC DE DF ( 1) and A = D Reprint 2025-26 TRIANGLES 91 Now, PQ || EF and ABC DPQ (How ?) So, A = D, B = P and C = Q Therefore, ABC ~ DEF (Why?) Example 4 : In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR. Solution : PQ || RS (Given) So, P = S (Alternate angles) and Q = R Also, POQ = SOR (Vertically opposite angles) Therefore, POQ ~ SOR (AAA similarity criterion) Example 5 : Observe Fig. 6.30 and then find P. We now take some examples to illustrate the use of these criteria. Fig. 6.29 Solution : In ABC and PQR, Reprint 2025-26 Fig. 6.30 92 MATHEMATICS That is, AB BC CA RQ QP PR So, ABC ~ RQP (SSS similarity) Therefore, C = P (Corresponding angles of similar triangles) But C = 180° – A – B (Angle sum property) So, P = 40° Example 6 : In Fig. 6.31, Show that A = C and B = D. Solution : OA . OB = OC . OD (Given) So, OA OC = OD OB (1) Also, we have AOD = COB (Vertically opposite angles) (2) Therefore, from (1) and (2), AOD ~ COB (SAS similarity criterion) So, A = C and D = B (Corresponding angles of similar triangles) OA . OB = OC . OD. AB 3.8 1 , RQ 7.6 2 BC 6 1 QP 12 2 and CA 3 3 1 PR 2 6 3 = 180° – 80° – 60° = 40° Fig. 6.31 Example 7 : A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. Solution : Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32). From the figure, you can see that DE is the shadow of the girl. Let DE be x metres. Reprint 2025-26 Fig. 6.32 TRIANGLES 93 Now, BD = 1.2 m × 4 = 4.8 m. Note that in ABE and CDE, and E = E (Same angle) So, ABE ~ CDE (AA similarity criterion) Therefore, BE DE = AB CD i.e., 4.8 + x x = 3.6 0.9 (90 cm = 90 100 m = 0.9 m) i.e., 4.8 + x =4x i.e., 3x = 4.8 i.e., x = 1.6 So, the shadow of the girl after walking for 4 seconds is 1.6 m long. Example 8 : In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR. If ABC ~ PQR, prove that : (ii) CM AB RN PQ (i) AMC ~ PNR B = D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground) (iii) CMB ~ RNQ Solution : (i) ABC ~ PQR (Given) So, AB PQ = BC CA QR RP (1) and A = P, B = Q and C = R (2) But AB = 2 AM and PQ = 2 PN (As CM and RN are medians) So, from (1), 2AM 2PN = CA RP Reprint 2025-26 Fig. 6.33 94 MATHEMATICS i.e., AM PN = CA RP (3) Also, MAC = NPR [From (2)] (4) So, from (3) and (4), (ii) From (5), CM RN = CA RP (6) But CA RP = AB PQ [From (1)] (7) Therefore, CM RN = AB PQ [From (6) and (7)] (8) (iii) Again, AB PQ = BC QR [From (1)] Therefore, CM RN = BC QR [From (8)] (9) Also, CM RN = AB 2 BM PQ 2 QN i.e., CM RN = BM QN (10) AMC ~ PNR (SAS similarity) (5) i.e., CM RN = BC BM QR QN [From (9) and (10)] Therefore, CMB ~ RNQ (SSS similarity) [Note : You can also prove part (iii) by following the same method as used for proving part (i).] 1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : EXERCISE 6.3 Reprint 2025-26 TRIANGLES 95 Fig. 6.34 2. In Fig. 6.35, ODC ~ OBA, BOC = 125° 3. Diagonals AC and BD of a trapezium ABCD and CDO = 70°. Find DOC, DCO and OAB. with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA OB OC OD Reprint 2025-26 Fig. 6.35 96 MATHEMATICS 4. In Fig. 6.36, QR QT QS PR and 1 = 2. Show that PQS ~ TQR. 5. S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS. 6. In Fig. 6.37, if ABE ACD, show that ADE ~ ABC. 7. In Fig. 6.38, altitudes AD and CE of ABC intersect each other at the point P. Show that: 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ~ CFB. 9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) AEP ~ CDP (ii) ABD ~ CBE (iii) AEP ~ ADB (iv) PDC ~ BEC Fig. 6.37 Fig. 6.36 10. CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively. If ABC ~ FEG, show that: (i) ABC ~ AMP (ii) CA BC PA MP (i) CD AC GH FG (ii) DCB ~ HGE (iii) DCA ~ HGF Reprint 2025-26 Fig. 6.38 Fig. 6.39 TRIANGLES 97 6.5 Summary 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of PQR (see Fig. 6.41). Show that ABC ~ PQR. 13. D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that CA2 = CB.CD. 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ABC ~ PQR. 15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. 16. If AD and PM are medians of triangles ABC and PQR, respectively where 11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD BC and EF AC, prove that ABD ~ ECF. ABC ~ PQR, prove that AB AD PQ PM Fig. 6.40 Fig. 6.41 In this chapter you have studied the following points : 1. Two figures having the same shape but not necessarily the same size are called similar figures. 2. All the congruent figures are similar but the converse is not true. 3. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion). 4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. 6. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion). 7. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion). Reprint 2025-26 98 MATHEMATICS 8. If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion). 9. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity criterion). If in two right triangles, hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle, then the two triangles are similar. This may be referred to as the RHS Similarity Criterion. If you use this criterion in Example 2, Chapter 8, the proof will become simpler. A NOTE TO THE READER Reprint 2025-26" class_10,7,Coordinate Geometry,ncert_books/class_10/jemh1dd/jemh107.pdf,"7.1 Introduction In Class IX, you have studied that to locate the position of a point on a plane, we require a pair of coordinate axes. The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y). Here is a play for you. Draw a set of a pair of perpendicular axes on a graph paper. Now plot the following points and join them as directed: Join the point A(4, 8) to B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle. Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points (0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got? COORDINATE GEOMETRY 99 COORDINATE GEOMETRY 7 ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically, gives a straight line. Further, in Chapter 2, you have seen the graph of y = ax 2 + bx + c (a ¹ 0), is a parabola. In fact, coordinate geometry has been developed as an algebraic tool for studying geometry of figures. It helps us to study geometry using algebra, and understand algebra with the help of geometry. Because of this, coordinate geometry is widely applied in various fields such as physics, engineering, navigation, seismology and art! whose coordinates are given. You will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio. Also, you have seen that a linear equation in two variables of the form In this chapter, you will learn how to find the distance between the two points Reprint 2025-26 100 MATHEMATICS 7.2 Distance Formula Let us consider the following situation: A town B is located 36 km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it. Let us see. This situation can be represented graphically as shown in Fig. 7.1. You may use the Pythagoras Theorem to calculate this distance. Now, suppose two points lie on the x-axis. Can we find the distance between them? For instance, consider two points A(4, 0) and B(6, 0) in Fig. 7.2. The points A and B lie on the x-axis. From the figure you can see that OA = 4 units and OB = 6 units. Therefore, the distance of B from A, i.e., AB = OB – OA = 6 – 4 = 2 units. So, if two points lie on the x-axis, we can easily find the distance between them. Now, suppose we take two points lying on the y-axis. Can you find the distance between them. If the points C(0, 3) and D(0, 8) lie on the y-axis, similarly we find that CD = 8 – 3 = 5 units (see Fig. 7.2). Fig. 7.2 Fig. 7.1 OC = 3 units, the distance of A from C, i.e., AC = 2 2 3 4 = 5 units. Similarly, you can find the distance of B from D = BD = 10 units. distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see an example. In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use Pythagoras theorem to find the distance between them? Let us draw PR and QS perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0), respectively. So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units. Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and Now, if we consider two points not lying on coordinate axis, can we find the Reprint 2025-26 COORDINATE GEOMETRY 101 Therefore, QT = 2 units and PT = RS = 2 units. Now, using the Pythagoras theorem, we have So, PQ = 2 2 units How will we find the distance between two points in two different quadrants? Consider the points P(6, 4) and Q(–5, –3) (see Fig. 7.4). Draw QS perpendicular to the x-axis. Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R. PQ2 = PT2 + QT2 = 22 + 22 = 8 Fig. 7.3 PQ = 2 2 11 7 = 170 units. Then PT = 11 units and QT = 7 units. (Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get Reprint 2025-26 Fig. 7.4 102 MATHEMATICS Let us now find the distance between any two points P(x1 , y1 ) and Q(x2 , y2). Draw PR and QS perpendicular to the x-axis. A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. 7.5). Then, OR = x1 , OS = x2 . So, RS = x2 – x1 = PT. Also, SQ = y2 , ST = PR = y1 . So, QT = y2 – y1 . Now, applying the Pythagoras theorem in PTQ, we get Therefore, PQ = 2 2 21 21 x x yy root. So, the distance between the points P(x1 , y1 ) and Q(x2 , y2 ) is which is called the distance formula. Remarks : 1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by 2. We can also write, PQ = 2 2 12 1 2 xx y y . (Why?) Note that since distance is always non-negative, we take only the positive square PQ2 = PT2 + QT2 PQ = 2 2 21 21 x – +– x yy , = (x2 – x1 ) 2 + (y2 – y1 )2 OP = 2 2 x y . Fig. 7.5 Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed. Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. PQ = 2 2 22 (3 2) (2 3) 5 5 50 = 7.07 (approx.) QR = 2 2 22 (–2 – 2) (–3 – 3) (– 4) (– 6) 52 = 7.21 (approx.) PR = 2 22 2 (3 – 2) (2 – 3) 1 ( 1) 2 = 1.41 (approx.) Reprint 2025-26 COORDINATE GEOMETRY 103 Also, PQ2 + PR2 = QR2 , by the converse of Pythagoras theorem, we have P = 90°. Therefore, PQR is a right triangle. Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square. Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now, Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square. Alternative Solution : We find the four sides and one diagonal, say, AC as above. Here AD2 + DC2 = 34 + 34 = 68 = AC2 . Therefore, by the converse of Pythagoras theorem, D = 90°. A quadrilateral with all four sides equal and one angle 90° is a square. So, ABCD is a square. AB = 2 2 (1 – 4) (7 2) 9 25 34 BC = 2 2 (4 1) (2 1) 25 9 34 CD = 2 2 (–1 4) (–1 – 4) 9 25 34 DA = 2 2 (1 4) (7 – 4) 25 9 34 AC = 2 2 (1 1) (7 1) 4 64 68 BD = 2 2 (4 4) (2 4) 64 4 68 Example 3 : Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. Fig. 7.6 Reprint 2025-26 104 MATHEMATICS Solution : Using the distance formula, we have Since, AB + BC = 3 2 2 2 5 2 AC, we can say that the points A, B and C are collinear. Therefore, they are seated in a line. Example 4 : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5). Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). We are given that AP = BP. So, AP2 = BP2 i.e., (x – 7)2 + (y – 1)2 =(x – 3)2 + (y – 5)2 i.e., x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25 i.e., x – y =2 which is the required relation. Remark : Note that the graph of the equation x – y = 2 is a line. From your earlier studies, you know that a point which is equidistant from A and B lies on the perpendicular bisector of AB. Therefore, the graph of x – y = 2 is the perpendicular bisector of AB (see Fig. 7.7). AB = 2 2 (6 3) (4 1) 9 9 18 3 2 BC = 2 2 (8 – 6) (6 – 4) 4 4 8 2 2 AC = 2 2 (8 – 3) (6 – 1) 25 25 50 5 2 Example 5 : Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3). Solution : We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B. Then (6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2 i.e., 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y i.e., 4y = 36 i.e., y =9 Reprint 2025-26 Fig. 7.7 COORDINATE GEOMETRY 105 So, the required point is (0, 9). Let us check our solution : AP = 2 2 (6 – 0) (5 – 9) 36 16 52 Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB. 1. Find the distance between the following pairs of points : 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2. 3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. 4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct. (i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b) BP = 2 2 (– 4 – 0) (3 – 9) 16 36 52 EXERCISE 7.1 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: 7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). 8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. (iii) (4, 5), (7, 6), (4, 3), (1, 2) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) Reprint 2025-26 Fig. 7.8 106 MATHEMATICS 7.3 Section Formula Let us recall the situation in Section 7.2. Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1 : 2 (see Fig. 7.9). If we take A as the origin O, and 1 km as one unit on both the axis, the coordinates of B will be (36, 15). In order to know the position of the tower, we must know the coordinates of P. How do we find these coordinates? x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion, studied in Chapter 6, POD and BPC are similar. Therefore , OD OP 1 PC PB 2 , and PD OP 1 BC PB 2 So, 1 36 2 x x and 1 15 2 y y These equations give x = 12 and y = 5. 10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). 9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR. Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the Fig. 7.9 You can check that P(12, 5) meets the condition that OP : PB = 1 : 2. Now let us use the understanding that you may have developed through this example to obtain the general formula. Consider any two points A(x1 , y1 ) and B(x2 , y2 ) and assume that P (x, y) divides AB internally in the ratio m1 : m2 , i.e., 2 PA PB m m (see Fig. 7.10). 1 Reprint 2025-26 Fig. 7.10 COORDINATE GEOMETRY 107 the x-axis. Then, by the AA similarity criterion, Therefore, PA AQ BP PC = PQ BC (1) Now, AQ = RS = OS – OR = x – x1 PC = ST = OT – OS = x2 – x PQ = PS – QS = PS – AR = y – y1 BC = BT– CT = BT – PS = y2 – y Substituting these values in (1), we get Taking 1 2 m m = 1 2 x x x x , we get x = 12 21 1 2 mx m x m m Similarly, taking 1 2 m m = 1 2 y y y y , we get y = 12 21 So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x1 , y1 ) and B(x2 , y2 ), internally, in the ratio m1 : m2 are Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to This is known as the section formula. 2 m m = 1 1 2 2 x x yy x xy y 1 12 12 , mx m x my m y mm mm (2) PAQ ~ BPC 12 21 1 2 2 1 1 2 my my m m y-axis and proceeding as above. 21 2 1 , 1 1 kx x ky y k k Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x1 , y1 ) and B(x2 , y2 ) is This can also be derived by drawing perpendiculars from A, P and B on the If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be 1 2 1 2 1 21 2 1 11 1 , , 11 11 2 2 x x y y x xyy . Let us solve a few examples based on the section formula. Reprint 2025-26 108 MATHEMATICS Example 6 : Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. Solution : Let P(x, y) be the required point. Using the section formula, we get Therefore, (7, 3) is the required point. Example 7 : In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)? Solution : Let (– 4, 6) divide AB internally in the ratio m1 : m2 . Using the section formula, we get Recall that if (x, y) = (a, b) then x = a and y = b. So, – 4 = 1 2 1 2 3 6 m m m m and 1 2 1 2 8 10 6 m m m m Now, – 4 = 1 2 1 2 3 6 m m m m gives us – 4m1 – 4m2 =3m1 – 6m2 i.e., 7m1 =2m2 i.e., m1 : m2 = 2 : 7 x = 3(8) 1(4) 7 3 1 , y = 3(5) 1(–3) 3 3 1 (– 4, 6) = 12 1 2 12 1 2 3 6 –8 10 , mm m m mm m m (1) You should verify that the ratio satisfies the y-coordinate also. Now, 1 2 1 2 8 10 m m m m = = 2 8 10 7 6 2 1 7 8 10 Reprint 2025-26 m m m m 1 2 1 2 (Dividing throughout by m2 ) 1 COORDINATE GEOMETRY 109 Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7. Alternatively : The ratio m1 : m2 can also be written as 1 2 :1, m m or k : 1. Let (– 4, 6) divide AB internally in the ratio k : 1. Using the section formula, we get So, – 4 = 3 6 1 k k i.e., – 4k – 4 = 3k – 6 i.e., 7k =2 i.e., k : 1 = 2 : 7 You can check for the y-coordinate also. So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7. Note : You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear. Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4). (– 4, 6) = 3 6 8 10 , 1 1 k k k k (2) Solution : Let P and Q be the points of trisection of AB i.e., AP = PQ = QB (see Fig. 7.11). Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are 1( 7) 2(2) 1(4) 2( 2) , 12 12 , i.e., (–1, 0) 2( 7) 1(2) 2(4) 1( 2) , 21 21 , i.e., (– 4, 2) Fig. 7.11 Reprint 2025-26 110 MATHEMATICS Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2). Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we could have obtained its coordinates using the mid-point formula. Example 9 : Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection. Solution : Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are 54 6 , 1 1 k k k k This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. Therefore, 5 1 k k =0 So, k =5 That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. 13 0, 3 . Solution : We know that diagonals of a parallelogram bisect each other. So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD i.e., 6 91 4 , 2 2 = 8 23 , 2 2 p i.e., 15 5, 2 2 = 8 5, 2 2 p so, 15 2 = 8 2 p i.e., p =7 Reprint 2025-26 COORDINATE GEOMETRY 111 1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. 2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). 3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown 4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6). 5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. 6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). 8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that in Fig. 7.12. Niharika runs 1 4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1 5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? EXERCISE 7.2 Fig. 7.12 AP = 3 AB 7 and P lies on the line segment AB. 9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts. 10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus = 1 2 (product of its diagonals)] Reprint 2025-26 112 MATHEMATICS 7.4 Summary In this chapter, you have studied the following points : 1. The distance between P(x1 , y1 ) and Q(x2 , y2 ) is 2 2 21 21 ( ) ( ). xx yy 2. The distance of a point P(x, y) from the origin is 2 2 x y . 3. The coordinates of the point P(x, y) which divides the line segment joining the points A(x 1, y 1) and B(x 2, y 2) internally in the ratio m1 : m2 are 12 1 2 , mx m x my m y mm m m 4. The mid-point of the line segment joining the points P(x1 , y1 ) and Q(x2 , y2 ) is Section 7.3 discusses the Section Formula for the coordinates (x, y) of a point P which divides internally the line segment joining the points A(x1 , y1 ) and B(x2 , y2 ) in the ratio m1 : m2 as follows : Note that, here, PA : PB = m1 : m2 . 1 21 2 , 2 2 x xy y . 12 21 1 2 21 x = 12 21 1 2 mx mx m m , y = 12 21 1 2 my my m m A NOTETOTHE READER However, if P does not lie between A and B but lies on the line AB, outside the line segment AB, and PA : PB = m1 : m2 , we say that P divides externally the line segment joining the points A and B. You will study Section Formula for such case in higher classes. Reprint 2025-26" class_10,8,Introduction to Trigonometry,ncert_books/class_10/jemh1dd/jemh108.pdf,"8.1 Introduction You have already studied about triangles, and in particular, right triangles, in your earlier classes. Let us take some examples from our surroundings where right triangles can be imagined to be formed. For instance : INTRODUCTION TO TRIGONOMETRY 113 1. Suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, as shown in Fig 8.1. Can the student find out the height of the Minar, without actually measuring it? There is perhaps nothing which so occupies the middle position of mathematics as trigonometry. INTRODUCTION TO TRIGONOMETRY – J.F. Herbart (1890) 8 2. Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation as shown in Fig.8.2. If you know the height at which the person is sitting, can you find the width of the river? Reprint 2025-26 Fig. 8.1 Fig. 8.2 114 MATHEMATICS some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. respect to its acute angles, called trigonometric ratios of the angle. We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also. We will also define the trigonometric ratios for angles of measure 0° and 90°. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities. 3. Suppose a hot air balloon is flying in the air. A girl happens to spot the balloon in the sky and runs to her mother to tell her about it. Her mother rushes out of the house to look at the balloon.Now when the girl had spotted the balloon intially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of B from the ground? In all the situations given above, the distances or heights can be found by using In this chapter, we will study some ratios of the sides of a right triangle with Fig. 8.3 8.2 Trigonometric Ratios In Section 8.1, you have seen some right triangles imagined to be formed in different situations. Let us take a right triangle ABC as shown in Fig. 8.4. Here, CAB (or, in brief, angle A) is an acute angle. Note the position of the side BC with respect to angle A. It faces A. We call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of A. So, we call it the side adjacent to angle A. Fig. 8.4 Reprint 2025-26 INTRODUCTION TO TRIGONOMETRY 115 Note that the position of sides change when you consider angle C in place of A (see Fig. 8.5). You have studied the concept of ‘ratio’ in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The trigonometric ratios of the angle A in right triangle ABC (see Fig. 8.4) are defined as follows : sine of A = side opposite to angle A BC hypotenuse AC cosine of A = side adjacent to angle A AB hypotenuse AC tangent of A = side opposite to angle A BC side adjacent to angle A AB cosecant of A = 1 hypotenuse AC sine of A side opposite to angle A BC secant of A = 1 hypotenuse AC cosine of A side adjacent to angle A AB Fig. 8.5 and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A. relationship between the angle and the length of its sides. triangle? (See Fig. 8.5) cotangent of A = 1 side adjacent to angle A AB tangent of A side opposite to angle A BC The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A Also, observe that tan A = So, the trigonometric ratios of an acute angle in a right triangle express the Why don’t you try to define the trigonometric ratios for angle C in the right BC BC sin A AC AB cos A AB AC Reprint 2025-26 and cot A = cosA sin A . 116 MATHEMATICS Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning. Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also. The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’. The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’. Aryabhata C.E. 476 – 550 Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in QAN? Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the triangles PAM and CAB are similar. Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional. So, we have AM AB = AP MP AC BC To answer this, first look at these triangles. Is PAM similar to CAB? From Reprint 2025-26 Fig. 8.6 INTRODUCTION TO TRIGONOMETRY 117 From this, we find MP AP = BC sin A AC . Similarly, AM AB AP AC = cos A, MP BC tan A AM AB and so on. This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB. trigonometric ratios) remains the same in QAN also. ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same. Note : For the sake of convenience, we may write sin2 A, cos2 A, etc., in place of (sin A)2 , (cos A)2 , etc., respectively. But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A). sin–1 A has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Sometimes, the Greek letter (theta) is also used to denote an angle. of the ratios, can we obtain the other ratios? Let us see. If in a right triangle ABC, sin A = 1 , 3 then this means that BC 1 AC 3 , i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1 : 3 (see Fig. 8.7). So if BC is equal to k, then AC will be 3k, where k is any positive number. To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB. Do you remember the Pythagoras theorem? Let us use it to determine the required length AB. AB2 = AC2 – BC2 = (3k) 2 – (k)2 = 8k2 = (2 2 k)2 In the same way, you should check that the value of sin A (and also of other From our observations, it is now clear that the values of the trigonometric We have defined six trigonometric ratios of an acute angle. If we know any one Therefore, AB = 2 2 k So, we get AB = 2 2 k (Why is AB not – 2 2 k ?) Now, cos A = AB 2 2 2 2 AC 3 3 k k Similarly, you can obtain the other trigonometric ratios of the angle A. Reprint 2025-26 Fig. 8.7 118 MATHEMATICS Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1). Let us consider some examples. Example 1 : Given tan A = 4 3 , find the other trigonometric ratios of the angle A. Solution : Let us first draw a right ABC (see Fig 8.8). Now, we know that tan A = BC 4 AB 3 . Therefore, if BC = 4k, then AB = 3k, where k is a positive number. Now, by using the Pythagoras Theorem, we have So, AC = 5k Now, we can write all the trigonometric ratios using their definitions. cos A = AB 3 3 AC 5 5 k k sin A = BC 4 4 AC 5 5 k k AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 Fig. 8.8 Therefore, cot A = 13 15 , cosec A = tan A 4 sin A 4 and sec A = 1 5 cos A 3 Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that B = Q. Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9). We have sin B = AC AB and sin Q = PR PQ Reprint 2025-26 Fig. 8.9 INTRODUCTION TO TRIGONOMETRY 119 Then AC AB = PR PQ Therefore, AC PR = AB , say PQ k (1) Now, using Pythagoras theorem, and QR = 2 2 PQ – PR So, BC QR = From (1) and (2), we have Then, by using Theorem 6.4, ACB ~ PRQ and therefore, B = Q. Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10). Determine the values of (ii) cos2 – sin2 (i) cos2 + sin2 , 2 2 22 22 AB AC PQ PR PQ PR PQ PR PQ PR PQ PR kk k k (2) 2 2 22 22 2 2 AC PR = AB BC PQ QR BC = 2 2 AB AC Solution : In ACB, we have So, sin = AC 20 BC 21 , cos = AB 29 AB 29 Now, (i) cos2 + sin2 = 2 2 2 2 and (ii) cos2 – sin2 = 2 2 AC = 2 2 AB BC = 2 2 (29) (21) = (29 21)(29 21) (8)(50) 400 20units 2 21 20 (21 20)(21 20) 41 29 29 841 29 . 2 20 21 20 21 400 441 1, 29 29 841 29 Reprint 2025-26 Fig. 8.10 120 MATHEMATICS Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1. Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11) i.e., BC = AB Let AB = BC = k, where k is a positive number. Now, AC = 2 2 AB BC Therefore, sin A = BC 1 AC 2 and cos A = AB 1 AC 2 So, 2 sin A cos A = 1 1 2 1, 2 2 which is the required value. Example 5 : In OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q. = 2 2 () () 2 k kk Fig. 8.11 Solution : In OPQ, we have i.e., (1 + PQ)2 = OP2 + PQ2 (Why?) i.e., 1 + PQ2 + 2PQ = OP2 + PQ2 i.e., 1 + 2PQ = 72 (Why?) i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm So, sin Q = 7 25 and cos Q = 24 25 OQ2 = OP2 + PQ2 Reprint 2025-26 Fig. 8.12 INTRODUCTION TO TRIGONOMETRY 121 10. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. 1. In ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A 2. In Fig. 8.13, find tan P – cot R. 3. If sin A = 3 , 4 calculate cos A and tan A. 4. Given 15 cot A = 8, find sin A and sec A. 5. Given sec = 13 , 12 calculate all other trigonometric ratios. 6. If A and B are acute angles such that cos A = cos B, then show that A = B. 7. If cot = 7 , 8 evaluate : (i) (1 sin )(1 sin ) , (1 cos )(1 cos ) (ii) cot2 8. If 3 cot A = 4, check whether 2 2 1 tan A 1 + tan A = cos2 A – sin2 A or not. 9. In triangle ABC, right-angled at B, if tan A = 1 , 3 find the value of: (ii) sin C, cos C (ii) cos A cos C – sin A sin C (i) sin A cos C + cos A sin C EXERCISE 8.1 Fig. 8.13 8.3 Trigonometric Ratios of Some Specific Angles From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°. In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°. 11. State whether the following are true or false. Justify your answer. (iv) cot A is the product of cot and A. (ii) sec A = 12 5 for some value of angle A. (iii) cos A is the abbreviation used for the cosecant of angle A. (v) sin = 4 3 for some angle . (i) The value of tan A is always less than 1. Reprint 2025-26 122 MATHEMATICS Trigonometric Ratios of 45° In ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., A = C = 45° (see Fig. 8.14). So, BC = AB (Why?) Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2 , and, therefore, AC = a 2 Using the definitions of the trigonometric ratios, we have : Also, cosec 45° = 1 2 sin 45 , sec 45° = 1 2 cos 45 , cot 45° = 1 1 tan 45 . cos 45° = side adjacent toangle 45° AB 1 hypotenuse AC 2 2 a a tan 45° = side opposite to angle 45° BC 1 side adjacent to angle 45° AB sin 45° = side opposite to angle 45° BC 1 hypotenuse AC 2 2 a a a a Fig. 8.14 Trigonometric Ratios of 30° and 60° Let us now calculate the trigonometric ratios of 30° and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, A = B = C = 60°. Draw the perpendicular AD from A to the side BC (see Fig. 8.15). Now ABD ACD (Why?) Therefore, BD = DC and BAD = CAD (CPCT) Now observe that: ABD is a right triangle, right-angled at D with BAD = 30° and ABD = 60° (see Fig. 8.15). Reprint 2025-26 Fig. 8.15 INTRODUCTION TO TRIGONOMETRY 123 As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a. Then, BD = 1 BC = 2 a and AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2 , Therefore, AD = a 3 Now, we have : Also, cosec 30° = 1 2, sin 30 sec 30° = 1 2 cos 30 3 Similarly, Trigonometric Ratios of 0° and 90° cosec 60° = 2 , 3 sec 60° = 2 and cot 60° = 1 3 tan 30° = BD 1 AD 3 3 a cot 30° = 1 3 tan 30 . sin 30° = BD 1 AB 2 2 a a , cos 30° = AD 3 3 AB 2 2 a a sin 60° = AD 3 3 AB 2 2 a a , cos 60° = 1 2 , tan 60° = 3 , a . Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC (see Fig. 8.16), till it becomes zero. As A gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B, and finally when A becomes very close to 0°, AC becomes almost the same as AB (see Fig. 8.17). Reprint 2025-26 Fig. 8.17 Fig. 8.16 124 MATHEMATICS When A is very close to 0°, BC gets very close to 0 and so the value of sin A = BC AC is very close to 0. Also, when A is very close to 0°, AC is nearly the same as AB and so the value of cos A = AB AC is very close to 1. This helps us to see how we can define the values of sin A and cos A when A = 0°. We define : sin 0° = 0 and cos 0° = 1. Now, let us see what happens to the trigonometric ratios of A, when it is made larger and larger in ABC till it becomes 90°. As A gets larger and larger, C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when A is very close to 90°, C becomes very close to 0° and the side AC almost coincides with side BC (see Fig. 8.18). Using these, we have : tan 0° = sin 0° cos 0° = 0, cot 0° = 1 , tan 0° which is not defined. (Why?) sec 0° = 1 cos 0 = 1 and cosec 0° = 1 , sin 0 which is again not defined.(Why?) When C is very close to 0°, A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when A is very close to 90°, C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0. We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60° and 90° in Table 8.1, for ready reference. So, we define : sin 90° = 1 and cos 90° = 0. Now, why don’t you find the other trigonometric ratios of 90°? Reprint 2025-26 Fig. 8.18 INTRODUCTION TO TRIGONOMETRY 125 Remark : From the table above you can observe that as A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. A 0° 30° 45° 60° 90° sin A 0 1 2 cos A 1 3 2 tan A 0 1 3 1 3 Not defined cosec A Not defined 2 2 2 3 1 sec A 1 2 3 2 2 Not defined cot A Not defined 3 1 1 3 0 Let us illustrate the use of the values in the table above through some examples. Table 8.1 1 2 3 2 1 1 2 1 2 0 Example 6 : In ABC, right-angled at B, AB = 5 cm and ACB = 30° (see Fig. 8.19). Determine the lengths of the sides BC and AC. Solution : To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore i.e., 5 BC = tan 30° = 1 which gives BC = 5 3 cm AB BC = tan C Reprint 2025-26 3 Fig. 8.19 126 MATHEMATICS To find the length of the side AC, we consider i.e., 1 2 = 5 AC i.e., AC = 10 cm Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above, i.e., AC = 222 2 AB BC 5 (5 3) cm = 10cm. Example 7 : In PQR, right-angled at Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. Determine QPR and PRQ. Solution : Given PQ = 3 cm and PR = 6 cm. Therefore, PQ PR = sin R or sin R = 3 1 6 2 So, PRQ = 30° sin 30° = AB AC (Why?) Fig. 8.20 and therefore, QPR = 60°. (Why?) You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined. Example 8 : If sin (A – B) = 1 , 2 cos (A + B) = 1 , 2 0° < A + B 90°, A > B, find A and B. Solution : Since, sin (A – B) = 1 2 , therefore, A – B = 30° (Why?) (1) Also, since cos (A + B) = 1 2 , therefore, A + B = 60° (Why?) (2) Solving (1) and (2), we get : A = 45° and B = 15°. Reprint 2025-26 INTRODUCTION TO TRIGONOMETRY 127 1. Evaluate the following : 2. Choose the correct option and justify your choice : (iv) 2 2 tan 30 1 tan 30 (iii) cos 45° sec 30° + cosec 30° (iv) sin 30° + tan 45° – cosec 60° sec 30° + cos 60° + cot 45° (iii) sin 2A = 2 sin A is true when A = (v) 2 22 2 2 5 cos 60 4 sec 30 tan 45 sin 30 cos 30 (ii) 2 2 1 tan 45 1 tan 45 (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° – sin2 60° (i) 2 2 tan 30 1 tan 30 (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (A) tan 90° (B) 1 (C) sin 45° (D) 0 (A) 0° (B) 30° (C) 45° (D) 60° EXERCISE 8.2 3. If tan (A + B) = 3 and tan (A – B) = 1 3 ; 0° < A + B 90°; A > B, find A and B. 4. State whether the following are true or false. Justify your answer. (iv) sin = cos for all values of . (iii) The value of cos increases as increases. (v) cot A is not defined for A = 0°. (ii) The value of sin increases as increases. (i) sin (A + B) = sin A + sin B. (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30° Reprint 2025-26 128 MATHEMATICS 8.4 Trigonometric Identities You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities. i.e., 2 2 AB BC AC AC = 2 AC AC i.e., (cos A)2 + (sin A)2 =1 i.e., cos2 A + sin2 A = 1 (2) In ABC, right-angled at B (see Fig. 8.21), we have: Dividing each term of (1) by AC2 , we get This is true for all A such that 0° A 90°. So, this is a trigonometric identity. Let us now divide (1) by AB2 . We get 2 2 AB BC AC AC = 2 2 AC AC AB2 + BC2 = AC2 (1) 2 2 Fig. 8.21 or, 2 2 AB BC AB AB = i.e., 1 + tan2 A = sec2 A (3) sec A are not defined for A = 90°. So, (3) is true for all A such that 0° A 90°. Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and Let us see what we get on dividing (1) by BC2 . We get 2 2 AB BC BC BC = 2 2 2 AB BC AB AB = 2 2 2 2 2 Reprint 2025-26 2 AC AB 2 AC AB 2 AC BC INTRODUCTION TO TRIGONOMETRY 129 i.e., 2 2 AB BC BC BC = i.e., cot2 A + 1 = cosec2 A (4) all A such that 0° < A 90°. trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios. tan A = 1 3 Then, cot A = 3 . Since, sec2 A = 1 + tan2 A = 1 4 1 , 3 3 sec A = 2 Again, sin A = 2 3 1 1 cos A 1 4 2 . Therefore, cosec A = 2. Example 9 : Express the ratios cos A, tan A and sec A in terms of sin A. Solution : Since cos2 A + sin2 A = 1, therefore, Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for Using these identities, we can express each trigonometric ratio in terms of other Let us see how we can do this using these identities. Suppose we know that 2 AC BC 3 , and cos A = 3 2 This gives cos A = 2 1 sin A (Why?) Hence, tan A = sin A cos A = 2 2 sin A 1 1 and sec A = 1 – sin A cos A 1 sin A Example 10 : Prove that sec A (1 – sin A)(sec A + tan A) = 1. Solution : LHS = sec A (1 – sin A)(sec A + tan A) = 1 1 sin A (1 sin A) cos A cos A cos A cos2 A = 1 – sin2 A, i.e., cos A = 2 1 sin A Reprint 2025-26 130 MATHEMATICS Example 11 : Prove that cot A – cos A cosec A – 1 cot A + cos A cosec A + 1 Solution : LHS = Example 12 : Prove that sin cos 1 1 , sin cos 1 sec tan using the identity sec2 = 1 + tan2 . Solution : Since we will apply the identity involving sec and tan , let us first convert the LHS (of the identity we need to prove) in terms of sec and tan by dividing numerator and denominator by cos cos A cos A cot A – cos A sin A cot A + cos A cos A cos A sin A = 1 1 cos A 1 1 sin A sin A cosec A – 1 1 1 cosec A + 1 cos A 1 1 sin A sin A = 2 = 2 2 2 (1 sin A)(1 + sin A) 1 sin A cos A cos A 2 cos A 1 cos A = RHS = RHS LHS = sin – cos + 1 tan 1 sec sin + cos – 1 tan 1 sec = (tan sec ) 1 {(tan sec ) 1} (tan sec ) (tan sec ) 1 {(tan sec ) 1} (tan sec ) = 2 2 (tan sec ) (tan sec ) {tan sec 1} (tan sec ) = – 1 tan sec (tan sec 1) (tan sec ) Reprint 2025-26 INTRODUCTION TO TRIGONOMETRY 131 which is the RHS of the identity, we are required to prove. 1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. 2. Write all the other trigonometric ratios of A in terms of sec A. 3. Choose the correct option. Justify your choice. 4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iv) 2 2 1 tan A 1 + cot A (iii) (sec A + tan A) (1 – sin A) = (ii) (1 + tan + sec ) (1 + cot – cosec ) = (A) 0 (B) 1 (C) 2 (D) –1 (i) 9 sec2 A – 9 tan2 A = (i) (cosec – cot )2 = 1 cos 1 cos (ii) cos A 1 sin A 2 sec A 1 + sin A cos A (A) 1 (B) 9 (C) 8 (D) 0 (A) sec A (B) sin A (C) cosec A (D) cos A (A) sec2 A (B) –1 (C) cot2 A (D) tan2 A = –1 1 , tan sec sec tan EXERCISE 8.3 (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A (iv) 2 1 sec A sin A sec A 1 – cos A [Hint : Simplify LHS and RHS separately] (vi) 1 sin A sec A + tan A 1 – sin A (vii) 3 3 sin 2 sin tan 2 cos cos (iii) tan cot 1 sec cosec 1 cot 1 tan (v) cos A – sin A + 1 cosec A + cot A, cos A + sin A – 1 using the identity cosec2 A = 1 + cot2 A. [Hint : Write the expression in terms of sin and cos ] Reprint 2025-26 132 MATHEMATICS 8.5 Summary In this chapter, you have studied the following points : 1. In a right triangle ABC, right-angled at B, 2. 1 1 1 sin A , cosec A = ; sec A = ; tan A = tan A = sin A cos A cot A cos A . 3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined. 4. The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°. 5. The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° £ A < 90°) or cosec A (0° < A £ 90º) is always greater than or equal to 1. 6. sin2 A + cos2 A = 1, sec2 A – tan2 A = 1 for 0° £ A < 90°, cosec2 A = 1 + cot2 A for 0° < A £ 90º. sin A = side opposite to angle A side adjacent to angle A , cos A = hypotenuse hypotenuse tan A = side opposite toangle A side adjacent to angle A . (ix) 1 (cosec A – sin A)(sec A – cos A) tan A + cot A (x) [Hint : Simplify LHS and RHS separately] 2 2 2 1 tan A 1 tan A 1 + cot A 1 – cot A = tan2 A Reprint 2025-26" class_10,9,Some Applications of Trigonometry,ncert_books/class_10/jemh1dd/jemh109.pdf,"SOME APPLICATIONS OF TRIGONOMETRY 133 9.1 Heights and Distances In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying about some ways in which trigonometry is used in the life around you. Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1. SOME APPLICATIONS OF TRIGONOMETRY 9 minar is called the line of sight. The student is looking at the top of the minar. The angle BAC, so formed by the line of sight with the horizontal, is called the angle of elevation of the top of the minar from the eye of the student. in the object viewed by the observer. The angle of elevation of the point viewed is In this figure, the line AC drawn from the eye of the student to the top of the Thus, the line of sight is the line drawn from the eye of an observer to the point Reprint 2025-26 Fig. 9.1 134 MATHEMATICS the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object (see Fig. 9.2). looking down at a flower pot placed on a stair of the temple. In this case, the line of sight is below the horizontal level. The angle so formed by the line of sight with the horizontal is called the angle of depression. formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed (see Fig. 9.3). Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is Thus, the angle of depression of a point on the object being viewed is the angle Fig. 9.2 Are they angles of elevation or angles of depression? without actually measuring it, what information do you need? You would need to know the following: Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3. Let us refer to Fig. 9.1 again. If you want to find the height CD of the minar (i) the distance DE at which the student is standing from the foot of the minar Reprint 2025-26 Fig. 9.3 SOME APPLICATIONS OF TRIGONOMETRY 135 height of the minar? which of the trigonometric ratios can we use? Which one of them has the two values that we have and the one we need to determine? Our search narrows down to using either tan A or cot A, as these ratios involve AB and BC. Therefore, tan A = BC AB or cot A = AB, BC which on solving would give us BC. By adding AE to BC, you will get the height of the minar. Example 1 : A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower. Solution : First let us draw a simple diagram to represent the problem (see Fig. 9.4). Here AB represents the tower, CB is the distance of the point from the tower and ACB is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angled at B. Assuming that the above three conditions are known, how can we determine the In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student. To find BC, we will use trigonometric ratios of BAC or A. In ABC, the side BC is the opposite side in relation to the known A. Now, Now let us explain the process, we have just discussed, by solving some problems. (iii) the height AE of the student. (ii) the angle of elevation, BAC, of the top of the minar To solve the problem, we choose the trigonometric ratio tan 60° (or cot 60°), as the ratio involves AB and BC. Now, tan 60° = AB BC i.e., 3 = AB 15 i.e., AB = 15 3 Hence, the height of the tower is 15 3 m. Reprint 2025-26 Fig. 9.4 136 MATHEMATICS Example 2 : An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take 3 = 1.73) Solution : In Fig. 9.5, the electrician is required to reach the point B on the pole AD. So, BD = AD – AB = (5 – 1.3)m = 3.7 m. Here, BC represents the ladder. We need to find its length, i.e., the hypotenuse of the right triangle BDC. Now, can you think which trigonometic ratio should we consider? It should be sin 60°. So, BD BC = sin 60° or 3.7 BC = 3 2 Therefore, BC = 3.7 2 3 = 4.28 m (approx.) Fig. 9.5 i.e., the length of the ladder should be 4.28 m. Now, DC BD = cot 60° = 1 i.e., DC = 3.7 Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole. 3 = 2.14 m (approx.) Reprint 2025-26 3 SOME APPLICATIONS OF TRIGONOMETRY 137 Example 3 : An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney? Solution : Here, AB is the chimney, CD the observer and ADE the angle of elevation (see Fig. 9.6). In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney. We have AB = AE + BE = AE + 1.5 and DE = CB = 28.5 m To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation. Now, tan 45° = AE DE i.e., 1 = AE 28.5 Therefore, AE = 28.5 So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m. Fig. 9.6 Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732) Solution : In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., DB and the distance of the building from the point P, i.e., PA. Reprint 2025-26 138 MATHEMATICS Since, we know the height of the building AB, we will first consider the right PAB. We have tan 30° = AB AP i.e., 1 3 = 10 AP Therefore, AP = 10 3 i.e., the distance of the building from P is 10 3 m = 17.32 m. Next, let us suppose DB = x m. Then AD = (10 + x) m. Now, in right PAD, tan 45° = AD 10 AP 10 3 x Therefore, 1 = 10 i.e., x = 10 3 1 = 7.32 So, the length of the flagstaff is 7.32 m. 10 3 x Fig. 9.7 Example 5 : The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Solution : In Fig. 9.8, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°. Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC. Reprint 2025-26 Fig. 9.8 SOME APPLICATIONS OF TRIGONOMETRY 139 So, DB = (40 + x) m Now, we have two right triangles ABC and ABD. In ABC, tan 60° = AB BC or, 3 = h x (1) In ABD, tan 30° = AB BD i.e., 1 3 = 40 h x (2) From (1), we have h = x 3 Putting this value in (2), we get x 3 3 = x + 40, i.e., 3x = x + 40 i.e., x = 20 So, h = 20 3 [From (1)] Therefore, the height of the tower is 20 3 m. Example 6 : The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multistoreyed building and the distance between the two buildings. Solution : In Fig. 9.9, PC denotes the multistoryed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore, QPB and PBD are alternate angles, and so are equal. So PBD = 30°. Similarly, PAC = 45°. In right PBD, we have Reprint 2025-26 Fig. 9.9 140 MATHEMATICS In right PAC, we have i.e., PC = AC Also, PC = PD + DC, therefore, PD + DC = AC. Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD 3 (Why?) This gives PD = So, the height of the multi-storeyed building is 4 3 1 8 m=4 3+ 3 m and the distance between the two buildings is also 4 3 3 m. Example 7 : From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river. Solution : In Fig 9.10, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3 m. We are interested to determine the width of the river, which is the length of the side AB of the D APB. PD BD = tan 30° = 1 PC AC = tan 45° = 1 8 31 8 4 3 1 m. 3 1 31 31 3 or BD = PD 3 Now, AB = AD + DB In right APD, A = 30°. So, tan 30° = PD AD Reprint 2025-26 Fig. 9.10 SOME APPLICATIONS OF TRIGONOMETRY 141 i.e., 1 Also, in right PBD, B = 45°. So, BD = PD = 3 m. Now, AB = BD + AD = 3 + 3 3 = 3 (1 + 3 ) m. Therefore, the width of the river is 3 3 1m . 1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11). 2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. 3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? 3 = 3 AD or AD = 3 3 m EXERCISE 9.1 Fig. 9.11 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. 7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Reprint 2025-26 142 MATHEMATICS 10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. 12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. 13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. 11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal. 8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. 9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Fig. 9.12 14. A 1.2 m tall girl spots a balloon moving 15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval. Reprint 2025-26 Fig. 9.13 SOME APPLICATIONS OF TRIGONOMETRY 143 9.2 Summary In this chapter, you have studied the following points : 1. (i) The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. (ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level, i.e., the case when we raise our head to look at the object. 2. The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios. (iii) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level, i.e., the case when we lower our head to look at the object. tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. Reprint 2025-26" class_10,10,Circles,ncert_books/class_10/jemh1dd/jemh110.pdf,"10.1 Introduction You have studied in Class IX that a circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre). You have also studied various terms related to a circle like chord, segment, sector, arc etc. Let us now examine the different situations that can arise when a circle and a line are given in a plane. in Fig. 10.1 below: 144 MATHEMATICS So, let us consider a circle and a line PQ. There can be three possibilities given CIRCLES 10 PQ is called a non-intersecting line with respect to the circle. In Fig. 10.1 (ii), there are two common points A and B that the line PQ and the circle have. In this case, we call the line PQ a secant of the circle. In Fig. 10.1 (iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a tangent to the circle. In Fig. 10.1 (i), the line PQ and the circle have no common point. In this case, Reprint 2025-26 Fig. 10.1 CIRCLES 145 You might have seen a pulley fitted over a well which is used in taking out water from the well. Look at Fig. 10.2. Here the rope on both sides of the pulley, if considered as a ray, is like a tangent to the circle representing the pulley. Is there any position of the line with respect to the circle other than the types given above? You can see that there cannot be any other type of position of the line with respect to the circle. In this chapter, we will study about the existence of the tangents to a circle and also study some of their properties. 10.2 Tangent to a Circle In the previous section, you have seen that a tangent* to a circle is a line that intersects the circle at only one point. the following activities: Activity 1 : Take a circular wire and attach a straight wire AB at a point P of the circular wire so that it can rotate about the point P in a plane. Put the system on a table and gently rotate the wire AB about the point P to get different positions of the straight wire [see Fig. 10.3(i)]. In various positions, the wire intersects the circular wire at P and at another point Q1 or Q2 or Q3, etc. In one position, you will see that it will intersect the circle at the point P only (see position AB of AB). This shows that a tangent exists at the point P of the circle. On rotating further, you can observe that in all other positions of AB, it will intersect the circle at P and at another point, say R1 or R2 or R3, etc. So, you can observe that there is only one tangent at a point of the circle. To understand the existence of the tangent to a circle at a point, let us perform Fig. 10.2 While doing activity above, you must have observed that as the position AB moves towards the position A B, the common point, say Q1, of the line AB and the circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides with the point P in the position AB of AB. Again note, what happens if ‘AB’ is rotated rightwards about P? The common point R3 gradually comes nearer and nearer to P and ultimately coincides with P. So, what we see is: The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide. *The word ‘tangent’ comes from the Latin word ‘tangere’, which means to touch and was introduced by the Danish mathematician Thomas Fineke in 1583. Reprint 2025-26 Fig. 10.3 (i) 146 MATHEMATICS Activity 2 : On a paper, draw a circle and a secant PQ of the circle. Draw various lines parallel to the secant on both sides of it. You will find that after some steps, the length of the chord cut by the lines will gradually decrease, i.e., the two points of intersection of the line and the circle are coming closer and closer [see Fig. 10.3(ii)]. In one case, it becomes zero on one side of the secant and in another case, it becomes zero on the other side of the secant. See the positions PQ and PQ of the secant in Fig. 10.3 (ii). These are the tangents to the circle parallel to the given secant PQ. This also helps you to see that there cannot be more than two tangents parallel to a given secant. Activity 1, namely, a tangent is the secant when both of the end points of the corresponding chord coincide. [the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the common point. Now look around you. Have you seen a bicycle or a cart moving? Look at its wheels. All the spokes of a wheel are along its radii. Now note the position of the wheel with respect to its movement on the ground. Do you see any tangent anywhere? (See Fig. 10.4). In fact, the wheel moves along a line which is a tangent to the circle representing the wheel. Also, notice that in all positions, the radius through the point of contact with the ground appears to be at right angles to the tangent (see Fig. 10.4). We shall now prove this property of the tangent. This activity also establishes, what you must have observed, while doing The common point of the tangent and the circle is called the point of contact Fig. 10.3 (ii) Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact. Proof : We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY. Reprint 2025-26 Fig. 10.4 CIRCLES 147 Take a point Q on XY other than P and join OQ (see Fig. 10.5). The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP. Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. (as shown in Theorem A1.7.) Remarks 1. By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent. 2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point. 1. How many tangents can a circle have? 2. Fill in the blanks : (i) A tangent to a circle intersects it in point (s). EXERCISE 10.1 Fig. 10.5 10.3 Number of Tangents from a Point on a Circle To get an idea of the number of tangents from a point on a circle, let us perform the following activity: 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119 cm. 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. (iv) The common point of a tangent to a circle and the circle is called . (iii) A circle can have parallel tangents at the most. (ii) A line intersecting a circle in two points is called a . Reprint 2025-26 148 MATHEMATICS Activity 3 : Draw a circle on a paper. Take a point P inside it. Can you draw a tangent to the circle through this point? You will find that all the lines through this point intersect the circle in two points. So, it is not possible to draw any tangent to a circle through a point inside it [see Fig. 10.6 (i)]. Next take a point P on the circle and draw tangents through this point. You have already observed that there is only one tangent to the circle at such a point [see Fig. 10.6 (ii)]. Finally, take a point P outside the circle and try to draw tangents to the circle from this point. What do you observe? You will find that you can draw exactly two tangents to the circle through this point [see Fig. 10.6 (iii)]. Case 1 : There is no tangent to a circle passing through a point lying inside the circle. Case 2 : There is one and only one tangent to a circle passing through a point lying on the circle. Case 3 : There are exactly two tangents to a circle through a point lying outside the circle. We can summarise these facts as follows: (i) (ii) In Fig. 10.6 (iii), T1 and T2 are the points of contact of the tangents PT1 and PT2 respectively. The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle. the circle. The lengths PT1 and PT2 have a common property. Can you find this? Measure PT1 and PT2 . Are these equal? In fact, this is always so. Let us give a proof of this fact in the following theorem. Note that in Fig. 10.6 (iii), PT1 and PT2 are the lengths of the tangents from P to Reprint 2025-26 Fig. 10.6 (iii) CIRCLES 149 Theorem 10.2 : The lengths of tangents drawn from an external point to a circle are equal. Proof : We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 10.7). We are required to prove that PQ = PR. For this, we join OP, OQ and OR. Then ∠ OQP and ∠ ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP, OQ = OR (Radii of the same circle) Therefore, ∆ OQP ≅ ∆ ORP (RHS) This gives PQ = PR (CPCT) Remarks 1. The theorem can also be proved by using the Pythagoras Theorem as follows: which gives PQ = PR. 2. Note also that ∠ OPQ = ∠ OPR. Therefore, OP is the angle bisector of ∠ QPR, i.e., the centre lies on the bisector of the angle between the two tangents. PQ2 = OP2 – OQ2 = OP2 – OR2 = PR2 (As OQ = OR) OP = OP (Common) Fig. 10.7 Let us take some examples. Example 1 : Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. Solution : We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1 which touches the smaller circle C2 at the point P (see Fig. 10.8). We need to prove that AP = BP. Let us join OP. Then, AB is a tangent to C2 at P and OP is its radius. Therefore, by Theorem 10.1, OP ⊥ AB Reprint 2025-26 Fig. 10.8 150 MATHEMATICS Now AB is a chord of the circle C1 and OP AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord, i.e., AP = BP Example 2 : Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = 2 OPQ. Solution : We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Fig. 10.9). We need to prove that Let PTQ = Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle. Therefore, TPQ = TQP = 1 1 (180° ) 90° 2 2 Also, by Theorem 10.1, OPT = 90° So, OPQ = OPT – TPQ = 1 90° 90° – 2 PTQ = 2 OPQ = 1 1 PTQ 2 2 Fig. 10.9 This gives PTQ = 2 OPQ Example 3 : PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10). Find the length TP. Solution : Join OT. Let it intersect PQ at the point R. Then TPQ is isosceles and TO is the angle bisector of PTQ. So, OT PQ and therefore, OT bisects PQ which gives PR = RQ = 4 cm. Also, OR = 2 2 22 OP PR 5 4 cm 3 c m. Reprint 2025-26 Fig. 10.10 CIRCLES 151 Now, TPR + RPO = 90° = TPR + PTR (Why?) So, RPO = PTR This gives TP PO = RP RO , i.e., TP 5 = 4 3 or TP = 20 3 cm. Note : TP can also be found by using the Pythagoras Theorem, as follows: Let TP = x and TR = y. Then Subtracting (1) from (2), we get Therefore, x2 = 2 16 16 16 25 16 (16 9) 39 9 [From (1)] or x = 20 3 In Q.1 to 3, choose the correct option and give justification. Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity. x2 + 52 =(y + 3)2 (Taking right OPT) (2) 25 = 6y – 7 or y = 32 16 6 3 x2 = y2 + 16 (Taking right PRT) (1) EXERCISE 10.2 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is 2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that POQ = 110°, then PTQ is equal to (A) 60° (B) 70° 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to (A) 50° (B) 60° (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm (C) 80° (D) 90° (C) 70° (D) 80° Reprint 2025-26 Fig. 10.11 152 MATHEMATICS 10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. 4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. 5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. 6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. 8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that 9. In Fig. 10.13, XY and XY are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XY at B. Prove that AOB = 90°. AB + CD = AD + BC Fig. 10.12 Fig. 10.13 12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC. 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Fig. 10.14 11. Prove that the parallelogram circumscribing a circle is a rhombus. Reprint 2025-26 CIRCLES 153 10.4 Summary In this chapter, you have studied the following points : 1. The meaning of a tangent to a circle. 2. The tangent to a circle is perpendicular to the radius through the point of contact. 3. The lengths of the two tangents from an external point to a circle are equal. Reprint 2025-26" class_10,11,Areas Related to Circles,ncert_books/class_10/jemh1dd/jemh111.pdf,"154 MATHEMATICS 11.1 Areas of Sector and Segment of a Circle You have already come across the terms sector and segment of a circle in your earlier classes. Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, in Fig. 11.1, shaded region OAPB is a sector of the circle with centre O. ∠ AOB is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is 360° – ∠ AOB. AREAS RELATED TO CIRCLES Fig. 11.1 11 Now, look at Fig. 11.2 in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment. Remark : When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise. Reprint 2025-26 Fig. 11.2 AREAS RELATED TO CIRCLES 155 Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas. Let OAPB be a sector of a circle with centre O and radius r (see Fig. 11.3). Let the degree measure of Ð AOB be q. You know that area of a circle (in fact of a circular region or disc) is pr 2 . In a way, we can consider this circular region to be a sector forming an angle of 360° (i.e., of degree measure 360) at the centre O. Now by applying the Unitary Method, we can arrive at the area of the sector OAPB as follows: sector = pr 2 sector = 2 sector = 2 circle: When degree measure of the angle at the centre is 360, area of the So, when the degree measure of the angle at the centre is 1, area of the Therefore, when the degree measure of the angle at the centre is q, area of the Thus, we obtain the following relation (or formula) for area of a sector of a Area of the sector of angle q = r 2 360 , 360 r = 2 360 r . 360 r Fig. 11.3 where r is the radius of the circle and q the angle of the sector in degrees. Now, a natural question arises : Can we find the length of the arc APB corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2pr, we can obtain the required length of the arc APB as 2 360 r . So, length of an arc of a sector of angle q = 2 360 r . Reprint 2025-26 Fig. 11.4 156 MATHEMATICS Now let us take the case of the area of the segment APB of a circle with centre O and radius r (see Fig. 11.4). You can see that : Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that: and Area of major segment AQB = r2 – Area of the minor segment APB Example 1 : Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use = 3.14). Solution : Given sector is OAPB (see Fig. 11.5). = 12.56 2 2 cm 4.19cm 3 (approx.) Area of the corresponding major sector Area of the segment APB = Area of the sector OAPB – Area of OAB = 2 – area of OAB 360 r Let us now take some examples to understand these concepts (or results). Area of the major sector OAQB = r2 – Area of the minor sector OAPB Area of the sector = 2 360 r = 30 2 3.14 4 4 cm 360 Fig. 11.5 Alternatively, area of the major sector = 2 (360 – ) 360 r = r2 – area of sector OAPB = (3.14 × 16 – 4.19) cm2 = 46.05 cm2 = 46.1 cm2 (approx.) Reprint 2025-26 = 360 30 2 3.14 16 cm 360 = 330 2 2 3.14 16cm 46.05 cm 360 = 46.1 cm2 (approx.) AREAS RELATED TO CIRCLES 157 Example 2 : Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and AOB = 120°. (Use = 22 7 ) Solution : Area of the segment AYB Now, area of the sector OAYB = 120 22 21 21 360 7 cm2 = 462 cm2 (2) For finding the area of OAB, draw OM AB as shown in Fig. 11.7. Note that OA = OB. Therefore, by RHS congruence, AMO BMO. So, M is the mid-point of AB and AOM = BOM = 1 120 60 2 . Let OM = x cm So, from OMA, OM OA = cos 60° = Area of sector OAYB – Area of OAB (1) Fig. 11.6 or, 21 x = 1 1 cos 60° = 2 2 or, x = 21 2 So, OM = 21 2 cm Also, AM OA = sin 60° = 3 2 So, AM = 21 3 2 cm Therefore, AB = 2 AM = 2 21 3 cm = 21 3 cm 2 Reprint 2025-26 Fig. 11.7 158 MATHEMATICS So, area of OAB = 1 AB × OM 2 = 1 21 2 21 3 cm 2 2 Therefore, area of the segment AYB = 441 2 462 3 cm 4 [From (1), (2) and (3)] Unless stated otherwise, use = 22 7 . 1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. 2. Find the area of a quadrant of a circle whose circumference is 22 cm. 3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. 4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use = 3.14) 5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord = 441 2 3 cm 4 (3) EXERCISE 11.1 = 21 2 (88 – 21 3)cm 4 6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. 7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. 8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find Fig. 11.8 (Use = 3.14 and 3 = 1.73) (Use = 3.14 and 3 = 1.73) Reprint 2025-26 AREAS RELATED TO CIRCLES 159 10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. 12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use = 3.14) 11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. 9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find : (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use = 3.14) (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. Fig. 11.10 Fig. 11.9 13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ` 0.35 per cm2 . (Use 3 = 1.7) 14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is (A) 2 R 180 p (B) 2 R 180 p (C) 2 R 360 p (D) 2 2 R 720 p Reprint 2025-26 Fig. 11.11 160 MATHEMATICS 11.2 Summary In this chapter, you have studied the following points : 1. Length of an arc of a sector of a circle with radius r and angle with degree measure is 2. Area of a sector of a circle with radius r and angle with degree measure is 2 360 r 3. Area of segment of a circle 2 360 r = Area of the corresponding sector – Area of the corresponding triangle. Reprint 2025-26" class_10,12,Surface Areas and Volumes,ncert_books/class_10/jemh1dd/jemh112.pdf,"SURFACE AREAS AND VOLUMES 161 12.1 Introduction From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes. SURFACE AREAS AND VOLUMES 12 of two or more of the basic solids as shown above. You must have seen a truck with a container fitted on its back (see Fig. 12.2), carrying oil or water from one place to another. Is it in the shape of any of the four basic solids mentioned above? You may guess that it is made of a cylinder with two hemispheres as its ends. In our day-to-day life, we come across a number of solids made up of combinations Reprint 2025-26 Fig. 12.1 Fig. 12.2 162 MATHEMATICS Again, you may have seen an object like the one in Fig. 12.3. Can you name it? A test tube, right! You would have used one in your science laboratory. This tube is also a combination of a cylinder and a hemisphere. Similarly, while travelling, you may have seen some big and beautiful buildings or monuments made up of a combination of solids mentioned above. If for some reason you wanted to find the surface areas, or volumes, or capacities of such objects, how would you do it? We cannot classify these under any of the solids you have already studied. objects. 12.2 Surface Area of a Combination of Solids Let us consider the container seen in Fig. 12.2. How do we find the surface area of such a solid? Now, whenever we come across a new problem, we first try to see, if we can break it down into smaller problems, we have earlier solved. We can see that this solid is made up of a cylinder with two hemispheres stuck at either end. It would look like what we have in Fig. 12.4, after we put the pieces all together. In this chapter, you will see how to find surface areas and volumes of such Fig. 12.4 Fig. 12.3 only the curved surfaces of the two hemispheres and the curved surface of the cylinder. areas of each of the individual parts. This gives, where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ respectively. together a hemisphere and a cone. Let us see the steps that we would be going through. If we consider the surface of the newly formed object, we would be able to see So, the total surface area of the new solid is the sum of the curved surface TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere Let us now consider another situation. Suppose we are making a toy by putting Reprint 2025-26 SURFACE AREAS AND VOLUMES 163 Here, of course, we would take the base radius of the cone equal to the radius of the hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown in Fig. 12.5. we want to find how much paint we would require to colour the surface of this toy, what would we need to know? We would need to know the surface area of the toy, which consists of the CSA of the hemisphere and the CSA of the cone. So, we can say: Example 1 : Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he First, we would take a cone and a hemisphere and bring their flat faces together. At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if Now, let us consider some examples. Total surface area of the toy = CSA of hemisphere + CSA of cone Fig. 12.5 has to colour. (Take = 22 7 ) Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we can conveniently use the result we have arrived at there. That is : Now, the curved surface area of the hemisphere = 1 2 2 (4 ) 2 2 r r TSA of the toy = CSA of hemisphere + CSA of cone Reprint 2025-26 = 22 3.5 3.5 2 2 cm 722 Fig. 12.6 . 164 MATHEMATICS Also, the height of the cone = height of the top – height (radius) of the hemispherical part So, the slant height of the cone (l ) = 2 22 2 3.5 (3.25) cm 2 r h = 3.7 cm (approx.) Therefore, CSA of cone = rl = 22 3.5 2 3.7 cm 7 2 This gives the surface area of the top as surface areas of the cone and hemisphere. Example 2 : The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. You may note that ‘total surface area of the top’ is not the sum of the total = 22 3.5 3.5 22 3.5 2 2 2 cm 3.7 cm 722 72 = 22 3.5 2 3.5 3.7 cm 7 2 = 11 2 2 (3.5 3.7) cm 39.6 cm (approx.) 2 = 3.5 5 cm 2 = 3.25 cm (Take = 22 7 ) Solution : The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2 . Note that the part of the cube where the hemisphere is attached is not included in the surface area. So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – r2 + 2 r2 = (150 + r2 ) cm2 Reprint 2025-26 = 2 2 22 4.2 4.2 150 cm cm 722 = (150 + 13.86) cm2 = 163.86 cm2 Fig. 12.7 SURFACE AREAS AND VOLUMES 165 Example 3 : A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take = 3.14) Solution : Denote radius of cone by r, slant height of cone by l, height of cone by h, radius of cylinder by r and height of cylinder by h. Then r = 2.5 cm, h = 6 cm, r = 1.5 cm, h = 26 – 6 = 20 cm and Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted. So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder l = 2 2 r h = 2 2 2.5 6 cm = 6.5 cm Fig. 12.8 Now, the area to be painted yellow = CSA of the cylinder + area of one base of the cylinder Reprint 2025-26 = rl + r2 – r2 = [(2.5 × 6.5) + (2.5)2 – (1.5)2 ] cm2 = [20.25] cm2 = 3.14 × 20.25 cm2 = 63.585 cm2 = 2rh + (r)2 = r (2h + r) = (3.14 × 1.5) (2 × 20 + 1.5) cm2 = 4.71 × 41.5 cm2 = 195.465 cm2 166 MATHEMATICS Example 4 : Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take = 22 7 ) Solution : Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, Unless stated otherwise, take = 22 7 1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. EXERCISE 12.1 = 22 2 2 30(145 30) cm 7 = 33000 cm2 = 3.3 m2 = 2rh + 2r2 = 2 r(h + r) Fig. 12.9 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. Fig. 12.10 Reprint 2025-26 SURFACE AREAS AND VOLUMES 167 12.3 Volume of a Combination of Solids In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their volumes. It may be noted that in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them. However, this will not be the case when we calculate the volume. The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below. Example 5 : Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m3 , and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ` 500 per m2 . (Note that the base of the tent will not be covered with canvas.) 8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2 . 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article. Fig. 12.11 shed? (Take = 22 7 ) Reprint 2025-26 Fig. 12.12 168 MATHEMATICS Solution : The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together. Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m. So, the required volume = volume of the cuboid + 1 2 volume of the cylinder Next, the total space occupied by the machinery = 300 m3 And the total space occupied by the workers = 20 × 0.08 m3 = 1.6 m3 Therefore, the volume of the air, when there are machinery and workers Example 6 : A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use = 3.14.) = 1 22 7 7 3 15 7 8 15 m 2 7 22 = 1128.75 m3 = 1128.75 – (300.00 + 1.60) = 827.15 m3 Fig. 12.13 Solution : Since the inner diameter of the glass = 5 cm and height = 10 cm, base of the glass. i.e., it is less by 2 3 r3 = 2 3 3.14 2.5 2.5 2.5 cm 3 = 32.71 cm3 So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere = (196.25 – 32.71) cm3 the apparent capacity of the glass = r2 h But the actual capacity of the glass is less by the volume of the hemisphere at the Reprint 2025-26 = 163.54 cm3 = 3.14 × 2.5 × 2.5 × 10 cm3 = 196.25 cm3 SURFACE AREAS AND VOLUMES 169 Example 7 : A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take = 3.14) Solution : Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as of the cone) = 1 2 × 4 cm = 2 cm. So, volume of the toy = 2 1 3 2 3 3 r rh Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm So, the volume required = volume of the right circular cylinder – volume of the toy = (3.14 × 22 × 4 – 25.12) cm3 = 25.12 cm3 Hence, the required difference of the two volumes = 25.12 cm3 . = 2 1 3 23 3.14 (2) 3.14 (2) 2 cm 3 3 = 25.12 cm3 Fig. 12.14 Unless stated otherwise, take = 22 7 . A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of . 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) EXERCISE 12.2 Reprint 2025-26 170 MATHEMATICS 3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15). 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16). 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use = 3.14) 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. Fig. 12.16 Fig. 12.15 12.4 Summary In this chapter, you have studied the following points: 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3 . Check whether she is correct, taking the above as the inside measurements, and = 3.14. 1. To determine the surface area of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere. 2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder, sphere and hemisphere. Reprint 2025-26" class_10,13,Statistics,ncert_books/class_10/jemh1dd/jemh113.pdf,"13.1 Introduction In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions. You have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths) and frequency polygons. In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data. We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives. 13.2 Mean of Grouped Data STATISTICS 171 STATISTICS 13 The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if x1 , x2 ,. . ., xn are observations with respective frequencies f 1 , f2 , . . ., fn , then this means observation x1 occurs f 1 times, x2 occurs f 2 times, and so on. the number of observations = f 1 + f2 + . . . + f n . sigma) which means summation. That is, Now, the sum of the values of all the observations = f 1 x1 + f2 x2 + . . . + f n xn , and So, the mean x of the data is given by Recall that we can write this in short form by using the Greek letter (capital x = 11 2 2 1 2 n fx fx fx ff f Reprint 2025-26 n n 172 MATHEMATICS which, more briefly, is written as x = i i i f x f , if it is understood that i varies from 1 to n. Let us apply this formula to find the mean in the following example. Example 1 : The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. Solution: Recall that to find the mean marks, we require the product of each xi with the corresponding frequency f i . So, let us put them in a column as shown in Table 13.1. 10 1 10 20 1 20 . 36 3 108 40 4 160 50 3 150 56 2 112 60 4 240 70 4 280 72 1 72 80 1 80 88 2 176 92 3 276 95 1 95 Marks obtained 10 20 36 40 50 56 60 70 72 80 88 92 95 (xi ) Number of 1 1 3 4 3 2 4 4 1 1 2 31 students ( fi ) Marks obtained (xi ) Number of students ( f i ) fi xi x = 1 i i i n i i f n 1 f x Table 13.1 Total f i = 30 f i xi = 1779 Reprint 2025-26 STATISTICS 173 Now, i i Therefore, the mean marks obtained is 59.3. study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean. class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table (see Table 13.2). representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each Class interval 10 - 25 25 - 40 40 - 55 55 - 70 70 - 85 85 - 100 Number of students 2 3 7 6 6 6 In most of our real life situations, data is usually so large that to make a meaningful Let us convert the ungrouped data of Example 1 into grouped data by forming Now, for each class-interval, we require a point which would serve as the i f x x f = 1779 30 = 59.3 Table 13.2 class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is, 17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table 13.3. These class marks serve as our xi ’s. Now, in general, for the ith class interval, we have the frequency f i corresponding to the class mark xi . We can now proceed to compute the mean in the same manner as in Example 1. With reference to Table 13.2, for the class 10-25, the class mark is 10 25 2 + , i.e., Class mark = Upper class limit + Lower class limit 2 Reprint 2025-26 174 MATHEMATICS given data is given by x = 1860.0 62 30 i i same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact mean, while 62 an approximate mean. of xi and fi becomes tedious and time consuming. So, for such situations, let us think of Class interval Number of students ( f i ) Class mark (xi ) f i xi The sum of the values in the last column gives us Σ f i xi . So, the mean x of the i f x f Σ = = Σ This new method of finding the mean is known as the Direct Method. We observe that Tables 13.1 and 13.3 are using the same data and employing the Sometimes when the numerical values of x i and f i are large, finding the product 10 - 25 2 17.5 35.0 25 - 40 3 32.5 97.5 40 - 55 7 47.5 332.5 55 - 70 6 62.5 375.0 70 - 85 6 77.5 465.0 85 - 100 6 92.5 555.0 Total Σ f i = 30 Σ f i x i = 1860.0 Table 13.3 a method of reducing these calculations. so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these xi ’s? Let us try this method. it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi which lies in the centre of x1 , x2 , . . ., xn . So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5. the deviation of ‘a’ from each of the xi ’s. i.e., di = xi – a = xi – 47.5 We can do nothing with the f i ’s, but we can change each x i to a smaller number The first step is to choose one among the xi ’s as the assumed mean, and denote The next step is to find the difference di between a and each of the x i ’s, that is, Reprint 2025-26 STATISTICS 175 The third step is to find the product of di with the corresponding f i , and take the sum of all the f i di ’s. The calculations are shown in Table 13.4. x , we need to add ‘a’ to d . This can be explained mathematically as: Class interval Number of Class mark di = xi – 47.5 f i di So, from Table 13.4, the mean of the deviations, d = i i i f d f Σ Σ . Now, let us find the relation between d and x . Since in obtaining di , we subtracted ‘a’ from each xi , so, in order to get the mean Mean of deviations, d = i i i f d f Σ Σ 10 - 25 2 17.5 –30 –60 25 - 40 3 32.5 –15 –45 40 - 55 7 47.5 0 0 55 - 70 6 62.5 15 90 70 - 85 6 77.5 30 180 85 - 100 6 92.5 45 270 Total Σf i = 30 Σf i di = 435 students ( f i ) (xi ) Table 13.4 So, x = a + d i.e., x = i i i f d a f Σ + Σ So, d = ( ) i i Reprint 2025-26 = i i i i i f x f a f f Σ Σ − Σ Σ = i i f x a f Σ − Σ = x a − i f x a f Σ − Σ 176 MATHEMATICS Therefore, the mean of the marks obtained by the students is 62. Activity 1 : From the Table 13.3 find the mean by taking each of x i (i.e., 17.5, 32.5, and so on) as ‘a’. What do you observe? You will find that the mean determined in each case is the same, i.e., 62. (Why ?) choice of ‘a’. we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with f i . (Here, 15 is the class size of each class interval.) Now, we calculate ui in this way and continue as before (i.e., find f i ui and then Σ f i ui ). Taking h = 15, let us form Table 13.5. Class interval f i xi di = xi – a ui = x – a i h f i ui Substituting the values of a, Σf i di and Σf i from Table 13.4, we get The method discussed above is called the Assumed Mean Method. So, we can say that the value of the mean obtained does not depend on the Observe that in Table 13.4, the values in Column 4 are all multiples of 15. So, if So, let ui = i x a h − , where a is the assumed mean and h is the class size. x = 435 47.5 47.5 14.5 62 30 + = + = . Table 13.5 Let u = i i i f u f Σ Σ Here, again let us find the relation between u and x . 10 - 25 2 17.5 –30 –2 –4 25 - 40 3 32.5 –15 –1 –3 40 - 55 7 47.5 0 0 0 55 - 70 6 62.5 15 1 6 70 - 85 6 77.5 30 2 12 85 - 100 6 92.5 45 3 18 Total Σf i = 30 Σf i ui = 29 Reprint 2025-26 STATISTICS 177 We have, ui = i x a h Therefore, u = So, hu = x a i.e., x = a + hu So, x = i i i f u a h f Now, substituting the values of a, h, f i ui and f i from Table 14.5, we get x = 29 47.5 15 30 = 1 ii i = 1 x a h = 47.5 + 14.5 = 62 i i fx f a hf f x a f h fx a f fhf ( ) 1 i i ii i i i So, the mean marks obtained by a student is 62. The method discussed above is called the Step-deviation method. We note that : Let us apply these methods in another example. the step-deviation method will be convenient to apply if all the di ’s have a The mean obtained by all the three methods is the same. The assumed mean method and step-deviation method are just simplified forms of the direct method. The formula x = a + hu still holds if a and h are not as given above, but are common factor. any non-zero numbers such that ui = i x a h . Reprint 2025-26 178 MATHEMATICS Example 2 : The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section. Source : Seventh All India School Education Survey conducted by NCERT Solution : Let us find the class marks, xi , of each class, and put them in a column (see Table 13.6): Percentage of 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 75 - 85 female teachers Number of 6 11 7 4 4 2 1 States/U.T. Percentage of female Number of xi teachers States /U.T. ( fi ) 15 - 25 6 20 25 - 35 11 30 35 - 45 7 40 Table 13.6 Here we take a = 50, h = 10, then di = xi – 50 and 50 10 i i x u . We now find di and ui and put them in Table 13.7. 45 - 55 4 50 55 - 65 4 60 65 - 75 2 70 75 - 85 1 80 Reprint 2025-26 STATISTICS 179 Percentage of Number of xi di = xi – 50 −50 = 10 i i x u f i xi f i di f i ui female states/U.T. teachers ( f i ) 15 - 25 6 20 –30 –3 120 –180 –18 25 - 35 11 30 –20 –2 330 –220 –22 35 - 45 7 40 –10 –1 280 –70 –7 45 - 55 4 50 0 0 200 0 0 55 - 65 4 60 10 1 240 40 4 65 - 75 2 70 20 2 140 40 4 75 - 85 1 80 30 3 80 30 3 Total 35 1390 –360 –36 From the table above, we obtain Σf i = 35, Σf i xi = 1390, Using the direct method, 1390 39.71 35 Σ = = = Σ i i i f x x f Using the assumed mean method, x = i i i f d a f Σ + Σ = ( 360) 50 39.71 35 − + = Σf i di = – 360, Σf i ui = –36. Table 13.7 rural areas is 39.71. Remark : The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of xi and f i . If xi and f i are sufficiently small, then the direct method is an appropriate choice. If xi and f i are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and xi are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the di ’s. Using the step-deviation method, Therefore, the mean percentage of female teachers in the primary schools of x = – 36 50 10 39.71 35 i i i f u a h f Σ + × = + × = Σ Reprint 2025-26 180 MATHEMATICS Example 3 : The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Solution : Here, the class size varies, and the xi , s are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8. Number of 20 - 60 60 - 100 100 - 150 150 - 250 250 - 350 350 - 450 wickets Number of 7 5 16 12 2 3 bowlers Number of Number of xi di = xi – 200 = 20 i i d u ui f i wickets bowlers taken ( f i ) 100 - 150 16 125 –75 –3.75 –60 150 - 250 12 200 0 0 0 60 - 100 5 80 –120 –6 –30 20 - 60 7 40 –160 –8 –56 Table 13.8 This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89. 250 - 350 2 300 100 5 10 350 - 450 3 400 200 10 30 Total 45 –106 So, 106 45 − u = ⋅ Therefore, x = 200 + 106 20 45 − = 200 – 47.11 = 152.89. Now, let us see how well you can apply the concepts discussed in this section! Reprint 2025-26 STATISTICS 181 Activity 2 : Divide the students of your class into three groups and ask each group to do one of the following activities. distribution tables, the groups should find the mean in each case by the method which they find appropriate. 1. Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution of the data obtained. 2. Collect the daily maximum temperatures recorded for a period of 30 days in your city. Present this data as a grouped frequency table. 3. Measure the heights of all the students of your class (in cm) and form a grouped frequency distribution table of this data. 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. After all the groups have collected the data and formed grouped frequency Which method did you use for finding the mean, and why? Number of plants 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14 Number of houses 121562 3 EXERCISE 13.1 2. Consider the following distribution of daily wages of 50 workers of a factory. 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Find the mean daily wages of the workers of the factory by using an appropriate method. Daily wages (in `) 500 - 520 520 -540 540 - 560 560 - 580 580 -600 Number of workers 12 14 8 6 10 Daily pocket 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 allowance (in `) Number of children 7 6 9 13 f 5 4 Reprint 2025-26 182 MATHEMATICS 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. 6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? Number of heartbeats 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 - 86 per minute Number of women 2438742 Number of mangoes 50 - 52 53 - 55 56 - 58 59 - 61 62 - 64 Number of boxes 15 110 135 115 25 Daily expenditure 100 - 150 150 - 200 200 - 250 250 - 300 300 - 350 (in `) Number of 4 5 12 2 2 households Find the mean daily expenditure on food by a suitable method. 7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: Find the mean concentration of SO2 in the air. Concentration of SO2 (in ppm) Frequency 0.00 - 0.04 4 0.04 - 0.08 9 0.08 - 0.12 9 0.12 - 0.16 2 0.16 - 0.20 4 0.20 - 0.24 2 Reprint 2025-26 STATISTICS 183 13.3 Mode of Grouped Data Recall from Class IX, a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only. 8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Let us first recall how we found the mode for ungrouped data through the following Number of 0 - 6 6 - 10 10 - 14 14 - 20 20 - 28 28 - 38 38 - 40 days Number of 11 10 7 4 4 3 1 students Literacy rate (in %) 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95 Number of cities 3 10 11 8 3 example. Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows: Find the mode of the data. Solution : Let us form the frequency distribution table of the given data as follows: Number of 0123456 wickets Number of 1132111 matches 2645 0 2 1323 Reprint 2025-26 184 MATHEMATICS (i.e., 3) of matches. So, the mode of this data is 2. looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula: Example 5 : A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household: Clearly, 2 is the number of wickets taken by the bowler in the maximum number In a grouped frequency distribution, it is not possible to determine the mode by where l = lower limit of the modal class, Let us consider the following examples to illustrate the use of this formula. Mode = 1 0 10 2 2 f f l h ff f Family size 1 - 3 3 - 5 5 - 7 7 - 9 9 - 11 Number of 78221 families f 1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class. h = size of the class interval (assuming all class sizes to be equal), Solution : Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 – 5. Now frequency ( f 1 ) of the modal class = 8, frequency ( f 0 ) of class preceding the modal class = 7, frequency ( f 2 ) of class succeeding the modal class = 2. Now, let us substitute these values in the formula : Find the mode of this data. modal class = 3 – 5, lower limit (l) of modal class = 3, class size (h) = 2 Reprint 2025-26 STATISTICS 185 Therefore, the mode of the data above is 3.286. Example 6 : The marks distribution of 30 students in a mathematics examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean. Solution : Refer to Table 13.3 of Example 1. Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore, the frequency ( f 1 ) of modal class = 7, the frequency ( f 0 ) of the class preceding the modal class = 3, the frequency ( f 2 ) of the class succeeding the modal class = 6. Now, using the formula: the lower limit (l) of the modal class = 40, the class size ( h) = 15, Mode = 1 0 10 2 2 f f l h ff f = 8 7 2 3 2 3 3.286 287 2 7 Mode = 1 0 102 2 f f l h ff f , we get Mode = 7 3 40 15 14 6 3 = 52 So, the mode marks is 52. Now, from Example 1, you know that the mean marks is 62. student obtained 62 marks. Remarks : 1. In Example 6, the mode is less than the mean. But for some other problems it may be equal or more than the mean also. 2. It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most So, the maximum number of students obtained 52 marks, while on an average a Reprint 2025-26 186 MATHEMATICS of the students. In the first situation, the mean is required and in the second situation, the mode is required. Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations assigned to the groups. Ask each group to find the mode of the data. They should also compare this with the mean, and interpret the meaning of both. Remark : The mode can also be calculated for grouped data with unequal class sizes. However, we shall not be discussing it. 1. The following table shows the ages of the patients admitted in a hospital during a year: 2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Determine the modal lifetimes of the components. Age (in years) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 Number of patients 6 11 21 23 14 5 Lifetimes (in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Frequency 10 35 52 61 38 29 EXERCISE 13.2 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : Expenditure (in `) Number of families 1000 - 1500 24 1500 - 2000 40 2000 - 2500 33 2500 - 3000 28 3000 - 3500 30 3500 - 4000 22 4000 - 4500 16 4500 - 5000 7 Reprint 2025-26 STATISTICS 187 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. 5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Number of students per teacher Number of states / U.T. Runs scored Number of batsmen 15 - 20 3 20 - 25 8 25 - 30 9 30 - 35 10 35 - 40 3 40 - 45 0 45 - 50 0 50 - 55 2 3000 - 4000 4 4000 - 5000 18 5000 - 6000 9 6000 - 7000 7 7000 - 8000 6 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data : Find the mode of the data. Number of cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 7 14 13 12 20 11 15 8 10000 - 11000 1 9000 - 10000 1 8000 - 9000 3 Reprint 2025-26 188 MATHEMATICS 13.4 Median of Grouped Data As you have studied in Class IX, the median is a measure of central tendency which gives the value of the middle-most observation in the data. Recall that for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the 1 2 n th observation. And, if n is even, then the median will be the average of the th 2 n and the 1 th 2 n observations. marks, out of 50, obtained by 100 students in a test : follows : Marks obtained 20 29 28 33 42 38 43 25 Number of students 6 28 24 15 2 4 1 20 Suppose, we have to find the median of the following data, which gives the First, we arrange the marks in ascending order and prepare a frequency table as Marks obtained Number of students (Frequency) 20 6 Table 13.9 Total 100 25 20 28 24 29 28 33 15 38 4 42 2 43 1 Reprint 2025-26 STATISTICS 189 observations, we proceed as follows: above and name it as cumulative frequency column. 1 2 n th observations, i.e., the 50th and 51st observations. To find these Here n = 100, which is even. The median will be the average of the 2 n th and the Now we add another column depicting this information to the frequency table Marks obtained Number of students upto 25 6 + 20 = 26 upto 28 26 + 24 = 50 upto 29 50 + 28 = 78 upto 33 78 + 15 = 93 upto 38 93 + 4 = 97 upto 42 97 + 2 = 99 upto 43 99 + 1 = 100 20 6 Table 13.10 Table 13.11 Marks obtained Number of students Cumulative frequency 20 6 6 25 20 26 28 24 50 29 28 78 33 15 93 38 4 97 42 2 99 43 1 100 Reprint 2025-26 190 MATHEMATICS From the table above, we see that: So, Median = 28 29 28.5 2 + = Remark : The part of Table 13.11 consisting Column 1 and Column 3 is known as Cumulative Frequency Table. The median marks 28.5 conveys the information that about 50% students obtained marks less than 28.5 and another 50% students obtained marks more than 28.5. Now, let us see how to obtain the median of grouped data, through the following situation. students, in a certain examination, as follows: Consider a grouped frequency distribution of marks obtained, out of 100, by 53 Marks Number of students 10 - 20 3 20 - 30 4 0 - 10 5 50th observaton is 28 (Why?) 51st observation is 29 Table 13.12 From the table above, try to answer the following questions: How many students have scored marks less than 10? The answer is clearly 5. 90 - 100 8 30 - 40 3 40 - 50 3 50 - 60 4 60 - 70 7 70 - 80 9 80 - 90 7 Reprint 2025-26 STATISTICS 191 of students who have scored less than 20 include the number of students who have scored marks from 0 - 10 as well as the number of students who have scored marks from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8. We say that the cumulative frequency of the class 10 -20 is 8. the number of students with marks less than 30, less than 40, . . ., less than 100. We give them in Table 13.13 given below: How many students have scored less than 20 marks? Observe that the number Similarly, we can compute the cumulative frequencies of the other classes, i.e., Marks obtained Number of students (Cumulative frequency) Less than 10 5 Less than 20 5 + 3 = 8 Less than 30 8 + 4 = 12 Less than 40 12 + 3 = 15 Less than 50 15 + 3 = 18 Less than 60 18 + 4 = 22 Less than 70 22 + 7 = 29 Less than 80 29 + 9 = 38 Less than 90 38 + 7 = 45 Table 13.13 the less than type. Here 10, 20, 30, . . . 100, are the upper limits of the respective class intervals. than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From Table 13.12, we observe that all 53 students have scored marks more than or equal to 0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing in the same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or above as 45 – 4 = 41, and so on, as shown in Table 13.14. The distribution given above is called the cumulative frequency distribution of We can similarly make the table for the number of students with scores, more Less than 100 45 + 8 = 53 Reprint 2025-26 192 MATHEMATICS than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective class intervals. cumulative frequency distributions. Marks Number of students ( f ) Cumulative frequency (cf ) The table above is called a cumulative frequency distribution of the more Now, to find the median of grouped data, we can make use of any of these Let us combine Tables 13.12 and 13.13 to get Table 13.15 given below: More than or equal to 0 53 More than or equal to 10 53 – 5 = 48 More than or equal to 20 48 – 3 = 45 More than or equal to 30 45 – 4 = 41 More than or equal to 40 41 – 3 = 38 More than or equal to 50 38 – 3 = 35 More than or equal to 60 35 – 4 = 31 More than or equal to 70 31 – 7 = 24 More than or equal to 80 24 – 9 = 15 More than or equal to 90 15 – 7 = 8 Marks obtained Number of students (Cumulative frequency) Table 13.14 Table 13.15 looking at the cumulative frequencies as the middle observation will be some value in 0 - 10 5 5 10 - 20 3 8 20 - 30 4 12 30 - 40 3 15 40 - 50 3 18 50 - 60 4 22 60 - 70 7 29 70 - 80 9 38 80 - 90 7 45 90 - 100 8 53 Now in a grouped data, we may not be able to find the middle observation by Reprint 2025-26 STATISTICS 193 a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves. But which class should this be? We now locate the class whose cumulative frequency is greater than (and nearest to) Now 60 – 70 is the class whose cumulative frequency 29 is greater than (and nearest to) 2 n , i.e., 26.5. Therefore, 60 – 70 is the median class. median. where l = lower limit of median class, 2 n This is called the median class. In the distribution above, n = 53. So, 2 n = 26.5. To find this class, we find the cumulative frequencies of all the classes and 2 n . After finding the median class, we use the following formula for calculating the cf = cumulative frequency of class preceding the median class, n = number of observations, f = frequency of median class, Median = cf 2 + , n l h f Substituting the values 26.5, 2 n l = 60, cf = 22, f = 7, h = 10 in the formula above, we get So, about half the students have scored marks less than 66.4, and the other half have scored marks more than 66.4. h = class size (assuming class size to be equal). Median = 26.5 22 60 10 7 = 60 + 45 7 = 66.4 Reprint 2025-26 194 MATHEMATICS Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained: Find the median height. Solution : To calculate the median height, we need to find the class intervals and their corresponding frequencies. upper limits of the corresponding class intervals. So, the classes should be below 140, 140 - 145, 145 - 150, . . ., 160 - 165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 - 145 is 11 – 4 = 7. Similarly, the frequency of 145 - 150 is 29 – 11 = 18, for 150 - 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the given cumulative frequencies becomes: The given distribution being of the less than type, 140, 145, 150, . . ., 165 give the Height (in cm) Number of girls Less than 140 4 Less than 145 11 Less than 150 29 Less than 155 40 Less than 160 46 Less than 165 51 Class intervals Frequency Cumulative frequency Below 140 4 4 140 - 145 7 11 145 - 150 18 29 150 - 155 11 40 155 - 160 6 46 160 - 165 5 51 Table 13.16 Reprint 2025-26 STATISTICS 195 Now n = 51. So, 51 25.5 2 2 n = = . This observation lies in the class 145 - 150. Then, = 145 + 72.5 18 = 149.03. So, the median height of the girls is 149.03 cm. 50% are taller than this height. Example 8 : The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Using the formula, Median = l + cf 2 n This means that the height of about 50% of the girls is less than this height, and cf (the cumulative frequency of the class preceding 145 - 150) = 11, h (the class size) = 5. f (the frequency of the median class 145 - 150) = 18, l (the lower limit) = 145, Median = 25.5 11 145 5 18 − + × − × h f , we have Class intervals Frequency 0 - 100 2 100 - 200 5 200 - 300 x 300 - 400 12 400 - 500 17 500 - 600 20 600 - 700 y 700 - 800 9 800 - 900 7 900 - 1000 4 Reprint 2025-26 196 MATHEMATICS Solution : It is given that n = 100 So, 76 + x + y = 100, i.e., x + y = 24 (1) The median is 525, which lies in the class 500 – 600 So, l = 500, f = 20, cf = 36 + x, h = 100 Using the formula : Median = cf 2 , n Class intervals Frequency Cumulative frequency 900 - 1000 4 76 + x + y 100 - 200 5 7 200 - 300 x 7 + x 300 - 400 12 19 + x 400 - 500 17 36 + x 500 - 600 20 56 + x 600 - 700 y 56 + x + y 700 - 800 9 65 + x + y 800 - 900 7 72 + x + y 0 - 100 2 2 l h f we get i.e., 525 – 500 = (14 – x) × 5 i.e., 25 = 70 – 5x i.e., 5x = 70 – 25 = 45 So, x =9 i.e., y = 15 Therefore, from (1), we get 9 + y = 24 525 = 50 36 500 100 20 x Reprint 2025-26 STATISTICS 197 Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement. The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance. However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data. In problems where individual observations are not important, and we wish to find out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency. In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc. Remarks : 1. There is a empirical relationship between the three measures of central tendency : 2. The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here. 3 Median = Mode + 2 Mean Reprint 2025-26 198 MATHEMATICS 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. 2. If the median of the distribution given below is 28.5, find the values of x and y. Monthly consumption (in units) Number of consumers Class interval Frequency 105 - 125 13 125 - 145 20 145 - 165 14 165 - 185 8 185 - 205 4 85 - 105 5 65 - 85 4 0 - 10 5 EXERCISE 13.3 3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. 50 - 60 5 10 - 20 x 20 - 30 20 30 - 40 15 40 - 50 y Total 60 Reprint 2025-26 STATISTICS 199 4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in mm) Number of leaves Age (in years) Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100 118 - 126 3 127 - 135 5 Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.) 136 - 144 9 145 - 153 12 154 - 162 5 163 - 171 4 172 - 180 2 Reprint 2025-26 200 MATHEMATICS 5. The following table gives the distribution of the life time of 400 neon lamps : 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: 7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Find the median life time of a lamp. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. Number of letters 1 - 4 4 - 7 7 - 10 10 - 13 13 - 16 16 - 19 Number of surnames 6 30 40 16 4 4 Life time (in hours) Number of lamps 1500 - 2000 14 2000 - 2500 56 2500 - 3000 60 3000 - 3500 86 3500 - 4000 74 4000 - 4500 62 4500 - 5000 48 13.5 Summary In this chapter, you have studied the following points: 1. The mean for grouped data can be found by : (ii) the assumed mean method : i i i f d x a f Σ = + Σ Weight (in kg) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 Number of students 2 3 8 6 6 3 2 (i) the direct method : i i i f x x f Σ = Σ Reprint 2025-26 STATISTICS 201 2. The mode for grouped data can be found by using the formula: 3. The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class. 4. The median for grouped data is formed by using the formula: For calculating mode and median for grouped data, it should be ensured that the class intervals are continuous before applying the formulae. Same condition also apply for construction of an ogive. Further, in case of ogives, the scale may not be the same on both the axes. with the assumption that the frequency of a class is centred at its mid-point, called its class mark. Mode = 1 0 102 2 f f l h ff f where symbols have their usual meanings. where symbols have their usual meanings. (iii) the step deviation method : i i i f u x a h f , A NOTE TO THE READER Median = cf 2 n l h f , Reprint 2025-26" class_10,14,Probability,ncert_books/class_10/jemh1dd/jemh114.pdf,"202 MATHEMATICS 14.1 Probability — ATheoretical Approach Let us consider the following situation : When we speak of a coin, we assume it to be ‘fair’, that is, it is symmetrical so that there is no reason for it to come down more often on one side than the other. We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’, we mean that the coin is allowed to fall freely without any bias or interference. Suppose a coin is tossed at random. The theory of probabilities and the theory of errors now constitute a formidable body of great mathematical interest and of great practical importance. PROBABILITY – R.S. Woodward 14 either head up or tail up (we dismiss the possibility of its ‘landing’ on its edge, which may be possible, for example, if it falls on sand). We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely. once. For us, a die will always mean a fair die. What are the possible outcomes? They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6. We know, in advance, that the coin can only land in one of two possible ways — For another example of equally likely outcomes, suppose we throw a die Reprint 2025-26 PROBABILITY 203 without looking into the bag. What are the outcomes? Are the outcomes — a red ball and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you would agree that you are more likely to get a red ball than a blue ball. So, the outcomes (a red ball or a blue ball) are not equally likely. However, the outcome of drawing a ball of any colour from the bag is equally likely. So, all experiments do not necessarily have equally likely outcomes. have equally likely outcomes. event E as P(E) = Number of trials in which the event happened Total number of trials The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing experiments. But how about repeating the experiment of launching a satellite in order to compute the empirical probability of its failure during launching, or the repetition of the phenomenon of an earthquake to compute the empirical probability of a multistoreyed building getting destroyed in an earthquake? Are the outcomes of every experiment equally likely? Let us see. Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball However, in this chapter, from now on, we will assume that all the experiments In Class IX, we defined the experimental or empirical probability P(E) of an In experiments where we are prepared to make certain assumptions, the repetition of an experiment can be avoided, as the assumptions help in directly calculating the exact (theoretical) probability. The assumption of equally likely outcomes (which is valid in many experiments, as in the two examples above, of a coin and of a die) is one such assumption that leads us to the following definition of probability of an event. written as P(E), is defined as where we assume that the outcomes of the experiment are equally likely. The theoretical probability (also called classical probability) of an event E, We will briefly refer to theoretical probability as probability. This definition of probability was given by Pierre Simon Laplace in 1795. P(E) = Number of outcomes favourable to E Number of all possible outcomes of the experiment , Reprint 2025-26 204 MATHEMATICS where the equally likely assumption holds. Example 1 : Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. Solution : In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of outcomes favourable to E, (i.e., of getting a head) is 1. Therefore, Probability theory had its origin in the 16th century when an Italian physician and mathematician J.Cardan wrote the first book on the subject, The Book on Games of Chance. Since its inception, the study of probability has attracted the attention of great mathematicians. James Bernoulli (1654 – 1705), A. de Moivre (1667 – 1754), and Pierre Simon Laplace are among those who made significant contributions to this field. Laplace’s Theorie Analytique des Probabilités, 1812, is considered to be the greatest contribution by a single person to the theory of probability. In recent years, probability has been used extensively in many areas such as biology, economics, genetics, physics, sociology etc. Let us find the probability for some of the events associated with experiments P(E) = P (head) = Number of outcomes favourable to E Number of all possible outcomes = 1 2 Pierre Simon Laplace (1749 – 1827) Similarly, if F is the event ‘getting a tail’, then Example 2 : A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them. (i) yellow ball? (ii) red ball? (iii) blue ball? P(F) = P(tail) = 1 2 (Why ?) Reprint 2025-26 PROBABILITY 205 out is blue’, and R be the event ‘the ball taken out is red’. Now, the number of possible outcomes = 3. (i) The number of outcomes favourable to the event Y = 1. So, P(Y) = 1 3 Similarly, (ii) P(R) = 1 3 and (iii) P(B) = 1 3 Remarks : 1. An event having only one outcome of the experiment is called an elementary event. In Example 1, both the events E and F are elementary events. Similarly, in Example 2, all the three events, Y, B and R are elementary events. 2. In Example 1, we note that : P(E) + P(F) = 1 In Example 2, we note that : P(Y) + P(R) + P(B) = 1 an experiment is 1. This is true in general also. Example 3 : Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ? Solution : (i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So, Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken Observe that the sum of the probabilities of all the elementary events of P(E) = P(number greater than 4) = 2 6 = 1 3 (ii) Let F be the event ‘getting a number less than or equal to 4’. Number of possible outcomes = 6 Outcomes favourable to the event F are 1, 2, 3, 4. So, the number of outcomes favourable to F is 4. Therefore, P(F) = 4 6 = 2 3 Reprint 2025-26 206 MATHEMATICS not because the event E has 2 outcomes and the event F has 4 outcomes. Remarks : From Example 1, we note that where E is the event ‘getting a head’ and F is the event ‘getting a tail’. where E is the event ‘getting a number >4’ and F is the event ‘getting a number 4’. than or equal to 4, and vice versa. ‘not E’ by E . So, P(E) + P(not E) = 1 i.e., P(E) + P( E ) = 1, which gives us P( E ) = 1 – P(E). Are the events E and F in the example above elementary events? No, they are From (i) and (ii) of Example 3, we also get Note that getting a number not greater than 4 is same as getting a number less In (1) and (2) above, is F not the same as ‘not E’? Yes, it is. We denote the event In general, it is true that for an event E, The event E , representing ‘not E’, is called the complement of the event E. P(E) + P(F) = 1 1 1 2 2 (1) P(E) + P(F) = 1 2 1 3 3 (2) P( E ) = 1 – P(E) We also say that E and E are complementary events. Let us answer (i) : We know that there are only six possible outcomes in a single throw of a die. These outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no outcome favourable to 8, i.e., the number of such outcomes is zero. In other words, getting 8 in a single throw of a die, is impossible. So, P(getting 8) = 0 6 = 0 (ii) What is the probability of getting a number less than 7 in a single throw of a die? (i) What is the probability of getting a number 8 in a single throw of a die? Before proceeding further, let us try to find the answers to the following questions: Reprint 2025-26 PROBABILITY 207 event is called an impossible event. Let us answer (ii) : will always get a number less than 7 when it is thrown once. So, the number of favourable outcomes is the same as the number of all possible outcomes, which is 6. Therefore, P(E) = P(getting a number less than 7) = 6 6 = 1 So, the probability of an event which is sure (or certain) to occur is 1. Such an event is called a sure event or a certain event. Note : From the definition of the probability P(E), we see that the numerator (number of outcomes favourable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore, playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each— spades (), hearts (), diamonds () and clubs (). Clubs and spades are of black colour, while hearts and diamonds are of red colour. The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face cards. Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will That is, the probability of an event which is impossible to occur is 0. Such an Since every face of a die is marked with a number less than 7, it is sure that we Now, let us take an example related to playing cards. Have you seen a deck of 0 P(E) 1 Solution : Well-shuffling ensures equally likely outcomes. (ii) not be an ace. (ii) Let F be the event ‘card drawn is not an ace’. (i) be an ace, (i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’. The number of outcomes favourable to E = 4 The number of possible outcomes = 52 (Why ?) Therefore, P(E) = 4 1 52 13 The number of outcomes favourable to the event F = 52 – 4 = 48 (Why?) Reprint 2025-26 208 MATHEMATICS Remark : Note that F is nothing but E . Therefore, we can also calculate P(F) as follows: P(F) = P( E ) = 1 – P(E) = 1 12 1 13 13 Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? Solution : Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively. Example 6 : Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). Solution : Out of the two friends, one girl, say, Savita’s birthday can be any day of the year. Now, Hamida’s birthday can also be any day of 365 days in the year. The probability of Sangeeta’s winning = P(S) = 0.62 (given) The probability of Reshma’s winning = P(R) = 1 – P(S) The number of possible outcomes = 52 Therefore, P(F) = 48 12 52 13 [As the events R and S are complementary] = 1 – 0.62 = 0.38 We assume that these 365 outcomes are equally likely. So, P (Hamida’s birthday is different from Savita’s birthday) = 364 365 (ii) P(Savita and Hamida have the same birthday) (i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes for her birthday is 365 – 1 = 364 = 1 – P (both have different birthdays) = 364 1 365 [Using P( E ) = 1 – P(E)] = 1 365 Reprint 2025-26 PROBABILITY 209 Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy? Solution : There are 40 students, and only one name card has to be chosen. Note : We can also determine P(Boy), by taking Example 8 : A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be Solution : Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the (ii) The number of outcomes favourable for a card with the name of a boy = 15 (Why?) (i) The number of all possible outcomes is 40 (i) white? (ii) blue? (iii) red? The number of outcomes favourable for a card with the name of a girl = 25 (Why?) Therefore, P (card with name of a girl) = P(Girl) = 25 5 40 8 Therefore, P(card with name of a boy) = P(Boy) = 15 3 40 8 P(Boy) = 1 – P(not Boy) = 1 – P(Girl) = 5 3 1 8 8 Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’ and R denote the event ‘marble is red’. (i) The number of outcomes favourable to the event W = 2 So, P(W) = 2 9 Similarly, (ii) P(B) = 3 9 = 1 3 and (iii) P(R) = 4 9 Note that P(W) + P(B) + P(R) = 1. number of possible outcomes = 3 +2 + 4 = 9 (Why?) Reprint 2025-26 210 MATHEMATICS Example 9 : Harpreet tosses two different coins simultaneously (say, one is of ` 1 and other of ` 2). What is the probability that she gets at least one head? Solution : We write H for ‘head’ and T for ‘tail’. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. Here (H, H) means head up on the first coin (say on ` 1) and head up on the second coin (` 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on. and (T, H). (Why?) So, the number of outcomes favourable to E is 3. Therefore, P(E) = 3 4 i.e., the probability that Harpreet gets at least one head is 3 4 Note : You can also find P(E) as follows: P (E) = 1 3 1 – P(E) = 1 – 4 4 1 Since P(E) = P(no head) = 4 Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now. given numbers, or in which the outcome is every point within a circle or rectangle, etc. Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle. So, the definition of (theoretical) probability which you have learnt so far cannot be applied in the present form. What is the way out? To answer this, let us consider the following example : Example 10* : In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting? The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) There are many experiments in which the outcome is any number between two Solution : Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2 (see Fig. 14.1). * Not from the examination point of view. Fig. 14.1 Reprint 2025-26 PROBABILITY 211 Let E be the event that ‘the music is stopped within the first half-minute’. The outcomes favourable to E are points on the number line from 0 to 1 2 . The distance from 0 to 2 is 2, while the distance from 0 to 1 2 is 1 2 . Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is 1 2 . So, P(E) = Distance favourable to the event E Total distance in which outcomes can lie = 1 2 1 2 4 Can we now extend the idea of Example 10 for finding the probability as the ratio of the favourable area to the total area? Example 11* : A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the lake shown in the figure? Solution : The helicopter is equally likely to crash anywhere in the region. Area of the entire region where the helicopter can crash * Not from the examination point of view. = (4.5 × 9) km2 = 40.5 km2 Reprint 2025-26 Fig. 14.2 212 MATHEMATICS Area of the lake = (2.5 × 3) km2 = 7.5 km2 Therefore, P (helicopter crashed in the lake) = 7.5 5 40.5 405 27 Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes. Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (ii) it is acceptable to Sujatha? (ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96 (Why?) (i) it is acceptable to Jimmy? (i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 (Why?) So, P (shirt is acceptable to Sujatha) = 96 0.96 100 Therefore, P (shirt is acceptable to Jimmy) = 88 0.88 100 Solution : When the blue die shows ‘1’, the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die. (i) 8? (ii) 13? (iii) less than or equal to 12? Reprint 2025-26 PROBABILITY 213 Note that the pair (1, 4) is different from (4, 1). (Why?) So, the number of possible outcomes = 6 × 6 = 36. (i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see Fig. 14.3) i.e., the number of outcomes favourable to E = 5. Hence, P(E) = 5 36 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) 1 2 3 4 5 6 Fig. 14.3 (iii) As you can see from Fig. 14.3, all the outcomes are favourable to the event G, ‘sum of two numbers £ 12’. (ii) As you can see from Fig. 14.3, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’. So, P(F) = 0 0 36 So, P(G) = 36 1 36 Reprint 2025-26 214 MATHEMATICS 1. Complete the following statements: 2. Which of the following experiments have equally likely outcomes? Explain. 3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? 4. Which of the following cannot be the probability of an event? (A) 2 3 (B) –1.5 (C) 15% (D) 0.7 (iv) The sum of the probabilities of all the elementary events of an experiment is . (iv) A baby is born. It is a boy or a girl. (iii) The probability of an event that is certain to happen is . Such an event is called . (iii) A trial is made to answer a true-false question. The answer is right or wrong. (v) The probability of an event is greater than or equal to and less than or equal to . (ii) The probability of an event that cannot happen is . Such an event is called . (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. (i) Probability of an event E + Probability of the event ‘not E’ = . (i) A driver attempts to start a car. The car starts or does not start. EXERCISE 14.1 5. If P(E) = 0.05, what is the probability of ‘not E’? 6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out 7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday? 8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red? 9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green? (ii) a lemon flavoured candy? (i) an orange flavoured candy? Reprint 2025-26 PROBABILITY 215 10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ` 5 coin? 12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9? 13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number. 14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds 15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? 11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish? Fig. 14.4 Fig. 14.5 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. 17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ? 18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5. Reprint 2025-26 216 MATHEMATICS 20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m? 19. A child has a die whose six faces show the letters as given below: 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that 22. Refer to Example 13. (i) Complete the following table: The die is thrown once. What is the probability of getting (i) A? (ii) D? Event : ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1 36 5 36 1 36 (ii) She will not buy it ? (i) She will buy it ? ABCDEA Fig. 14.6 3 m 2 m * Not from the examination point of view. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. 24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment] (ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1 11 . Do you agree with this argument? Justify your answer. Reprint 2025-26 PROBABILITY 217 14.2 Summary In this chapter, you have studied the following points : 25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. 1. The theoretical (classical) probability of an event E, written as P(E), is defined as 2. The probability of a sure event (or certain event) is 1. 3. The probability of an impossible event is 0. 4. The probability of an event E is a number P(E) such that 5. An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1. where we assume that the outcomes of the experiment are equally likely. (ii) If a die is thrown, there are two possible outcomes—an odd number or an even (i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1 3 number. Therefore, the probability of getting an odd number is 1 2 . P (E) = Number of outcomes favourable to E Number of all possible outcomes of the experiment 0 P (E) 1 6. For any event E, P (E) + P ( E ) = 1, where E stands for ‘not E’. E and E are called complementary events. The experimental or empirical probability of an event is based on what has actually happened while the theoretical probability of the event attempts to predict what will happen on the basis of certain assumptions. As the number of trials in an experiment, go on increasing we may expect the experimental and theoretical probabilities to be nearly the same. A NOTE TO THE READER Reprint 2025-26" class_11,1,Sets,ncert_books/class_11/kemh1dd/kemh101.pdf,"1.1 Introduction The concept of set serves as a fundamental part of the present day mathematics. Today this concept is being used in almost every branch of mathematics. Sets are used to define the concepts of relations and functions. The study of geometry, sequences, probability, etc. requires the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”. In this Chapter, we discuss some basic definitions and operations involving sets. vIn these days of conflict between ancient and modern studies; there must surely be something to be said for a study which did not begin with Pythagoras and will not end with Einstein; but is the oldest and the youngest. — G.H. HARDY v SETS Chapter 1 Georg Cantor (1845-1918) 1.2 Sets and their Representations In everyday life, we often speak of collections of objects of a particular kind, such as, a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come across collections, for example, of natural numbers, points, prime numbers, etc. More specially, we examine the following collections: (i) Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9 (ii) The rivers of India (iii) The vowels in the English alphabet, namely, a, e, i, o, u (iv) Various kinds of triangles (v) Prime factors of 210, namely, 2,3,5 and 7 (vi) The solution of the equation: x 2 – 5x + 6 = 0, viz, 2 and 3. We note that each of the above example is a well-defined collection of objects in Reprint 2025-26 2 MATHEMATICS the sense that we can definitely decide whether a given particular object belongs to a given collection or not. For example, we can say that the river Nile does not belong to the collection of rivers of India. On the other hand, the river Ganga does belong to this colleciton. R+ : the set of positive real numbers. The symbols for the special sets given above will be referred to throughout this text. Again the collection of five most renowned mathematicians of the world is not well-defined, because the criterion for determining a mathematician as most renowned may vary from person to person. Thus, it is not a well-defined collection. We shall say that a set is a well-defined collection of objects. The following points may be noted : We give below a few more examples of sets used particularly in mathematics, viz. N : the set of all natural numbers Z : the set of all integers Q : the set of all rational numbers R : the set of real numbers Z + : the set of positive integers Q+ : the set of positive rational numbers, and If a is an element of a set A, we say that “ a belongs to A” the Greek symbol ∈ (i) Objects, elements and members of a set are synonymous terms. (ii) Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc. (iii) The elements of a set are represented by small letters a, b, c, x, y, z, etc. (epsilon) is used to denote the phrase ‘belongs to’. Thus, we write a ∈ A. If ‘b’ is not an element of a set A, we write b ∉ A and read “b does not belong to A”. Thus, in the set V of vowels in the English alphabet, a ∈ V but b ∉ V. In the set P of prime factors of 30, 3 ∈ P but 15 ∉ P. There are two methods of representing a set : (i) Roster or tabular form (ii) Set-builder form. (i) In roster form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within braces { }. For example, the set of all even positive integers less than 7 is described in roster form as {2, 4, 6}. Some more examples of representing a set in roster form are given below : (a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}. Reprint 2025-26 (ii) In set-builder form, all the elements of a set possess a single common property which is not possessed by any element outside the set. For example, in the set {a, e, i, o, u}, all the elements possess a common property, namely, each of them is a vowel in the English alphabet, and no other letter possess this property. Denoting this set by V, we write V = {x : x is a vowel in English alphabet} It may be observed that we describe the element of the set by using a symbol x (any other symbol like the letters y, z, etc. could be used) which is followed by a colon “ : ”. After the sign of colon, we write the characteristic property possessed by the elements of the set and then enclose the whole description within braces. The above description of the set V is read as “the set of all x such that x is a vowel of the English alphabet”. In this description the braces stand for “the set of all”, the colon stands for “such that”. For example, the set A = {x : x is a natural number and 3 < x < 10} is read as “the set of all x such that x is a natural number and x lies between 3 and 10.” Hence, the numbers 4, 5, 6, 7, 8 and 9 are the elements of the set A. If we denote the sets described in (a), (b) and (c) above in roster form by A, B, C, respectively, then A, B, C can also be represented in set-builder form as follows: A= {x : x is a natural number which divides 42} B= {y : y is a vowel in the English alphabet} C= {z : z is an odd natural number} ANote In roster form, the order in which the elements are listed is immaterial. Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}. (b) The set of all vowels in the English alphabet is {a, e, i, o, u}. (c) The set of odd natural numbers is represented by {1, 3, 5, . . .}. The dots tell us that the list of odd numbers continue indefinitely. ANote It may be noted that while writing the set in roster form an element is not generally repeated, i.e., all the elements are taken as distinct. For example, the set of letters forming the word ‘SCHOOL’ is { S, C, H, O, L} or {H, O, L, C, S}. Here, the order of listing elements has no relevance. SETS 3 Example 1 Write the solution set of the equation x 2 + x – 2 = 0 in roster form. Solution The given equation can be written as (x – 1) (x + 2) = 0, i. e., x = 1, – 2 Therefore, the solution set of the given equation can be written in roster form as {1, – 2}. Example 2 Write the set {x : x is a positive integer and x 2 < 40} in the roster form. Reprint 2025-26 4 MATHEMATICS Solution The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3, 4, 5, 6}. Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form. Solution We may write the set A as A = {x : x is the square of a natural number} Alternatively, we can write A = {x : x = n 2 , where n ∈ N} Example 4 Write the set 1 2 3 4 5 6 { } 2 3 4 5 6 7 , , , , , in the set-builder form. Solution We see that each member in the given set has the numerator one less than the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the set-builder form the given set is Example 5 Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form : (i) {P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18} (ii) { 0 } (b) { x : x is an integer and x 2 – 9 = 0} (iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1= 1} (iv) {3, –3} (d) {x : x is a letter of the word PRINCIPAL} Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x 2 – 9 = 0 implies x = 3, –3 and so (iv) matches (b). where is a natural number and 1 6 1 n x : x , n n n = ≤ ≤ + 1. Which of the following are sets ? Justify your answer. (i) The collection of all the months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. (v) The collection of all natural numbers less than 100. (vi) A collection of novels written by the writer Munshi Prem Chand. (vii) The collection of all even integers. EXERCISE 1.1 Reprint 2025-26 (viii) The collection of questions in this Chapter. (ix) A collection of most dangerous animals of the world. 2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces: (i) 5. . .A (ii) 8 . . . A (iii) 0. . .A (iv) 4. . . A (v) 2. . .A (vi) 10. . .A 3. Write the following sets in roster form: (i) A = {x : x is an integer and –3 ≤ x < 7} (ii) B = {x : x is a natural number less than 6} (iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8} (iv) D = {x : x is a prime number which is divisor of 60} (v) E = The set of all letters in the word TRIGONOMETRY (vi) F = The set of all letters in the word BETTER 4. Write the following sets in the set-builder form : (i) (3, 6, 9, 12} (ii) {2,4,8,16,32} (iii) {5, 25, 125, 625} (iv) {2, 4, 6, . . .} (v) {1,4,9, . . .,100} 5. List all the elements of the following sets : (i) A = {x : x is an odd natural number} (iii) C = {x : x is an integer, x 2 ≤ 4} (iv) D = {x : x is a letter in the word “LOYAL”} (v) E = {x : x is a month of a year not having 31 days} (vi) F = {x : x is a consonant in the English alphabet which precedes k }. 6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form: (i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6} (ii) {2, 3} (b) {x : x is an odd natural number less than 10} (iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6} (iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}. (ii) B = {x : x is an integer, 1 2 – < x < 9 2 } SETS 5 1.3 The Empty Set Consider the set A = { x : x is a student of Class XI presently studying in a school } We can go to the school and count the number of students presently studying in Class XI in the school. Thus, the set A contains a finite number of elements. We now write another set B as follows: Reprint 2025-26 6 MATHEMATICS B = { x : x is a student presently studying in both Classes X and XI } We observe that a student cannot study simultaneously in both Classes X and XI. Thus, the set B contains no element at all. Definition 1 A set which does not contain any element is called the empty set or the null set or the void set. According to this definition, B is an empty set while A is not an empty set. The empty set is denoted by the symbol φ or { }. We give below a few examples of empty sets. (i) Let A = {x : 1 < x < 2, x is a natural number}. Then A is the empty set, because there is no natural number between 1 and 2. (ii) B = {x : x 2 – 2 = 0 and x is rational number}. Then B is the empty set because the equation x 2 – 2 = 0 is not satisfied by any rational value of x. (iii) C = {x : x is an even prime number greater than 2}.Then C is the empty set, because 2 is the only even prime number. (iv) D = { x : x 2 = 4, x is odd }. Then D is the empty set, because the equation x 2 = 4 is not satisfied by any odd value of x. 1.4 Finite and Infinite Sets Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e, g} and C = { men living presently in different parts of the world} We observe that A contains 5 elements and B contains 6 elements. How many elements does C contain? As it is, we do not know the number of elements in C, but it is some natural number which may be quite a big number. By number of elements of a set S, we mean the number of distinct elements of the set and we denote it by n (S). If n (S) is a natural number, then S is non-empty finite set. Consider the set of natural numbers. We see that the number of elements of this set is not finite since there are infinite number of natural numbers. We say that the set of natural numbers is an infinite set. The sets A, B and C given above are finite sets and n(A) = 5, n(B) = 6 and n(C) = some finite number. Definition 2 A set which is empty or consists of a definite number of elements is called finite otherwise, the set is called infinite. Consider some examples : (i) Let W be the set of the days of the week. Then W is finite. (ii) Let S be the set of solutions of the equation x 2 –16 = 0. Then S is finite. (iii) Let G be the set of points on a line. Then G is infinite. When we represent a set in the roster form, we write all the elements of the set within braces { }. It is not possible to write all the elements of an infinite set within braces { } because the numbers of elements of such a set is not finite. So, we represent Reprint 2025-26 some infinite set in the roster form by writing a few elements which clearly indicate the structure of the set followed ( or preceded ) by three dots. For example, {1, 2, 3 . . .} is the set of natural numbers, {1, 3, 5, 7, . . .} is the set of odd natural numbers, {. . .,–3, –2, –1, 0,1, 2 ,3, . . .} is the set of integers. All these sets are infinite. ANote All infinite sets cannot be described in the roster form. For example, the set of real numbers cannot be described in this form, because the elements of this set do not follow any particular pattern. Example 6 State which of the following sets are finite or infinite : (i) {x : x ∈ N and (x – 1) (x –2) = 0} (ii) {x : x ∈ N and x 2 = 4} (iii) {x : x ∈ N and 2x –1 = 0} (iv) {x : x ∈ N and x is prime} (v) {x : x ∈ N and x is odd} Solution (i) Given set = {1, 2}. Hence, it is finite. (ii) Given set = {2}. Hence, it is finite. (iii) Given set = φ. Hence, it is finite. (iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite (v) Since there are infinite number of odd numbers, hence, the given set is infinite. 1.5 Equal Sets Given two sets A and B, if every element of A is also an element of B and if every element of B is also an element of A, then the sets A and B are said to be equal. Clearly, the two sets have exactly the same elements. SETS 7 Definition 3 Two sets A and B are said to be equal if they have exactly the same elements and we write A = B. Otherwise, the sets are said to be unequal and we write A ≠ B. We consider the following examples : (i) Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B. (ii) Let A be the set of prime numbers less than 6 and P the set of prime factors of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of 30 and also these are less than 6. ANote A set does not change if one or more elements of the set are repeated. For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each Reprint 2025-26 8 MATHEMATICS Example 7 Find the pairs of equal sets, if any, give reasons: A = {0}, B = {x : x > 15 and x < 5}, C = {x : x – 5 = 0 }, D = {x: x 2 = 25}, E = {x : x is an integral positive root of the equation x 2 – 2x –15 = 0}. Solution Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E. Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D. Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E. Thus, the only pair of equal sets is C and E. Example 8 Which of the following pairs of sets are equal? Justify your answer. (i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”. (ii) A = {n : n ∈ Z and n 2 ≤ 4} and B = {x : x ∈ R and x 2 – 3x + 2 = 0}. Solution (i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are equal sets as repetition of elements in a set do not change a set. Thus, X = {A, L, O, Y} = B (ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 ∈ A and 0 ∉ B, A and B are not equal sets. 1. Which of the following are examples of the null set (i) Set of odd natural numbers divisible by 2 (ii) Set of even prime numbers (iii) { x : x is a natural numbers, x < 5 and x > 7 } (iv) { y : y is a point common to any two parallel lines} 2. Which of the following sets are finite or infinite (i) The set of months of a year (ii) {1, 2, 3, . . .} (iii) {1, 2, 3, . . .99, 100} (iv) The set of positive integers greater than 100 (v) The set of prime numbers less than 99 3. State whether each of the following set is finite or infinite: (i) The set of lines which are parallel to the x-axis (ii) The set of letters in the English alphabet (iii) The set of numbers which are multiple of 5 element of A is in B and vice-versa. That is why we generally do not repeat any element in describing a set. EXERCISE 1.2 Reprint 2025-26 (iv) The set of animals living on the earth (v) The set of circles passing through the origin (0,0) 4. In the following, state whether A = B or not: (i) A = { a, b, c, d } B = { d, c, b, a } (ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18} (iii) A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x ≤ 10} (iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . } 5. Are the following pair of sets equal ? Give reasons. (i) A = {2, 3}, B = {x : x is solution of x 2 + 5x + 6 = 0} (ii) A = { x : x is a letter in the word FOLLOW} B = { y : y is a letter in the word WOLF} 6. From the sets given below, select equal sets : A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2} E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1} 1.6 Subsets Consider the sets : X = set of all students in your school, Y = set of all students in your class. We note that every element of Y is also an element of X; we say that Y is a subset of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂ stands for ‘is a subset of’ or ‘is contained in’. Definition 4 A set A is said to be a subset of a set B if every element of A is also an element of B. In other words, A ⊂ B if whenever a ∈ A, then a ∈ B. It is often convenient to use the symbol “⇒” which means implies. Using this symbol, we can write the definiton of subset as follows: A ⊂ B if a ∈ A ⇒ a ∈ B We read the above statement as “A is a subset of B if a is an element of A implies that a is also an element of B”. If A is not a subset of B, we write A ⊄ B. We may note that for A to be a subset of B, all that is needed is that every element of A is in B. It is possible that every element of B may or may not be in A. If it so happens that every element of B is also in A, then we shall also have B ⊂A. In this case, A and B are the same sets so that we have A ⊂ B and B ⊂ A ⇔ A = B, where “⇔” is a symbol for two way implications, and is usually read as if and only if (briefly written as “iff”). It follows from the above definition that every set A is a subset of itself, i.e., A ⊂ A. Since the empty set φ has no elements, we agree to say that φ is a subset of every set. We now consider some examples : SETS 9 Reprint 2025-26 10 MATHEMATICS (i) The set Q of rational numbers is a subset of the set R of real numbes, and we write Q ⊂ R. (ii) If A is the set of all divisors of 56 and B the set of all prime divisors of 56, then B is a subset of A and we write B ⊂ A. (iii) Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Then A ⊂ B and B ⊂ A and hence A = B. (iv) Let A = { a, e, i, o, u} and B = { a, b, c, d}. Then A is not a subset of B, also B is not a subset of A. Let A and B be two sets. If A ⊂ B and A ≠ B , then A is called a proper subset of B and B is called superset of A. For example, A = {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}. If a set A has only one element, we call it a singleton set. Thus,{ a } is a singleton set. Example 9 Consider the sets φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄ between each of the following pair of sets: Solution (i) φ ⊂ B as φ is a subset of every set. (ii) A ⊄ B as 3 ∈ A and 3 ∉ B (iii) A ⊂ C as 1, 3 ∈ A also belongs to C (iv) B ⊂ C as each element of B is also an element of C. Example 10 Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B ? No. (Why?). Is B a subset of A? No. (Why?) Example 11 Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example. (i) φ . . . B (ii) A . . . B (iii) A . . . C (iv) B . . . C Solution No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1} and B ⊂ C. But A ⊄ C as 1 ∈ A and 1 ∉ C. Note that an element of a set can never be a subset of itself. 1.6.1 Subsets of set of real numbers As noted in Section 1.6, there are many important subsets of R. We give below the names of some of these subsets. The set of natural numbers N = {1, 2, 3, 4, 5, . . .} The set of integers Z = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .} The set of rational numbers Q = { x : x = p q , p, q ∈ Z and q ≠ 0} Reprint 2025-26 which is read “ Q is the set of all numbers x such that x equals the quotient p q , where p and q are integers and q is not zero”. Members of Q include –5 (which can be expressed as 5 1 – ) , 7 5 , 1 3 2 (which can be expressed as 7 2 ) and 11 3 – . The set of irrational numbers, denoted by T, is composed of all other real numbers. Thus T = {x : x ∈ R and x ∉ Q}, i.e., all real numbers that are not rational. Members of T include 2 , 5 and π . Some of the obvious relations among these subsets are: N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊄ T. 1.6.2 Intervals as subsets of R Let a, b ∈ R and a < b. Then the set of real numbers { y : a < y < b} is called an open interval and is denoted by (a, b). All the points between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval. The interval which contains the end points also is called closed interval and is denoted by [ a, b ]. Thus [ a, b ] = {x : a ≤ x ≤ b} We can also have intervals closed at one end and open at the other, i.e., [ a, b ) = {x : a ≤ x < b} is an open interval from a to b, including a but excluding b. ( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding a. These notations provide an alternative way of designating the subsets of set of real numbers. For example , if A = (–3, 5) and B = [–7, 9], then A ⊂ B. The set [ 0, ∞) defines the set of non-negative real numbers, while set ( – ∞, 0 ) defines the set of negative real numbers. The set ( – ∞, ∞ ) describes the set of real numbers in relation to a line extending from – ∞ to ∞. On real number line, various types of intervals described above as subsets of R, are shown in the Fig 1.1. SETS 11 Here, we note that an interval contains infinitely many points. For example, the set {x : x ∈ R, –5 < x ≤ 7}, written in set-builder form, can be written in the form of interval as (–5, 7] and the interval [–3, 5) can be written in setbuilder form as {x : –3 ≤ x < 5}. Reprint 2025-26 Fig 1.1 12 MATHEMATICS [a, b) or (a, b]. 1.7 Universal Set Usually, in a particular context, we have to deal with the elements and subsets of a basic set which is relevant to that particular context. For example, while studying the system of numbers, we are interested in the set of natural numbers and its subsets such as the set of all prime numbers, the set of all even numbers, and so forth. This basic set is called the “Universal Set”. The universal set is usually denoted by U, and all its subsets by the letters A, B, C, etc. For example, for the set of all integers, the universal set can be the set of rational numbers or, for that matter, the set R of real numbers. For another example, in human population studies, the universal set consists of all the people in the world. 1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces : (i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 } (ii) { a, b, c } . . . { b, c, d } (iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school} (iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with radius 1 unit} (v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane} (vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane} (vii) {x : x is an even natural number} . . . {x : x is an integer} 2. Examine whether the following statements are true or false: (i) { a, b } ⊄ { b, c, a } (ii) { a, e } ⊂ { x : x is a vowel in the English alphabet} (iii) { 1, 2, 3 } ⊂ { 1, 3, 5 } (iv) { a } ⊂ { a, b, c } (v) { a } ∈ { a, b, c } (vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36} 3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why? (i) {3, 4} ⊂ A (ii) {3, 4} ∈ A (iii) {{3, 4}} ⊂ A (iv) 1 ∈ A (v) 1 ⊂ A (vi) {1, 2, 5} ⊂ A (vii) {1, 2, 5} ∈ A (viii) {1, 2, 3} ⊂ A (ix) φ ∈ A (x) φ ⊂ A (xi) {φ} ⊂ A 4. Write down all the subsets of the following sets (i) {a} (ii) {a, b} (iii) {1, 2, 3} (iv) φ The number (b – a) is called the length of any of the intervals (a, b), [a, b], EXERCISE 1.3 Reprint 2025-26 5. Write the following as intervals : (i) {x : x ∈ R, – 4 < x ≤ 6} (ii) {x : x ∈ R, – 12 < x < –10} (iii) {x : x ∈ R, 0 ≤ x < 7} (iv) {x : x ∈ R, 3 ≤ x ≤ 4} 6. Write the following intervals in set-builder form : (i) (– 3, 0) (ii) [6 , 12] (iii) (6, 12] (iv) [–23, 5) 7. What universal set(s) would you propose for each of the following : (i) The set of right triangles. (ii) The set of isosceles triangles. 8. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C (i) {0, 1, 2, 3, 4, 5, 6} (ii) φ (iii) {0,1,2,3,4,5,6,7,8,9,10} (iv) {1,2,3,4,5,6,7,8} 1.8 Venn Diagrams Most of the relationships between sets can be represented by means of diagrams which are known as Venn diagrams. Venn diagrams are named after the English logician, John Venn (1834-1883). These diagrams consist of rectangles and closed curves usually circles. The universal set is represented usually by a rectangle and its subsets by circles. In Venn diagrams, the elements of the sets are written in their respective circles (Figs 1.2 and 1.3) Illustration 1 In Fig 1.2, U = {1,2,3, ..., 10} is the universal set of which A = {2,4,6,8,10} is a subset. Fig 1.2 SETS 13 Illustration 2 In Fig 1.3, U = {1,2,3, ..., 10} is the universal set of which A = {2,4,6,8,10} and B = {4, 6} are subsets, and also B ⊂ A. The reader will see an extensive use of the Venn diagrams when we discuss the union, intersection and difference of sets. 1.9 Operations on Sets In earlier classes, we have learnt how to perform the operations of addition, subtraction, multiplication and division on numbers. Each one of these operations was performed on a pair of numbers to get another number. For example, when we perform the operation of addition on the pair of numbers 5 and 13, we get the number 18. Again, performing the operation of multiplication on the pair of numbers 5 and 13, we get 65. Reprint 2025-26 Fig 1.3 14 MATHEMATICS Similarly, there are some operations which when performed on two sets give rise to another set. We will now define certain operations on sets and examine their properties. Henceforth, we will refer all our sets as subsets of some universal set. 1.9.1 Union of sets Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once. The symbol ‘∪’ is used to denote the union. Symbolically, we write A ∪ B and usually read as ‘A union B’. Example 12 Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B. Solution We have A ∪ B = { 2, 4, 6, 8, 10, 12} Note that the common elements 6 and 8 have been taken only once while writing A ∪ B. Example 13 Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A Solution We have, A ∪ B = { a, e, i, o, u } = A. This example illustrates that union of sets A and its subset B is the set A itself, i.e., if B ⊂ A, then A ∪ B = A. Example 14 Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X ∪ Y and interpret the set. Solution We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of students from Class XI who are in the hockey team or the football team or both. Thus, we can define the union of two sets as follows: Definition 5 The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both). In symbols, we write. A ∪ B = { x : x ∈A or x ∈B } The union of two sets can be represented by a Venn diagram as shown in Fig 1.4. The shaded portion in Fig 1.4 represents A ∪ B. Some Properties of the Operation of Union (i) A ∪ B = B ∪ A (Commutative law) (iii) A ∪ φ = A (Law of identity element, φ is the identity of ∪) (ii) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C) (Associative law ) Reprint 2025-26 Fig 1.4 1.9.2 Intersection of sets The intersection of sets A and B is the set of all elements which are common to both A and B. The symbol ‘∩’ is used to denote the intersection. The intersection of two sets A and B is the set of all those elements which belong to both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}. Example 15 Consider the sets A and B of Example 12. Find A ∩ B. Solution We see that 6, 8 are the only elements which are common to both A and B. Hence A ∩ B = { 6, 8 }. Example 16 Consider the sets X and Y of Example 14. Find X ∩ Y. Solution We see that element ‘Geeta’ is the only element common to both. Hence, X ∩ Y = {Geeta}. Example 17 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and hence show that A ∩ B = B. Solution We have A ∩ B = { 2, 3, 5, 7 } = B. We note that B ⊂ A and that A ∩ B = B. Definition 6 The intersection of two sets A and B is the set of all those elements which belong to both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B} The shaded portion in Fig 1.5 indicates the intersection of A and B. If A and B are two sets such that A ∩ B = φ, then A and B are called disjoint sets. For example, let A = { 2, 4, 6, 8 } and B = { 1, 3, 5, 7 }. Then A and B are disjoint sets, because there are no elements which are common to A and B. The disjoint sets can be represented by means of Venn diagram as shown in the Fig 1.6 In the above diagram, A and B are disjoint sets. Some Properties of Operation of Intersection (i) A ∩ B = B ∩ A (Commutative law). (ii) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) (Associative law). (iii) φ ∩ A = φ, U ∩ A = A (Law of φ and U). (iv) A ∩ A = A (Idempotent law) Fig 1.5 (iv) A ∪ A = A (Idempotent law) (v) U ∪ A = U (Law of U) U SETS 15 Reprint 2025-26 A B Fig 1.6 16 MATHEMATICS (v) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (Distributive law ) i. e., ∩ distributes over ∪ This can be seen easily from the following Venn diagrams [Figs 1.7 (i) to (v)]. (i) (iii) (ii) (iv) 1.9.3 Difference of sets The difference of the sets A and B in this order is the set of elements which belong to A but not to B. Symbolically, we write A – B and read as “A minus B”. Example 18 Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A. Solution We have, A – B = { 1, 3, 5 }, since the elements 1, 3, 5 belong to A but not to B and B – A = { 8 }, since the element 8 belongs to B and not to A. We note that A – B ≠ B – A. (v) Figs 1.7 (i) to (v) Reprint 2025-26 Example 19 Let V = { a, e, i, o, u } and B = { a, i, k, u}. Find V – B and B – V Solution We have, V – B = { e, o }, since the elements e, o belong to V but not to B and B – V = { k }, since the element k belongs to B but not to V. We note that V – B ≠ B – V. Using the setbuilder notation, we can rewrite the definition of difference as A – B = { x : x ∈ A and x ∉ B } The difference of two sets A and B can be represented by Venn diagram as shown in Fig 1.8. The shaded portion represents the difference of the two sets A and B. Remark The sets A – B, A ∩ B and B – A are mutually disjoint sets, i.e., the intersection of any of these two sets is the null set as shown in Fig 1.9. 1. Find the union of each of the following pairs of sets : (i) X = {1, 3, 5} Y = {1, 2, 3} (ii) A = [ a, e, i, o, u} B = {a, b, c} (iii) A = {x : x is a natural number and multiple of 3} B = {x : x is a natural number less than 6} (iv) A = {x : x is a natural number and 1 < x ≤6 } B = {x : x is a natural number and 6 < x < 10 } (v) A = {1, 2, 3}, B = φ 2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ? 3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ? 4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find (i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D (v) A ∪ B ∪ C (vi) A ∪ B ∪ D (vii) B ∪ C ∪ D 5. Find the intersection of each pair of sets of question 1 above. 6. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find (i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D (iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C) (vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) ( A ∩ B ) ∩ ( B ∪ C ) (x) ( A ∪ D) ∩ ( B ∪ C) EXERCISE 1.4 Fig 1.8 Fig 1.9 SETS 17 Reprint 2025-26 18 MATHEMATICS 7. If A = {x : x is a natural number }, B = {x : x is an even natural number} C = {x : x is an odd natural number}andD = {x : x is a prime number }, find (i) A ∩ B (ii) A ∩ C (iii) A ∩ D (iv) B ∩ C (v) B ∩ D (vi) C ∩ D 8. Which of the following pairs of sets are disjoint (i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 } (ii) { a, e, i, o, u } and { c, d, e, f } (iii) {x : x is an even integer } and {x : x is an odd integer} 9. If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 }, C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find (i) A – B (ii) A – C (iii) A – D (iv) B – A (v) C – A (vi) D – A (vii) B – C (viii) B – D (ix) C – B (x) D – B (xi) C – D (xii) D – C 10. If X= { a, b, c, d } and Y = { f, b, d, g}, find (i) X – Y (ii) Y – X (iii) X ∩ Y 11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q? 12. State whether each of the following statement is true or false. Justify your answer. (i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets. (ii) { a, e, i, o, u } and { a, b, c, d }are disjoint sets. (iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets. (iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets. 1.10 Complement of a Set Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42. Thus, A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these three prime numbers, i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by A′. So we have A′ = {2, 3, 7}. Thus, we see that A′ = {x : x ∈ U and x ∉ A }. This leads to the following definition. Definition 7 Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A. Symbolically, we write A′ to denote the complement of A with respect to U. Thus, A′ = {x : x ∈ U and x ∉ A }. Obviously A′ = U – A We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A. Reprint 2025-26 Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′. Solution We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Hence A′ = { 2, 4, 6, 8,10 }. Example 21 Let U be universal set of all the students of Class XI of a coeducational school and A be the set of all girls in Class XI. Find A′. Solution Since A is the set of all girls, A′ is clearly the set of all boys in the class. Now, we want to find the results for ( A ∪ B )′ and A′ ∩ B′ in the followng example. Example 22 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′∩ B′. Solution Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′ = { 1, 6 } Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B)′ = { 1, 6 } ( A ∪ B )′ = { 1, 6 } = A′ ∩ B′ It can be shown that the above result is true in general. If A and B are any two subsets of the universal set U, then ANote If A is a subset of the universal set U, then its complement A′ is also a subset of U. Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 } Hence (A′)′= {x : x ∈ U and x ∉ A′} = {1, 3, 5, 7, 9} = A It is clear from the definition of the complement that for any subset of the universal set U, we have ( A′)′ = A SETS 19 ( A ∪ B )′ = A′ ∩ B′. Similarly, ( A ∩ B )′ = A′ ∪ B′ . These two results are stated in words as follows : The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgan’s laws. These are named after the mathematician De Morgan. The complement A′ of a set A can be represented by a Venn diagram as shown in Fig 1.10. The shaded portion represents the complement of the set A. Fig 1.10 Reprint 2025-26 20 MATHEMATICS Some Properties of Complement Sets 1. Complement laws: (i) A ∪ A′ = U (ii) A ∩ A′ = φ 3. Law of double complementation : (A′)′ = A 4. Laws of empty set and universal set φ′ = U and U′ = φ. These laws can be verified by using Venn diagrams. 1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′ 2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets : (i) A = {a, b, c} (ii) B = {d, e, f, g} (iii) C = {a, c, e, g} (iv) D = { f, g, h, a} 3. Taking the set of natural numbers as the universal set, write down the complements of the following sets: (i) {x : x is an even natural number} (ii) { x : x is an odd natural number } (iii) {x : x is a positive multiple of 3} (iv) { x : x is a prime number } (v) {x : x is a natural number divisible by 3 and 5} (vi) { x : x is a perfect square } (vii) { x : x is a perfect cube} (viii) { x : x + 5 = 8 } (ix) { x : 2x + 5 = 9} (x) { x : x ≥ 7 } (xi) { x : x ∈ N and 2x + 1 > 10 } 4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that (i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′ 5. Draw appropriate Venn diagram for each of the following : (i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′ 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′? 7. Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . . 2. De Morgan’s law: (i) (A ∪ B)´ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′ EXERCISE 1.5 Example 23 Show that the set of letters needed to spell “ CATARACT ” and the set of letters needed to spell “ TRACT” are equal. Solution Let X be the set of letters in “CATARACT”. Then X = { C, A, T, R } Miscellaneous Examples Reprint 2025-26 Let Y be the set of letters in “ TRACT”. Then Y = { T, R, A, C, T } = { T, R, A, C } Since every element in X is in Y and every element in Y is in X. It follows that X = Y. Example 24 List all the subsets of the set { –1, 0, 1 }. Solution Let A = { –1, 0, 1 }. The subset of A having no element is the empty set φ. The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three elements of A is A itself. So, all the subsets of A are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}. Example 25 Show that A ∪ B = A ∩ B implies A = B Solution Let a ∈ A. Then a ∈ A ∪ B. Since A ∪ B = A ∩ B , a ∈ A ∩ B. So a ∈ B. Therefore, A ⊂ B. Similarly, if b ∈ B, then b ∈ A ∪ B. Since A ∪ B = A ∩ B, b ∈ A ∩ B. So, b ∈ A. Therefore, B ⊂ A. Thus, A = B 1. Decide, among the following sets, which sets are subsets of one and another: A = { x : x ∈ R and x satisfy x 2 – 8x + 12 = 0 }, B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }. 2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If x ∈ A and A ∈ B , then x ∈ B (ii) If A ⊂ B and B ∈ C , then A ∈ C (iii) If A ⊂ B and B ⊂ C , then A ⊂ C (iv) If A ⊄ B and B ⊄ C , then A ⊄ C (v) If x ∈ A and A ⊄ B , then x ∈ B (vi) If A ⊂ B and x ∉ B , then x ∉ A 3. Let A, B, and C be the sets such that A ∪ B = A∪ C and A ∩ B = A ∩ C. Show that B = C. 4. Show that the following four conditions are equivalent : (i) A ⊂ B(ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A Miscellaneous Exercise on Chapter 1 SETS 21 5. Show that if A ⊂ B, then C – B ⊂ C – A. 6. Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B ) 7. Using properties of sets, show that (i) A ∪ ( A ∩ B ) = A (ii) A ∩ ( A ∪ B ) = A. 8. Show that A ∩ B = A ∩ C need not imply B = C. Reprint 2025-26 22 MATHEMATICS 9. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law ) 10. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ. This chapter deals with some basic definitions and operations involving sets. These are summarised below: ÆA set is a well-defined collection of objects. ÆA set which does not contain any element is called empty set. ÆA set which consists of a definite number of elements is called finite set, otherwise, the set is called infinite set. ÆTwo sets A and B are said to be equal if they have exactly the same elements. ÆA set A is said to be subset of a set B, if every element of A is also an element of B. Intervals are subsets of R. ÆThe union of two sets A and B is the set of all those elements which are either in A or in B. ÆThe intersection of two sets A and B is the set of all elements which are common. The difference of two sets A and B in this order is the set of elements which belong to A but not to B. ÆThe complement of a subset A of universal set U is the set of all elements of U which are not the elements of A. ÆFor any two sets A and B, (A ∪ B)′ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′ Summary The modern theory of sets is considered to have been originated largely by the German mathematician Georg Cantor (1845-1918). His papers on set theory appeared sometimes during 1874 to 1897. His study of set theory came when he was studying trigonometric series of the form a1 sin x + a2 sin 2x + a3 sin 3x + ... He published in a paper in 1874 that the set of real numbers could not be put into one-to-one correspondence wih the integers. From 1879 onwards, he publishd several papers showing various properties of abstract sets. Historical Note Reprint 2025-26 Cantor’s work was well received by another famous mathematician Richard Dedekind (1831-1916). But Kronecker (1810-1893) castigated him for regarding infinite set the same way as finite sets. Another German mathematician Gottlob Frege, at the turn of the century, presented the set theory as principles of logic. Till then the entire set theory was based on the assumption of the existence of the set of all sets. It was the famous Englih Philosopher Bertand Russell (1872- 1970 ) who showed in 1902 that the assumption of existence of a set of all sets leads to a contradiction. This led to the famous Russell’s Paradox. Paul R.Halmos writes about it in his book ‘Naïve Set Theory’ that “nothing contains everything”. The Russell’s Paradox was not the only one which arose in set theory. Many paradoxes were produced later by several mathematicians and logicians. As a consequence of all these paradoxes, the first axiomatisation of set theory was published in 1908 by Ernst Zermelo. Another one was proposed by Abraham Fraenkel in 1922. John Von Neumann in 1925 introduced explicitly the axiom of regularity. Later in 1937 Paul Bernays gave a set of more satisfactory axiomatisation. A modification of these axioms was done by Kurt Gödel in his monograph in 1940. This was known as Von Neumann-Bernays (VNB) or GödelBernays (GB) set theory. Despite all these difficulties, Cantor’s set theory is used in present day mathematics. In fact, these days most of the concepts and results in mathematics are expressed in the set theoretic language. SETS 23 Reprint 2025-26 — v —" class_11,2,Relations and Functions,ncert_books/class_11/kemh1dd/kemh102.pdf,"2.1 Introduction Much of mathematics is about finding a pattern – a recognisable link between quantities that change. In our daily life, we come across many patterns that characterise relations such as brother and sister, father and son, teacher and student. In mathematics also, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, we notice that a relation involves pairs of objects in certain order. In this Chapter, we will learn how to link pairs of objects from two sets and then introduce relations between the two objects in the pair. Finally, we will learn about special relations which will qualify to be functions. The concept of function is very important in mathematics since it captures the idea of a mathematically precise correspondence between one quantity with the other. RELATIONS AND FUNCTIONS vMathematics is the indispensable instrument of all physical research. – BERTHELOT v Chapter 2 G. W. Leibnitz (1646–1716) 2.2 Cartesian Products of Sets Suppose A is a set of 2 colours and B is a set of 3 objects, i.e., A = {red, blue}and B = {b, c, s}, where b, c and s represent a particular bag, coat and shirt, respectively. How many pairs of coloured objects can be made from these two sets? Proceeding in a very orderly manner, we can see that there will be 6 distinct pairs as given below: (red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s). Thus, we get 6 distinct objects (Fig 2.1). Let us recall from our earlier classes that an ordered pair of elements taken from any two sets P and Q is a pair of elements written in small Fig 2.1 Reprint 2025-26 brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This leads to the following definition: Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e., P × Q = { (p,q) : p ∈ P, q ∈ Q } If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φ From the illustration given above we note that A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}. Again, consider the two sets: A = {DL, MP, KA}, where DL, MP, KA represent Delhi, Madhya Pradesh and Karnataka, respectively and B = {01,02, 03}representing codes for the licence plates of vehicles issued by DL, MP and KA . If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the restriction that the code begins with an element from set A, which are the pairs available from these sets and how many such pairs will there be (Fig 2.2)? The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03) and the product of set A and set B is given by A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}. It can easily be seen that there will be 9 such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also note that the order in which these elements are paired is crucial. For example, the code (DL, 01) will not be the same as the code (01, DL). As a final illustration, consider the two sets A= {a1 , a2 } and B = {b1 , b2 , b3 , b4 } (Fig 2.3). RELATIONS AND FUNCTIONS 25 03 02 01 DL MP KA Fig 2.2 (a2 , b3 ), (a2 , b4 )}. The 8 ordered pairs thus formed can represent the position of points in the plane if A and B are subsets of the set of real numbers and it is obvious that the point in the position (a1 , b2 ) will be distinct from the point in the position (b2 , a1 ). Remarks (i) Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal. A × B = {( a1 , b1 ), (a1 , b2 ), (a1 , b3 ), (a1 , b4 ), (a2 , b1 ), (a2 , b2 ), Reprint 2025-26 Fig 2.3 26 MATHEMATICS Example 1 If (x + 1, y – 2) = (3,1), find the values of x and y. Solution Since the ordered pairs are equal, the corresponding elements are equal. Therefore x + 1 = 3 and y – 2 = 1. Solving we get x = 2 and y = 3. Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. Are these two products equal? Solution By the definition of the cartesian product, P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)} Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair (r, a), we conclude that P × Q ≠ Q × P. However, the number of elements in each set will be the same. Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find (i) A × (B ∩ C) (ii) (A × B) ∩ (A × C) (iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C) (iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered (iii) If A and B are non-empty sets and either A or B is an infinite set, then so is (ii) If there are p elements in A and q elements in B, then there will be pq elements in A × B, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq. A × B. triplet. Solution (i) By the definition of the intersection of two sets, (B ∩ C) = {4}. Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}. (ii) Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)} and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)} Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}. (iii) Since, (B ∪ C) = {3, 4, 5, 6}, we have A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)}. (iv) Using the sets A × B and A × C from part (ii) above, we obtain (A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)}. Reprint 2025-26 Example 4 If P = {1, 2}, form the set P × P × P. Solution We have, P× P× P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}. Example 5 If R is the set of all real numbers, what do the cartesian products R × R and R × R × R represent? Solution The Cartesian product R × R represents the set R × R={(x, y) : x, y ∈ R} which represents the coordinates of all the points in two dimensional space and the cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R} which represents the coordinates of all the points in three-dimensional space. Example 6 If A × B ={(p, q),(p, r), (m, q), (m, r)}, find A and B. Solution A = set of first elements = {p, m} B = set of second elements = {q, r}. 1. If 2 5 1 1 3 3 3 3 x ,y – , + = , find the values of x and y. 2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B). 3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G. 4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. (i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}. (ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B. (iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ. 5. If A = {–1, 1}, find A × A × A. 6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B. 7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that EXERCISE 2.1 RELATIONS AND FUNCTIONS 27 (i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D. 8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them. 9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements. Reprint 2025-26 28 MATHEMATICS 10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A. 2.3 Relations Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}. The cartesian product of P and Q has 15 ordered pairs which can be listed as P × Q = {(a, Ali), (a,Bhanu), (a, Binoy), ..., (c, Divya)}. We can now obtain a subset of P × Q by introducing a relation R between the first element x and the second element y of each ordered pair (x, y) as R= { (x,y): x is the first letter of the name y, x ∈ P, y ∈ Q}. Then R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)} A visual representation of this relation R (called an arrow diagram) is shown in Fig 2.4. Definition 2 A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element. Definition 3 The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R. Fig 2.4 Definition 4 The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range ⊂ codomain. Remarks (i) A relation may be represented algebraically either by the Roster method or by the Set-builder method. (ii) An arrow diagram is a visual representation of a relation. Example 7 Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 } (i) Depict this relation using an arrow diagram. (ii) Write down the domain, codomain and range of R. Solution (i) By the definition of the relation, R = {(1,2), (2,3), (3,4), (4,5), (5,6)}. Reprint 2025-26 The corresponding arrow diagram is shown in Fig 2.5. (ii) We can see that the domain ={1, 2, 3, 4, 5,} Similarly, the range = {2, 3, 4, 5, 6} and the codomain = {1, 2, 3, 4, 5, 6}. Example 8 The Fig 2.6 shows a relation between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form. What is its domain and range? Solution It is obvious that the relation R is “x is the square of y”. (i) In set-builder form, R = {(x, y): x is the square of y, x ∈ P, y ∈ Q} (ii) In roster form, R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)} The domain of this relation is {4, 9, 25}. The range of this relation is {– 2, 2, –3, 3, –5, 5}. Note that the element 1 is not related to any element in set P. The set Q is the codomain of this relation. ANote The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A ) = p and n(B) = q, then n (A × B) = pq and the total number of relations is 2 pq . RELATIONS AND FUNCTIONS 29 Fig 2.5 Fig 2.6 Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B. Solution We have, A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. Since n (A×B ) = 4, the number of subsets of A×B is 24 . Therefore, the number of relations from A into B will be 24 . Remark A relation R from A to A is also stated as a relation on A. 1. Let A = {1, 2, 3,...,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range. EXERCISE 2.2 Reprint 2025-26 30 MATHEMATICS 2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, 3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form. 4. The Fig2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range? 5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}. (iii) Find the range of R. 6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}. 7. Write the relation R = {(x, x3 ) : x is a prime number less than 10} in roster form. 8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B. 9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R. Definition 5 A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element. If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the image of a under f and a is called the preimage of b under f. Fig 2.7 x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range. (ii) Find the domain of R (i) Write R in roster form 2.4 Functions In this Section, we study a special type of relation called function. It is one of the most important concepts in mathematics. We can, visualise a function as a rule, which produces new elements out of some given elements. There are many terms such as ‘map’ or ‘mapping’ used to denote a function. Reprint 2025-26 The function f from A to B is denoted by f: A ‡ B. Looking at the previous examples, we can easily see that the relation in Example 7 is not a function because the element 6 has no image. Again, the relation in Example 8 is not a function because the elements in the domain are connected to more than one images. Similarly, the relation in Example 9 is also not a function. (Why?) In the examples given below, we will see many more relations some of which are functions and others are not. Example 10 Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y ∈ N}. What is the domain, codomain and range of R? Is this relation a function? Solution The domain of R is the set of natural numbers N. The codomain is also N. The range is the set of even natural numbers. Since every natural number n has one and only one image, this relation is a function. Example 11 Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not? (i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)} (iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} Solution (i) Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function. (ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function. (iii) Since every element has one and only one image, this relation is a function. RELATIONS AND FUNCTIONS 31 Definition 6 A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function. Example 12 Let N be the set of natural numbers. Define a real valued function f : N‡ N by f (x) = 2x + 1. Using this definition, complete the table given below. x 1 2 3 4 5 6 7 Solution The completed table is given by x 1 2 3 4 5 6 7 y f (1) = 3 f (2) = 5 f (3) = 7 f (4) = 9 f (5) = 11 f (6) = 13 f (7) =15 y f (1) = ... f (2) = ... f (3) = ... f (4) = ... f (5) = ... f (6) = ... f (7) = ... Reprint 2025-26 Fig 2.8 (ii) Constant function Define the function f: R → R by y = f (x) = c, x ∈ R where c is a constant and each x ∈ R. Here domain of f is R and its range is {c}. 32 MATHEMATICS 2.4.1 Some functions and their graphs (i) Identity function Let R be the set of real numbers. Define the real valued function f : R → R by y = f(x) = x for each x ∈ R. Such a function is called the identity function. Here the domain and range of f are R. The graph is a straight line as shown in Fig 2.8. It passes through the origin. Reprint 2025-26 Fig 2.9 The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its graph will be a line as shown in the Fig 2.9. (iii) Polynomial function A function f : R → R is said to be polynomial function if for each x in R, y = f (x) = a0 + a1 x + a2 x 2 + ...+ an x n , where n is a non-negative integer and a0 , a1 , a2 ,...,an∈R. The functions defined by f(x) = x3 – x2 + 2, and g(x) = x4 + 2 x are some examples of polynomial functions, whereas the function h defined by h(x) = 2 3 x + 2x is not a polynomial function.(Why?) Example 13 Define the function f: R → R by y = f(x) = x2 , x ∈ R. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f. Solution The completed Table is given below: Domain of f = {x : x∈R}. Range of f = {x 2 : x ∈ R}. The graph of f is given by Fig 2.10 x – 4 –3 –2 –1 0 1 2 3 4 y = f(x) = x2 x – 4 –3 –2 –1 0 1 2 3 4 y = f (x) = x2 16 9 4 1 0 1 4 9 16 RELATIONS AND FUNCTIONS 33 Reprint 2025-26 Fig 2.10 34 MATHEMATICS Example 14 Draw the graph of the function f :R → R defined by f (x) = x 3 , x∈R. Solution We have f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc. Therefore, f = {(x,x 3 ): x∈R}. The graph of f is given in Fig 2.11. (iv) Rational functions are functions of the type ( ) ( ) f x g x , where f(x) and g(x) are Example 15 Define the real valued function f : R – {0} → R defined by 1 f x( ) = x , polynomial functions of x defined in a domain, where g(x) ≠ 0. Fig 2.11 x ∈ R –{0}. Complete the Table given below using this definition. What is the domain and range of this function? Solution The completed Table is given by x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2 y = 1 x ... ... ... ... ... ... ... ... ... x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2 y = 1 x – 0.5 – 0.67 –1 – 2 4 2 1 0.67 0.5 Reprint 2025-26 except 0. The graph of f is given in Fig 2.12. (v) The Modulus function The function f: R→R defined by f(x) = |x| for each x ∈R is called modulus function. For each non-negative value of x, f(x) is equal to x. But for negative values of x, the value of f(x) is the negative of the value of x, i.e., The domain is all real numbers except 0 and its range is also all real numbers Fig 2.12 RELATIONS AND FUNCTIONS 35 The graph of the modulus function is given in Fig 2.13. (vi) Signum function The function f:R→R defined by 0 ( ) 0 x,x f x x,x ≥ = − < , x f x , x 1 if 0 ( ) 0 if 0 > = = − < 1 if 0 , x Reprint 2025-26 Fig 2.13 36 MATHEMATICS is called the signum function. The domain of the signum function is R and the range is the set {–1, 0, 1}. The graph of the signum function is given by the Fig 2.14. (vii) Greatest integer function The function f: R → R defined by f(x) = [x], x ∈R assumes the value of the greatest integer, less than or equal to x. Such a function is called the greatest integer function. From the definition of [x], we can see that [x] = –1 for –1 ≤ x < 0 [x] = 0 for 0 ≤ x < 1 [x] = 1 for 1 ≤ x < 2 [x] = 2 for 2 ≤ x < 3 and so on. The graph of the function is shown in Fig 2.15. Fig 2.14 2.4.2 Algebra of real functions In this Section, we shall learn how to add two real functions, subtract a real function from another, multiply a real function by a scalar (here by a scalar we mean a real number), multiply two real functions and divide one real function by another. (i) Addition of two real functions Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f + g): X → R by (f + g) (x) = f (x) + g (x), for all x ∈ X. Reprint 2025-26 Fig 2.15 (ii) Subtraction of a real function from another Let f : X → R and g: X → R be any two real functions, where X ⊂ R. Then, we define (f – g) : X→R by (f–g) (x) = f(x) –g(x), for all x ∈ X. (iii) Multiplication by a scalar Let f : X→R be a real valued function and α be a scalar. Here by scalar, we mean a real number. Then the product α f is a function from X to R defined by (α f ) (x) = α f (x), x ∈X. (iv) Multiplication of two real functions The product (or multiplication) of two real functions f:X→R and g:X→R is a function fg:X→R defined by (fg) (x) = f(x) g(x), for all x ∈ X. This is also called pointwise multiplication. X→R, where X ⊂ R. The quotient of f by g denoted by f g is a function defined by , Example 16 Let f(x) = x 2 and g(x) = 2x + 1 be two real functions.Find (f + g) (x), (f –g) (x), (fg) (x), ( ) f x g . Solution We have, (f + g) (x) = x 2 + 2x + 1, (f –g) (x) = x 2 – 2x – 1, (v) Quotient of two real functions Let f and g be two real functions defined from ( ) ( ) ( ) f f x x g g x = , provided g(x) ≠ 0, x ∈ X RELATIONS AND FUNCTIONS 37 Example 17 Let f(x) = x and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and f g (x). Solution We have (f + g) (x) = x + x, (f – g) (x) = x – x , (fg) (x) = x 2 (2x + 1) = 2x 3 + x 2 , ( ) f x g = (fg) x = 3 2 x( x ) x = and ( ) f x g Reprint 2025-26 1 2 0 x – x , x x = = ≠ 2 1 x x + , x ≠ 1 2 − 2 38 MATHEMATICS 1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. (i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} (ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)} (iii) {(1,3), (1,5), (2,5)}. 2. Find the domain and range of the following real functions: 3. A function f is defined by f(x) = 2x –5. Write down the values of (i) f (0), (ii) f (7), (iii) f (–3). 4. The function ‘t’ which maps temperature in degree Celsius into temperature in Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212. 5. Find the range of each of the following functions. (i) f (x) = 2 – 3x, x ∈ R, x > 0. (ii) f (x) = x 2 + 2, x is a real number. (iii) f (x) = x, x is a real number. Example 18 Let R be the set of real numbers. Define the real function degree Fahrenheit is defined by t(C) = 9C 5 + 32. (i) f(x) = – x (ii) f(x) = 2 9 − x . Miscellaneous Examples EXERCISE 2.3 and sketch the graph of this function. Solution Here f(0) = 10, f(1) = 11, f(2) = 12, ..., f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, ..., f(–10) = 0 and so on. Therefore, shape of the graph of the given function assumes the form as shown in Fig 2.16. Remark The function f defined by f(x) = mx + c , x ∈ R, is called linear function, where m and c are constants. Above function is an example of a linear function. Fig 2.16 f: R→R by f(x) = x + 10 Reprint 2025-26 Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and a – b ∈ Z}. Show that (i) (a,a) ∈ R for all a ∈ Q (ii) (a,b) ∈ R implies that (b, a) ∈ R (iii) (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈R Solution (i) Since, a – a = 0 ∈ Z, if follows that (a, a) ∈ R. (ii) (a,b) ∈ R implies that a – b ∈ Z. So, b – a ∈ Z. Therefore, (b, a) ∈ R (iii) (a, b) and (b, c) ∈ R implies that a – b ∈ Z. b – c ∈ Z. So, a – c = (a – b) + (b – c) ∈ Z. Therefore, (a,c) ∈ R Example 20 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x). Solution Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R, f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1. Example 21 Find the domain of the function 2 Solution Since x 2 –5x + 4 = (x – 4) (x –1), the function f is defined for all real numbers except at x = 4 and x = 1. Hence the domain of f is R – {1, 4}. Example 22 The function f is defined by − < = + > 1 0 1 0 1 0 , x x , x x, x 2 3 5 ( ) 5 4 x x f x x x + + = − + RELATIONS AND FUNCTIONS 39 Draw the graph of f (x). Solution Here, f(x) = 1 – x, x < 0, this gives f(– 4) = 1 – (– 4)= 5; and f(1) = 2, f (2) = 3, f (3) = 4 f(– 3) =1 – (– 3) = 4, f(– 2) = 1 – (– 2)= 3 f(–1) = 1 – (–1) = 2; etc, Thus, the graph of f is as shown in Fig 2.17 Fig 2.17 f(4) = 5 and so on for f(x) = x + 1, x > 0. f (x) = Reprint 2025-26 40 MATHEMATICS 1. The relation f is defined by 2 0 3 ( ) = 3 3 10 x , x f x x, x ≤ ≤ ≤ ≤ 2. If f (x) = x 2 , find (1 1) (1) (1 1 1) f . – f . – . 3. Find the domain of the function f (x) 2 4. Find the domain and the range of the real function f defined by f (x) = ( 1) x − . 5. Find the domain and the range of the real function f defined by f (x) = x –1 . 6. Let 2 of f. 7. Let f, g : R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find The relation g is defined by 2 , 0 2 ( ) 3 , 2 10 x x g x x x ≤ ≤ = ≤ ≤ Show that f is a function and g is not a function. f + g, f – g and f g . 2 , : 1 x f x x x = ∈ + R be a function from R into R. Determine the range Miscellaneous Exercise on Chapter 2 2 2 1 8 12 x x x – x + + = + . 9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈N and a = b2 }. Are the following true? (i) (a,a) ∈ R, for all a ∈ N (ii) (a,b) ∈ R, implies (b,a) ∈ R (iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R. Justify your answer in each case. 10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? (i) f is a relation from A to B (ii) f is a function from A to B. Justify your answer in each case. 8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b. Reprint 2025-26 11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer. 12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n. Find the range of f. In this Chapter, we studied about relations and functions.The main features of this Chapter are as follows: Æ Ordered pair A pair of elements grouped together in a particular order. Æ Cartesian product A × B of two sets A and B is given by and R × R × R = {(x, y, z): x, y, z ∈ R} Æ If (a, b) = (x, y), then a = x and b = y. Æ If n(A) = p and n(B) = q, then n(A × B) = pq. Æ A × φ = φ Æ In general, A × B ≠ B × A. Æ Relation A relation R from a set A to a set B is a subset of the cartesian product A × B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A × B. Æ The image of an element x under a relation R is given by y, where (x, y) ∈ R, Æ The domain of R is the set of all first elements of the ordered pairs in a relation R. Æ The range of the relation R is the set of all second elements of the ordered pairs in a relation R. Æ Function A function f from a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B. A × B = {(a, b): a ∈ A, b ∈ B} In particular R × R = {(x, y): x, y ∈ R} Summary RELATIONS AND FUNCTIONS 41 We write f: A→B, where f(x) = y. Æ A is the domain and B is the codomain of f. Reprint 2025-26 42 MATHEMATICS Æ The range of the function is the set of images. Æ A real function has the set of real numbers or one of its subsets both as its domain and as its range. Æ Algebra of functions For functions f : X → R and g : X → R, we have The word FUNCTION first appears in a Latin manuscript “Methodus tangentium inversa, seu de fuctionibus” written by Gottfried Wilhelm Leibnitz (1646-1716) in 1673; Leibnitz used the word in the non-analytical sense. He considered a function in terms of “mathematical job” – the “employee” being just a curve. On July 5, 1698, Johan Bernoulli, in a letter to Leibnitz, for the first time deliberately assigned a specialised use of the term function in the analytical sense. At the end of that month, Leibnitz replied showing his approval. Function is found in English in 1779 in Chambers’ Cyclopaedia: “The term function is used in algebra, for an analytical expression any way compounded of a variable quantity, and of numbers, or constant quantities”. (f + g) (x) = f (x) + g(x), x ∈ X (f – g) (x) = f (x) – g(x), x ∈ X (f.g) (x) = f (x) .g (x), x ∈ X (kf) (x) = k ( f (x) ), x ∈ X, where k is a real number. ( ) f x g = ( ) ( ) f x g x , x ∈ X, g(x) ≠ 0 Historical Note Reprint 2025-26 — v —" class_11,3,Trigonometric Functions,ncert_books/class_11/kemh1dd/kemh103.pdf,"3.1 Introduction The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. The subject was originally developed to solve geometric problems involving triangles. It was studied by sea captains for navigation, surveyor to map out the new lands, by engineers and others. Currently, trigonometry is used in many areas such as the science of seismology, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing a musical tone and in many other areas. In earlier classes, we have studied the trigonometric ratios of acute angles as the ratio of the sides of a right angled triangle. We have also studied the trigonometric identities and application of trigonometric ratios in solving the problems related to heights and distances. In this Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions and study their properties. TRIGONOMETRIC FUNCTIONS vA mathematician knows how to solve a problem, he can not solve it. – MILNE v Chapter 3 Arya Bhatt (476-550) 3.2 Angles Angle is a measure of rotation of a given ray about its initial point. The original ray is Reprint 2025-26 Fig 3.1 Vertex 44 MATHEMATICS called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative (Fig 3.1). The measure of an angle is the amount of rotation performed to get the terminal side from the initial side. There are several units for measuring angles. The definition of an angle suggests a unit, viz. one complete revolution from the position of the initial side as indicated in Fig 3.2. This is often convenient for large angles. For example, we can say that a rapidly spinning wheel is making an angle of say 15 revolution per second. We shall describe two other units of measurement of an angle which are most commonly used, viz. degree measure and radian measure. 3.2.1 Degree measure If a rotation from the initial side to terminal side is th 1 360 of a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″. Thus, 1° = 60′, 1′ = 60″ Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are shown in Fig 3.3. Fig 3.2 Reprint 2025-26 Fig 3.3 3.2.2 Radian measure There is another unit for measurement of an angle, called the radian measure. Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig 3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the angles whose measures are 1 radian, –1 radian, 1 1 2 radian and –1 1 2 radian. (i) (ii) (iii) Fig 3.4 (i) to (iv) (iv) TRIGONOMETRIC FUNCTIONS 45 complete revolution of the initial side subtends an angle of 2π radian. 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 radian, an arc of length l will subtend an angle whose measure is l r radian. Thus, if in a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have θ = l r or l = r θ. We know that the circumference of a circle of radius 1 unit is 2π. Thus, one More generally, in a circle of radius r, an arc of length r will subtend an angle of Reprint 2025-26 46 MATHEMATICS 3.2.3 Relation between radian and real numbers Consider the unit circle with centre O. Let A be any point on the circle. Consider OA as initial side of an angle. Then the length of an arc of the circle will give the radian measure of the angle which the arc will subtend at the centre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the real number zero, AP represents positive real number and AQ represents negative real numbers (Fig 3.5). If we rope the line AP in the anticlockwise direction along the circle, and AQ in the clockwise direction, then every real number will correspond to a radian measure and conversely. Thus, radian measures and real numbers can be considered as one and the same. 3.2.4 Relation between degree and radian Since a circle subtends at the centre an angle whose radian measure is 2π and its degree measure is 360°, it follows that The above relation enables us to express a radian measure in terms of degree measure and a degree measure in terms of radian measure. Using approximate value of π as 22 7 , we have 2π radian = 360° or π radian = 180° Fig 3.5 A O 1 0 P −2 Q 1 2 −1 Also 1° = π 180 radian = 0.01746 radian approximately. The relation between degree measures and radian measure of some common angles are given in the following table: Degree 30° 45° 60° 90° 180° 270° 360° Radian π 6 1 radian = 180 π ° = 57° 16′ approximately. π 4 Reprint 2025-26 π 3 π 2 π 3π 2 2π Notational Convention Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle θ°, we mean the angle whose degree measure is θ and whenever we write angle β, we mean the angle whose radian measure is β. Note that when an angle is expressed in radians, the word ‘radian’ is frequently omitted. Thus, π π 180 and 45 4 = ° = ° are written with the understanding that π and π 4 are radian measures. Thus, we can say that Example 1 Convert 40° 20′ into radian measure. Solution We know that 180° = π radian. Hence 40° 20′ = 40 1 3 degree = π 180 × 121 3 radian = 121π 540 radian. Therefore 40° 20′ = 121π 540 radian. Example 2 Convert 6 radians into degree measure. Solution We know that π radian = 180°. Radian measure = π 180 × Degree measure Degree measure = 180 π ×Radian measure TRIGONOMETRIC FUNCTIONS 47 Hence 6 radians = 180 π × 6 degree = 1080 7 = 343° + 38′ + 2 11 minute [as 1′ = 60″] = 343° + 38′ + 10.9″ = 343°38′ 11″ approximately. Hence 6 radians = 343° 38′ 11″ approximately. Example 3 Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use 22 π 7 = ). = 343 7 11degree = 343° + 7 60 11 × minute [as 1° = 60′] Reprint 2025-26 22 × degree 48 MATHEMATICS Solution Here l = 37.4 cm and θ = 60° = 60π π radian = 180 3 Hence, by r = θ l , we have Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14). Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through 2 or 4π 3 radian. Hence, the required distance travelled is given by Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii. Solution Let r 1 and r 2 be the radii of the two circles. Given that θ1 = 65° = π 65 180 × = 13π 36 radian l = r θ = 1.5 × 4π 3 cm = 2π cm = 2 × 3.14 cm = 6.28 cm. r = 37.4×3 37.4×3×7 = π 22 = 35.7 cm 3 of a revolution. Therefore, 2 θ = × 360° 3 and θ2 = 110° = π 110 180 × = 22π 36 radian Let l be the length of each of the arc. Then l = r 1 θ1 = r2 θ2 , which gives Hence r 1 : r 2 = 22 : 13. 1. Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47°30′ (iii) 240° (iv) 520° 13π 36 × r 1 = 22π 36 × r 2 , i.e., 1 2 r r = 22 13 EXERCISE 3.1 Reprint 2025-26 2 . Find the degree measures corresponding to the following radian measures (i) 11 16 (ii) – 4 (iii) 5π 3 (iv) 7π 6 3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? 4. Find the degree measure of the angle subtended at the centre of a circle of 5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. 6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. 7. Find the angle in radian through which a pendulum swings if its length is 75 cm and th e tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm 3.3 Trigonometric Functions In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions. Consider a unit circle with centre at origin of the coordinate axes. Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x (Fig 3.6). We define cos x = a and sin x = b Since ∆OMP is a right triangle, we have OM2 + MP2 = OP2 or a 2 + b 2 = 1 Thus, for every point on the unit circle, we have a 2 + b 2 = 1 or cos2 x + sin2 x = 1 Since one complete revolution subtends an angle of 2π radian at the (Use 22 π 7 = ). radius 100 cm by an arc of length 22 cm (Use 22 π 7 = ). TRIGONOMETRIC FUNCTIONS 49 centre of the circle, ∠AOB = π 2 , Fig 3.6 Reprint 2025-26 50 MATHEMATICS ∠AOC = π and ∠AOD = 3π 2 . All angles which are integral multiples of π 2 are called quadrantal angles. The coordinates of the points A, B, C and D are, respectively, (1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have cos 2π = 1 sin 2π = 0 Now, if we take one complete revolution from the point P, we again come back to same point P. Thus, we also observe that if x increases (or decreases) by any integral multiple of 2π, the values of sine and cosine functions do not change. Thus, sin (2nπ + x) = sin x, n ∈ Z , cos (2nπ + x) = cos x , n ∈ Z Further, sin x = 0, if x = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π and cos x = 0, if x = ± π 2 , ± 3π 2 , ± 5π 2 , ... i.e., cos x vanishes when x is an odd multiple of π 2 . Thus sin x = 0 implies x = nπ, where n is any integer cos x = 0 implies x = (2n + 1) π 2 , where n is any integer. cos 3π 2 = 0 sin 3π 2 = –1 cos π 2 = 0 sin π 2 = 1 cos 0° = 1 sin 0° = 0, cosπ = − 1 sinπ = 0 We now define other trigonometric functions in terms of sine and cosine functions: cosec x = 1 sin x , x ≠ nπ, where n is any integer. sec x = 1 cos x , x ≠ (2n + 1) π 2 , where n is any integer. tan x = sin cot x = cos sin x x , x ≠ n π, where n is any integer. cos x x , x ≠ (2n +1) π 2 , where n is any integer. Reprint 2025-26 We have shown that for all real x, sin2 x + cos2 x = 1 It follows that 1 + cot2 x = cosec2 x (why?) In earlier classes, we have discussed the values of trigonometric ratios for 0°, 30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table: The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectively. cos 1 3 2 tan 0 1 3 1 3 0 0 sin 0 1 2 0° π 6 1 + tan2 x = sec2 x (why?) π 4 1 2 3 2 1 0 – 1 0 1 2 π 3 1 2 0 – 1 0 1 not defined not defined π 2 π 3π 2 2π TRIGONOMETRIC FUNCTIONS 51 3.3.1 Sign of trigonometric functions Let P (a, b) be a point on the unit circle with centre at the origin such that ∠AOP = x. If ∠AOQ = – x, then the coordinates of the point Q will be (a, –b) (Fig 3.7). Therefore cos (– x) = cos x and sin (– x) = – sin x Since for every point P (a, b) on the unit circle, – 1 ≤ a ≤ 1 and Fig 3.7 Reprint 2025-26 52 MATHEMATICS – 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in previous classes that in the first quadrant (0 < x < π 2 ) a and b are both positive, in the second quadrant ( π 2 < x <π) a is negative and b is positive, in the third quadrant (π < x < 3π 2 ) a and b are both negative and in the fourth quadrant ( 3π 2 < x < 2π) a is positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for π < x < 2π. Similarly, cos x is positive for 0 < x < π 2 , negative for π 2 < x < 3π 2 and also positive for 3π 2 < x < 2π. Likewise, we can find the signs of other trigonometric functions in different quadrants. In fact, we have the following table. sin x + + – – cos x + – – + tan x + – + – I II III IV 3.3.2 Domain and range of trigonometric functions From the definition of sine and cosine functions, we observe that they are defined for all real numbers. Further, we observe that for each real number x, – 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1 is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1. Thus, domain of y = sin x and y = cos x is the set of all real numbers and range cosec x + + – – sec x + – – + cot x + – + – Reprint 2025-26 x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) π 2 , n ∈ Z} and range is the set {y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and x ≠ (2n + 1) π 2 , n ∈ Z} and range is the set of all real numbers. The domain of y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers. increases from 0 to 1, as x increases from π 2 to π, sin x decreases from 1 to 0. In the third quadrant, as x increases from π to 3π 2 , sin x decreases from 0 to –1and finally, in the fourth quadrant, sin x increases from –1 to 0 as x increases from 3π 2 to 2π. Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table: sin increases from 0 to 1 decreases from 1 to 0 decreases from 0 to –1 increases from –1 to 0 Since cosec x = 1 sin x , the domain of y = cosec x is the set { x : x ∈ R and We further observe that in the first quadrant, as x increases from 0 to π 2 , sin x I quadrant II quadrant III quadrant IV quadrant TRIGONOMETRIC FUNCTIONS 53 Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for 0 < x < π 2 simply means that tan x increases as x increases for 0 < x < π 2 and cos decreases from 1 to 0 decreases from 0 to – 1 increases from –1 to 0 increases from 0 to 1 tan increases from 0 to ∞ increases from –∞to 0 increases from 0 to ∞ increases from –∞to 0 cot decreases from ∞ to 0 decreases from 0 to–∞ decreases from ∞ to 0 decreases from 0to –∞ sec increases from 1 to ∞ increases from –∞to–1 decreases from –1to–∞ decreases from ∞ to 1 cosec decreases from ∞ to 1 increases from 1 to ∞ increases from –∞to–1 decreases from–1to–∞ Reprint 2025-26 54 MATHEMATICS assumes arbitraily large positive values as x approaches to π 2 . Similarly, to say that cosec x decreases from –1 to – ∞ (minus infinity) in the fourth quadrant means that cosec x decreases for x ∈ ( 3π 2 , 2π) and assumes arbitrarily large negative values as x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour of functions and variables. We have already seen that values of sin x and cos x repeats after an interval of 2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We Fig 3.8 Fig 3.10 Fig 3.11 Reprint 2025-26 Fig 3.9 shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after an interval of π. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above: Example 6 If cos x = – 3 5 , x lies in the third quadrant, find the values of other five trigonometric functions. Solution Since cos x = 3 5 − , we have sec x = 5 3 − Fig 3.12 Fig 3.13 TRIGONOMETRIC FUNCTIONS 55 Now sin2 x + cos2 x = 1, i.e., sin2 x = 1 – cos2 x or sin2 x = 1 – 9 25 = 16 25 Hence sin x = ± 4 5 Since x lies in third quadrant, sin x is negative. Therefore sin x = – 4 5 which also gives cosec x = – 5 4 Reprint 2025-26 56 MATHEMATICS Further, we have Example 7 If cot x = – 5 12 , x lies in second quadrant, find the values of other five trigonometric functions. Solution Since cot x = – 5 12 , we have tan x = – 12 5 Now sec2 x = 1 + tan2 x = 1 + 144 25 = 169 25 Hence sec x = ± 13 5 Since x lies in second quadrant, sec x will be negative. Therefore which also gives 5 cos 13 x = − Further, we have tan x = sin sec x = – 13 5 , sin x = tan x cos x = (– 12 5 ) ×(– 5 13 ) = 12 13 cos x x = 4 3 and cot x = cos sin x x = 3 4 . and cosec x = 1 sin x = 13 12 . Example 8 Find the value of sin 31π 3 . Solution We know that values of sin x repeats after an interval of 2π. Therefore sin 31π 3 = sin (10π + π 3 ) = sin π 3 = 3 2 . Reprint 2025-26 Example 9 Find the value of cos (–1710°). Solution We know that values of cos x repeats after an interval of 2π or 360°. Therefore, cos (–1710°) = cos (–1710° + 5 × 360°) = cos (–1710° + 1800°) = cos 90° = 0. Find the values of other five trigonometric functions in Exercises 1 to 5. Find the values of the trigonometric functions in Exercises 6 to 10. 1. cos x = – 1 2 , x lies in third quadrant. 2. sin x = 3 5 , x lies in second quadrant. 3. cot x = 4 3 , x lies in third quadrant. 4. sec x = 13 5 , x lies in fourth quadrant. 5. tan x = – 5 6. sin 765° 7. cosec (– 1410°) 12 , x lies in second quadrant. EXERCISE 3.2 TRIGONOMETRIC FUNCTIONS 57 10. cot (– 15π 4 ) 3.4 Trigonometric Functions of Sum and Difference of Two Angles In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions. The basic results in this connection are called trigonometric identities. We have seen that We shall now prove some more results: 8. tan 19π 3 9. sin (– 11π 3 ) 1. sin (– x) = – sin x 2. cos (– x) = cos x Reprint 2025-26 58 MATHEMATICS 3. cos (x + y) = cos x cos y – sin x sin y Consider the unit circle with centre at the origin. Let x be the angle P4OP1 and y be the angle P1OP2 . Then (x + y) is the angle P4OP2 . Also let (– y) be the angle P4OP3 . Therefore, P1 , P2 , P3 and P4 will have the coordinates P1 (cos x, sin x), P2 [cos (x + y), sin (x + y)], P3 [cos (– y), sin (– y)] and P4 (1, 0) (Fig 3.14). P1 P3 and P2 P4 are equal. By using distance formula, we get Also, P2 P4 2 = [1 – cos (x + y)] 2 + [0 – sin (x + y)]2 Consider the triangles P1OP3 and P2OP4 . They are congruent (Why?). Therefore, P1 P3 2 = [cos x – cos (– y)]2 + [sin x – sin(–y] 2 = (cos x – cos y) 2 + (sin x + sin y) 2 = cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y + 2sin x sin y = 2 – 2 (cos x cos y – sin x sin y) (Why?) = 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y) = 2 – 2 cos (x + y) Reprint 2025-26 Fig 3.14 Since P1 P3 = P2 P4 , we have P1 P3 2 = P2 P4 2 . Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y). Hence cos (x + y) = cos x cos y – sin x sin y 4 . cos (x – y) = cos x cos y + sin x sin y Replacing y by – y in identity 3, we get cos (x + (– y)) = cos x cos (– y) – sin x sin (– y) or cos (x – y) = cos x cos y + sin x sin y 5. cos ( x π – 2 ) = sin x If we replace x by π 2 and y by x in Identity (4), we get 6. sin ( x π – 2 ) = cos x Using the Identity 5, we have sin ( π 2 − x ) = cos π π 2 2 x − − = cos x. 7. sin (x + y) = sin x cos y + cos x sin y We know that cos ( π 2 − x ) = cos π 2 cos x + sin π 2 sin x = sin x. TRIGONOMETRIC FUNCTIONS 59 = sin x cos y + cos x sin y 8. sin (x – y) = sin x cos y – cos x sin y If we replace y by –y, in the Identity 7, we get the result. 9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the following results: sin (x + y) = cos π ( ) 2 x y − + = cos π ( ) 2 x y − − cos x π ( + ) 2 = – sin x sin x π ( + ) 2 = cos x cos (π – x) = – cos x sin (π – x) = sin x = cos ( π 2 − x ) cos y + sin π ( ) 2 − x sin y Reprint 2025-26 60 MATHEMATICS Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin x and cos x. 10. If none of the angles x, y and (x + y) is an odd multiple of π 2 , then Since none of the x, y and (x + y) is an odd multiple of π 2 , it follows that cos x, cos y and cos (x + y) are non-zero. Now Dividing numerator and denominator by cos x cos y, we have cos (π + x) = – cos x sin (π + x) = – sin x cos (2π – x) = cos x sin (2π – x) = – sin x tan (x + y) = x y x y tan + tan 1 – tan tan tan (x + y) = sin( ) cos( ) x y x y + + = sin cos cos sin cos cos sin sin x y x y x y x y + − . tan (x + y) = = tan tan 1 – tan tan x y x y + coscos sinsin coscos coscos coscos sincos coscos cossin yx yx yx yx yx yx yx yx − + 11. tan ( x – y) = x y x y tan – tan 1 + tan tan If we replace y by – y in Identity 10, we get tan (x – y) = tan [x + (– y)] 12. If none of the angles x, y and (x + y) is a multiple of π, then cot ( x + y) = x y y x cot cot – 1 cot +cot = tan tan ( ) 1 tan tan ( ) x y x y + − − − = tan tan 1 tan tan x y x y − + Reprint 2025-26 Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and sin (x + y) are non-zero. Now, 13. cot (x – y)= x y y x cot cot +1 cot – cot if none of angles x, y and x–y is a multiple of π If we replace y by –y in identity 12, we get the result 14. cos 2x = cos2x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin 2 x = x x 2 We know that cos (x + y) = cos x cos y – sin x sin y Replacing y by x, we get cos 2x = cos2x – sin2 x = cos2 x – (1 – cos2 x) = 2 cos2x – 1 Again, cos 2x = cos2 x – sin2 x = 1 – sin2 x – sin2 x = 1 – 2 sin2 x. We have cos 2x = cos2 x – sin 2 x = 2 2 cot ( x + y)= cos ( ) cos cos – sin sin sin ( ) sin cos cos sin x y x y x y x y x y x y + = + + Dividing numerator and denominator by sin x sin y, we have cot (x + y) = cot cot –1 cot cot x y y x + 2 2 cos sin cos sin x x x x − + TRIGONOMETRIC FUNCTIONS 61 2 1 – tan 1 + tan 15. sin 2x = 2 sinx cos x = x x 2 2tan 1 + tan π π 2 x n ≠ + , where n is an integer We have sin (x + y) = sin x cos y + cos x sin y Replacing y by x, we get sin 2x = 2 sin x cos x. Again sin 2x = 2 2 2sin cos cos sin x x x x + Dividing numerator and denominator by cos2 x, we get cos 2x = 2 2 1 – tan 1+ tan x x , π π 2 x n ≠ + , where n is an integer Reprint 2025-26 62 MATHEMATICS 16. tan 2x = x x 2 2tan 1 – tan if π 2 π 2 x n ≠ + , where n is an integer We know that 17. sin 3x = 3 sin x – 4 sin3 x We have, sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x = 2 sin x cos x cos x + (1 – 2sin2 x) sin x = 2 sin x (1 – sin2 x) + sin x – 2 sin3 x = 2 sin x – 2 sin3 x + sin x – 2 sin3 x = 3 sin x – 4 sin3 x 18. cos 3x= 4 cos3 x – 3 cos x We have, cos 3x = cos (2x +x) = cos 2x cos x – sin 2x sin x = (2cos2 x – 1) cos x – 2sin x cos x sin x = (2cos2 x – 1) cos x – 2cos x (1 – cos2 x) = 2cos3 x – cos x – 2cos x + 2 cos3 x = 4cos3 x – 3cos x. Dividing each term by cos2 x, we get Replacing y by x , we get 2 2 tan tan 2 1 tan x x x = − sin 2x = 2 2tan 1 tan x + x tan (x + y) = tan tan 1 tan tan x y – x y + 19. = x x x x 3 We have tan 3x =tan (2x + x) 2 3 tan – tan tan3 1– 3tan if π 3 π 2 x n ≠ + , where n is an integer = tan 2 tan 1 tan 2 tan x x – x x + 2 Reprint 2025-26 + = 2tan tan 1 tan 2tan tan 1 1 tan x x – x x . x – – x 2 20. (i) cos x + cos y = x y x y + – 2cos cos 2 2 (iv) sin x – sin y = x y x y + – 2cos sin 2 2 We know that cos (x + y) = cos x cos y – sin x sin y ... (1) and cos (x – y) = cos x cos y + sin x sin y ... (2) Adding and subtracting (1) and (2), we get cos (x + y) + cos(x – y) = 2 cos x cos y ... (3) and cos (x + y) – cos (x – y) = – 2 sin x sin y ... (4) Further sin (x + y) = sin x cos y + cos x sin y ... (5) and sin (x – y) = sin x cos y – cos x sin y ... (6) Adding and subtracting (5) and (6), we get sin (x + y) + sin (x – y) = 2 sin x cos y ... (7) sin (x + y) – sin (x – y) = 2cos x sin y ... (8) Let x + y = θ and x – y = φ. Therefore (iii) sin x + sin y = x y x y + – 2sin cos 2 2 (ii) cos x – cos y = – x y x y + – 2sin sin 2 2 2 2 2 2tan tan tan 3 tan tan 1 tan 2tan 1 3tan x x – x x – x – x – x – x + = = 3 3 TRIGONOMETRIC FUNCTIONS 63 θ θ and 2 2 x y +φ −φ = = Substituting the values of x and y in (3), (4), (7) and (8), we get cos θ + cos φ = 2 cos θ θ cos 2 2 +φ −φ cos θ – cos φ = – 2 sin θ θ sin 2 + φ φ – 2 sin θ + sin φ = 2 sin θ θ cos 2 2 +φ −φ Reprint 2025-26 64 MATHEMATICS sin θ – sin φ = 2 cos θ θ sin 2 2 +φ −φ Since θ and φ can take any real values, we can replace θ by x and φ by y. Thus, we get Remark As a part of identities given in 20, we can prove the following results: 21. (i) 2 cos x cos y = cos (x + y) + cos (x – y) (ii) –2 sin x sin y = cos (x + y) – cos (x – y) (iii) 2 sin x cos y = sin (x + y) + sin (x – y) (iv) 2 cos x sin y = sin (x + y) – sin (x – y). Example 10 Prove that Solution We have cos x + cos y = 2 cos cos 2 2 x y x y + − ; cos x – cos y = – 2 sin sin 2 2 x y x y + − , sin x + sin y = 2 sin cos 2 2 x y x y + − ; sin x – sin y = 2 cos sin 2 2 x y x y + − . L.H.S. = 5 3sin sec 4sin cot 6 3 6 4 π π π π − 5 3sin sec 4sin cot 1 6 3 6 4 π π π π − = = 3 × 1 2 × 2 – 4 sin 6 π π− × 1 = 3 – 4 sin 6 π Example 11 Find the value of sin 15°. Solution We have sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° Example 12 Find the value of tan 13 12 π . = 3 – 4 × 1 2 = 1 = R.H.S. = 1 3 1 1 3 1 2 2 2 2 2 2 – × − × = . Reprint 2025-26 Solution We have Example 13 Prove that Solution We have Example 14 Show that tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x Solution We know that 3x = 2x + x tan 13 12 π = tan 12 π π + = tan tan 12 4 6 π π π = − L.H.S. sin ( ) sin cos cos sin sin( ) sin cos cos sin x y x y x y x y x y x y + + = = − − Dividing the numerator and denominator by cos x cos y, we get = tan tan 4 6 1 tan tan 4 6 π π + = sin ( ) tan tan sin ( ) tan tan x y x y x y x y + + = − − . π π − sin ( ) tan tan sin ( ) tan tan x y x y x y x y + + = − − . 1 1 3 3 1 2 3 1 3 1 1 3 − − = = − + + TRIGONOMETRIC FUNCTIONS 65 Therefore, tan 3x = tan (2x + x) or tan 2 tan tan3 1– tan 2 tan x x x x x + = or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x. Example 15 Prove that Solution Using the Identity 20(i), we have cos cos 2 cos 4 4 x x x π π + + − = Reprint 2025-26 66 MATHEMATICS Example 16 Prove that cos 7 cos 5 cot sin 7 – sin 5 x x x x x + = Solution Using the Identities 20 (i) and 20 (iv), we get Example 17 Prove that sin 5 2sin 3 sin tan cos5 cos x x x x x x − + = = − L.H.S. = L.H.S. cos cos 4 4 x x π π = + + − = 2 cos 4 π cos x = 2 × 1 2 cos x = 2 cos x = R.H.S. ( ) 4 4 4 4 2cos cos 2 2 x x x – x π π π π + + − + − = 7 5 7 5 2cos cos 2 2 7 5 7 5 2cos sin 2 2 x x x x x x x x + − + − = cos sin cot x x = x = R.H.S. Solution We have L.H.S. sin 5 2sin3 sin cos5 cos x x x x x − + = − sin 5 sin 2sin 3 cos5 cos x x x x x + − = − 2 1 cos 2 2sin sin 2 2sin cos x x x x x − = = = tan x = R.H.S. 2sin3 cos 2 2sin3 – 2sin3 sin 2 x x x x x − = sin 3 (cos2 1) sin 3 sin 2 x x – x x − = Reprint 2025-26 EXERCISE 3.3 Prove that: Prove the following: 9. 3π 3π cos cos (2π ) cot cot (2π ) 1 2 2 x x x x + + − + + = 10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x 11. 3 3 cos cos 2 sin 4 4 x x x π π + − − = − 1. sin2 π 6 + cos2 3 π – tan2 1 – 4 2 π = 2. 2sin2 6 π + cosec2 7 3 2 cos 6 3 2 π π = 3. 2 5 2 cot cosec 3tan 6 6 6 6 π π π + + = 4. 2 3 2 2 2sin 2cos 2sec 10 4 4 3 π π π + + = 5. Find the value of: (i) sin 75° (ii) tan 15° 6. cos cos sin sin sin( ) 4 4 4 4 x y x y x y π π π π − − − − − = + 7. 2 π tan 4 1 tan π 1 tan tan 4 x x x x + + = − − 8. cos ( ) cos ( ) 2 cot sin ( ) cos 2 TRIGONOMETRIC FUNCTIONS 67 x x x x x π + − = π π − + 12. sin2 6x – sin2 4x = sin 2x sin 10x 13. cos2 2x – cos2 6x = sin 4x sin 8x 14. sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x 15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x) 16. cos cos 18. sin sin 20. sin sin sin sin cos cos tan x y sin cos sin x x 17 3 − = 3 2 2 2 21. cos cos cos cos 9 5 x y − x y x x x − 10 x x + = − x x x − − = − 17. sin sin sin 2 19. sin sin 2 x Reprint 2025-26 cos cos tan 5 3 cos cos tan x x sin sin sin cot 4 3 2 5 3 4 x x x x x + 4 3 2 3 x x x x x x + x x x x + + + = 3 + = + + = 3 2 68 MATHEMATICS 22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1 23. 2 25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1 Example 18 If sin x = 3 find the value of sin (x + y). Solution We know that sin (x + y) = sin x cos y + cos x sin y ... (1) Now cos2 x = 1 – sin2 x = 1 – 9 Therefore cos x = ± 4 Since x lies in second quadrant, cos x is negative. Hence cos x = − 4 Now sin2 y = 1 – cos2 y = 1 – 144 i.e. sin y = ± 5 2 4 4tan (1 tan ) tan 4 1 6 tan tan x x x x x − = − + 24. cos 4x = 1 – 8sin2 x cos2 x 5 . 5 5 , cos y = − 12 Miscellaneous Examples 169 13 , where x and y both lie in second quadrant, 25 = 16 169 = 25 25 Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = 5 13 . Substituting the values of sin x, sin y, cos x and cos y in (1), we get Example 19 Prove that Solution We have 9 5 cos 2 cos cos 3 cos sin 5 sin 2 2 2 x x x x − x = x . sin( )x y + = × − + − × 3 13 . 5 Reprint 2025-26 12 13 4 5 13 = 36 20 56 65 65 65 − − = − . 5 Example 20 Find the value of tan π 8 . Solution Let π 8 x = . Then π 2 4 x = . Now tan tan L.H.S. = 1 9 2cos 2 cos 2cos cos 3 2 2 2 x x x x − = 1 9 9 cos 2 cos 2 cos 3 cos 3 2 2 2 2 2 x x x x x x x x + + − − + − − = 1 = = − − sin sin sin sin 5 = 5 2 5 15 5 15 1 2 2 2 2 2sin sin 2 2 2 tan 2 2 2 cos cos cos cos x x x x + − − = 1 x x x x + − − 1 2 x x x = − 2 x x x x = R.H.S. 5 2 2 5 5 3 2 15 2 3 TRIGONOMETRIC FUNCTIONS 69 2 2 cos cos x x − 5 2 15 or 2 π 2tan π 8 tan 4 π 1 tan 8 Let y = tan π 8 . Then 1 = 2 1 2 y y − or y2 + 2y – 1 = 0 Therefore y = − ± = − ± 2 2 2 = − 2 1 2 Reprint 2025-26 70 MATHEMATICS Since π 8 lies in the first quadrant, y = tan π 8 is positve. Hence Example 21 If 3 3π tan = , π < < 4 2 x x , find the value of sin x Solution Since 3π π 2 < 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0? Let us examine. Note that Again, the symbol −3 is meant to represent 3 i only, i.e., −3 = 3 i . Generally, if a is a positive real number, −a = a −1 = a i , We already know that a b × = ab for all positive real number a and b. This ( ) 2 − 3i = ( ) 2 − 3 i 2 = – 3 Reprint 2025-26 80 MATHEMATICS Further, if any of a and b is zero, then, clearly, a b ab × = = 0. 4.3.7 Identities We prove the following identity Proof We have, (z 1 + z 2 ) 2 = (z 1 + z 2 ) (z 1 + z 2 ), (iv) ( ) ( ) 2 2 1 2 1 2 1 2 z – z z z z – z = + In fact, many other identities which are true for all real numbers, can be proved to be true for all complex numbers. ( ) ( ) 2 i = − − = − − 1 1 1 1 (by assuming a b × = ab for all real numbers) = 1 = 1, which is a contradiction to the fact that = − 2 i 1. Therefore, a b ab × ≠ if both a and b are negative real numbers. = 2 2 1 1 2 2 z z z z + + 2 Similarly, we can prove the following identities: ( )2 2 2 z z z z z z 1 2 1 2 1 2 + = + + 2 , for all complex numbers z 1 and z 2 . (iii) ( )3 3 2 2 3 z z z z z z z z 1 2 1 1 2 1 2 2 − = − + − 3 3 (ii) ( )3 3 2 2 3 z z z z z z z z 1 2 1 1 2 1 2 2 + = + + + 3 3 (i) ( )2 2 2 z z z z z z 1 2 1 1 2 2 − = − + 2 = 2 2 1 1 2 1 2 2 z z z z z z + + + (Commutative law of multiplication) = (z 1 + z 2 ) z 1 + (z 1 + z 2 ) z 2 (Distributive law) = 2 2 1 2 1 1 2 2 z z z z z z + + + (Distributive law) Example 2 Express the following in the form of a + bi: Solution (i) ( ) 1 5 8 i i − = 5 2 8 i − = ( ) 5 1 8 − − = 5 8 = 5 0 8 + i (ii) ( ) ( ) 3 1 2 8 i i i − − = 1 5 2 8 8 8 × ×i × × = ( )2 1 2 256 i 1 256 i i = . (i) ( ) 1 5 8 i i − (ii) (−i i ) (2 ) Reprint 2025-26 3 1 8 i − Example 3 Express (5 – 3i) 3 in the form a + ib. Solution We have, (5 – 3i) 3 = 5 3 – 3 × 52 × (3i) + 3 × 5 (3i) 2 – (3i) 3 Example 4 Express (− + − − 3 2 2 3 )( i)in the form of a + ib Solution We have, (− + − − 3 2 2 3 ) ( i) = (− + − 3 2 2 3 i i ) ( ) 4.4 The Modulus and the Conjugate of a Complex Number Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined to be the non-negative real number 2 2 a b + , i.e., | z | = 2 2 a b + and the conjugate of z, denoted as z , is the complex number a – ib, i.e., z = a – ib. For example, 2 2 3 3 1 10 + = + = i , 2 2 2 5 2 ( 5) 29 − = + − = i , and 3 3 + = − i i , 2 5 2 5 − = + i i , − − 3 5 i = 3i – 5 Observe that the multiplicative inverse of the non-zero complex number z is given by z –1 = 1 a ib + = 2 2 2 2 a b i a b a b − + + + = 2 2 a ib a b − + = 2 z = 2 − + + − 6 3 2 6 2 i i i = (− + + + 6 2 3 1 2 2 ) ( )i = 125 – 225i – 135 + 27i = – 10 – 198i. COMPLEX NUMBERS AND QUADRATIC EQUATIONS 81 or z 2 z z = Furthermore, the following results can easily be derived. For any two compex numbers z 1 and z 2 , we have (iii) 1 2 1 2 z z z z = (iv) 1 2 1 2 z z z z ± = ± (v) 1 1 2 2 z z z z = provided z 2 ≠ 0. (i) 1 2 1 2 z z z z = (ii) 1 1 Reprint 2025-26 2 2 z z z z = provided 2 z ≠ 0 z 82 MATHEMATICS Example 5 Find the multiplicative inverse of 2 – 3i. Solution Let z = 2 – 3i Then z = 2 + 3i and 2 2 2 z = + − = 2 ( 3) 13 Therefore, the multiplicative inverse of 2 3 − i is given by The above working can be reproduced in the following manner also, Example 6 Express the following in the form a + ib Solution (i) We have, 5 2 5 2 1 2 (i) 5 2 z –1 = 1 2 3 2 3 (2 3 )(2 3 ) i i i i + = − − + z –1 2 2 3 2 3 13 13 13 z i i z 1 2 i = 2 2 2 3 2 3 2 3 2 (3 ) 13 13 13 i i i i + + = = + − i + − (ii) i–35 + = = = + 1 2 1 2 1 2 i i i = 3 6 2 3(1 2 2 ) 1 2 3 + + i i = + = 1 2 2 + i . i i i + + + = × − − + ( )2 5 5 2 2 2 i + + − = − 1 2 i i Express each of the complex number given in the Exercises 1 to 10 in the form a + ib. 1. ( ) 3 5 5 i i − 2. i i 9 19 + 3. i −39 (ii) ( ) 35 35 17 2 1 1 1 i i i i i i i − = = = × − = 2 i i i = − EXERCISE 4.1 Reprint 2025-26 Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13. 11. 4 – 3i 12. 5 3 + i 13. – i 14. Express the following expression in the form of a + ib : 4.5 Argand Plane and Polar Representation We already know that corresponding to each ordered pair of real numbers (x, y), we get a unique point in the XYplane and vice-versa with reference to a set of mutually perpendicular lines known as the x-axis and the y-axis. The complex number x + iy which corresponds to the ordered pair (x, y) can be represented geometrically as the unique point P(x, y) in the XY-plane and vice-versa. Some complex numbers such as 2 + 4i, – 2 + 3i, 0 + 1i, 2 + 0i, – 5 –2i and 1 – 2i which correspond to the ordered pairs (2, 4), ( – 2, 3), (0, 1), (2, 0), ( –5, –2), and (1, – 2), respectively, have been represented geometrically by the points A, B, C, D, E, and F, respectively in the Fig 4.1. The plane having a complex number assigned to each of its point is called the complex plane or the Argand plane. 4. 3(7 + i7) + i (7 + i7) 5. (1 – i) – ( –1 + i6) 6. 1 2 5 4 5 5 2 i i + − + 7. 1 7 1 4 4 3 3 3 3 i i i + + + − − + 8. (1 – i) 4 9. 3 1 3 3 i + 10. 3 1 2 3 i − − COMPLEX NUMBERS AND QUADRATIC EQUATIONS 83 ( ) ( ) 3 2 3 2 i i ( ) ( ) 3 5 3 5 + − − i i + − Reprint 2025-26 Fig 4.1 84 MATHEMATICS 0 + i b. The x-axis and y-axis in the Argand plane are called, respectively, the real axis and the imaginary axis. The representation of a complex number z = x + iy and its conjugate z = x – iy in the Argand plane are, respectively, the points P (x, y) and Q (x, – y). Geometrically, the point (x, – y) is the mirror image of the point (x, y) on the real axis (Fig 4.3). x + iy = 2 2 x y + is the distance between the point P(x, y) and the origin O (0, 0) (Fig 4.2). The points on the x-axis corresponds to the complex numbers of the form a + i 0 and the points on the y-axis corresponds to the complex numbers of the form Obviously, in the Argand plane, the modulus of the complex number Fig 4.2 Reprint 2025-26 Fig 4.3 Example 7 Find the conjugate of (3 2 )(2 3 ) (1 2 )(2 ) i i i i − + + − . Solution We have , (3 2 )(2 3 ) (1 2 )(2 ) i i i i − + + − Therefore, conjugate of (3 2 )(2 3 ) 63 16 is (1 2 )(2 ) 25 25 i i i i i − + + + − . Example 8 If x + iy = a ib a ib + − , prove that x 2 + y 2 = 1. Solution We have, = 6 9 4 6 2 4 2 i i i i + − + − + + = 12 5 4 3 4 3 4 3 i i i i + − × + − = 48 36 20 15 63 16 16 9 25 − + + − i i i = + = 63 16 25 25 − i x + iy = ( )( ) ( )( ) a ib a ib a ib a ib + + − + = Miscellaneous Examples COMPLEX NUMBERS AND QUADRATIC EQUATIONS 85 2 2 a b abi 2 a b − + + = 2 2 2 2 2 2 a b ab 2 i a b a b − + + + 2 2 So that, x – iy = 2 2 Therefore, x 2 + y 2 = (x + iy) (x – iy) = 2 2 2 2 2 1. Evaluate: 2. For any two complex numbers z 1 and z 2 , prove that Re (z 1 z 2 ) = Re z 1 Re z 2 – Imz 1 Imz 2 . 3 25 18 1 i i + . 2 2 2 2 a b ab 2 i a b a b − − + + Miscellaneous Exercise on Chapter 4 2 2 2 2 2 2 ( ) 4 ( ) ( ) a b a b a b a b − + + + = 2 2 2 Reprint 2025-26 2 2 2 ( ) ( ) a b a b + + = 1 86 MATHEMATICS 6. If a + ib = 2 7. Let z 1 = 2 – i, z 2 = –2 + i. Find 8. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i. 9. Find the modulus of 1 1 1 1 i i i i + − − − + . 10. If (x + iy) 3 = u + iv, then show that 2 2 4( – ) u v x y x y + = . 3. Reduce 1 2 3 4 1 4 1 5 4. If a ib x iy c id − − = − prove that ( ) 2 2 2 2 2 2 2 a b x y c d + + = + . 5. If z 1 = 2 – i, z 2 = 1 + i, find 1 2 1 2 (i) 1 2 1 Re z z z , (ii) 1 1 1 Im z z . i i i i − − − + + to the standard form . 2 ( ) 2 1 x i x + + , prove that a2 + b 2 = ( ) 1 – 1 z z z z + + + . 2 2 ( 1) 2 1 x x + 2 2 + . 11. If α and β are different complex numbers with β 1 = , then find β α 1 αβ – – . 12. Find the number of non-zero integral solutions of the equation 1 2 x x – i = . 13. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that 14. If 1 1 1 (a 2 + b 2 ) (c 2 + d 2 ) (e 2 + f 2 ) (g 2 + h 2 ) = A2 + B2 m i – i + = , then find the least positive integral value of m. Reprint 2025-26 ÆA number of the form a + ib, where a and b are real numbers, is called a complex number, a is called the real part and b is called the imaginary part of the complex number. ÆLet z 1 = a + ib and z 2 = c + id. Then (i) z 1 + z 2 = (a + c) + i (b + d) (ii) z 1 z 2 = (ac – bd) + i (ad + bc) ÆFor any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists the =1 ÆFor any integer k, i 4k = 1, i 4k + 1 = i, i 4k + 2 = – 1, i 4k + 3 = – i ÆThe conjugate of the complex number z = a + ib, denoted by z , is given by z = a – ib. complex number 2 2 2 2 a b i a b a b − + + + , denoted by 1 z or z –1, called the multiplicative inverse of z such that (a + ib) 2 2 2 2 − + + + a b i a b a b = 1 + i0 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 87 Historical Note Summary The fact that square root of a negative number does not exist in the real number system was recognised by the Greeks. But the credit goes to the Indian mathematician Mahavira (850) who first stated this difficulty clearly. “He mentions in his work ‘Ganitasara Sangraha’ as in the nature of things a negative (quantity) is not a square (quantity)’, it has, therefore, no square root”. Bhaskara, another Indian mathematician, also writes in his work Bijaganita, written in 1150. “There is no square root of a negative quantity, for it is not a square.” Cardan (1545) considered the problem of solving x + y = 10, xy = 40. Reprint 2025-26 88 MATHEMATICS He obtained x = 5 + −15 and y = 5 – −15 as the solution of it, which was discarded by him by saying that these numbers are ‘useless’. Albert Girard (about 1625) accepted square root of negative numbers and said that this will enable us to get as many roots as the degree of the polynomial equation. Euler was the first to introduce the symbol i for −1 and W.R. Hamilton (about 1830) regarded the complex number a + ib as an ordered pair of real numbers (a, b) thus giving it a purely mathematical definition and avoiding use of the so called ‘imaginary numbers’. — v — Reprint 2025-26" class_11,5,Linear Inequalities,ncert_books/class_11/kemh1dd/kemh105.pdf,"5.1 Introduction In earlier classes, we have studied equations in one variable and two variables and also solved some statement problems by translating them in the form of equations. Now a natural question arises: ‘Is it always possible to translate a statement problem in the form of an equation? For example, the height of all the students in your class is less than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than or equal) and ≥ (greater than or equal) which are known as inequalities. In this Chapter, we will study linear inequalities in one and two variables. The study of inequalities is very useful in solving problems in the field of science, mathematics, statistics, economics, psychology, etc. 5.2 Inequalities Let us consider the following situations: LINEAR INEQUALITIES vMathematics is the art of saying many things in many different ways. – MAXWELLv Chapter 5 (i) Ravi goes to market with `200 to buy rice, which is available in packets of 1kg. The price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets only, he may not be able to spend the entire amount of ` 200. (Why?) Hence 30x < 200 ... (1) Clearly the statement (i) is not an equation as it does not involve the sign of equality. (ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of registers and y, the number of pens which Reshma buys, then the total amount spent by her is ` (40x + 20y) and we have 40x + 20y ≤ 120 ... (2) Reprint 2025-26 90 MATHEMATICS Since in this case the total amount spent may be upto ` 120. Note that the statement (2) consists of two statements Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation. Definition 1 Two real numbers or two algebraic expressions related by the symbol ‘<’, ‘>’, ‘≤’ or ‘≥’ form an inequality. Statements such as (1), (2) and (3) above are inequalities. 3 < 5; 7 > 5 are the examples of numerical inequalities while 3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities. Some more examples of inequalities are: x < 5; y > 2; x ≥ 3, y ≤ 4 are some examples of literal inequalities. and 40x + 20y = 120 ... (4) ax + b < 0 ... (5) ax + b > 0 ... (6) ax + b ≤ 0 ... (7) ax + b ≥ 0 ... (8) ax + by < c ... (9) ax + by > c ... (10) 40x + 20y < 120 ... (3) Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8), (11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear inequalities in one variable x when a ≠ 0, while inequalities from (9) to (12) are linear inequalities in two variables x and y when a ≠ 0, b ≠ 0. Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities in one variable x when a ≠ 0). In this Chapter, we shall confine ourselves to the study of linear inequalities in one and two variables only. ax + by ≤ c ... (11) ax + by ≥ c ... (12) ax2 + bx + c ≤ 0 ... (13) ax2 + bx + c > 0 ... (14) Reprint 2025-26 5.3 Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation Let us consider the inequality (1) of Section 6.2, viz, 30x < 200 Note that here x denotes the number of packets of rice. Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this inequality is 30x and right hand side (RHS) is 200. Therefore, we have For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true. For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true. For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true. inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6} is called its solution set. Thus, any solution of an inequality in one variable is a value of the variable which makes it a true statement. We have found the solutions of the above inequality by trial and error method which is not very efficient. Obviously, this method is time consuming and sometimes not feasible. We must have some better or systematic techniques for solving inequalities. Before that we should go through some more properties of numerical inequalities and follow them as rules while solving the inequalities. You will recall that while solving linear equations, we followed the following rules: For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true. For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true. For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true. For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true. For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false. In the above situation, we find that the values of x, which makes the above LINEAR INEQUALITIES 91 Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation. Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero number. In the case of solving inequalities, we again follow the same rules except with a difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, ≤’ becomes ‘≥’ and so on) whenever we multiply (or divide) both sides of an inequality by a negative number. It is evident from the facts that 3 > 2 while – 3 < – 2, – 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14. Reprint 2025-26 92 MATHEMATICS Thus, we state the following rules for solving an inequality: Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality. Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed. Example 1 Solve 30 x < 200 when (i) x is a natural number, (ii) x is an integer. Solution We are given 30 x < 200 or 30 200 30 30 x < (Rule 2), i.e., x < 20 / 3. (i) When x is a natural number, in this case the following values of x make the statement true. 1, 2, 3, 4, 5, 6. The solution set of the inequality is {1,2,3,4,5,6}. (ii) When x is an integer, the solutions of the given inequality are ..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6 The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6} Example 2 Solve 5x – 3 < 3x +1 when (i) x is an integer, (ii) x is a real number. Solution We have, 5x –3 < 3x + 1 Now, let us consider some examples. or 5x –3 + 3 < 3x +1 +3 (Rule 1) or 5x < 3x +4 or 5x – 3x < 3x + 4 – 3x (Rule 1) or 2x < 4 or x < 2 (Rule 2) (i) When x is an integer, the solutions of the given inequality are ..., – 4, – 3, – 2, – 1, 0, 1 (ii) When x is a real number, the solutions of the inequality are given by x < 2, i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x ∈ (– ∞, 2). We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers. Reprint 2025-26 Example 3 Solve 4x + 3 < 6x +7. Solution We have, 4x + 3 < 6x + 7 or 4x – 6x < 6x + 4 – 6x or – 2x < 4 or x > – 2 i.e., all the real numbers which are greater than –2, are the solutions of the given inequality. Hence, the solution set is (–2, ∞). Example 4 Solve 5 2 5 3 6 – x x ≤ – . Solution We have or 2 (5 – 2x) ≤ x – 30. or 10 – 4x ≤ x – 30 or – 5x ≤ – 40, i.e., x ≥ 8 Thus, all real numbers x which are greater than or equal to 8 are the solutions of the given inequality, i.e., x ∈ [8, ∞). Example 5 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line. Solution We have 7x + 3 < 5x + 9 or 2x < 6 or x < 3 The graphical representation of the solutions are given in Fig 5.1. 5 2 5 3 6 – x x ≤ – LINEAR INEQUALITIES 93 Example 6 Solve 3 4 1 1 2 4 x x − + ≥ − . Show the graph of the solutions on number line. Solution We have or 3 4 3 2 4 x x − − ≥ or 2 (3x – 4) ≥ (x – 3) 3 4 1 1 2 4 x x − + ≥ − Reprint 2025-26 Fig 5.1 94 MATHEMATICS The graphical representation of solutions is given in Fig 5.2. Example 7 The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks. Solution Let x be the marks obtained by student in the annual examination. Then or 110 + x ≥ 180 or x ≥ 70 Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks. Example 8 Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40. Solution Let x be the smaller of the two consecutive odd natural number, so that the other one is x +2. Then, we should have 62 48 60 3 + + x ≥ or 6x – 8 ≥ x – 3 or 5x ≥ 5 or x ≥ 1 Fig 5.2 x > 10 ... (1) and x + ( x + 2) < 40 ... (2) Solving (2), we get 2x + 2 < 40 i.e., x < 19 ... (3) From (1) and (3), we get 10 < x < 19 Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be (11, 13), (13, 15), (15, 17), (17, 19) Reprint 2025-26 Solve the inequalities in Exercises 5 to 16 for real x. 11. 3( 2) 5(2 ) 5 3 x − − x ≤ 12. 1 3 1 4 ( 6) 2 5 3 x x + ≥ − 13. 2 (2x + 3) – 10 < 6 (x – 2) 14. 37 – (3x + 5) > 9x – 8 (x – 3) 15. (5 2) (7 3) 4 3 5 x x x − − < − 16. (2 1) (3 2) (2 ) 3 4 5 x x x − − − ≥ − Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line 1. Solve 24x < 100, when (i) x is a natural number. (ii) x is an integer. 2. Solve – 12x > 30, when (i) x is a natural number. (ii) x is an integer. 3. Solve 5x – 3 < 7, when (i) x is an integer. (ii) x is a real number. 4. Solve 3x + 8 >2, when (i) x is an integer. (ii) x is a real number. 5. 4x + 3 < 5x + 7 6. 3x – 7 > 5x – 1 7. 3(x – 1) ≤ 2 (x – 3) 8. 3 (2 – x) ≥ 2 (1 – x) 9. 11 2 3 x x x + + < 10. 1 3 2 x x > + EXERCISE 5.1 LINEAR INEQUALITIES 95 17. 3x – 2 < 2x + 1 18. 5x – 3 > 3x – 5 19. 3 (1 – x) < 2 (x + 4) 20. (5 – 2) (7 –3) – 2 3 5 x x x ≥ 21. Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks. 22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course. 23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11. 24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23. Reprint 2025-26 96 MATHEMATICS 25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side. 26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5]. Example 9 Solve – 8 ≤ 5x – 3 < 7. Solution In this case, we have two inequalities, – 8 ≤ 5x – 3 and 5x – 3 < 7, which we will solve simultaneously. We have – 8 ≤ 5x –3 < 7 or –5 ≤ 5x < 10 or –1 ≤ x < 2 Example 10 Solve – 5 ≤ 5 3 2 – x ≤ 8. Solution We have – 5 ≤ 5 3 2 – x ≤ 8 or –10 ≤ 5 – 3x ≤ 16 or – 15 ≤ – 3x ≤ 11 or 5 ≥ x ≥ – 11 3 Miscellaneous Examples which can be written as –11 3 ≤ x ≤ 5 Example 11 Solve the system of inequalities: 3x – 7 < 5 + x ... (1) 11 – 5 x ≤ 1 ... (2) and represent the solutions on the number line. Solution From inequality (1), we have 3x – 7 < 5 + x or x < 6 ... (3) Also, from inequality (2), we have 11 – 5 x ≤ 1 or – 5 x ≤ – 10 i.e., x ≥ 2 ... (4) Reprint 2025-26 If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of x, which are common to both, are shown by bold line in Fig 5.3. Thus, solution of the system are real numbers x lying between 2 and 6 including 2, i.e., 2 ≤ x < 6 Example 12 In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by C = 5 9 (F – 32), where C and F represent temperature in degree Celsius and degree Fahrenheit, respectively. Solution It is given that 30 < C < 35. Putting C = 5 9 (F – 32), we get or 9 5 × (30) < (F – 32) < 9 5 × (35) 30 < 5 9 (F – 32) < 35, Fig 5.3 LINEAR INEQUALITIES 97 or 54 < (F – 32) < 63 or 86 < F < 95. Thus, the required range of temperature is between 86° F and 95° F. Example 13 A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%? Solution Let x litres of 30% acid solution is required to be added. Then Total mixture = (x + 600) litres Therefore 30% x + 12% of 600 > 15% of (x + 600) and 30% x + 12% of 600 < 18% of (x + 600) or 30 100 x + 12 100 (600) > 15 100 (x + 600) Reprint 2025-26 98 MATHEMATICS and 30 100 x + 12 100 (600) < 18 100 (x + 600) or 30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800 or 15x > 1800 and 12x < 3600 or x > 120 and x < 300, i.e. 120 < x < 300 Thus, the number of litres of the 30% solution of acid will have to be more than 120 litres but less than 300 litres. Solve the inequalities in Exercises 1 to 6. 1. 2 ≤ 3x – 4 ≤ 5 2. 6 ≤ – 3 (2x – 4) < 12 Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line. 7. 5x + 1 > – 24, 5x – 1 < 24 8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x 9. 3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x 10. 5 (2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47 . 3. 7 3 4 18 2 x – ≤ − ≤ 4. 3 2 15 0 5 ( x ) − − < ≤ 5. 3 12 4 2 5 x − < − ≤ − 6. 3 11 7 11 2 ( x ) + ≤ ≤ . Miscellaneous Exercise on Chapter 5 11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by 12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added? F = 9 5 C + 32 ? Reprint 2025-26 13. How many litres of water will have to be added to 1125 litres of the 45% solution 14. IQ of a person is given by the formula where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age. Summary ÆTwo real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality. ÆEqual numbers may be added to (or subtracted from ) both sides of an inequality. ÆBoth sides of an inequality can be multiplied (or divided ) by the same positive number. But when both sides are multiplied (or divided) by a negative number, then the inequality is reversed. ÆThe values of x, which make an inequality a true statement, are called solutions of the inequality. ÆTo represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a. ÆTo represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and dark the line to the left (or right) of the number x. of acid so that the resulting mixture will contain more than 25% but less than 30% acid content? IQ = MA CA × 100, LINEAR INEQUALITIES 99 Reprint 2025-26 — v —" class_11,6,Permutations and Combinations,ncert_books/class_11/kemh1dd/kemh106.pdf,"6.1 Introduction Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9. The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Some how, you have forgotten this specific sequence of digits. You remember only the first digit which is 7. In order to open the lock, how many sequences of 3-digits you may have to check with? To answer this question, you may, immediately, start listing all possible arrangements of 9 remaining digits taken 3 at a time. But, this method will be tedious, because the number of possible sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing 3-digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examine a principle which is most fundamental to the learning of these techniques. 100 MATHEMATICS PERMUTATIONS AND COMBINATIONS vEvery body of discovery is mathematical in form because there is no other guidance we can have – DARWINv Chapter 6 Jacob Bernoulli (1654-1705) 6.2 Fundamental Principle of Counting Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are 3 × 2 = 6 pairs of a pant and a shirt. Reprint 2025-26 these six possibilities can be illustrated in the Fig. 6.1. Let us consider another problem of the same type. Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles. In how many ways can she carry these items (choosing one each). A school bag can be chosen in 2 different ways. After a school bag is chosen, a tiffin box can be chosen in 3 different ways. Hence, there are 2 × 3 = 6 pairs of school bag and a tiffin box. For each of these pairs a water bottle can be chosen in 2 different ways. Hence, there are 6 × 2 = 12 different ways in which, Sabnam can carry these items to school. If we name the 2 school bags as B1 , B2 , the three tiffin boxes as T1 , T2 , T3 and the two water bottles as W1 , W2 , these possibilities can be illustrated in the Fig. 6.2. Let us name the three pants as P1 , P2 , P3 and the two shirts as S1 , S2 . Then, PERMUTATIONS AND COMBINATIONS 101 Fig 6.1 Reprint 2025-26 Fig 6.2 principle known as the fundamental principle of counting, or, simply, the multiplication principle, which states that “If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m×n.” The above principle can be generalised for any finite number of events. For example, for 3 events, the principle is as follows: ‘If an event can occur in m different ways, following which another event can occur in n different ways, following which a third event can occur in p different ways, then the total number of occurrence to ‘the events in the given order is m × n × p.” In the first problem, the required number of ways of wearing a pant and a shirt was the number of different ways of the occurence of the following events in succession: ways of the occurence of the following events in succession: Here, in both the cases, the events in each problem could occur in various possible orders. But, we have to choose any one of the possible orders and count the number of different ways of the occurence of the events in this chosen order. 102 MATHEMATICS In fact, the problems of the above types are solved by applying the following In the second problem, the required number of ways was the number of different (i) the event of choosing a school bag (ii) the event of choosing a tiffin box (iii) the event of choosing a water bottle. (i) the event of choosing a pant (ii) the event of choosing a shirt. Example 1 Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed. Solution There are as many words as there are ways of filling in 4 vacant places first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the required number of words is 24. by the 4 letters, keeping in mind that the repetition is not allowed. The Reprint 2025-26 Example 2 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other? Solution There will be as many signals as there are ways of filling in 2 vacant places be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12. Example 3 How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated? Solution There will be as many ways as there are ways of filling 2 vacant places place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10. Example 4 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available. ANote If the repetition of the letters was allowed, how many words can be formed? One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256. in succession by the five given digits. Here, in this case, we start filling in unit’s in succession by the 4 flags of different colours. The upper vacant place can PERMUTATIONS AND COMBINATIONS 103 Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers. There will be as many 2 flag signals as there are ways of filling in 2 vacant places ways is 5 × 4 = 20. Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places in succession by the 5 flags. in succession by the 5 flags available. By Multiplication rule, the number of Reprint 2025-26 The number of ways is 5 × 4 × 3 = 60. Continuing the same way, we find that The number of 4 flag signals = 5 × 4 × 3 × 2 = 120 and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120 Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320. 1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that 2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? 3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated? 4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once? 5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there? 6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? 6.3 Permutations 104 MATHEMATICS (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed? EXERCISE 6.1 In Example 1 of the previous Section, we are actually counting the different possible arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each arrangement is different from other. In other words, the order of writing the letters is important. Each arrangement is called a permutation of 4 different letters taken all at a time. Now, if we have to determine the number of 3-letter words, with or without meaning, which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using multiplication principle). If the repetition of the letters was allowed, the required number of words would be 6 × 6 × 6 = 216. Reprint 2025-26 Definition 1 A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. In the following sub-section, we shall obtain the formula needed to answer these questions immediately. 6.3.1 Permutations when all the objects are distinct Theorem 1 The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by nPr . Proof There will be as many permutations as there are ways of filling in r vacant places . . . by ← r vacant places → the n objects. The first place can be filled in n ways; following which, the second place can be filled in (n – 1) ways, following which the third place can be filled in (n – 2) ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or n ( n – 1) (n – 2) ... (n – r + 1) This expression for nPr is cumbersome and we need a notation which will help to reduce the size of this expression. The symbol n! (read as factorial n or n factorial ) comes to our rescue. In the following text we will learn what actually n! means. 6.3.2 Factorial notation The notation n! represents the product of first n natural numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n ! 1 = 1 ! 1 × 2 = 2 ! 1× 2 × 3 = 3 ! 1 × 2 × 3 × 4 = 4 ! and so on. PERMUTATIONS AND COMBINATIONS 105 We define 0 ! = 1 We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 ! = 5 × 4 × 3 × 2 × 1! Clearly, for a natural number n n ! = n (n – 1) ! = n (n – 1) (n – 2) ! [provided (n ≥ 2)] = n (n – 1) (n – 2) (n – 3) ! [provided (n ≥ 3)] and so on. Reprint 2025-26 Example 5 Evaluate (i) 5 ! (ii) 7 ! (iii) 7 ! – 5! Solution (i) 5 ! = 1 × 2 × 3 × 4 × 5 = 120 (ii) 7 ! = 1 × 2 × 3 × 4 × 5 × 6 ×7 = 5040 and (iii) 7 ! – 5! = 5040 – 120 = 4920. Example 6 Compute (i) 7! 5! (ii) ( ) 12! 10! (2!) Solution (i) We have 7! 5! = 7 6 5! 5! × × = 7 × 6 = 42 and (ii) ( ) ( ) 12! 10! 2! = ( ) Example 7 Evaluate ( ) ! ! ! n r n r − , when n = 5, r = 2. Solution We have to evaluate ( ) 5! 2! 5 2 ! − (since n = 5, r = 2) We have ( ) 5! 2 ! 5 2 ! − = 5! 5 4 10 2! 3! 2 × = = × . Example 8 If 1 1 8! 9! 10! x + = , find x. 106 MATHEMATICS ( ) ( ) 12 11 10! 10! 2 × × × = 6 × 11 = 66. Solution We have 1 1 8! 9 8! 10 9 8! x + = × × × Therefore 1 1 9 10 9 x + = × or 10 9 10 9 x = × So x = 100. 1. Evaluate (i) 8 ! (ii) 4 ! – 3 ! EXERCISE 6.2 Reprint 2025-26 2. Is 3 ! + 4 ! = 7 ! ? 3. Compute 8! 6! 2! × 4. If 1 1 6! 7! 8! x + = , find x 5. Evaluate ( ) ! ! n n r − , when 6.3.3 Derivation of the formula for nPr Let us now go back to the stage where we had determined the following formula: nPr = n (n – 1) (n – 2) . . . (n – r + 1) Multiplying numerator and denomirator by (n – r) (n – r – 1) . . . 3 × 2 × 1, we get Thus ( ) ! P ! n r n n r = − , where 0 < r ≤n This is a much more convenient expression for nPr than the previous one. In particular, when r = n, ! P ! 0! n n n = = n (i) n = 6, r = 2 (ii) n = 9, r = 5. ( )( ) 1 2 1 1 3 2 1 P 1 3 2 1 n r n n n ... n r n r n r ... n r n r ... − − − + − − − × × = − − − × × = ( ) ! ! n n r − , ( ) ( ) ( )( )( ) ( ) ! P ! n r n n r − = , 0 ≤ r ≤ n PERMUTATIONS AND COMBINATIONS 107 all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have Therefore, the formula (1) is applicable for r = 0 also. Thus ( ) ! P 0 ! n r n , r n n r = ≤ ≤ − . Counting permutations is merely counting the number of ways in which some or n P0 = 1 = ! ! ! ( 0)! = − n n n n ... (1) Reprint 2025-26 Theorem 2 The number of permutations of n different objects taken r at a time, where repetition is allowed, is n r . Proof is very similar to that of Theorem 1 and is left for the reader to arrive at. Here, we are solving some of the problems of the pervious Section using the formula for nPr to illustrate its usefulness. In Example 1, the required number of words = 4P4 = 4! = 24. Here repetition is not allowed. If repetition is allowed, the required number of words would be 44 = 256. The number of 3-letter words which can be formed by the letters of the word NUMBER = 6 3 6! P 3! = = 4 × 5 × 6 = 120. Here, in this case also, the repetition is not allowed. If the repetition is allowed,the required number of words would be 63 = 216. The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly 12 2 12! P 11 12 10! = = × = 132. 6.3.4 Permutations when all the objects are not distinct objects Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O1 and O2 . The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO1O2 T. Corresponding to this permutation,we have 2 ! permutations RO1O2 T and RO2O1 T which will be exactly the same permutation if O1 and O2 are not treated as different, i.e., if O1 and O2 are the same O at both places. 108 MATHEMATICS Therefore, the required number of permutations = 4! 3 4 12 2! = × = . Permutations when O1 , O2 are Permutations when O1 , O2 are different. the same O. 2 1 RO O T RO O T R O O T 2 1 T O O R T O O R T O O R 1 2 1 2 Reprint 2025-26 2 1 O T O R O T O R O T O R 2 1 O O R T O O T R O O R T 2 1 O R O T O R O T O R O T 2 1 O R T O O R T O O R T O 2 1 T O R O T O R O T O R O 2 1 T R O O T R O O T R O O 2 1 R O T O R O T O R O T O 2 1 R T O O R T O O R T O O 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 PERMUTATIONS AND COMBINATIONS 109 INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times. Temporarily, let us treat these letters different and name them as I1 , I2 , T1 , T2 , T3 . The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, I1 NT1 SI2 T2 U E T3 . Here if I1 , I2 are not same Let us now find the number of ways of rearranging the letters of the word 2 1 O T R O O T R O O T R O O O T R O O T R 2 1 O O T R 1 2 1 2 Reprint 2025-26 and T1 , T2 , T3 are not same, then I1 , I2 can be arranged in 2! ways and T1 , T2 , T3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I1NT1 SI2 T2UET3 . Hence, total number of different permutations will be 9! 2! 3! Theorem 3 The number of permutations of n objects, where p objects are of the same kind and rest are all different = ! ! n p . Theorem 4 The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ..., pk are of k th kind and the rest, if any, are of different kind is 1 2 ! ! ! !k n p p ... p . Example 9 Find the number of permutations of the letters of the word ALLAHABAD. Solution Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different. Therefore, the required number of arrangements = 9! 5 6 7 8 9 4!2! 2 × × × × = = 7560 110 MATHEMATICS We can state (without proof) the following theorems: In fact, we have a more general theorem. Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed? Solution Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time. Therefore, the required 4 digit numbers ( ) 9 4 9! 9! = P = = 9 – 4 ! 5! = 9 × 8 × 7 × 6 = 3024. Example 11 How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to Reprint 2025-26 count the permutations of 6 digits taken 3 at a time. This number would be 6P3 . But, these permutations will include those also where 0 is at the 100’s place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from 6P3 to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is 5P2 . So The required number 6 5 3 2 6! 5! = P P 3! 3! − = − Example 12 Find the value of n such that Solution (i) Given that P 42 P 5 3 n n = or n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2) Since n > 4 so n(n – 1) (n – 2) ≠ 0 Therefore, by dividing both sides by n(n – 1) (n – 2), we get (n – 3 (n – 4) = 42 or n 2 – 7n – 30 = 0 or n 2 – 10n + 3n – 30 or (n – 10) (n + 3) = 0 (i) P 42 P 4 5 3 n n = > , n (ii) 4 –1 4 P 5 = P 3 n = 4 × 5 × 6 – 4 ×5 = 100 PERMUTATIONS AND COMBINATIONS 111 n , n > 4 or n – 10 = 0 or n + 3 = 0 or n = 10 or n = – 3 As n cannot be negative, so n = 10. Therefore 3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4) or 3n = 5 (n – 4) [as (n – 1) (n – 2) (n – 3) ≠ 0, n > 4] or n = 10. (ii) Given that 4 –1 4 P 5 P 3 n n = Reprint 2025-26 Example 13 Find r, if 5 4Pr = 6 5Pr–1 . Solution We have 4 5 1 5 P 6 P r r = − or ( ) ( ) 4! 5! 5 6 4 ! 5 1 ! r r × = × − − + or ( ) ( ) ( )( ) 5! 6 5! 4 ! 5 1 5 5 1 ! r r r r × = − − + − − − or (6 – r) (5 – r) = 6 or r 2 – 11r + 24 = 0 or r 2 – 8r – 3r + 24 = 0 or (r – 8) (r – 3) = 0 or r = 8 or r = 3. Hence r = 8, 3. Example 14 Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together (ii) all vowels do not occur together. Solution (i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320. (ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. 112 MATHEMATICS Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000 Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ? Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind Reprint 2025-26 (red), 3 are of the second kind (yellow) and 2 are of the third kind (green). Therefore, the number of arrangements 9! =1260 4! 3! 2! . Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels never occur together (iv) do the words begin with I and end in P? Solution There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore The required number of arrangements 12! 1663200 3! 4! 2! = = (i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required number of words starting with P (ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in 11! 138600 3! 2! 4! = = . PERMUTATIONS AND COMBINATIONS 113 (iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together. E and I can be rearranged in 5! 4! ways. Therefore, by multiplication principle, the required number of arrangements 8! 3! 2! ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, 8! 5! = 16800 3! 2! 4! × = Reprint 2025-26 = 1663200 – 16800 = 1646400 (iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters. Hence, the required number of arrangements 8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once? 9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel? 10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? 11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S? 114 MATHEMATICS 1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? 2. How many 4-digit numbers are there with no digit repeated? 3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? 4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even? 5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position? 6. Find n if n – 1P3 : nP4 = 1 : 9. 7. Find r if (i) 5 6 P 2 P r r = −1 (ii) 5 6 P P r r = −1 . = 10! 3! 2! 4! = 12600 EXERCISE 6.3 6.4 Combinations Let us now assume that there is a group of 3 lawn tennis players X, Y, Z. A team consisting of 2 players is to be formed. In how many ways can we do so? Is the team of X and Y different from the team of Y and X ? Here, order is not important. In fact, there are only 3 possible ways in which the team could be constructed. Reprint 2025-26 These are XY, YZ and ZX (Fig 6.3). Here, each selection is called a combination of 3 different objects taken 2 at a time. In a combination, the order is not important. Now consider some more illustrations. Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time. Seven points lie on a circle. How many chords can be drawn by joining these points pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time. Now, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by nCr .. Suppose we have 4 different objects A, B, C and D. Taking 2 at a time, if we have to make combinations, these will be AB, AC, AD, BC, BD, CD. Here, AB and BA are the same combination as order does not alter the combination. This is why we have not included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations of 4 different objects taken 2 at a time, i.e., 4C2 = 6. Corresponding to each combination in the list, we can arrive at 2! permutations as 2 objects in each combination can be rearranged in 2! ways. Hence, the number of permutations = 4C2 × 2!. Fig. 6.3 PERMUTATIONS AND COMBINATIONS 115 On the other hand, the number of permutations of 4 different things taken 2 at a time = 4P2 . Now, let us suppose that we have 5 different objects A, B, C, D, E. Taking 3 at a time, if we have to make combinations, these will be ABC, ABD, ABE, BCD, BCE, CDE, ACE, ACD, ADE, BDE. Corresponding to each of these 5C3 combinations, there are 3! permutations, because, the three objects in each combination can be Therefore 4P2 = 4C2 × 2! or ( ) 4 2 4! C 4 2 ! 2! = − Reprint 2025-26 rearranged in 3 ! ways. Therefore, the total of permutations = 5C 3! 3 × permutaion and combination: Theorem 5 P C ! n n r r = r , 0 < r ≤ n. Proof Corresponding to each combination of nCr , we have r ! permutations, because r objects in every combination can be rearranged in r ! ways. Hence, the total number of permutations of n different things taken r at a time is nCr × r!. On the other hand, it is P n r . Thus Remarks 1. From above ( ) ! C ! ! n r n r n r = × − , i.e., ( ) ! C ! ! n r n r n r = − . In particular, if r n= , ! C 1 ! 0! n n n n = = . 2. We define nC0 = 1, i.e., the number of combinations of n different things taken 116 MATHEMATICS Therefore 5P3 = 5C3 × 3! or ( ) 5 3 5! C 5 3 ! 3! = − These examples suggest the following theorem showing relationship between nothing at all is considered to be 1. Counting combinations is merely counting the number of ways in which some or all objects at a time are selected. Selecting nothing at all is the same as leaving behind all the objects and we know that there is only one way of doing so. This way we define nC0 = 1. P C ! n n r r = × r , 0 < ≤r n . 3. As ( ) 0 ! 1 C 0! 0 ! n n n = = − , the formula ( ) ! C ! ! n r n r n r = − is applicable for r = 0 also. 4. ( ) ( ) ( ) ! C ! ! n n r n n r n n r − = − − − = ( ) ! ! ! n n r r − = C n r , Hence ( ) ! C ! ! n r n r n r = − , 0 ≤ r ≤ n. Reprint 2025-26 i.e., selecting r objects out of n objects is same as rejecting (n – r) objects. 5. nCa = nCb ⇒ a = b or a = n – b, i.e., n = a + b Theorem 6 1 C C C 1 n n n r r r + + = − Proof We have ( ) ( ) ( ) 1 ! ! C C ! ! 1 ! 1 ! n n r r n n r n r r n r + = + − − − − + Example 17 If C C 9 8 n n = , find C17 n . Solution We have C C 9 8 n n = i.e., ( ) ( ) ! ! 9! 9 ! 8 ! 8! n n n n = − − or 1 1 9 8 n = − or n – 8 = 9 or n = 17 = ( ) ( ) ! 1 ! ! n r r n r × − − + ( ) ( ) ( ) ! 1 ! 1 ! n r n r n r − − + − = ( ) ( ) ! 1 ! ! n r n r − − = ( ) ( ) ( ) ! 1 1 ! ! 1 n n r r r n r r n r − + + × − − − + = ( ) 1 1 r n r 1 + − + PERMUTATIONS AND COMBINATIONS 117 ( ) 1 1 ! C ! 1 ! n r n r n r + + = + − Therefore 17 C C 1 17 17 n = = . Example 18 A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Solution Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways = 5 3 5! 4 5 C 10 3! 2! 2 × = = = . selected from 3 women in 3C2 ways. Therefore, the required number of committees Now, 1 man can be selected from 2 men in 2C1 ways and 2 women can be Reprint 2025-26 Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these Solution There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore The required number of ways = 52 4 52! 49 50 51 52 C 4! 48! 2 3 4 × × × = = × × (i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are 13C4 ways of choosing 4 diamonds. Similarly, there are 13C4 ways of choosing 4 clubs, 13C4 ways of choosing 4 spades and 13C4 ways of choosing 4 hearts. Therefore The required number of ways = 13C4 + 13C4 + 13C4 + 13C4 . 118 MATHEMATICS (i) four cards are of the same suit, (ii) four cards belong to four different suits, (iii) are face cards, (iv) two are red cards and two are black cards, (v) cards are of the same colour? = 2 3 1 2 2! 3! C C 6 1! 1! 2! 1! × = × = . = 13! 4 2860 4! 9! × = = 270725 (iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be (ii) There are13 cards in each suit. Therefore, there are 13C1 ways of choosing 1 card from 13 cards of diamond, 13C1 ways of choosing 1 card from 13 cards of hearts, 13C1 ways of choosing 1 card from 13 cards of clubs, 13C1 ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways done in 12C4 ways. Therefore, the required number of ways = 12! 495 4! 8! = . = 13C1 × 13C1 × 13C1 × 13C1 = 134 Reprint 2025-26 (iv) There are 26 red cards and 26 black cards. Therefore, the required number of ways = 26C2 × 26C2 (v) 4 red cards can be selected out of 26 red cards in 26C4 ways. 4 black cards can be selected out of 26 black cards in 26C4ways. Therefore, the required number of ways = 26C4 + 26C4 1. If nC8 = nC2 , find nC2 . 2. Determine n if (i) 2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11 : 1 3. How many chords can be drawn through 21 points on a circle? 4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? 5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. 6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination. 7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? 8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected. 9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student? = ( ) 2 26! 2 325 2! 24! = = 105625 EXERCISE 6.4 = 26! 2 4! 22! × = 29900. PERMUTATIONS AND COMBINATIONS 119 Example 20 How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ? Solution In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 consonants, namely, N, V, L and T. Miscellaneous Examples Reprint 2025-26 4 × 6 = 24. themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880. Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ? (iii) at least 3 girls ? Solution (i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in 7C5 ways. Therefore, the required number of ways = 7 5 7! 6 7 C 21 5! 2! 2 × = = = (ii) Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of 120 MATHEMATICS The number of ways of selecting 3 vowels out of 4 = 4C3 = 4. The number of ways of selecting 2 consonants out of 4 = 4C2 = 6. Therefore, the number of combinations of 3 vowels and 2 consonants is Now, each of these 24 combinations has 5 letters which can be arranged among (a) 1 boy and 4 girls (b) 2 boys and 3 girls (c) 3 boys and 2 girls (d) 4 boys and 1 girl. 1 boy and 4 girls can be selected in 7C1 × 4C4 ways. 2 boys and 3 girls can be selected in 7C2 × 4C3 ways. 3 boys and 2 girls can be selected in 7C3 × 4C2 ways. = 7C1 × 4C4 + 7C2 × 4C3 + 7C3 × 4C2 + 7C4 × 4C1 = 7 + 84 + 210 + 140 = 441 (iii) Since, the team has to consist of at least 3 girls, the team can consist of (a) 3 girls and 2 boys, or (b) 4 girls and 1 boy. Note that the team cannot have all 5 girls, because, the group has only 4 girls. Therefore, the required number of ways Therefore, the required number of ways 3 girls and 2 boys can be selected in 4C3 × 7C2 ways. 4 girls and 1 boy can be selected in 4C4 × 7C1 ways. 4 boys and 1 girl can be selected in 7C4 × 4C1 ways. = 4C3 × 7C2 + 4C4 × 7C1 = 84 + 7 = 91 Reprint 2025-26 Example 22 Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word? Solution There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, the required number of words = 5! 60 2! = . position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with A = 4! = 24. Then, starting with G, the number of words 4! 2! = = 12 as after placing G at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained = 24 + 12 + 12 =48. Example 23 How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? Solution Since, 1000000 is a 7-digit number and the number of digits to be used is also 7. Therefore, the numbers to be counted will be 7-digit only. Also, the numbers have to be greater than 1000000, so they can begin either with 1, 2 or 4. To get the number of words starting with A, we fix the letter A at the extreme left The 49th word is NAAGI. The 50th word is NAAIG. PERMUTATIONS AND COMBINATIONS 121 fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2s and 2, 4s. The number of numbers beginning with 1 = 6! 4 5 6 3! 2! 2 × × = = 60, as when 1 is Total numbers begining with 2 = 6! 3 4 5 6 2! 2! 2 × × × = = 180 and total numbers begining with 4 6! 4 5 6 3! = = × × = 120 Reprint 2025-26 Alternative Method The number of 7-digit arrangements, clearly, 7! 420 3! 2! = . But, this will include those numbers also, which have 0 at the extreme left position. The number of such arrangements 6! 3! 2! (by fixing 0 at the extreme left position) = 60. Therefore, the required number of numbers = 420 – 60 = 360. ANote If one or more than one digits given in the list is repeated, it will be understood that in any number, the digits can be used as many times as is given in the list, e.g., in the above example 1 and 0 can be used only once whereas 2 and 4 can be used 3 times and 2 times, respectively. Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together? Solution Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can be seated only at the cross marked places. × G × G × G × G × G ×. There are 6 cross marked places and the three boys can be seated in 6P3 ways. Hence, by multiplication principle, the total number of ways 122 MATHEMATICS Therefore, the required number of numbers = 60 + 180 + 120 = 360. = 5! × 6P3 = 6! 5!× 3! = 4 × 5 × 2 × 3 × 4 × 5 × 6 = 14400. 1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ? 2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? 3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: 4. If the different permutations of all the letter of the word EXAMINATION are (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ? Miscellaneous Exercise on Chapter 6 Reprint 2025-26 listed as in a dictionary, how many words are there in this list before the first word starting with E ? 5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ? 6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ? 7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ? 8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king. 9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ? 10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ? 11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ? Summary ÆFundamental principle of counting If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m × n. ÆThe number of permutations of n different things taken r at a time, where PERMUTATIONS AND COMBINATIONS 123 where 0 ≤ r ≤ n. Æn! = 1 × 2 × 3 × ...×n Æn! = n × (n – 1) ! ÆThe number of permutations of n different things, taken r at a time, where repeatition is allowed, is n r . ÆThe number of permutations of n objects taken all at a time, where p1 objects repetition is not allowed, is denoted by nPr and is given by nPr = ! ( )! n n r − , Reprint 2025-26 124 MATHEMATICS kind and rest, if any, are all different is 1 2 ! ! ! !k n p p ... p . ÆThe number of combinations of n different things taken r at a time, denoted by The concepts of permutations and combinations can be traced back to the advent of Jainism in India and perhaps even earlier. The credit, however, goes to the Jains who treated its subject matter as a self-contained topic in mathematics, under the name Vikalpa. Among the Jains, Mahavira, (around 850) is perhaps the world’s first mathematician credited with providing the general formulae for permutations and combinations. In the 6th century B.C., Sushruta, in his medicinal work, Sushruta Samhita, asserts that 63 combinations can be made out of 6 different tastes, taken one at a time, two at a time, etc. Pingala, a Sanskrit scholar around third century B.C., gives the method of determining the number of combinations of a given number of letters, taken one at a time, two at a time, etc. in his work Chhanda Sutra. Bhaskaracharya (born 1114) treated the subject matter of permutations and combinations under the name Anka Pasha in his famous work Lilavati. In addition to the general formulae for nCr and nPr already provided by Mahavira, Bhaskaracharya gives several important theorems and results concerning the subject. Outside India, the subject matter of permutations and combinations had its humble beginnings in China in the famous book I–King (Book of changes). It is difficult to give the approximate time of this work, since in 213 B.C., the emperor had ordered all books and manuscripts in the country to be burnt which fortunately was not completely carried out. Greeks and later Latin writers also did some scattered work on the theory of permutations and combinations. Some Arabic and Hebrew writers used the concepts of permutations and combinations in studying astronomy. Rabbi ben Ezra, for instance, determined the number of combinations of known planets taken two at a time, three at a time and so on. This was around 1140. It appears that Rabbi ben Ezra did not know are of first kind, p2 objects are of the second kind, ..., pk objects are of the k th nCr , is given by nCr = ! ! ! n r ( n r ) = − , 0 ≤ r ≤ n. Historical Note Reprint 2025-26 the formula for nCr . However, he was aware that nCr = nCn–r for specific values n and r. In 1321, Levi Ben Gerson, another Hebrew writer came up with the formulae for nPr , nPn and the general formula for nCr . The first book which gives a complete treatment of the subject matter of permutations and combinations is Ars Conjectandi written by a Swiss, Jacob Bernoulli (1654 – 1705), posthumously published in 1713. This book contains essentially the theory of permutations and combinations as is known today. PERMUTATIONS AND COMBINATIONS 125 Reprint 2025-26 — v —" class_11,7,Binomial Theorem,ncert_books/class_11/kemh1dd/kemh107.pdf,"126 MATHEMATICS 7.1 Introduction In earlier classes, we have learnt how to find the squares and cubes of binomials like a + b and a – b. Using them, we could evaluate the numerical values of numbers like (98)2 = (100 – 2)2 , (999)3 = (1000 – 1)3 , etc. However, for higher powers like (98)5 , (101)6 , etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem. It gives an easier way to expand (a + b) n , where n is an integer or a rational number. In this Chapter, we study binomial theorem for positive integral indices only. 7.2 Binomial Theorem for Positive Integral Indices vMathematics is a most exact science and its conclusions are capable of absolute proofs. – C.P. STEINMETZv BINOMIALTHEOREM Chapter 7 Blaise Pascal (1623-1662) Let us have a look at the following identities done earlier: (a+ b) 0 = 1 a + b ≠ 0 (a+ b) 1 = a + b (a+ b) 2 = a 2 + 2ab + b 2 In these expansions, we observe that (i) The total number of terms in the expansion is one more than the index. For example, in the expansion of (a + b) 2 , number of terms is 3 whereas the index of (a + b) 2 is 2. (ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the second quantity ‘b’ increase by 1, in the successive terms. (iii) In each term of the expansion, the sum of the indices of a and b is the same and is equal to the index of a + b. (a+ b) 3 = a 3 + 3a 2b + 3ab2 + b 3 (a+ b) 4 = (a + b) 3 (a + b) = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4 Reprint 2025-26 Do we observe any pattern in this table that will help us to write the next row? Yes we do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each row. This can be continued till any index of our interest. We can extend the pattern given in Fig 7.2 by writing a few more rows. We now arrange the coefficients in these expansions as follows (Fig 7.1): Fig 7.1 BINOMIAL THEOREM 127 Pascal’s Triangle The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running down the two slanting sides. This array of numbers is known as Pascal’s triangle, after the name of French mathematician Blaise Pascal. It is also known as Meru Prastara by Pingla. Expansions for the higher powers of a binomial are also possible by using Pascal’s triangle. Let us expand (2x + 3y) 5 by using Pascal’s triangle. The row for index 5 is 1 5 10 10 5 1 Using this row and our observations (i), (ii) and (iii), we get (2x + 3y) 5 = (2x) 5 + 5(2x) 4 (3y) + 10(2x) 3 (3y) 2 +10 (2x) 2 (3y) 3 + 5(2x)(3y) 4 +(3y) 5 = 32x 5 + 240x 4y + 720x 3y 2 + 1080x 2y 3 + 810xy4 + 243y 5 . Reprint 2025-26 Fig 7.2 128 MATHEMATICS the row for index 12. This can be done by writing all the rows of the Pascal’s triangle till index 12. This is a slightly lengthy process. The process, as you observe, will become more difficult, if we need the expansions involving still larger powers. We thus try to find a rule that will help us to find the expansion of the binomial for any power without writing all the rows of the Pascal’s triangle, that come before the row of the desired index. For this, we make use of the concept of combinations studied earlier to rewrite the numbers in the Pascal’s triangle. We know that ! C !( )! n r n r n – r = , 0 ≤ r ≤ n and n is a non-negative integer. Also, nC0 = 1 = nCn The Pascal’s triangle can now be rewritten as (Fig 7.3) Now, if we want to find the expansion of (2x + 3y) 12, we are first required to get Observing this pattern, we can now write the row of the Pascal’s triangle for any index without writing the earlier rows. For example, for the index 7 the row would be Thus, using this row and the observations (i), (ii) and (iii), we have (a + b) 7 = 7C0 a 7 + 7C1 a 6b + 7C2 a 5b 2 + 7C3 a 4b 3 + 7C4 a 3b 4 + 7C5 a 2b 5 + 7C6 ab6 + 7C7 b 7 An expansion of a binomial to any positive integral index say n can now be visualised using these observations. We are now in a position to write the expansion of a binomial to any positive integral index. 7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7. Fig 7.3 Pascal’s triangle Reprint 2025-26 7.2.1 Binomial theorem for any positive integer n, Proof The proof is obtained by applying principle of mathematical induction. Let the given statement be Thus, P (1) is true. (1) We shall prove that P(k + 1) is also true, i.e., Now, (a + b) k + 1 = (a + b) (a + b) k Suppose P (k) is true for some positive integer k, i.e. (a + b) k + 1 = k + 1C0 a k + 1 + k + 1C1 a kb + k + 1C2 a k – 1b 2 + ...+ k + 1Ck+1 b k + 1 P(n) : (a + b) n = nC0 a n + nC1 a n – 1b + nC2 a n – 2b 2 + ...+ nCn–1a.b n – 1 + nCn b n For n = 1, we have P (1) : (a + b) 1 = 1C0 a 1 + 1C1 b 1 = a + b (a + b) k = kC0 a k + kC1 a k – 1b + kC2 a k – 2b 2 + ...+ kCk b k ... (a + b) n = nC0 a n + nC1 a n–1b + nC2 a n–2 b 2 + ...+ nCn – 1a.b n–1 + nCn b n = (a + b) (kC0 a k + kC1 a k – 1 b + kC2 a k – 2 b 2 +...+ kCk – 1 abk – 1 + kCk b k ) [from (1)] = kC0 a k + 1 + kC1 a kb + kC2 a k – 1b 2 +...+ kCk – 1 a 2b k – 1 + kCk abk +kC0 a kb + kC1 a k – 1b 2 + kC2 a k – 2b 3+...+ kCk-1abk + kCk b k + 1 BINOMIAL THEOREM 129 [by actual multiplication] Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by principle of mathematical induction, P(n) is true for every positive integer n. Thus (x + 2)6 = x 6 + 12x 5 + 60x 4 + 160x 3 + 240x 2 + 192x + 64. We illustrate this theorem by expanding (x + 2)6 : (x + 2)6 = 6C0 x 6 + 6C1 x 5 .2 + 6C2 x 42 2 + 6C3 x 3 .23 + 6C4 x 2 .24 + 6C5 x.25 + 6C6 .26 . (by using k + 1C0 =1, kCr + kCr–1 = k + 1Cr and kCk = 1= k + 1Ck + 1) = x 6 + 12x 5 + 60x 4 + 160x 3 + 240x 2 + 192x + 64 = kC0 a k + 1 + (kC1 + kC0 ) akb + (kC2 + kC1 )a k – 1b 2 + ... + (kCk + kCk–1) abk + kCk b k + 1 [grouping like terms] = k + 1C0 a k + 1 + k + 1C1 a kb + k + 1C2 a k – 1b 2 +...+ k + 1Ck abk + k + 1Ck + 1 b k +1 Reprint 2025-26 130 MATHEMATICS Observations 1. The notation ∑= − n Hence the theorem can also be stated as 2. The coefficients nCr occuring in the binomial theorem are known as binomial coefficients. 3. There are (n+1) terms in the expansion of (a+b) n , i.e., one more than the index. 4. In the successive terms of the expansion the index of a goes on decreasing by unity. It is n in the first term, (n–1) in the second term, and so on ending with zero in the last term. At the same time the index of b increases by unity, starting with zero in the first term, 1 in the second and so on ending with n in the last term. 5. In the expansion of (a+b) n , the sum of the indices of a and b is n + 0 = n in the first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion. 7.2.2 Some special cases In the expansion of (a + b) n , (i) Taking a = x and b = – y, we obtain nC0 a nb 0 + nC1 a n–1b 1 + ...+ nCr a n–rb r + ...+nCn a n–nb n , where b 0 = 1 = an–n . (x – y)n = [x + (–y)]n = nC0 x n + nC1 x n – 1(–y) + nC2 x n–2(–y) 2 + nC3 x n–3(–y) 3 + ... + nCn (–y) n k n kk k n a b 0 C stands for kkn k n n ba a b 0 C)( . ∑= − =+ n k Thus (x–y) n = nC0 x n – nC1 x n – 1 y + nC2 x n – 2 y 2 + ... + (–1)n nCn y n Using this, we have (x–2y) 5 = 5C0 x 5 – 5C1 x 4 (2y) + 5C2 x 3 (2y) 2 – 5C3 x 2 (2y) 3 + (ii) Taking a = 1, b = x, we obtain Thus (1 + x) n = nC0 + nC1 x + nC2 x 2 + nC3 x 3 + ... + nCn x n (1 + x) n = nC0 (1)n + nC1 (1)n – 1x + nC2 (1)n – 2 x 2 + ... + nCn x n = nC0 x n – nC1 x n – 1y + nC2 x n – 2y 2 – nC3 x n – 3y 3 + ... + (–1)n nCn y n = nC0 + nC1 x + nC2 x 2 + nC3 x 3 + ... + nCn x n 5C4 x(2y) 4 – 5C5 (2y) 5 = x 5 –10x 4y + 40x 3y 2 – 80x 2y 3 + 80xy4 – 32y 5 . Reprint 2025-26 In particular, for x = 1, we have (iii) Taking a = 1, b = – x, we obtain Example 1 Expand 4 2 3 x x + , x ≠ 0 Solution By using binomial theorem, we have Example 2 Compute (98)5 . Solution We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem. Write 98 = 100 – 2 Therefore, (98)5 = (100 – 2)5 4 2 3 x x + = 4C0 (x 2 ) 4 + 4C1 (x 2 ) 3 x 3 + 4C2 (x 2 ) 2 2 3 x + 4C3 (x 2 ) In particular, for x = 1, we get (1– x) n = nC0 – nC1 x + nC2 x 2 – ... + (– 1)n nCn x n = x 8 + 4.x 6 . x 3 + 6.x 4 . 2 9 = x 8 + 12x 5 + 54x 2 + 4 81108 x x + . 2 n = nC0 + nC1 + nC2 + ... + nCn . 0 = nC0 – nC1 + nC2 – ... + (–1)n nCn x + 4.x 2 . 3 27 x + 4 81 x BINOMIAL THEOREM 131 3 3 x + 4C4 4 3 x Example 3 Which is larger (1.01)1000000 or 10,000? Solution Splitting 1.01 and using binomial theorem to write the first few terms we have = 5C0 (100)5 – 5C1 (100)4 .2 + 5C2 (100)32 2 – 5C3 (100)2 (2)3 + 5C4 (100) (2)4 – 5C5 (2)5 = 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32 = 10040008000 – 1000800032 = 9039207968. Reprint 2025-26 132 MATHEMATICS (1.01)1000000 = (1 + 0.01)1000000 Hence (1.01)1000000 > 10000 Example 4 Using binomial theorem, prove that 6n–5n always leaves remainder 1 when divided by 25. Solution For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6n – 5n leaves remainder 1 when divided by 25, we prove that 6 n – 5n = 25k + 1, where k is some natural number. We have For a = 5, we get i.e. (6)n = 1 + 5n + 52 . nC2 + 53 . nC3 + ... + 5n (1 + 5)n = nC0 + nC1 5 + nC2 5 2 + ... + nCn 5 n = 1000000C0 + 1000000C1 (0.01) + other positive terms = 1 + 1000000 × 0.01 + other positive terms = 1 + 10000 + other positive terms (1 + a) n = nC0 + nC1 a + nC2 a 2 + ... + nCn a n > 10000 i.e. 6 n – 5n = 1+52 (nC2 + nC3 5 + ... + 5n-2) or 6 n – 5n = 1+ 25 (nC2 + 5 . nC3 + ... + 5n-2) or 6 n – 5n = 25k+1 where k = nC2 + 5 . nC3 + ... + 5n–2 . This shows that when divided by 25, 6n – 5n leaves remainder 1. Expand each of the expressions in Exercises 1 to 5. 1. (1–2x) 5 2. 5 2 2 x – x 3. (2x – 3)6 EXERCISE 7.1 Reprint 2025-26 Using binomial theorem, evaluate each of the following: 6. (96)3 7. (102)5 8. (101)4 9. (99)5 10. Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000. 12. Find (x + 1)6 + (x – 1)6 . Hence or otherwise evaluate ( 2 + 1)6 + ( 2 – 1)6 . 13. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. 14. Prove that ∑ = = n 11. Find (a + b) 4 – (a – b) 4 . Hence, evaluate 4 + )23( – 4 )2–3( . 4. 5 1 3 x x + 5. 1. If a and b are distinct integers, prove that a – b is a factor of a n – b n , whenever 2. Evaluate ( ) ( ) 6 6 3 2 3 2 + − − . 3. Find the value of ( ) ( ) 4 4 2 2 2 2 a a a a + − + − − 1 1 . n is a positive integer. [Hint write a n = (a – b + b) n and expand] r nr n r 0 3 4C . Miscellaneous Exercise on Chapter 7 6 1 + x x BINOMIAL THEOREM 133 4. Find an approximation of (0.99)5 using the first three terms of its expansion. 5. Expand using Binomial Theorem 4 2 1 0 2 x , x x + − ≠ . 6. Find the expansion of (3x 2 – 2ax + 3a 2 ) 3 using binomial theorem. Summary ÆThe expansion of a binomial for any positive integral n is given by Binomial Theorem, which is (a + b) n = nC0 a n + nC1 a n – 1b + nC2 a n – 2b 2 + ...+ nCn – 1a.b n – 1 + nCn b n . ÆThe coefficients of the expansions are arranged in an array. This array is called Pascal’s triangle. Reprint 2025-26 134 MATHEMATICS The ancient Indian mathematicians knew about the coefficients in the expansions of (x + y) n , 0 ≤ n ≤ 7. The arrangement of these coefficients was in the form of a diagram called Meru-Prastara, provided by Pingla in his book Chhanda shastra (200B.C.). This triangular arrangement is also found in the work of Chinese mathematician Chu-shi-kie in 1303. The term binomial coefficients was first introduced by the German mathematician, Michael Stipel (1486-1567) in approximately 1544. Bombelli (1572) also gave the coefficients in the expansion of (a + b) n , for n = 1,2 ...,7 and Oughtred (1631) gave them for n = 1, 2,..., 10. The arithmetic triangle, popularly known as Pascal’s triangle and similar to the MeruPrastara of Pingla was constructed by the French mathematician Blaise Pascal (1623-1662) in 1665. The present form of the binomial theorem for integral values of n appeared in Trate du triange arithmetic, written by Pascal and published posthumously in 1665. Historical Note — v — Reprint 2025-26" class_11,8,Sequences and Series,ncert_books/class_11/kemh1dd/kemh108.pdf,"8.1 Introduction In mathematics, the word, “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on. For example, population of human beings or bacteria at different times form a sequence. The amount of money deposited in a bank, over a number of years form a sequence. Depreciated values of certain commodity occur in a sequence. Sequences have important applications in several spheres of human activities. Sequences, following specific patterns are called progressions. In previous class, we have studied about arithmetic progression (A.P). In this Chapter, besides discussing more about A.P.; arithmetic mean, geometric mean, relationship between A.M. and G.M., special series in forms of sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers will also be studied. vNatural numbers are the product of human spirit. – DEDEKINDv SEQUENCES AND SERIES 8 Chapter Fibonacci (1175-1250) 8.2 Sequences Let us consider the following examples: Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years. Here, the total number of generations = 300 10 30 = Reprint 2025-26 136 MATHEMATICS The number of person’s ancestors for the first, second, third, …, tenth generations are 2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence. Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by a1 , a2 , a3 , …, an , …, etc., the subscripts denote the position of the term. The n th term is the number at the n th position of the sequence and is denoted by an. The n th term is also called the generalterm of the sequence. Thus, the terms of the sequence of person’s ancestors mentioned above are: a1 = 2, a2 = 4, a3 = 8, …, a10 = 1024. Similarly, in the example of successive quotients a1 = 3, a2 = 3.3, a3 = 3.33, …, a6 = 3.33333, etc. A sequence containing finite number of terms is called a finite sequence. For example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed number). A sequence is called infinite, if it is not a finite sequence. For example, the sequence of successive quotients mentioned above is an infinite sequence, infinite in the sense that it never ends. Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers 2, 4, 6, … Here a1 = 2 = 2 × 1 a2 = 4 = 2 × 2 a3 = 6 = 2 × 3 a4 = 8 = 2 × 4 .... .... .... .... .... .... where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …, the n th term is given by the formula, an = 2n – 1, where n is a natural number. In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible pattern, but the sequence is generated by the recurrence relation given by a1 = a2 = 1 a3 = a1 + a2 an = an – 2 + an – 1, n > 2 This sequence is called Fibonacci sequence. In fact, we see that the n th term of this sequence can be written as an = 2n, .... .... .... .... .... .... a23 = 46 = 2 × 23, a24 = 48 = 2 × 24, and so on. Reprint 2025-26 prime. Such sequence can only be described by verbal description. In every sequence, we should not expect that its terms will necessarily be given by a specific formula. However, we expect a theoretical scheme or a rule for generating the terms a1 , a2 , a3 ,…,an ,… in succession. In view of the above, a sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for an . 8.3 Series Let a1 , a2 , a3 ,…,an , be a given sequence. Then, the expression is called the series associated with the given sequence .The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter ∑ (sigma) as means of indicating the summation involved. Thus, the series a1+ a2 + a3 + ... + an is abbreviated as 1 n k k a = ∑ . Remark When the series is used, it refers to the indicated sum not to the sum itself. For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase “sum of a series,” we will mean the number that results from adding the terms, the sum of the series is 16. In the sequence of primes 2,3,5,7,…, we find that there is no formula for the n th We now consider some examples. a1 + a2 + a3 +,…+ an + ... SEQUENCES AND SERIES 137 Example 1 Write the first three terms in each of the following sequences defined by the following: Solution (i) Here an = 2n + 5 Substituting n = 1, 2, 3, we get a1 = 2(1) + 5 = 7, a2 = 9, a3 = 11 Therefore, the required terms are 7, 9 and 11. (ii) Here an = 3 4 n − . Thus, 1 2 3 1 3 1 1 0 4 2 4 a , a , a − = = − = − = (i) an = 2n + 5, (ii) an = 3 4 n − . Reprint 2025-26 138 MATHEMATICS Hence, the first three terms are 1 1 2 4 – , – and 0. Example 2 What is the 20th term of the sequence defined by an = (n – 1) (2 – n) (3 + n) ? Solution Putting n = 20 , we obtain Example 3 Let the sequence an be defined as follows: Find first five terms and write corresponding series. Solution We have Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series is 1 + 3 + 5 + 7 + 9 +... Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5, a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9. a20 = (20 – 1) (2 – 20) (3 + 20) = 19 × (– 18) × (23) = – 7866. a1 = 1, an = an – 1 + 2 for n ≥ 2. EXERCISE 8.1 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n th terms are: 1. an = n (n + 2) 2. an = 1 n n + 3. an = 2 n 4. an = 2 3 6 n − 5. an = (–1)n–1 5n+1 6. an 2 5 4 n n + = . 7. an = 4n – 3; a17, a24 8. an = 9. an = (–1)n – 1n 3 ; a9 10. 20 ( – 2) ; 3 n n n a a n = + . Reprint 2025-26 7 ; 2 n n a 2 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 13. a1 = a2 = 2, an = an – 1–1, n > 2 14. The Fibonacci sequence is defined by 8.4 Geometric Progression (G. P.) Let us consider the following sequences: In each of these sequences, how their terms progress? We note that each term, except the first progresses in a definite order. 11. a1 = 3, an = 3an – 1 + 2 for all n > 1 12. a1 = – 1, an = n 1 a n − , n ≥ 2 Find n 1 In (i), we have a a a In (ii), we observe, a a a (i) 2,4,8,16,..., (ii) 1 1 1 1 9 27 81 243 – – , , , ... (iii) .01,.0001,.000001,... n a a + , for n = 1, 2, 3, 4, 5 1 = a1 = a2 and an = an – 1 + an – 2, n > 2. a a 1 2 3 = = = = 2 2 2 2 , , , and so on. a a 1 2 1 3 = = = = , , , and so on. 1 3 1 9 a a 3 2 1 3 a a 3 4 1 3 4 SEQUENCES AND SERIES 139 Similarly, state how do the terms in (iii) progress? It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it. In (i), this constant ratio is 2; in (ii), it is – 1 3 and in (iii), the constant ratio is 0.01. Such sequences are called geometric sequence or geometric progression abbreviated as G.P. A sequence a1 , a2 , a3 , …, an , … is called geometric progression, if each term is non-zero and a a k k + 1 = r (constant), for k ≥ 1. is called the first term and r is called the common ratio of the G.P. Common ratio in geometric progression (i), (ii) and (iii) above are 2, – 1 3 and 0.01, respectively. By letting a1 = a, we obtain a geometric progression, a, ar, ar2 , ar3 ,…., where a 1 Reprint 2025-26 2 140 MATHEMATICS As in case of arithmetic progression, the problem of finding the n th term or sum of n terms of a geometric progression containing a large number of terms would be difficult without the use of the formulae which we shall develop in the next Section. We shall use the following notations with these formulae: 8.4.1 General term of a G.P. Let us consider a G.P. with first non-zero term ‘a’ and common ratio ‘r’. Write a few terms of it. The second term is obtained by multiplying a by r, thus a2 = ar. Similarly, third term is obtained by multiplying a2 by r. Thus, a3 = a2 r = ar2 , and so on. We write below these and few more terms. 1 st term = a1 = a = ar1–1, 2nd term = a2 = ar = ar2–1, 3rd term = a3 = ar2 = ar3–1 4 th term = a4 = ar3 = ar4–1, 5th term = a5 = ar4 = ar5–1 Do you see a pattern? What will be 16th term? a16 = ar16–1 = ar15 Therefore, the pattern suggests that the n th term of a G.P. is given by an = arn–1 . Thus, a, G.P. can be written as a, ar, ar2 , ar3 , … arn – 1; a, ar, ar2 ,...,arn – 1 ...;according as G.P. is finite or infinite, respectively. The series a + ar + ar2 + ... + arn–1 or a + ar + ar2 + ... + arn–1 +...are called finite or infinite geometric series, respectively. 8.4.2. Sum to n terms of a G.P. Let the first term of a G.P. be a and the common ratio be r. Let us denote by Sn the sum to first n terms of G.P. Then Sn = a + ar + ar2 +...+ arn–1 ... (1) Case 1 If r = 1, we have Sn = a + a + a + ... + a (n terms) = na a = the first term, r = the common ratio, l = the last term, n = the numbers of terms, Sn = the sum of first n terms. S Case 2 If r ≠ 1, multiplying (1) by r, we have rSn = ar + ar2 + ar3 + ... + arn ... (2) Subtracting (2) from (1), we get (1 – r) Sn = a – arn = a(1 – r n ) This gives Example 4 Find the 10th and n th terms of the G.P. 5, 25,125,… . Solution Here a = 5 and r = 5. Thus, a10 = 5(5)10–1 = 5(5)9 = 510 and an = arn–1 = 5(5)n–1 = 5n . Reprint 2025-26 or ( 1) S 1 n n a r r − = − Example 5 Which term of the G.P., 2,8,32, ... up to n terms is 131072? Solution Let 131072 be the n th term of the given G.P. Here a = 2 and r = 4. Therefore 131072 = an = 2(4)n – 1 or 65536 = 4n – 1 This gives 4 8 = 4n – 1. So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P. Example 6 In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term. Solution Here, a ar 3 2 = = 24 ... (1) and a ar 6 5 = = 192 ... (2) Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6. Hence a10 = 6 (2)9 = 3072. Example 7 Find the sum of first n terms and the sum of first 5 terms of the geometric series 2 4 1 3 9 + + +... Solution Here a = 1 and r = 2 3 . Therefore Sn = n a r r 2 1 3 (1 ) 1 2 1 3 n − − = − − = 2 3 1 3 n − SEQUENCES AND SERIES 141 In particular, 5 Example 8 How many terms of the G.P. 3 3 3 2 4 , , ,... are needed to give the sum 3069 512 ? Solution Let n be the number of terms needed. Given that a = 3, r = 1 2 and 3069 S 512 n = Since (1 ) S 1 n n a – r r = − 5 2 S 3 1 3 = − = 211 3 243 × = 211 81 . Reprint 2025-26 142 MATHEMATICS Therefore or 3069 3072 = 1 1 2 n − or 1 2 n = 3069 1 3072 − 3 1 3072 1024 = = or 2 n = 1024 = 210 , which gives n = 10. Example 9 The sum of first three terms of a G.P. is 13 12 and their product is – 1. Find the common ratio and the terms. Solution Let a r , a, ar be the first three terms of the G.P. Then and ( ) ( ) 1 a a ar – r = ... (2) From (2), we get a 3 = – 1, i.e., a = – 1 (considering only real roots) a ar a r + + = 13 12 ... (1) 1 3(1 ) 3069 1 2 6 1 512 1 2 1 2 n n − = = − − Substituting a = –1 in (1), we have This is a quadratic in r, solving, we get 3 4 or 4 3 r – – = . Thus, the three terms of G.P. are : 4 3 –3 3 4 –4 , 1, for = and , 1, for = 3 4 4 4 3 3 – r – r , Example10 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms. Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as 1 13 1 12 – – – r r = or 12r 2 + 25r + 12 = 0. Sn = 7 + 77 + 777 + 7777 + ... to n terms Reprint 2025-26 Example 11 A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. Solution Here a = 2, r = 2 and n = 10 Using the sum formula Sn = ( 1) 1 n a r r We have S10 = 2(210 – 1) = 2046 Hence, the number of ancestors preceding the person is 2046. 8.4.3 Geometric Mean (G.M.) The geometric mean of two positive numbers a and b is the number ab . Therefore, the geometric mean of 2 and 8 is 4. We observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers. Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P. Let G1 , G2 ,…, Gn be n numbers between positive numbers a and b such that a,G1 ,G2 ,G3 ,…,Gn ,b is a G.P. Thus, b being the (n + 2)th term,we have = 7 [9 99 999 9999 to term] 9 + + + + ... n = 7 2 3 4 [(10 1) (10 1) (10 1) (10 1) terms] 9 − + − + − + − + ...n = 7 2 3 [(10 10 10 terms) (1+1+1+... terms)] 9 + + +...n – n = 7 10(10 1) 7 10 (10 1) 9 10 1 9 9 n n n n − − − = − − . − − SEQUENCES AND SERIES 143 Hence 1 1 G1 b n ar a a + = = , 2 1 2 G2 b n ar a a + = = , n 1 b ar + = , or 1 G n n n n b ar a a + = = 1 b n 1 r a + = Reprint 2025-26 . 3 1 3 G3 b n ar a a + = = , 144 MATHEMATICS Example12 Insert three numbers between 1 and 256 so that the resulting sequence is a G.P. Solution Let G1 , G2 ,G3 be three numbers between 1 and 256 such that 1, G1 ,G2 ,G3 ,256 is a G.P. Therefore 256 = r 4 giving r = ± 4 (Taking real roots only) For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64 Similarly, for r = – 4, numbers are – 4,16 and – 64. Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are in G.P. 8.5 Relationship Between A.M. and G.M. Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then Thus, we have From (1), we obtain the relationship A ≥ G. Example 13 If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers. A – G = 2 a b ab + − = 2 2 a b ab + − A and G 2 a b ab + = = = ( ) 2 0 2 a b − ≥ ... (1) Solution Given that A.M. 10 2 a b + = = ... (1) and G.M. 8 = = ab ... (2) From (1) and (2), we get a + b = 20 ... (3) ab = 64 ... (4) Putting the value of a and b from (3), (4) in the identity (a – b) 2 = (a + b) 2 – 4ab, we get (a – b) 2 = 400 – 256 = 144 or a – b = ± 12 ... (5) Reprint 2025-26 Solving (3) and (5), we obtain a = 4, b = 16 or a = 16, b = 4 Thus, the numbers a and b are 4, 16 or 16, 4 respectively. Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7. 0.15, 0.015, 0.0015, ... 20 terms. 1. Find the 20th and n th terms of the G.P. 5 5 5 2 4 8 , , , ... 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. 3. The 5th, 8th and 11 th terms of a G.P. are p, q and s, respectively. Show that q 2 = ps. 4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term. 5. Which term of the following sequences: 6. For what values of x, the numbers 2 7 7 2 – , x, – are in G.P.? (a) 2 2 2 4 is 128 ? , , ,... (b) 3 3 3 3 is729 ? , , ,... (c) 1 1 1 1 is 3 9 27 19683 , , ,... ? EXERCISE 8.2 SEQUENCES AND SERIES 145 8. 7 , 21 , 3 7 , ... n terms. 9. 1, – a, a 2 , – a3 , ... n terms (if a ≠ – 1). 10. x 3 , x 5 , x 7 , ... n terms (if x ≠ ± 1). 11. Evaluate 11 12. The sum of first three terms of a G.P. is 39 10 and their product is 1. Find the common ratio and the terms. 13. How many terms of G.P. 3, 32 , 33 , … are needed to give the sum 120? 14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. 15. Given a G.P. with a = 729 and 7th term 64, determine S7 . 1 (2 3 ) k k= ∑ + . Reprint 2025-26 146 MATHEMATICS 16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term. 17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. 18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… . 19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 20. Show that the products of the corresponding terms of the sequences a, ar, ar2 , …arn – 1 and A, AR, AR2 , … ARn – 1 form a G.P, and find the common ratio. 21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18. 22. If the p th , q th and r th terms of a G.P. are a, b and c, respectively. Prove that a q – r br – pc P – q = 1. 23. If the first and the n th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n . 24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from 25. If a, b, c and d are in G.P. show that (a 2 + b 2 + c 2 ) (b 2 + c 2 + d 2 ) = (ab + bc + cd) 2 . 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P. 16, 32 and 128, 32, 8, 2, 1 2 . (n + 1)th to (2n) th term is 1 n r . 27. Find the value of n so that a b a b n n a and b. 28. The sum of two numbers is 6 times their geometric mean, show that numbers 29. If A and G be A.M. and G.M., respectively between two positive numbers, 30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2 nd hour, 4th hour and n th hour ? are in the ratio (3 2 2 : 3 2 2 + − ) ( ). prove that the numbers are A A G A G ± + − ( )( ) . Reprint 2025-26 n n + + + + 1 1 may be the geometric mean between 31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? 32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. Example 14 If a, b, c, d and p are different real numbers such that (a 2 + b 2 + c 2 )p 2 – 2(ab + bc + cd) p + (b 2 + c2 + d2 ) ≤ 0, then show that a, b, c and d are in G.P. Solution Given that (a 2 + b 2 + c 2 ) p 2 – 2 (ab + bc + cd) p + (b 2 + c2 + d2 ) ≤ 0 ... (1) But L.H.S. which gives (ap – b) 2 + (bp – c) 2 + (cp – d) 2 ≥ 0 ... (2) Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), we have, (ap – b) 2 + (bp – c) 2 + (cp – d) 2 = 0 or ap – b = 0, bp – c = 0, cp – d = 0 This implies that b c d p a b c = = = Hence a, b, c and d are in G.P. = (a 2p 2 – 2abp + b2 ) + (b 2p 2 – 2bcp + c2 ) + (c 2p 2 – 2cdp + d 2 ), Miscellaneous Examples SEQUENCES AND SERIES 147 5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. 1. If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that 2. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. 3. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. 4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. f(1) = 3 and 1 ( ) 120 n x f x = ∑ = , find the value of n. Miscellaneous Exercise On Chapter 8 Reprint 2025-26 148 MATHEMATICS 6. If a bx a bx 7. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2R n = Sn . 8. If a, b, c, d are in G.P, prove that (a n + b n ), (b n + c n ), (c n + d n ) are in G.P. 9. If a and b are the roots of x 2 – 3x + p = 0 and c, d are roots of x 2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15. 10. The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that ( ) ( ) 2 2 2 2 a b m m – n : m – m – n : = + . 11. Find the sum of the following series up to n terms: (i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666+… 12. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms. 13. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him? 14. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? 15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed. c dx c dx x + − = + − = + − ( ) ,0 ≠ then show that a, b, c and d are in G.P. b cx b cx 16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years. 17. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. 18. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. Reprint 2025-26 ÆBy a sequence, we mean an arrangement of number in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, ....k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence. ÆLet a1 , a2 , a3 , ... be the sequence, then the sum expressed as a1 + a2 + a3 + ... is called series. A series is called finite series if it has got finite number of terms. ÆA sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called ÆThe geometric mean (G.M.) of any two positive numbers a and b is given by the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the n th term of G.P. is given by an = arn – 1 . The sum Sn of the first n terms of G.P. is given by ( – 1 1– ) ( ) S 1 1 1 – n n n a r a r = or , if r r – r ≠ ab i.e., the sequence a, G, b is G.P. Summary SEQUENCES AND SERIES 149 Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and geometric sequences. According to Boethius (510), arithmetic and geometric sequences were known to early Greek writers. Among the Indian mathematician, Aryabhatta (476) was the first to give the formula for the sum of squares and cubes of natural numbers in his famous work Aryabhatiyam, written around 499. He also gave the formula for finding the sum to n terms of an arithmetic sequence starting with p th term. Noted Indian mathematicians Brahmgupta Historical Note Reprint 2025-26 150 MATHEMATICS (598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares and cubes. Another specific type of sequence having important applications in mathematics, called Fibonacci sequence, was discovered by Italian mathematician Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification of series into specific forms. In 1671 James Gregory used the term infinite series in connection with infinite sequence. It was only through the rigorous development of algebraic and set theoretic tools that the concepts related to sequence and series could be formulated suitably. — v — Reprint 2025-26" class_11,9,Straight Lines,ncert_books/class_11/kemh1dd/kemh109.pdf,"9.1 Introduction We are familiar with two-dimensional coordinate geometry from earlier classes. Mainly, it is a combination of algebra and geometry. A systematic study of geometry by the use of algebra was first carried out by celebrated French philosopher and mathematician René Descartes, in his book ‘La Géométry, published in 1637. This book introduced the notion of the equation of a curve and related analytical methods into the study of geometry. The resulting combination of analysis and geometry is referred now as analytical geometry. In the earlier classes, we initiated the study of coordinate geometry, where we studied about coordinate axes, coordinate plane, plotting of points in a plane, distance between two points, section formulae, etc. All these concepts are the basics of coordinate geometry. Let us have a brief recall of coordinate geometry done in earlier classes. To recapitulate, the location of the points (6, – 4) and (3, 0) in the XY-plane is shown in Fig 9.1. We may note that the point (6, – 4) is at 6 units distance from the y-axis measured along the positive x-axis and at 4 units distance from the x-axis measured along the negative y-axis. Similarly, the point (3, 0) is at 3 units distance from the y-axis measured along the positive x-axis and has zero distance from the x-axis. We also studied there following important formulae: vG eometry, as a logical system, is a means and even the most powerful means to make children feel the strength of the human spirit that is of their own spirit. – H. FREUDENTHALv STRAIGHT LINES Chapter 9 René Descartes (1596 -1650) Reprint 2025-26 Fig 9.1 152 MATHEMATICS For example, distance between the points (6, – 4) and (3, 0) is For example, the coordinates of the point which divides the line segment joining A (1, –3) and B (–3, 9) internally, in the ratio 1: 3 are given by 1 ( 3) 3 1 0 1 3 . . x − + = = + III. In particular, if m = n, the coordinates of the mid-point of the line segment IV. Area of the triangle whose vertices are (x1, y1 ), (x2 , y2 ) and (x3 , y3 ) is II. The coordinates of a point dividing the line segment joining the points (x1, y1 ) I. Distance between the points P (x1, y1 ) and Q (x2 , y2 ) is and (x2 , y2 ) internally, in the ratio m: n are + + and 1.9 + 3. –3 ( ) = = 0. 1 + 3 y joining the points (x1, y1 ) and (x2 , y2 ) are + + 1( ) ( ) 2 3 2 3 1 3 ( ) 1 2 1 2 x x x y y y y y y − + − + − . ( ) 1 ( ) 2 2 PQ 2 2 1 = + x – x y – y ( ) ( ) 2 2 3 6 0 4 9 16 5 − + + = + = units. 2 , 2 21 21 xx yy . nm m x n x12 12 , . + + nm m y n y For example, the area of the triangle, whose vertices are (4, 4), (3, – 2) and (– 3, 16) is Remark If the area of the triangle ABC is zero, then three points A, B and C lie on a line, i.e., they are collinear. In the this Chapter, we shall continue the study of coordinate geometry to study properties of the simplest geometric figure – straight line. Despite its simplicity, the line is a vital concept of geometry and enters into our daily experiences in numerous interesting and useful ways. Main focus is on representing the line algebraically, for which slope is most essential. 9.2 Slope of a Line A line in a coordinate plane forms two angles with the x-axis, which are supplementary. 1 54 4( 2 16) 3(16 4) ( 3)(4 2) 27. 2 2 − − − + − + − + = = Reprint 2025-26 The angle (say) θ made by the line l with positive direction of x-axis and measured anti clockwise is called the inclination of the line. Obviously 0° ≤ θ ≤ 180° (Fig 9.2). We observe that lines parallel to x-axis, or coinciding with x-axis, have inclination of 0°. The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°. Definition 1 If θ is the inclination of a line l, then tan θ is called the slope or gradient of the line l. The slope of a line whose inclination is 90° is not defined. The slope of a line is denoted by m. Thus, m = tan θ, θ ≠ 90° It may be observed that the slope of x-axis is zero and slope of y-axis is not defined. 9.2.1 Slope of a line when coordinates of any two points on the line are given We know that a line is completely determined when we are given two points on it. Hence, we proceed to find the slope of a line in terms of the coordinates of two points on the line. Let P(x 1 , y 1 ) and Q(x 2 , y 2 ) be two points on non-vertical line l whose inclination is θ. Obviously, x 1 ≠ x 2 , otherwise the line will become perpendicular to x-axis and its slope will not be defined. The inclination of the line l may be acute or obtuse. Let us take these two cases. Draw perpendicular QR to x-axis and PM perpendicular to RQ as shown in Figs. 9.3 (i) and (ii). STRAIGHT LINES 153 Fig 9.2 Case 1 When angle θ is acute: In Fig 9.3 (i), ∠MPQ = θ. ... (1) Therefore, slope of line l = m = tan θ. But in ∆MPQ, we have 2 1 2 1 MQ tanθ . MP y y x x − = = − ... (2) Reprint 2025-26 Fig 9. 3 (i) 154 MATHEMATICS From equations (1) and (2), we have Case II When angle θ is obtuse: In Fig 9.3 (ii), we have ∠MPQ = 180° – θ. Therefore, θ = 180° – ∠MPQ. Now, slope of the line l m = tan θ = tan ( 180° – ∠MPQ) = – tan ∠MPQ Consequently, we see that in both the cases the slope m of the line through the points (x 1 , y 1 ) and (x 2 , y 2 ) is given by 2 1 2 1 y y m x x − = − . 9.2.2 Conditions for parallelism and perpendicularity of lines in terms of their slopes In a coordinate plane, suppose that non-vertical lines l1 and l2 have slopes m1 and m2 , respectively. Let their inclinations be α and β, respectively. If the line l 1 is parallel to l 2 (Fig 9.4), then their inclinations are equal, i.e., α = β, and hence, tan α = tan β Therefore m1 = m2 , i.e., their slopes are equal. Conversely, if the slope of two lines l 1 and l 2 is same, i.e., 2 1 . y y m x x − = − 2 1 = 2 1 1 2 MQ MP y y x x − − = − − = 2 1 2 1 y y . x x − − Fig 9. 3 (ii) Then tan α = tan β. By the property of tangent function (between 0° and 180°), α = β. Therefore, the lines are parallel. m1 = m2 . Reprint 2025-26 Fig 9. 4 are equal. If the lines l1 and l 2 are perpendicular (Fig 9.5), then β = α + 90°. Therefore, tan β = tan (α + 90°) i.e., m2 = 1 1 m − or m1 m2 = – 1 Conversely, if m1 m2 = – 1, i.e., tan α tan β = – 1. Then tan α = – cot β = tan (β + 90°) or tan (β – 90°) Therefore, α and β differ by 90°. Thus, lines l1 and l 2 are perpendicular to each other. Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other, i.e., m2 = 1 1 m − or, m1 m2 = – 1. Example 1 Find the slope of the lines: (a) Passing through the points (3, – 2) and (–1, 4), (b) Passing through the points (3, – 2) and (7, – 2), Hence, two non vertical lines l1 and l2 are parallel if and only if their slopes Let us consider the following example. = – cot α = 1 tanα − STRAIGHT LINES 155 Fig 9. 5 Solution (a) The slope of the line through (3, – 2) and (– 1, 4) is 4 ( 2) 6 3 1 3 4 2 m − − = = = − − − − . –2 – (–2) 0 = = = 0 7 – 3 4 m . (c) Passing through the points (3, – 2) and (3, 4), (d) Making inclination of 60° with the positive direction of x-axis. (b) The slope of the line through the points (3, – 2) and (7, – 2) is (c) The slope of the line through the points (3, – 2) and (3, 4) is Reprint 2025-26 156 MATHEMATICS 9.2.3 Angle between two lines When we think about more than one line in a plane, then we find that these lines are either intersecting or parallel. Here we will discuss the angle between two lines in terms of their slopes. Let L1 and L2 be two non-vertical lines with slopes m1 and m2 ,respectively. If α1 and α2 are the inclinations of lines L1 and L2 , respectively. Then vertically opposite angles such that sum of any two adjacent angles is 180°. Let θ and φ be the adjacent angles between the lines L1 and L2 (Fig 9.6). Then θ = α2 – α1 and α1 , α2 ≠ 90°. Therefore tan θ = tan (α2 – α1 ) 2 1 2 1 1 2 1 2 tan tan 1 tan tan 1 m m m m α α and φ = 180° – θ so that We know that when two lines intersect each other, they make two pairs of tan φ = tan (180° – θ ) = – tan θ = 2 1 1 2 – 1 m m m m − + , as 1 + m1m2 ≠ 0 (d) Here inclination of the line α = 60°. Therefore, slope of the line is m = tan 60° = 3 . 4 – (–2) 6 = = 3 – 3 0 m , which is not defined. m1 = 1 andαtan m2 = αtan 2 . α α − − = = + + (as 1 + m1m2 ≠ 0) Fig 9. 6 Now, there arise two cases: Reprint 2025-26 Case I If 2 1 1 1 2 m m– + m m is positive, then tan θ will be positive and tan φ will be negative, which means θ will be acute and φ will be obtuse. Case II If 2 1 1 1 2 m m– + m m is negative, then tan θ will be negative and tan φ will be positive, which means that θ will be obtuse and φ will be acute. Thus, the acute angle (say θ) between lines L1 and L2 with slopes m1 and m2 , respectively, is given by The obtuse angle (say φ) can be found by using φ =1800 – θ. Example 2 If the angle between two lines is π 4 and slope of one of the lines is 1 2 , find the slope of the other line. Solution We know that the acute angle θ between two lines with slopes m1 and m2 is given by 2 1 1 2 tan θ 1 m m m m − = + ... (1) Let m1 = 2 1 , m2 = m and θ = π 4 . 2 1 1 2 1 2 tan θ , as 1 0 1 m m m m m m − = + ≠ + ... (1) STRAIGHT LINES 157 Now, putting these values in (1), we get which gives 1 1 π 2 2 tan or 1 4 1 1 1 1 2 2 1 1 2 2 1 or 1 1 1 1 1 2 2 m m – . m m − − = = + + − − = = + + m m , m m Reprint 2025-26 158 MATHEMATICS Hence, slope of the other line is 3 or 1 3 − . Fig 9.7 explains the reason of two answers. Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. Solution Slope of the line through the points (– 2, 6) and (4, 8) is ( ) 1 8 6 2 1 4 2 6 3 m − = = = − − Slope of the line through the points (8, 12) and (x, 24) is 2 24 12 12 8 8 m x x − = = − − Since two lines are perpendicular, m1 m2 = –1, which gives 1 Therefore 3 or 3 m m . = = − Fig 9.7 1 12 1 or = 4 3 8 x x × = − − . 1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), 2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle. (5, – 5) and (– 4, –2). Also, find its area. EXERCISE 9.1 Reprint 2025-26 10. The slope of a line is double of the slope of another line. If tangent of the angle 11. A line passes through (x1 , y1 ) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1 ). 9.3 Various Forms of the Equation of a Line We know that every line in a plane contains infinitely many points on it. This relationship between line and points leads us to find the solution of the following problem: How can we say that a given point lies on the given line? Its answer may be that for a given line we should have a definite condition on the points lying on the line. Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the equation of L, we wish to construct a statement or condition for the point P that is true, when P is on L, otherwise false. Of course the statement is merely an algebraic equation involving the variables x and y. Now, we will discuss the equation of a line under different conditions. 3. Find the distance between P (x1 , y1 ) and Q (x2 , y2 ) when : (i) PQ is parallel to the 4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). 5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0). 6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle. 7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise. 8. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram. 9. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2). y-axis, (ii) PQ is parallel to the x-axis. between them is 3 1 , find the slopes of the lines. STRAIGHT LINES 159 9.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the xaxis then ordinate of every point lying on the line is either a or – a [Fig 9.8 (a)]. Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis. Similarly, the equation of a vertical line at a distance b from the y-axis is either x = b or x = – b [Fig 9.8(b)]. Reprint 2025-26 160 MATHEMATICS Example 4 Find the equations of the lines parallel to axes and passing through (– 2, 3). Solution Position of the lines is shown in the Fig 9.9. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through (– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3) is x = – 2. 9.3.2 Point-slope form Suppose that P0 (x0 , y0 ) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L (Fig 9.10). Then, by the definition, the slope of L is given by Fig 9.8 Since the point P0 (x0 , y0 ) along with all points (x, y) on L satisfies (1) and no other point in the plane satisfies (1). Equation (1) is indeed the equation for the given line L. y y m( ) x x x x y y m 0 0 0 0 ,i.e., −=− − − = ...(1) Reprint 2025-26 Fig 9.10 Fig 9.9 if and only if, its coordinates satisfy the equation y – y0 = m (x – x0 ) Example 5 Find the equation of the line through (– 2, 3) with slope – 4. Solution Here m = – 4 and given point (x0 , y0 ) is (– 2, 3). By slope-intercept form formula (1) above, equation of the given line is y – 3 = – 4 (x + 2) or 4x + y + 5 = 0, which is the required equation. 9.3.3 Two-point form Let the line L passes through two given points P1 (x1 , y1 ) and P2 (x2 , y2 ). Let P (x, y) be a general point on L (Fig 9.11). The three points P1 , P2 and P are collinear, therefore, we have slope of P1 P = slope of P1 P2 i.e., 1 2 1 2 1 1 1 1 2 1 2 1 or y y y y y y , y y ( x ). x x x x x x x − − − = − = − − − − Thus, equation of the line passing through the points (x1 , y1 ) and (x2 , y2 ) is given by Thus, the point (x, y) lies on the line with slope m through the fixed point (x0 , y0 ), Fig 9.11 STRAIGHT LINES 161 )( 1 12 12 1 xx xx yy y y − − − =− ... (2) Example 6 Write the equation of the line through the points (1, –1) and (3, 5). Solution Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form (2) above for the equation of the line, we have ( ) ( ) ( ) 5 – –1 – –1 = – 1 3 – 1 y x or –3x + y + 4 = 0, which is the required equation. 9.3.4 Slope-intercept form Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines. Reprint 2025-26 162 MATHEMATICS Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin (Fig 9.12). The distance c is called the yintercept of the line L. Obviously, coordinates of the point where the line meet the y-axis are (0, c). Thus, L has slope m and passes through a fixed point (0, c). Therefore, by point-slope form, the equation of L is y c m( x ) y mx c − = − = + 0 or Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx +c ...(3) Note that the value of c will be positive or negative according as the intercept is made on the positive or negative side of the y-axis, respectively. Case II Suppose line L with slope m makes x-intercept d. Then equation of L is y = m(x – d) ... (4) Students may derive this equation themselves by the same method as in Case I. Example 7 Write the equation of the lines for which tan θ = 2 1 , where θ is the inclination of the line and (i) y-intercept is 3 2 – (ii) x-intercept is 4. Solution (i) Here, slope of the line is m = tan θ = 2 1 and y - intercept c = – 2 3 . Fig 9.12 Therefore, by slope-intercept form (3) above, the equation of the line is which is the required equation. (ii) Here, we have m = tan θ = 2 1 and d = 4. Therefore, by slope-intercept form (4) above, the equation of the line is which is the required equation. 042or)4( 2 1 xyxy =+−−= , 032or 2 3 2 1 −= xyxy =+− , Reprint 2025-26 9.3.5 Intercept - form Suppose a line L makes x-intercept a and y-intercept b on the axes. Obviously L meets x-axis at the point (a, 0) and y-axis at the point (0, b) (Fig .9.13). By two-point form of the equation of the line, we have i.e., =+ 1 b y Thus, equation of the line making intercepts a and b on x-and y-axis, respectively, is Example 8 Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively. Solution Here a = –3 and b = 2. By intercept form (5) above, equation of the line is 1 or 2 3 6 0 3 2 x y + = − + = x y − . Any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously is called general linear equation or general equation of a line. 0 0 ( ) or 0 b y x a ay bx ab a − − = − = − + − , a x . a x ... (5) =+ 1 b y Fig 9.13 STRAIGHT LINES 163 In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: 1. Write the equations for the x-and y-axes. 2. Passing through the point (– 4, 3) with slope 2 1 . 3. Passing through (0, 0) with slope m. 4. Passing through ( 32,2 )and inclined with the x-axis at an angle of 75o . 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2. 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis. EXERCISE 9.2 Reprint 2025-26 164 MATHEMATICS 7. Passing through the points (–1, 1) and (2, – 4). 8. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R. 9. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 10. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line. 11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3). 12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9. 13. Find equation of the line through the point (0, 2) making an angle 2π 3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin. 14. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line. 15. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C. 16. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre? 17. P (a, b) is the mid-point of a line segment between axes. Show that equation 18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line. 19. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear. 9.4 Distance of a Point From a Line The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1 , y1 ) is d. Draw a perpendicular PM from the point P to the line L (Fig 9.14). If the of the line is =+ 2 b y a x . Reprint 2025-26 line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of the points are Q C 0 A , − and R C 0 B , − . Thus, the area of the triangle PQR is given by Also, area 1 ( ) 1 1 1 C C C (∆PQR) 0 0 0 2 B A B x y y = + + − − − + − area 1 ( PQR) PM.QR 2 ∆ = , which gives 2 area (∆PQR) PM = QR ... (1) Fig 9.14 STRAIGHT LINES 165 or 1 1 C 2 area (∆PQR) A B C and AB = + + . , x y Substituting the values of area (∆PQR) and QR in (1), we get 1 1 2 2 A B C PM A B x + + y = + ( ) 2 2 C C C 2 2 QR 0 0 A B A B AB = + = + + − 2 1 1 1 C C C 2 B A AB = + + x y Reprint 2025-26 166 MATHEMATICS or 1 1 2 2 A B C Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1 , y1 ) is given by 1 1 2 2 A B C 9.4.1 Distance between two parallel lines We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form y = mx + c1 ... (1) and y = mx + c 2 ... (2) Line (1) will intersect x-axis at the point A 1 0 c , m − as shown in Fig 9.15. A to line (2). Therefore, distance between the lines (1) and (2) is Distance between two lines is equal to the length of the perpendicular from point A B x y d + + = + . A B x y d + + = + . Fig 9.15 Thus, the distance d between two parallel lines 1 y mx c = + and 2 y mx c = + is given by then above formula will take the form 1 2 2 2 C C Students can derive it themselves. If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0, ( ) ( ) 1 2 1 2 2 2 or = 1 1 − − + − − + + . c m c m c c d m m 2 = 1 c c d m − Reprint 2025-26 1 2 A B d − = + + . Example 9 Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0. Solution Given line is 3x – 4y –26 = 0 ... (1) Comparing (1) with general equation of line Ax + By + C = 0, we get A = 3, B = – 4 and C = – 26. Given point is (x1 , y1 ) = (3, –5). The distance of the given point from given line is Example 10 Find the distance between the parallel lines 3x – 4y +7 = 0 and 3x – 4y + 5 = 0 Solution Here A = 3, B = –4, C1 = 7 and C2 = 5. Therefore, the required distance is 1. Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts. (i) x + 7y = 0, (ii) 6x + 3y – 5 = 0, (iii) y = 0. 2. Reduce the following equations into intercept form and find their intercepts on the axes. (i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0. 3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2). ( ) 1 1 2 2 2 2 A B C 3 3 4 5 26 3 . A B 5 3 4 x y . – – – d – + + + = = = + + ( )2 2 7 5 2 . 5 3 4 – d – = = + EXERCISE 9.3 ( )( ) STRAIGHT LINES 167 4. Find the points on the x-axis, whose distances from the line 1 3 4 x y + = are 4 units. 5. Find the distance between parallel lines (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0. 6. Find equation of the line parallel to the line 3 4 2 0 x y − + = and passing through the point (–2, 3). 7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. 8. Find angles between the lines yxyx =+=+ .13and13 9. The line through the points (h, 3) and (4, 1) intersects the line 7 9 19 0 x y . − − = at right angle. Find the value of h. Reprint 2025-26 168 MATHEMATICS 10. Prove that the line through the point (x1 , y1 ) and parallel to the line Ax + By + C = 0 is A (x –x1 ) + B (y – y1 ) = 0. 11. Two lines passing through the point (2, 3) intersects each other at an angle of 60o . If slope of one line is 2, find equation of the other line. 12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2). 13. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0. 14. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c. 15. If p and q are the lengths of perpendiculars from the origin to the lines − = kyx θ2cosθsinθcos and x sec θ + y cosec θ = k, respectively, prove that p 2 + 4q 2 = k 2 . 16. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A. 17. If p is the length of perpendicular from the origin to the line whose intercepts on Example 11 If the lines 2 3 0 5 3 0 x y , x ky + − = + − = and 3 2 0 x y − − = are concurrent, find the value of k. the axes are a and b, then show that . 111 2 22 p ba += Miscellaneous Examples Solution Three lines are said to be concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the third line. Here given lines are 2x + y – 3 = 0 ... (1) 5x + ky – 3 = 0 ... (2) 3x – y – 2 = 0 ... (3) Solving (1) and (3) by cross-multiplication method, we get 1 = = or = 1, = 1 –2 – 3 –9 + 4 –2 – 3 x y x y . Therefore, the point of intersection of two lines is (1, 1). Since above three lines are concurrent, the point (1, 1) will satisfy equation (2) so that 5.1 + k .1 – 3 = 0 or k = – 2. Reprint 2025-26 Example 12 Find the distance of the line 4x – y = 0 from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis. Solution Given line is 4x – y = 0 ... (1) In order to find the distance of the line (1) from the point P (4, 1) along another line, we have to find the point of intersection of both the lines. For this purpose, we will first find the equation of the second line (Fig 9.16). Slope of second line is tan 135° = –1. Equation of the line with slope – 1 through the point P (4, 1) is Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2) = the distance between the points P (4, 1) and Q (1, 4). Example 13 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x – 3y + 4 = 0. y – 1 = – 1 (x – 4) or x + y – 5 = 0 ... (2) = ( ) ( ) 2 2 1 4 4 1 3 2 units − + − = . (1, 4) Fig 9.16 STRAIGHT LINES 169 Solution Let Q (h, k) is the image of the point P (1, 2) in the line x – 3y + 4 = 0 ... (1) Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 9.17). Reprint 2025-26 Fig 9.17 170 MATHEMATICS Hence Slope of line PQ = 1 Slope of line 3 4 0 x y − − + = , so that 2 1 or 3 5 1 1 3 and the mid-point of PQ, i.e., point ++ 2 2 , 2 1 kh will satisfy the equation (1) so that Solving (2) and (3), we get h = 5 6 and k = 5 7 . Hence, the image of the point (1, 2) in the line (1) is 6 7 5 5 , . Example 14 Show that the area of the triangle formed by the lines y = m1 x + c1 , y = m2 x + c2 and x = 0 is ( ) 2 1 2 Solution Given lines are y = m1 x + c1 ... (1) y = m2 x + c 2 ... (2) x = 0 ... (3) We know that line y = mx + c meets the line x = 0 (y-axis) at the point (0, c). Therefore, two vertices of the triangle formed by lines (1) to (3) are P (0, c 1 ) and Q (0, c 2 ) (Fig 9.18). Third vertex can be obtained by solving equations (1) and (2). Solving (1) and (2), we get k h k h − − = + = − ... (2) 33or04 2 2 3 2 1 −=−=+ + − + kh kh ... (3) 2 1 2 cc m m – − . Reprint 2025-26 Fig 9.18 Therefore, third vertex of the triangle is R ( ) Now, the area of the triangle is Example 15 A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is bisected at the point (1, 5). Obtain its equation. Solution Given lines are 5x – y + 4 = 0 ... (1) 3x + 4y – 4 = 0 ... (2) Let the required line intersects the lines (1) and (2) at the points, (α1 , β1 ) and (α2 , β2 ), respectively (Fig 9.19). Therefore 5α1 – β1 + 4 = 0 and 3 α2 + 4 β2 – 4 = 0 or β1 = 5α1 + 4 and 2 2 4 – 3α β 4 = . ( ) ( ) 2 2 1 1 2 2 1 2 1 1 2 2 1 2 2 1 1 1 2 1 2 1 2 1 2 1 0 0 2 2 c c m c m c c c m c m c c c c c m m m m m m m m − − − = − + − + − = − − − − 1 2 1 2 and c c m c m c x y m m m m − − = = − − ( ) ( ) ( ) 2 1 1 2 2 1 1 2 1 2 c c m c m c , m m m m − − − − . ( ) ( ) 2 1 1 2 2 1 ( ) ( ) STRAIGHT LINES 171 − We are given that the mid point of the segment of the required line between (α1 , β1 ) and (α2 , β2 ) is (1, 5). Therefore or or α1 + α2 = 2 and 20α1 – 3α2 = 20 ... (3) Solving equations in (3) for α1 and α2 , we get 4 – 3α 5α + 4 + 4 α + = 2 and α = 5, 2 α1 + α2 2 β + β 1 = 1 and = 5 2 2 , 2 1 1 2 Reprint 2025-26 Fig 9.19 172 MATHEMATICS Equation of the required line passing through (1, 5) and (α1 , β1 ) is or 107x – 3y – 92 = 0, which is the equation of required line. Example 16 Show that the path of a moving point such that its distances from two lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line. Solution Given lines are 3x – 2y = 5 … (1) and 3x + 2y = 5 … (2) Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore which gives 3h – 2k – 5 = 3h + 2k – 5 or – (3h – 2k – 5) = 3h + 2k – 5. Solving these two relations we get k = 0 or h = 3 5 . Thus, the point (h, k) satisfies the 49 523 −+=−− + −+ = + −− khkh khkh , 1 26 α = 23 and 2 20 α = 23 and hence, 23 222 4 23 26 β .5 1 =+= . )1( α 1 β 5 5 1 1 − − − y =− x or 523523or 49 523 y x − − = − − 222 5 23 5 ( 1) 26 1 23 equations y = 0 or x = 3 5 , which represent straight lines. Hence, path of the point equidistant from the lines (1) and (2) is a straight line. 1. Find the values of k for which the line (k–3) x – (4 – k 2 ) y + k 2 –7k + 6 = 0 is (a) Parallel to the x-axis, (b) Parallel to the y-axis, (c) Passing through the origin. 2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively. Miscellaneous Exercise on Chapter 9 Reprint 2025-26 point, where it meets the y-axis. 7. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0. 8. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point. 9. If three lines whose equations are y = m1 x + c1 , y = m2 x + c2 and y = m3 x + c3 are concurrent, then show that m1 (c2 – c3 ) + m2 (c3 – c1 ) + m3 (c1 – c2 ) = 0. 10. Find the equation of the lines through the point (3, 2) which make an angle of 45o with the line x – 2y = 3. 11. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes. 12. Show that the equation of the line passing through the origin and making an angle 13. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? 14. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0. 15. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. 16. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes. 17. Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror. 18. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m. 19. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y +7 = 0 is always 10. Show that P must move on a line. 3. What are the points on the y-axis whose distance from the line 1 3 4 x y + = is 4 units. 4. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) and (cos φ, sin φ). 5. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0. 6. Find the equation of a line drawn perpendicular to the line 1 4 6 =+ yx through the θ with the line y mx c is y x m m = + = ± tan ‚ ∓ tan ‚1 . STRAIGHT LINES 173 Reprint 2025-26 174 MATHEMATICS 20. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0. 21. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A. 22. Prove that the product of the lengths of the perpendiculars drawn from the 23. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow. Summary ÆSlope (m) of a non-vertical line passing through the points (x 1 , y1 ) and (x 2 , y2 ) is given by 2 1 1 2 1 2 2 1 1 2 y y y y m , . x x x x x x − − = = ≠ − − Æ If a line makes an angle á with the positive direction of x-axis, then the slope of the line is given by m = tan α, α ≠ 90°. ÆSlope of horizontal line is zero and slope of vertical line is undefined. Æ An acute angle (say θ) between lines L1 and L2 with slopes m1 and m2 is given by 2 1 1 2 1 2 tanθ 1 0 1 m – m , m m m m = + ≠ + . ÆTwo lines are parallel if and only if their slopes are equal. ÆTwo lines are perpendicular if and only if product of their slopes is –1. ÆThree points A, B and C are collinear, if and only if slope of AB = slope of BC. ÆEquation of the horizontal line having distance a from the x-axis is either y = a or y = – a. ÆEquation of the vertical line having distance b from the y-axis is either x = b or x = – b. ÆThe point (x, y) lies on the line with slope m and through the fixed point (xo , yo ), if and only if its coordinates satisfy the equation y – yo = m (x – xo ). Æ Equation of the line passing through the points (x1 , y1 ) and (x2 , y2 ) is given by points ( ) 2 2 a b , − 0 and ( ) 2 2 − − a b ,0 to the line 2 cosθ sin θ 1is x y b a b + = . ).( 1 12 12 1 x x xx yy y y − − − =− Reprint 2025-26 ÆThe point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c. ÆIf a line with slope m makes x-intercept d. Then equation of the line is y = m (x – d). ÆEquation of a line making intercepts a and b on the x-and y-axis, ÆAny equation of the form Ax + By + C = 0, with A and B are not zero, simultaneously, is called the general linear equation or general equation of a line. ÆThe perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1 , y1 ) ÆDistance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, respectively, is =+ 1 b y a x . is given by 1 1 2 2 A B C is given by 1 2 2 2 C C A B d − = + . A B x y d + + = + . STRAIGHT LINES 175 Reprint 2025-26" class_11,10,Conic Sections,ncert_books/class_11/kemh1dd/kemh110.pdf,"176 MATHEMATICS 10.1 Introduction In the preceding Chapter 10, we have studied various forms of the equations of a line. In this Chapter, we shall study about some other curves, viz., circles, ellipses, parabolas and hyperbolas. The names parabola and hyperbola are given by Apollonius. These curves are in fact, known as conic sections or more commonly conics because they can be obtained as intersections of a plane with a double napped right circular cone. These curves have a very wide range of applications in fields such as planetary motion, design of telescopes and antennas, reflectors in flashlights and automobile headlights, etc. Now, in the subsequent sections we will see how the intersection of a plane with a double napped right circular cone results in different types of curves. vLet the relation of knowledge to real life be very visible to your pupils and let them understand how by knowledge the world could be transformed. – BERTRAND RUSSELL v CONIC SECTIONS Chapter 10 10.2 Sections of a Cone Let l be a fixed vertical line and m be another line intersecting it at a fixed point V and inclined to it at an angle α (Fig10.1). Suppose we rotate the line m around the line l in such a way that the angle α remains constant. Then the surface generated is a double-napped right circular hollow cone herein after referred as Reprint 2025-26 Apollonius (262 B.C. -190 B.C.) Fig 10. 1 cone and extending indefinitely far in both directions (Fig10.2). The point V is called the vertex; the line l is the axis of the cone. The rotating line m is called a generator of the cone. The vertex separates the cone into two parts called nappes. If we take the intersection of a plane with a cone, the section so obtained is called a conic section. Thus, conic sections are the curves obtained by intersecting a right circular cone by a plane. We obtain different kinds of conic sections depending on the position of the intersecting plane with respect to the cone and by the angle made by it with the vertical axis of the cone. Let β be the angle made by the intersecting plane with the vertical axis of the cone (Fig10.3). The intersection of the plane with the cone can take place either at the vertex of the cone or at any other part of the nappe either below or above the vertex. Fig 10. 2 Fig 10. 3 CONIC SECTIONS 177 10.2.1 Circle, ellipse, parabola and hyperbola When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations: (a) When β = 90o , the section is a circle (Fig10.4). (b) When α < β < 90o , the section is an ellipse (Fig10.5). (c) When β = α; the section is a parabola (Fig10.6). (In each of the above three situations, the plane cuts entirely across one nappe of the cone). (d) When 0 ≤ β < α; the plane cuts through both the nappes and the curves of intersection is a hyperbola (Fig10.7). Reprint 2025-26 178 MATHEMATICS Fig 10. 4 Fig 10. 5 10.2.2 Degenerated conic sections When the plane cuts at the vertex of the cone, we have the following different cases: (a) When α < β ≤ 90o , then the section is a point (Fig10.8). (b) When β = α, the plane contains a generator of the cone and the section is a straight line (Fig10.9). It is the degenerated case of a parabola. (c) When 0 ≤ β < α, the section is a pair of intersecting straight lines (Fig10.10). It is the degenerated case of a hyperbola. Fig 10. 6 Fig 10. 7 Reprint 2025-26 sections in standard form by defining them based on geometric properties. In the following sections, we shall obtain the equations of each of these conic Fig 10. 8 Fig 10. 9 CONIC SECTIONS 179 Fig 10. 10 10.3 Circle Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. to a point on the circle is called the radius of the circle (Fig 10.11). The fixed point is called the centre of the circle and the distance from the centre Reprint 2025-26 180 MATHEMATICS However, we derive below the equation of the circle with a given centre and radius (Fig 10.12). Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle (Fig10.12). Then, by the definition, | CP | = r . By the distance formula, we have i.e. (x – h) 2 + (y – k) 2 = r 2 This is the required equation of the circle with centre at (h,k) and radius r . Example 1 Find an equation of the circle with centre at (0,0) and radius r. Solution Here h = k = 0. Therefore, the equation of the circle is x 2 + y 2 = r 2 . The equation of the circle is simplest if the centre of the circle is at the origin. Fig 10. 11 Fig 10. 12 2 2 ( ) ( ) x – h y – k r + = Example 2 Find the equation of the circle with centre (–3, 2) and radius 4. Solution Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is (x + 3)2 + (y –2)2 = 16 Example 3 Find the centre and the radius of the circle x 2 + y2 + 8x + 10y – 8 = 0 Solution The given equation is (x 2 + 8x) + (y 2 + 10y) = 8 Now, completing the squares within the parenthesis, we get (x 2 + 8x + 16) + (y 2 + 10y + 25) = 8 + 16 + 25 i.e. (x + 4)2 + (y + 5)2 = 49 i.e. {x – (– 4)}2 + {y – (–5)}2 = 72 Therefore, the given circle has centre at (– 4, –5) and radius 7. Reprint 2025-26 Example 4 Find the equation of the circle which passes through the points (2, – 2), and (3,4) and whose centre lies on the line x + y = 2. Solution Let the equation of the circle be (x – h) 2 + (y – k) 2 = r 2 . Since the circle passes through (2, – 2) and (3,4), we have (2 – h) 2 + (–2 – k) 2 = r 2 ... (1) and (3 – h) 2 + (4 – k) 2 = r 2 ... (2) Also since the centre lies on the line x + y = 2, we have h + k = 2 ... (3) Solving the equations (1), (2) and (3), we get h = 0.7, k = 1.3 and r 2 = 12.58 Hence, the equation of the required circle is (x – 0.7)2 + (y – 1.3)2 = 12.58. In each of the following Exercises 1 to 5, find the equation of the circle with 5. centre (–a, –b) and radius 22 − ba . In each of the following Exercises 6 to 9, find the centre and radius of the circles. 1. centre (0,2) and radius 2 2. centre (–2,3) and radius 4 3. centre ( 4 1 , 2 1 ) and radius 12 1 4. centre (1,1) and radius 2 6. (x + 5)2 + (y – 3)2 = 36 7. x 2 + y 2 – 4x – 8y – 45 = 0 EXERCISE 10.1 CONIC SECTIONS 181 10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16. 11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0. 12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3). 13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes. 14. Find the equation of a circle with centre (2,2) and passes through the point (4,5). 15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25? 8. x 2 + y 2 – 8x + 10y – 12 = 0 9. 2x 2 + 2y 2 – x = 0 Reprint 2025-26 182 MATHEMATICS 10.4 Parabola Definition 2 A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. The fixed line is called the directrix of the parabola and the fixed point F is called the focus (Fig 10.13). (‘Para’ means ‘for’ and ‘bola’ means ‘throwing’, i.e., the shape described when you throw a ball in the air). ANote If the fixed point lies on the fixed line, then the set of points in the plane, which are equidistant from the fixed point and the fixed line is the straight line through the fixed point and perpendicular to the fixed line. We call this straight line as degenerate case of the parabola. A line through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola (Fig10.14). 10.4.1 Standard equations of parabola The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola are shown below in Fig10.15 (a) to (d). Fig 10. 13 Fig 10.14 Reprint 2025-26 focus at (a, 0) a > 0; and directricx x = – a as below: Let F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid-point O is on the parabola and is called the vertex of the parabola. Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in Fig10.16. Let P(x, y) be any point on the parabola such that PF = PB, ... (1) where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance formula, we have We will derive the equation for the parabola shown above in Fig 10.15 (a) with Fig 10.15 (a) to (d) CONIC SECTIONS 183 PF = 2 2 ( ) x – a y + and PB = 2 ( ) x a + Since PF = PB, we have 2 2 2 ( ) ( ) x – a y x a + = + i.e. (x – a) 2 + y 2 = (x + a) 2 or x 2 – 2ax + a 2 + y 2 = x 2 + 2ax + a 2 or y 2 = 4ax ( a > 0). Reprint 2025-26 Fig 10.16 184 MATHEMATICS Hence, any point on the parabola satisfies y 2 = 4ax. ... (2) and so P(x,y) lies on the parabola. vertex at the origin, focus at (a,0) and directrix x = – a is y 2 = 4ax. Discussion In equation (2), since a > 0, x can assume any positive value or zero but no negative value and the curve extends indefinitely far into the first and the fourth quadrants. The axis of the parabola is the positive x-axis. Similarly, we can derive the equations of the parabolas in: These four equations are known as standard equations of parabolas. ANote The standard equations of parabolas have focus on one of the coordinate axis; vertex at the origin and thereby the directrix is parallel to the other coordinate axis. However, the study of the equations of parabolas with focus at any point and any line as directrix is beyond the scope here. Conversely, let P(x, y) satisfy the equation (2) Thus, from (2) and (3) we have proved that the equation to the parabola with Fig 11.15 (b) as y 2 = – 4ax, Fig 11.15 (c) as x 2 = 4ay, Fig 11.15 (d) as x 2 = – 4ay, PF = 2 2 ( ) x – a y + = 2 ( ) 4 x – a ax + = 2 ( ) x a + = PB ... (3) observations: From the standard equations of the parabolas, Fig10.15, we have the following 1. Parabola is symmetric with respect to the axis of the parabola.If the equation has a y 2 term, then the axis of symmetry is along the x-axis and if the equation has an x 2 term, then the axis of symmetry is along the y-axis. 2. When the axis of symmetry is along the x-axis the parabola opens to the (a) right if the coefficient of x is positive, (b) left if the coefficient of x is negative. 3. When the axis of symmetry is along the y-axis the parabola opens (c) upwards if the coefficient of y is positive. (d) downwards if the coefficient of y is negative. Reprint 2025-26 10.4.2 Latus rectum Definition 3 Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola (Fig10.17). To find the Length of the latus rectum of the parabola y 2 = 4ax (Fig 10.18). By the definition of the parabola, AF = AC. But AC = FM = 2a Hence AF = 2a. And since the parabola is symmetric with respect to x-axis AF = FB and so AB = Length of the latus rectum = 4a. Fig 10.17 Fig 10.18 CONIC SECTIONS 185 Example 5 Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y 2 = 8x. Solution The given equation involves y 2 , so the axis of symmetry is along the x-axis. The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y 2 = 4ax, we find that a = 2. Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = – 2 (Fig 10.19). Length of the latus rectum is 4a = 4 × 2 = 8. Reprint 2025-26 Fig 10.19 186 MATHEMATICS Example 6 Find the equation of the parabola with focus (2,0) and directrix x = – 2. Solution Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either y 2 = 4ax or y 2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola is to be of the form y 2 = 4ax with a = 2. Hence the required equation is y2 = 4(2)x = 8x Example 7 Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2). Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x 2 = 4ay. thus, we have x 2 = 4(2)y, i.e., x2 = 8y. Example 8 Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3). Solution Since the parabola is symmetric about y-axis and has its vertex at the origin, the equation is of the form x 2 = 4ay or x 2 = – 4ay, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through (2,–3) which lies in the fourth quadrant, it must open downwards. Thus the equation is of the form x 2 = – 4ay. Since the parabola passes through ( 2,–3), we have 2 2 = – 4a (–3), i.e., a = 1 3 Therefore, the equation of the parabola is In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: 1. y 2 = 12x 2. x 2 = 6y 3. y 2 = – 8x 4. x 2 = – 16y 5. y 2 = 10x 6. x 2 = – 9y x 2 = 1 4 3 − y, i.e., 3x 2 = – 4y. EXERCISE 10.2 Reprint 2025-26 12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis. 10. 5 Ellipse Definition 4 An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse (Fig10.20). The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse(Fig 10.21). 11. Vertex (0,0) passing through (2,3) and axis is along x-axis. 7. Focus (6,0); directrix x = – 6 8. Focus (0,–3); directrix y = 3 9. Vertex (0,0); focus (3,0) 10. Vertex (0,0); focus (–2,0) ANote The constant which is the sum of the distances of a point on the ellipse from the two fixed points is always greater than the distance between the two fixed points. Fig 10.20 CONIC SECTIONS 187 We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b (Fig10.22). Fig 10.21 Fig 10.22 Reprint 2025-26 188 MATHEMATICS 10.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse (Fig 10.23). Take a point P at one end of the major axis. Sum of the distances of the point P to the foci is F1 P + F2 P = F1O + OP + F2 P (Since, F1 P = F1O + OP) = c + a + a – c = 2a Take a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is F1Q + F2Q = 22 22 +++ cbcb = 22 2 + cb Since both P and Q lies on the ellipse. By the definition of ellipse, we have or a 2 = b 2 + c 2 , i.e., c = 22 − ba . 10.5.2 Eccentricity Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is 2 22 + cb = 2a, i.e., a = 22 + cb Fig 10.23 denoted by e) i.e., c e a = . Then since the focus is at a distance of c from the centre, in terms of the eccentricity the focus is at a distance of ae from the centre. 10.5.3 Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two such possible orientations are shown in Fig 10.24. We will derive the equation for the ellipse shown above in Fig 10.24 (a) with foci on the x-axis. Reprint 2025-26 be the origin and the line from O through F2 be the positive x-axis and that through F1 as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) (Fig 10.25). Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a so given PF1 + PF2 = 2a. ... (1) Using the distance formula, we have Let F1 and F2 be the foci and O be the mid-point of the line segment F1 F2 . Let O Fig 10.24 (a) CONIC SECTIONS 189 22 22 )( )( +−+++ ycxycx = 2a i.e., 22 )( ++ ycx = 2a – 22 )( +− ycx Squaring both sides, we get (x + c) 2 + y 2 = 4a 2 – 4a 22 22 )()( +−++− ycxycx Fig 10.25 Reprint 2025-26 2 2 1 x y a b + = 2 2 190 MATHEMATICS which on simplification gives Squaring again and simplifying, we get i.e., 2 2 Hence any point on the ellipse satisfies Conversely, let P (x, y) satisfy the equation (2) with 0 < c < a. Then Therefore, PF1 = 2 2 ( ) x c y + + b y a x + = 1. ... (2) 2 2 ca y a x − + = 1 b y a x + = 1 (Since c 2 = a 2 – b 2 ) 2 2 2 2 x a c aycx −=+− 22 )( y 2 = b 2 2 2 = 22 2 − 2 2 1 a x − ++ 2 22 22 )( a xa bcx Similarly PF2 = c a x a − Hence PF1 + PF2 = 2 c c a x a – x a a a + + = ... (3) = 2 2 2 2 2 2 ( ) ( ) a x x c a c a − + + − (since b 2 = a 2 – c 2 ) = 2 cx c a a x a a + = + Reprint 2025-26 So, any point that satisfies 2 2 P(x, y) lies on the ellipse. Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is Discussion From the equation of the ellipse obtained above, it follows that for every point P (x, y) on the ellipse, we have Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines. Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines. These two equations are known as standard equations of the ellipses. ANote The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes. However, the study of the ellipses with centre at any other point, and any line through the centre as major and the minor axes passing through the centre and perpendicular to major axis are beyond the scope here. Similarly, we can derive the equation of the ellipse in Fig 10.24 (b) as 2 2 2 2 1 b y a x −= ≤ 1, i.e., x 2 ≤ a 2 , so – a ≤ x ≤ a. b y a x + = 1, satisfies the geometric condition and so 2 2 2 2 2 2 x y a b + = 1. 2 2 CONIC SECTIONS 191 2 2 1 x y b a + = . observations: (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse. finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x 2 has the larger denominator and it is along the y-axis if the coefficient of y 2 has the larger denominator. From the standard equations of the ellipses (Fig10.24), we have the following 1. Ellipse is symmetric with respect to both the coordinate axes since if 2. The foci always lie on the major axis. The major axis can be determined by Reprint 2025-26 192 MATHEMATICS 10.5.4 Latus rectum Definition 6 Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse (Fig 10.26). To find the length of the latus rectum of the ellipse 1 x y a b 2 2 Let the length of AF2 be l. Then the coordinates of A are (c, l ),i.e., (ae, l ) Since A lies on the ellipse 2 2 But 2 2 2 2 2 2 2 2 1 c a – b b e – a a a = = = Therefore l 2 = 4 2 2 + = ⇒ l2 = b 2 (1 – e 2 ) 2 2 ( ) 1 ae l a b + = 2 2 1 x y a b + = , we have 2 2 2 b a , i.e., 2 b l a = Fig 10. 26 Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 2 2b a . Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse Solution Since denominator of 2 2 2 1 25 9 x y + = 25 x is larger than the denominator of 2 Reprint 2025-26 9 y , the major axis is along the x-axis. Comparing the given equation with 2 2 Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and (5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the eccentricity is 4 5 and latus rectum is 2 2 18 5 b a = . Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x 2 + 4y 2 = 36. Solution The given equation of the ellipse can be written in standard form as Since the denominator of 2 along the y-axis. Comparing the given equation with the standard equation Also c = 2 2 a – b = 9 4 5 – = 2 2 1 4 9 x y + = a = 5 and b = 3. Also 2 2 c a – b – = = = 25 9 4 2 2 1 x y b a + = , we have b = 2 and a = 3. 2 2 9 y is larger than the denominator of 2 2 2 1 x y a b + = , we get CONIC SECTIONS 193 4 x , the major axis is and 5 3 c e a = = Hence the foci are (0, 5 ) and (0, – 5 ), vertices are (0,3) and (0, –3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the ellipse is 5 3 . Example 11 Find the equation of the ellipse whose vertices are (± 13, 0) and foci are (± 5, 0). Solution Since the vertices are on x-axis, the equation will be of the form 2 2 1 x y a b + = , where a is the semi-major axis. 2 2 Reprint 2025-26 194 MATHEMATICS Given that a = 13, c = ± 5. Therefore, from the relation c 2 = a 2 – b 2 , we get 25 = 169 – b 2 , i.e., b = 12 Hence the equation of the ellipse is 2 2 1 169 144 x y + = . Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5). Solution Since the foci are on y-axis, the major axis is along the y-axis. So, equation of the ellipse is of the form 2 2 Given that and the relation c 2 = a2 – b 2 gives 5 2 = 102 – b 2 i.e., b 2 = 75 Therefore, the equation of the ellipse is Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4). a = semi-major axis 20 10 2 = = 2 2 1 x y b a + = . 2 2 1 75 100 x y + = Solution The standard form of the ellipse is 2 2 and (–1, 4) lie on the ellipse, we have and 22 161 ba + = 1 ….(2) Solving equations (1) and (2), we find that 2 247 7 a = and 2 247 15 b = . Hence the required equation is 1 916 22 =+ ba ... (1) Reprint 2025-26 b y a x + = 1. Since the points (4, 3) 2 2 In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 10. Vertices (± 5, 0), foci (± 4, 0) 11. Vertices (0, ± 13), foci (0, ± 5) 12. Vertices (± 6, 0), foci (± 4, 0) 13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2) 1. 2 2 1 36 16 x y + = 2. 2 2 1 4 25 x y + = 3. 2 2 1 16 9 x y + = 4. 2 2 1 25 100 x y + = 5. 2 2 1 49 36 x y + = 6. 400100 2 2 yx + = 1 7. 36x 2 + 4y 2 = 144 8. 16x 2 + y 2 = 16 9. 4x 2 + 9y 2 = 36 x y + = 2 2 1 247 247 7 15 EXERCISE 10.3 , i.e., 7x 2 + 15y 2 = 247. CONIC SECTIONS 195 14. Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0) 15. Length of major axis 26, foci (± 5, 0) 16. Length of minor axis 16, foci (0, ± 6). 17. Foci (± 3, 0), a = 4 18. b = 3, c = 4, centre at the origin; foci on the x axis. 19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6). 20. Major axis on the x-axis and passes through the points (4,3) and (6,2). 10.6 Hyperbola Definition 7 A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. Reprint 2025-26 196 MATHEMATICS The term “difference” that is used in the definition means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola (Fig 10.27). We denote the distance between the two foci by 2c, the distance between two vertices (the length of the transverse axis) by 2a and we define the quantity b as Fig 10.27 b = 2 2 c – a Also 2b is the length of the conjugate axis (Fig 10.28). To find the constant P1 F2 – P1 F1 : By taking the point P at A and B in the Fig 10.28, we have BF1 – BF2 = AF2 – AF1 (by the definition of the hyperbola) BA +AF1 – BF2 = AB + BF2 – AF1 i.e., AF1 = BF2 So that, BF1 – BF2 = BA + AF1 – BF2 = BA = 2a Reprint 2025-26 Fig 10.28 10.6.1 Eccentricity Definition 8 Just like an ellipse, the ratio e = c a is called the eccentricity of the hyperbola. Since c ≥ a, the eccentricity is never less than one. In terms of the eccentricity, the foci are at a distance of ae from the centre. 10.6.2 Standard equation of Hyperbola The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two such possible orientations are shown in Fig10.29. We will derive the equation for the hyperbola shown in Fig 10.29(a) with foci on (a) (b) Fig 10.29 CONIC SECTIONS 197 the x-axis. Let F1 and F2 be the foci and O be the mid-point of the line segment F1 F2 . Let O be the origin and the line through O through F2 be the positive x-axis and that through F1 as the negative x-axis. The line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c,0) and F2 be (c,0) (Fig 10.30). Let P(x, y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a. So given, PF1 – PF2 = 2a Reprint 2025-26 Fig 10.30 198 MATHEMATICS i.e., 2 2 2 2 ( ) 2 ( ) x c y a x – c y + + = + + Squaring both side, we get and on simplifying, we get On squaring again and further simplifying, we get i.e., 2 2 Hence any point on the hyperbola satisfies 2 2 Using the distance formula, we have Conversely, let P(x, y) satisfy the above equation with 0 < a < c. Then (x + c) 2 + y2 = 4a 2 + 4a 2 2 ( ) x – c y + + (x – c) 2 + y2 2 2 2 1 x y – a c – a = 2 2 1 x y – a b = (Since c 2 – a 2 = b 2 ) 2 2 y 2 = b 2 2 2 a cx – a = 2 2 ( ) x – c y + 2 2 2 2 ( ) ( ) 2 x c y – x – c y a + + + = 2 x – a a 2 2 1 x y – a b = 1. Therefore, PF1 = + 2 2 ( ) x c y + + Similarly, PF2 = a – a c x In hyperbola c > a; and since P is to the right of the line x = a, x > a, c a x > a.Therefore, a – c a x becomes negative. Thus, PF2 = c a x – a. = + 2 2 2 2 2 ( ) x – a x c b a + + = a + x a c Reprint 2025-26 Therefore PF1 – PF2 = a + c a x – cx a + a = 2a Also, note that if P is to the left of the line x = – a, then In that case P F2 – PF1 = 2a. So, any point that satisfies 2 2 hyperbola. Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis along x-axis is 2 2 Discussion From the equation of the hyperbola we have obtained, it follows that, we have for every point (x, y) on the hyperbola, 2 2 i.e, a x ≥ 1, i.e., x ≤ – a or x ≥ a. Therefore, no portion of the curve lies between the lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis). ANote A hyperbola in which a = b is called an equilateral hyperbola. 2 2 1 x y – a b = . PF1 c – a x a = + , PF2 = a – c x a . 2 2 1 x y a b = + ≥ 1. 2 2 1 x y – a b = , lies on the CONIC SECTIONS 199 Similarly, we can derive the equation of the hyperbola in Fig 11.31 (b) as 2 2 These two equations are known as the standard equations of hyperbolas. ANote The standard equations of hyperbolas have transverse and conjugate axes as the coordinate axes and the centre at the origin. However, there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes. From the standard equations of hyperbolas (Fig10.27), we have the following observations: 1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola. Reprint 2025-26 2 2 y x a b − = 1 200 MATHEMATICS 2. The foci are always on the transverse axis. It is the positive term whose 10.6.3 Latus rectum Definition 9 Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is 2 2b a . Example 14 Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas: Solution (i) Comparing the equation 2 2 1 9 16 x y – = with the standard equation denominator gives the transverse axis. For example, 2 2 1 9 16 x y – = has transverse axis along x-axis of length 6, while 2 2 1 25 16 y x – = has transverse axis along y-axis of length 10. (i) 2 2 1 9 16 x y – = , (ii) y 2 – 16x 2 = 16 2 2 1 x y – a b = 2 2 Here, a = 3, b = 4 and c = 2 2 a b + = + = 9 16 5 Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 3, 0).Also, The eccentricity e = 5 3 c a = . The latus rectum 2 2 32 3 b a = = Comparing the equation with the standard equation 2 2 a = 4, b = 1 and 2 2 c a b = + = + = 16 1 17 . (ii) Dividing the equation by 16 on both sides, we have 2 2 1 16 1 y x – = Reprint 2025-26 2 2 1 y x – a b = , we find that Therefore, the coordinates of the foci are (0, ± 17 ) and that of the vertices are (0, ± 4). Also, The eccentricity 17 4 c e a = = . The latus rectum 2 2 1 Example 15 Find the equation of the hyperbola with foci (0, ± 3) and vertices (0, ± 11 2 ). Also, since foci are (0, ± 3); c = 3 and b 2 = c 2 – a 2 = 25 4 . Therefore, the equation of the hyperbola is Example 16 Find the equation of the hyperbola where foci are (0, ±12) and the length Solution Since the foci is on y-axis, the equation of the hyperbola is of the form Since vertices are (0, ± 11 2 ), a = 11 2 2 2 1 y x – a b = 2 2 y x – = 1, i.e., 100 y 2 – 44 x 2 = 275. 11 25 4 4 2 2 2 b a = = . CONIC SECTIONS 201 of the latus rectum is 36. Solution Since foci are (0, ± 12), it follows that c = 12. Length of the latus rectum = 36 2 2 = a b or b 2 = 18a Therefore c2 = a 2 + b 2 ; gives 144 = a 2 + 18a i.e., a 2 + 18a – 144 = 0, Therefore, the equation of the required hyperbola is 2 2 1 36 108 y x – = , i.e., 3y 2 – x 2 = 108 So a = – 24, 6. Since a cannot be negative, we take a = 6 and so b 2 = 108. Reprint 2025-26 202 MATHEMATICS EXERCISE 10.4 In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. 4. 16x 2 – 9y 2 = 576 5. 5y 2 – 9x 2 = 36 6. 49y 2 – 16x 2 = 784. In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions. 7. Vertices (± 2, 0), foci (± 3, 0) 8. Vertices (0, ± 5), foci (0, ± 8) 9. Vertices (0, ± 3), foci (0, ± 5) 10. Foci (± 5, 0), the transverse axis is of length 8. 11. Foci (0, ±13), the conjugate axis is of length 24. 12. Foci (± 3 5 , 0), the latus rectum is of length 8. 13. Foci (± 4, 0), the latus rectum is of length 12 14. vertices (± 7,0), e = 3 4 . 15. Foci (0, ± 10 ), passing through (2,3) Example 17 The focus of a parabolic mirror as shown in Fig 10.31 is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB (Fig 10.31). 1. 2 2 1 16 9 x y – = 2. 2 2 1 9 27 y x – = 3. 9y 2 – 4x 2 = 36 Miscellaneous Examples Solution Since the distance from the focus to the vertex is 5 cm. We have, a = 5. If the origin is taken at the vertex and the axis of the mirror lies along the positive x-axis, the equation of the parabolic section is y 2 = 4 (5) x = 20 x Note that x = 45. Thus y 2 = 900 Therefore y = ± 30 Hence AB = 2y = 2 × 30 = 60 cm. Example 18 A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at its centre, there Reprint 2025-26 Fig 10.31 is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm? Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate axis be chosen as shown in Fig 10.32. Let AB be the deflection of the beam which is 1 100 m. Coordinates of B are (x, 2 100 ). Therefore x 2 = 4 × 300 × 2 100 = 24 i.e. x = 24 = 2 6 metres 3 6 100 , , we have (6)2 = 4a 3 100 , i.e., a = 36 100 12 × = 300 m The equation of the parabola takes the form x 2 = 4ay. Since it passes through Fig 10.32 CONIC SECTIONS 203 Example 19 A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse. Solution Let AB be the rod making an angle θ with OX as shown in Fig 10.33 and P (x, y) the point on it such that AP = 6 cm. Since AB = 15 cm, we have PB = 9 cm. From P draw PQ and PR perpendiculars on y-axis and x-axis, respectively. Fig 10.33 Reprint 2025-26 204 MATHEMATICS From ∆ PBQ, cos θ = 9 x From ∆ PRA, sin θ = 6 y Since cos2 θ + sin2 θ = 1 or 2 2 1 81 36 x y + = Thus the locus of P is an ellipse. 1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. 2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? 3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle. Miscellaneous Exercise on Chapter 10 2 2 1 9 6 x y + = 4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end. 5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis. 6. Find the area of the triangle formed by the lines joining the vertex of the parabola x 2 = 12y to the ends of its latus rectum. 7. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man. 8. An equilateral triangle is inscribed in the parabola y 2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Reprint 2025-26 In this Chapter the following concepts and generalisations are studied. ÆA circle is the set of all points in a plane that are equidistant from a fixed point in the plane. ÆThe equation of a circle with centre (h, k) and the radius r is (x – h) 2 + (y – k) 2 = r 2 . ÆA parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane. ÆThe equation of the parabola with focus at (a, 0) a > 0 and directrix x = – a is y 2 = 4ax. ÆLatus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola. ÆLength of the latus rectum of the parabola y 2 = 4ax is 4a. ÆAn ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. ÆThe equation of an ellipse with foci on the x-axis is 2 2 ÆLatus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse. Summary 2 2 1 x y + = a b . CONIC SECTIONS 205 ÆLength of the latus rectum of the ellipse 2 2 ÆThe eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse. ÆA hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. ÆThe equation of a hyperbola with foci on the x-axis is : 2 2 Reprint 2025-26 2 2 + = 1 x y a b is 2 2b a . 2 2 1 x y a b − = 206 MATHEMATICS ÆLatus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. ÆLength of the latus rectum of the hyperbola : 2 2 ÆThe eccentricity of a hyperbola is the ratio of the distances from the centre of Geometry is one of the most ancient branches of mathematics. The Greek geometers investigated the properties of many curves that have theoretical and practical importance. Euclid wrote his treatise on geometry around 300 B.C. He was the first who organised the geometric figures based on certain axioms suggested by physical considerations. Geometry as initially studied by the ancient Indians and Greeks, who made essentially no use of the process of algebra. The synthetic approach to the subject of geometry as given by Euclid and in Sulbasutras, etc., was continued for some 1300 years. In the 200 B.C., Apollonius wrote a book called ‘The Conic’ which was all about conic sections with many important discoveries that have remained unsurpassed for eighteen centuries. Modern analytic geometry is called ‘Cartesian’ after the name of Rene Descartes (1596-1650) whose relevant ‘La Geometrie’ was published in 1637. But the fundamental principle and method of analytical geometry were already discovered by Pierre de Fermat (1601-1665). Unfortunately, Fermats treatise on the subject, entitled Ad Locus Planos et So LIDOS Isagoge (Introduction to Plane and Solid Loci) was published only posthumously in 1679. So, Descartes came to be regarded as the unique inventor of the analytical geometry. Isaac Barrow avoided using cartesian method. Newton used method of undetermined coefficients to find equations of curves. He used several types of coordinates including polar and bipolar. Leibnitz used the terms ‘abscissa’, ‘ordinate’ and ‘coordinate’. L’ Hospital (about 1700) wrote an important textbook on analytical geometry. Clairaut (1729) was the first to give the distance formula although in clumsy form. He also gave the intercept form of the linear equation. Cramer (1750) the hyperbola to one of the foci and to one of the vertices of the hyperbola. Historical Note 2 2 1 x y a b − = is : 2 2b a . Reprint 2025-26 made formal use of the two axes and gave the equation of a circle as ( y – a) 2 + (b – x) 2 = r He gave the best exposition of the analytical geometry of his time. Monge (1781) gave the modern ‘point-slope’ form of equation of a line as y – y′ = a (x – x′) and the condition of perpendicularity of two lines as aa′ + 1 = 0. S.F. Lacroix (1765–1843) was a prolific textbook writer, but his contributions to analytical geometry are found scattered. He gave the ‘two-point’ form of equation of a line as and the length of the perpendicular from (α, β) on y = ax + b as 2 (β ) His formula for finding angle between two lines was tan θ 1 a – a aa ′ = + ′ . It is, of course, surprising that one has to wait for more than 150 years after the invention of analytical geometry before finding such essential basic formula. In 1818, C. Lame, a civil engineer, gave mE + m′E′ = 0 as the curve passing through the points of intersection of two loci E = 0 and E′ = 0. Many important discoveries, both in Mathematics and Science, have been linked to the conic sections. The Greeks particularly Archimedes (287–212 B.C.) and Apollonius (200 B.C.) studied conic sections for their own beauty. These curves are important tools for present day exploration of outer space and also for research into behaviour of atomic particles. β β β = ( α) α α – y – x – – ′ ′ CONIC SECTIONS 207 1 – a – b + a . Reprint 2025-26 — v —" class_11,11,Introduction to Three Dimensional Geometry,ncert_books/class_11/kemh1dd/kemh111.pdf,"208 MATHEMATICS 11.1 Introduction You may recall that to locate the position of a point in a plane, we need two intersecting mutually perpendicular lines in the plane. These lines are called the coordinate axes and the two numbers are called the coordinates of the point with respect to the axes. In actual life, we do not have to deal with points lying in a plane only. For example, consider the position of a ball thrown in space at different points of time or the position of an aeroplane as it flies from one place to another at different times during its flight. Similarly, if we were to locate the position of the lowest tip of an electric bulb hanging from the ceiling of a room or the position of the central tip of the ceiling fan in a room, we will not only require the perpendicular distances of the point to be located from two perpendicular walls of the room but also the height of the point from the floor of the room. Therefore, we need not only two but three numbers representing the perpendicular distances of the point from three mutually perpendicular planes, namely the floor of the room and two adjacent walls of the room. The three numbers representing the three distances are called the coordinates of the point with reference to the three coordinate planes. So, a point in space has three coordinates. In this Chapter, we shall study the basic concepts of geometry in three dimensional space.* INTRODUCTION TO THREE DIMENSIONAL GEOMETRY vMathematics is both the queen and the hand-maiden of all sciences – E.T. BELLv Chapter 11 * For various activities in three dimensional geometry one may refer to the Book, “A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005. Reprint 2025-26 Leonhard Euler (1707-1783) 11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space Consider three planes intersecting at a point O such that these three planes are mutually perpendicular to each other (Fig 11.1). These three planes intersect along the lines X′OX, Y′OY and Z′OZ, called the x, y and z-axes, respectively. We may note that these lines are mutually perpendicular to each other. These lines constitute the rectangular coordinate system. The planes XOY, YOZ and ZOX, called, respectively the XY-plane, YZ-plane and the ZX-plane, are known as the three coordinate planes. We take the XOY plane as the plane of the paper and the line Z′OZ as perpendicular to the plane XOY. If the plane of the paper is considered as horizontal, then the line Z′OZ will be vertical. The distances measured from XY-plane upwards in the direction of OZ are taken as positive and those measured downwards in the direction of OZ′ are taken as negative. Similarly, the distance measured to the right of ZX-plane along OY are taken as positive, to the left of ZX-plane and along OY′ as negative, in front of the YZ-plane along OX as positive and to the back of it along OX′ as negative. The point O is called the origin of the coordinate system. The three coordinate planes divide the space into eight parts known as octants. These octants could be named as XOYZ, X′OYZ, X′OY′Z, XOY′Z, XOYZ′, X′OYZ′, X′OY′Z′ and XOY′Z′. and denoted by I, II, III, ..., VIII , respectively. 11.3 Coordinates of a Point in Space INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 209 Fig 11.1 Having chosen a fixed coordinate system in the space, consisting of coordinate axes, coordinate planes and the origin, we now explain, as to how, given a point in the space, we associate with it three coordinates (x,y,z) and conversely, given a triplet of three numbers (x, y, z), how, we locate a point in the space. Given a point P in space, we drop a perpendicular PM on the XY-plane with M as the foot of this perpendicular (Fig 11.2). Then, from the point M, we draw a perpendicular ML to the x-axis, meeting it at L. Let OL be x, LM be y and MP be z. Then x,y and z are called the x, y and z coordinates, respectively, of the point P in the space. In Fig 11.2, we may note that the point P (x, y, z) lies in the octant XOYZ and so all x, y, z are positive. If P was in any other octant, the signs of x, y and z would change Reprint 2025-26 Fig 11.2 210 MATHEMATICS accordingly. Thus, to each point P in the space there corresponds an ordered triplet (x, y, z) of real numbers. Conversely, given any triplet (x, y, z), we would first fix the point L on the x-axis corresponding to x, then locate the point M in the XY-plane such that (x, y) are the coordinates of the point M in the XY-plane. Note that LM is perpendicular to the x-axis or is parallel to the y-axis. Having reached the point M, we draw a perpendicular MP to the XY-plane and locate on it the point P corresponding to z. The point P so obtained has then the coordinates (x, y, z). Thus, there is a one to one correspondence between the points in space and ordered triplet (x, y, z) of real numbers. Alternatively, through the point P in the space, we draw three planes parallel to the coordinate planes, meeting the x-axis, y-axis and z-axis in the points A, B and C, respectively (Fig 11.3). Let OA = x, OB = y and OC = z. Then, the point P will have the coordinates x, y and z and we write P (x, y, z). Conversely, given x, y and z, we locate the three points A, B and C on the three coordinate axes. Through the points A, B and C we draw planes parallel to the YZ-plane, ZX-plane and XY-plane, respectively. The point of interesection of these three planes, namely, ADPF, BDPE and CEPF is obviously the point P, corresponding to the ordered triplet (x, y, z). We observe that if P (x, y, z) is any point in the space, then x, y and z are perpendicular distances from YZ, ZX and XY planes, respectively. ANote The coordinates of the origin O are (0,0,0). The coordinates of any point on the x-axis will be as (x,0,0) and the coordinates of any point in the YZ-plane will be as (0, y, z). Fig 11.3 Remark The sign of the coordinates of a point determine the octant in which the point lies. The following table shows the signs of the coordinates in eight octants. Coordinates Octants x + – – + + – – + y + + – – + + – – z + + + + – – – – I II III IV V VI VII VIII Reprint 2025-26 Table 11.1 Example 1 In Fig 11.3, if P is (2,4,5), find the coordinates of F. Solution For the point F, the distance measured along OY is zero. Therefore, the coordinates of F are (2,0,5). Example 2 Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie. Solution From the Table 11.1, the point (–3,1, 2) lies in second octant and the point (–3, 1, – 2) lies in octant VI. 1. A point is on the x-axis. What are its y-coordinate and z-coordinates? 2. A point is in the XZ-plane. What can you say about its y-coordinate? 3. Name the octants in which the following points lie: 4. Fill in the blanks: 11.4 Distance between Two Points We have studied about the distance between two points in two-dimensional coordinate system. Let us now extend this study to three-dimensional system. Let P(x1 , y1 , z 1 ) and Q ( x2 , y2 , z 2 ) be two points referred to a system of rectangular axes OX, OY and OZ. Through the points P and Q draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig 11.4). Now, since ∠PAQ is a right angle, it follows that, in triangle PAQ, PQ2 = PA2 + AQ2 ... (1) (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (– 2, – 4, –7). (iii) Coordinate planes divide the space into ______ octants. (ii) The coordinates of points in the XY-plane are of the form _______. (i) The x-axis and y-axis taken together determine a plane known as_______. INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 211 EXERCISE 11.1 Also, triangle ANQ is right angle triangle with ∠ANQ a right angle. Reprint 2025-26 Fig 11.4 212 MATHEMATICS Therefore AQ2 = AN2 + NQ2 ... (2) From (1) and (2), we have PQ2 = PA2 + AN2 + NQ2 Now PA = y2 – y1 , AN = x2 – x1 and NQ = z 2 – z 1 Hence PQ2 = (x2 – x1 ) 2 + (y2 – y1 ) 2 + (z 2 – z 1 ) 2 Therefore PQ = 2 12 2 12 2 12 −+−+− zzyyxx )()()( This gives us the distance between two points (x1 , y1 , z 1 ) and (x2 , y2 , z 2 ). In particular, if x1 = y1 = z 1 = 0, i.e., point P is origin O, then OQ = 2 2 2 2 2 2 ++ zyx , which gives the distance between the origin O and any point Q (x2 , y2 , z 2 ). Example 3 Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2). Solution The distance PQ between the points P (1,–3, 4) and Q (– 4, 1, 2) is Example 4 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear. Solution We know that points are said to be collinear if they lie on a line. Now, PQ = 14419)53()32()21( 2 2 2 =++=−+−++ PQ = 2 2 2 −+++−− )42()31()14( = ++ 41625 = 45 = 3 5 units and PR = 14312636981)51()30()27( 2 2 2 ==++=−−+−++ Thus, PQ + QR = PR. Hence, P, Q and R are collinear. Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices of a right angled triangle? Solution By the distance formula, we have AB2 = (10 – 3)2 + (20 – 6)2 + (30 – 9)2 BC2 = (25 – 10)2 + (– 41 – 20)2 + (5 – 30)2 = 225 + 3721 + 625 = 4571 = 49 + 196 + 441 = 686 QR = 1425616436)31()20()17( 2 2 2 ==++=−−+−+− Reprint 2025-26 We find that CA2 + AB2 ≠ BC2 . Hence, the triangle ABC is not a right angled triangle. Example 6 Find the equation of set of points P such that PA2 + PB2 = 2k 2 , where A and B are the points (3, 4, 5) and (–1, 3, –7), respectively. Solution Let the coordinates of point P be (x, y, z). Here PA2 = (x – 3)2 + (y – 4)2 + ( z – 5)2 By the given condition PA2 + PB2 = 2k 2 , we have i.e., 2x 2 + 2y 2 + 2z 2 – 4x – 14y + 4z = 2k 2 – 109. 1. Find the distance between the following pairs of points: 2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear. 3. Verify the following: (iii) (–1, 3, – 4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3). (x – 3)2 + (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k 2 (i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1) (i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle. CA2 = (3 – 25)2 + (6 + 41)2 + (9 – 5)2 PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2 = 484 + 2209 + 16 = 2709 INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 213 EXERCISE 11.2 Example 7 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle. Solution To show ABCD is a parallelogram we need to show opposite side are equal Note that. 4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). 5. Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10. (iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram. (ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle. Miscellaneous Examples Reprint 2025-26 214 MATHEMATICS Since AB = CD and BC = AD, ABCD is a parallelogram. Now, it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal. We have Since AC ≠ BD, ABCD is not a rectangle. Example 8 Find the equation of the set of the points P such that its distances from the points A (3, 4, –5) and B (– 2, 1, 4) are equal. Solution If P (x, y, z) be any point such that PA = PB. ANote We can also show that ABCD is a parallelogram, using the property that diagonals AC and BD bisect each other. DA = 2 2 2 −+−+− )63()72()41( = =++ 439259 AB = 2 2 2 −−+−−+−− )31()22()11( = 4 16 16 + + = 6 CD = 2 2 2 −+−+− )26()37()24( = =++ 616164 BC = 2 2 2 +++++ )12()23()12( = ++ 9259 = 43 AC = 3111)32()23()12( 2 2 2 =++=−+−+− BD = 155498125)16()27()14( 2 2 2 =++=+++++ . Now 2 2 2 2 2 2 zyx zyx −+−++=++−+− )4()1()2()5()4()3( or 2 2 2 2 2 2 zyx zyx −+−++=++−+− )4()1()2()5()4()3( or 10 x+ 6y – 18z – 29 = 0. Example 9 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, – 6), respectively, find the coordinates of the point C. Solution Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be (1, 1, 1). Then Reprint 2025-26 1. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex. 2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0). 3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c. 4. If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k 2 , where k is a constant. Summary ÆIn three dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the x, y and z-axes. ÆThe three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes. ÆThe three coordinate planes divide the space into eight parts known as octants. ÆThe coordinates of a point P in three dimensional geometry is always written in the form of triplet like (x, y, z). Here x, y and z are the distances from the Hence, coordinates of C are (1, 1, 2). x + − = 3 1 3 1, i.e., x = 1; y − + = 5 7 3 1, i.e., y = 1; z + − = 7 6 3 1, i.e., z = 2. Miscellaneous Exercise on Chapter 11 INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 215 YZ, ZX and XY-planes. Æ (i) Any point on x-axis is of the form (x, 0, 0) (iii) Any point on z-axis is of the form (0, 0, z). ÆDistance between two points P(x1 , y1 , z 1 ) and Q (x2 , y2 , z 2 ) is given by (ii) Any point on y-axis is of the form (0, y, 0) 2 2 2 PQ 2 1 2 1 2 1 = − + − + − ( x x ) ( y y ) ( z z ) Reprint 2025-26 216 MATHEMATICS Rene’ Descartes (1596–1650), the father of analytical geometry, essentially dealt with plane geometry only in 1637. The same is true of his co-inventor Pierre Fermat (1601-1665) and La Hire (1640-1718). Although suggestions for the three dimensional coordinate geometry can be found in their works but no details. Descartes had the idea of coordinates in three dimensions but did not develop it. J.Bernoulli (1667-1748) in a letter of 1715 to Leibnitz introduced the three coordinate planes which we use today. It was Antoinne Parent (1666-1716), who gave a systematic development of analytical solid geometry for the first time in a paper presented to the French Academy in 1700. L.Euler (1707-1783) took up systematically the three dimensional coordinate geometry, in Chapter 5 of the appendix to the second volume of his “Introduction to Geometry” in 1748. It was not until the middle of the nineteenth century that geometry was extended to more than three dimensions, the well-known application of which is in the Space-Time Continuum of Einstein’s Theory of Relativity. Historical Note — v — Reprint 2025-26" class_11,12,Limits and Derivatives,ncert_books/class_11/kemh1dd/kemh112.pdf,"12.1 Introduction This chapter is an introduction to Calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. First, we give an intuitive idea of derivative (without actually defining it). Then we give a naive definition of limit and study some algebra of limits. Then we come back to a definition of derivative and study some algebra of derivatives. We also obtain derivatives of certain standard functions. 12.2 Intuitive Idea of Derivatives vWith the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature – WHITEHEAD v LIMITS AND DERIVATIVES Chapter 12 Sir Issac Newton (1642-1727) Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9t 2 metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by s = 4.9t 2 . The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff. The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds. Average velocity between t = t 1 and t = t2 equals distance travelled between t = t1 and t = t2 seconds divided by (t 2 – t 1 ). Hence the average velocity in the first two seconds Reprint 2025-26 218 MATHEMATICS Likewise we compute the average velocitiy between t = t 1 and t = 2 for various t 1 . The following Table 13.2 gives the average velocity (v), t = t 1 seconds and t = 2 seconds. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551m/s. This conclusion is somewhat strengthened by the following set of computation. Compute the average velocities for various time intervals starting at t = 2 seconds. As before the average velocity v between t = 2 seconds and t = t 2 seconds is = 2 1 2 1 Distance travelled between 2 0 Time interval ( ) t and t t t = = − = ( ) Similarly, the average velocity between t = 1 and t = 2 is t 1 0 1 1.5 1.8 1.9 1.95 1.99 v 9.8 14.7 17.15 18.62 19.11 19.355 19.551 From Table 12.2, we observe that the average velocity is gradually increasing. ( ) ( ) 19.6 – 4.9 ( ) 19.6 0 9.8 / 2 0 2 1 m m s s − = − . m − s = 14.7 m/s Table 12.2 t s 0 0 1 4.9 1.5 11.025 1.8 15.876 1.9 17.689 1.95 18.63225 2 19.6 2.05 20.59225 2.1 21.609 2.2 23.716 2.5 30.625 3 44.1 4 78.4 Table 12.1 = 2 2 Distance travelled between 2 seconds and seconds 2 = 2 2 Distance travelled in seconds Distance travelled in 2 seconds 2 t t Reprint 2025-26 t t − − − between t = 2 seconds and t 2 seconds. better idea of the velocity at t = 2. In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens just before t = 2. In the second set of computations, we have found the average velocities decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens just after t = 2. Purely on the physical grounds, both these sequences of average velocities must approach a common limit. We can safely conclude that the velocity of the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is well-known, velocity is the rate of change of displacement. Hence what we have accomplished is the following. From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function s = 4.9t 2 at t = 2 is between 19.551 and 19.649. An alternate way of viewing this limiting process is shown in Fig 12.1. This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals h1 , h2 , ..., approaches zero, the sequence of average velocities approaches the same limit as does the sequence of ratios Fig 12.1 Here again we note that if we take smaller time intervals starting at t = 2, we get The following Table 12.3 gives the average velocity v in metres per second t 2 4 3 2.5 2.2 2.1 2.05 2.01 v 29.4 24.5 22.05 20.58 20.09 19.845 19.649 = 2 2 Distance travelled in seconds 19.6 2 t t Table 12.3 − − LIMITS AND DERIVATIVES 219 Reprint 2025-26 220 MATHEMATICS where C1B1 = s 1 – s 0 is the distance travelled by the body in the time interval h1 = AC1 , etc. From the Fig 12.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t 2 at t = 2. 12.3 Limits The above discussion clearly points towards the fact that we need to understand limiting process in greater clarity. We study a few illustrative examples to gain some familiarity with the concept of limits. Consider the function f(x) = x 2 . Observe that as x takes values very close to 0, the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say (to be read as limit of f(x) as x tends to zero equals zero). The limit of f(x) as x tends to zero is to be thought of as the value f (x) should assume at x = 0. In general as x → a, f (x) → l, then l is called limit of the function f (x) which is symbolically written as lim ( ) x a f x l → = . Computing the value of g(x) for values of x very near to 0, we see that the value of g(x) moves Consider the following function g(x) = |x|, x ≠ 0. Observe that g(0) is not defined. 1 2 3 C B C B C B , , AC AC AC , ... 1 1 2 2 3 3 ( ) 0 lim 0 x f x → = towards 0. So, 0 lim x→ g(x) = 0. This is intuitively clear from the graph of y = |x| for x ≠ 0. (See Fig 2.13, Chapter 2). Consider the following function. Compute the value of h(x) for values of x very near to 2 (but not at 2). Convince yourself that all these values are near to 4. This is somewhat strengthened by considering the graph of the function y = h(x) given here (Fig 12.2). Fig 12.2 ( ) 2 4 , 2 2 x h x x x − = ≠ − . Reprint 2025-26 point x = a did not really depend on how is x tending to a. Note that there are essentially two ways x could approach a number a either from left or from right, i.e., all the values of x near a could be less than a or could be greater than a. This naturally leads to two limits – the right hand limit and the left hand limit. Right hand limit of a function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends to a from the right. Similarly, the left hand limit. To illustrate this, consider the function Graph of this function is shown in the Fig 12.3. It is clear that the value of f at 0 dictated by values of f(x) with x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is Similarly, the value of f at 0 dictated by values of f (x) with x > 0 equals 2, i.e., the right hand limit of f (x) at 0 is limit of f (x) as x tends to zero does not exist (even though the function is defined at 0). In this case the right and left hand limits are different, and hence we say that the In all these illustrations the value which the function should assume at a given ( ) 1, 0 2, 0 x f x x ≤ = > 0 lim ( ) 2 x f x → + = . 0 lim ( ) 1 x f x → = . Summary LIMITS AND DERIVATIVES 221 Fig 12.3 Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this function at x = 5. Let us compute the value of the function f(x) for x very near to 5. Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values of the function at these points are tabulated below. Similarly, the real number 5.001, We say lim x a → – f(x) is the expected value of f at x = a given the values of f near x to the left of a. This value is called the left hand limit of f at a. f near x to the right of a. This value is called the right hand limit of f(x) at a. If the right and left hand limits coincide, we call that common value as the limit of f(x) at x = a and denote it by lim x a → f(x). We say lim ( ) x a f x → + is the expected value of f at x = a given the values of Reprint 2025-26 222 MATHEMATICS 5.01, 5.1 are also points near and to the right of 5. Values of the function at these points are also given in the Table 12.4. Table 12.4 14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995 and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by the numbers to the left of 5 is 15, i.e., both equal to 15. Thus, seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we note that as x approaches 5 from either right or left, the graph of the function f(x) = x +10 approaches the point (5, 15). We observe that the value of the function at x = 5 also happens to be equal to 15. From the Table 12.4, we deduce that value of f(x) at x = 5 should be greater than Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e., Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are This conclusion about the limit being equal to 15 is somewhat strengthened by f(x) 14.9 14.95 14.99 14.995 15.001 15.01 15.1 x 4.9 4.95 4.99 4.995 5.001 5.01 5.1 ( ) ( ) ( ) 5 5 5 lim lim lim 15 x x x f x f x f x → → − + → = = = . ( ) – 5 lim 15 x f x → = . ( ) 5 lim 15 x f x → + = . Illustration 2 Consider the function f(x) = x3 . Let us try to find the limit of this function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at x near 1. This is given in the Table 12.5. Table 12.5 From this table, we deduce that value of f(x) at x = 1 should be greater than 0.997002999 and less than 1.003003001 assuming nothing dramatic happens between f(x) 0.729 0.970299 0.997002999 1.003003001 1.030301 1.331 x 0.9 0.99 0.999 1.001 1.01 1.1 Reprint 2025-26 x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as dictated by the numbers to the left of 1 is 1, i.e., both equal to 1. Thus, seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we note that as x approaches 1 from either right or left, the graph of the function f(x) = x3 approaches the point (1, 1). We observe, again, that the value of the function at x = 1 also happens to be equal to 1. Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this function at x = 2. The following Table 12.6 is now self-explanatory. Similarly, when x approaches 1 from the right, f(x)should be taking value 1, i.e., Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are This conclusion about the limit being equal to 1 is somewhat strengthened by f(x) 5.7 5.85 5.97 5.997 6.003 6.03 6.3 x 1.9 1.95 1.99 1.999 2.001 2.01 2.1 ( ) ( ) ( ) 1 1 1 lim lim lim 1 x x x f x f x f x → → − + → = = = . ( ) 1 lim 1 x f x → − = . ( ) 1 lim 1 x f x → + = . Table 12.6 LIMITS AND DERIVATIVES 223 As before we observe that as x approaches 2 from either left or right, the value of f(x) seem to approach 6. We record this as Its graph shown in Fig 12.4 strengthens this fact. Here again we note that the value of the function at x = 2 coincides with the limit at x = 2. Illustration 4 Consider the constant function f(x) = 3. Let us try to find its limit at x = 2. This function being the constant function takes the same Fig 12.4 ( ) ( ) ( ) 2 2 2 lim lim lim 6 x x x f x f x f x → → − + → = = = Reprint 2025-26 224 MATHEMATICS value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In fact, it is easily observed that lim 3 ( ) x a f x → = for any real number a. Illustration 5 Consider the function f(x) = x2 + x. We want to find ( ) 1 lim x f x → . We tabulate the values of f(x) near x = 1 in Table 12.7. From this it is reasonable to deduce that From the graph of f(x) = x2 + x shown in the Fig 12.5, it is clear that as x approaches 1, the graph approaches (1, 2). Here, again we observe that the ( ) ( ) ( ) 1 1 1 lim lim lim 2 x x x f x f x f x → → − + → = = = . Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and f(x) 1.71 1.9701 1.997001 2.0301 2.31 2.64 x 0.9 0.99 0.999 1.01 1.1 1.2 ( ) ( ) ( ) 2 2 2 lim lim lim 3 x x x f x f x f x → → + → = = = Table 12.7 Now, convince yourself of the following three facts: Then 2 2 1 1 1 lim lim 1 1 2 lim x x x x x x x → → → + = + = = + . Also ( ) ( ) 2 1 1 1 1 lim . lim 1 1.2 2 lim 1 lim x x x x x x x x x x → → → → + = = = + = + . 1 lim x→ f (x) = f (1) 1 1 1 lim 1, lim 1 and lim 1 2 x x x x x x → → → = = + = 2 Reprint 2025-26 Fig 12.5 Illustration 6 Consider the function f(x) = sin x. We are interested in where the angle is measured in radians. this, we may deduce that (Chapter 3). In this case too, we observe that Illustration 7 Consider the function f(x) = x + cos x. We want to find the 0 lim x→ f (x). Here, we tabulate the (approximate) value of f(x) near 2 π (Table 12.8). From Further, this is supported by the graph of f(x) = sin x which is given in the Fig 3.8 Here we tabulate the (approximate) value of f(x) near 0 (Table 12.9). x 0.1 2 π − 0.01 2 π − 0.01 2 π + 0.1 2 π + f(x) 0.9950 0.9999 0.9999 0.9950 2 2 2 lim lim lim 1 x x x f x f x f x π − π + π → → → = = = . ( ) ( ) ( ) Table 12.8 2 lim x π → sin x = 1. LIMITS AND DERIVATIVES 225 2 lim sin x x π → , From the Table 13.9, we may deduce that f(x) 0.9850 0.98995 0.9989995 1.0009995 1.00995 1.0950 x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1 In this case too, we observe that 0 lim x→ f (x) = f (0) = 1. Now, can you convince yourself that [ ] 0 0 0 lim cos lim lim cos x x x x x x x → → → + = + is indeed true? ( ) ( ) ( ) 0 0 0 lim lim lim 1 x x x f x f x f x → → − + → = = = Reprint 2025-26 Table 12.9 226 MATHEMATICS Illustration 8 Consider the function ( ) 2 1 f x x = for x > 0 . We want to know 0 lim x→ f (x). numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of x approaching 0 from the left. Below we tabulate the values of the function for positive x close to 0 (in this table n denotes any positive integer). From the Table 12.10 given below, we see that as x tends to 0, f(x) becomes larger and larger. What we mean here is that the value of f(x) may be made larger than any given number. Table 12.10 Mathematically, we say We also remark that we will not come across such limits in this course. Illustration 9 We want to find ( ) 0 lim x f x → , where Here, observe that the domain of the function is given to be all positive real x 1 0.1 0.01 10–n f(x) 1 100 10000 102n ( ) 0 lim x f x → = +∞ f x x x x ( ) − < = = + > x x 0 , 0 2, 0 2, 0 As usual we make a table of x near 0 with f(x). Observe that for negative values of x we need to evaluate x – 2 and for positive values, we need to evaluate x + 2. function is decreasing to –2 and hence. f(x) – 2.1 – 2.01 – 2.001 2.001 2.01 2.1 From the first three entries of the Table 12.11, we deduce that the value of the x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1 ( ) 0 lim 2 x f x → − = − Reprint 2025-26 Table 12.11 is increasing from 2 and hence Since the left and right hand limits at 0 do not coincide, we say that the limit of the function at 0 does not exist. Graph of this function is given in the Fig12.6. Here, we remark that the value of the function at x = 0 is well defined and is, indeed, equal to 0, but the limit of the function at x = 0 is not even defined. Illustration 10 As a final illustration, we find ( ) 1 lim x f x → , where x less than 1, it seems that the function should take value 3 at x = 1., i.e., From the last three entires of the table we deduce that the value of the function f(x) 2.9 2.99 2.999 3.001 3.01 3.1 As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for x 0.9 0.99 0.999 1.001 1.01 1.1 ( ) 0 lim 2 x f x → + = ( ) 2 1 0 1 x x f x x + ≠ = = Table 12.12 LIMITS AND DERIVATIVES 227 Fig 12.6 Similarly, the value of f(x)should be 3 as dictated by values of f(x) at x greater than 1. i.e. But then the left and right hand limits coincide and hence Graph of function given in Fig 12.7 strengthens our deduction about the limit. Here, we ( ) ( ) ( ) 1 1 1 lim lim lim 3 x x x f x f x f x → → − + → = = = . ( ) 1 lim 3 x f x → − = . ( ) 1 lim 3 x f x → + = . Reprint 2025-26 Fig 12.7 228 MATHEMATICS note that in general, at a given point the value of the function and its limit may be different (even when both are defined). 12.3.1 Algebra of limits In the above illustrations, we have observed that the limiting process respects addition, subtraction, multiplication and division as long as the limits and functions under consideration are well defined. This is not a coincidence. In fact, below we formalise these as a theorem without proof. Theorem 1 Let f and g be two functions such that both lim x a → f(x) and lim x a → g(x) exist. Then (i) Limit of sum of two functions is sum of the limits of the functions, i.e., (iii) Limit of product of two functions is product of the limits of the functions, i.e., (iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e., (ii) Limit of difference of two functions is difference of the limits of the functions, i.e., lim x a → [f(x) + g (x)] = lim x a → f(x) + lim x a → g(x). lim x a → [f(x) – g(x)] = lim x a → f(x) – lim x a → g(x). lim x a → [f(x) . g(x)] = lim x a → f(x). lim x a → g(x). ( ) lim lim lim x a x a x a f x f x g x g x → → → = ( ) ( ) ( ) limits of special types of functions. 12.3.2 Limits of polynomials and rational functions A function f is said to be a polynomial function of degree n f(x) = a0 + a1 x + a2 x 2 +. . . + an x n , where ai s are real numbers such that an ≠ 0 for some natural number n. ANote In particular as a special case of (iii), when g is the constant function such that g(x) = λ , for some real number λ , we have In the next two subsections, we illustrate how to exploit this theorem to evaluate We know that lim x a → x = a. Hence lim . .lim ( ) ( ) ( ) x a x a f x f x → → λ = λ . Reprint 2025-26 An easy exercise in induction on n tells us that of each of 2 0 1 2 , , ,..., n n a a x a x a x as a function, we have Now, let ( ) 2 0 1 2 ... n n f x a a x a x a x = + + + + be a polynomial function. Thinking (Make sure that you understand the justification for each step in the above!) A function f is said to be a rational function, if f(x) = ( ) lim n n x a x a → = lim ( ) x a f x → = 2 0 1 2 lim ... n n x a a a x a x a x → + + + + ( ) 2 2 lim lim . lim .lim . x a x a x a x a x x x x x a a a → → → → = = = = = 2 0 1 2 lim lim lim ... lim n n x a x a x a x a a a x a x a x → → → → + + + + = 2 0 1 2 lim lim ... lim n n x a x a x a a a x a x a x → → → + + + + = 2 0 1 2 ... n n a a a a a a a + + + + = f a( ) LIMITS AND DERIVATIVES 229 ( ) g x h x , where g(x) and h(x) are polynomials such that h(x) ≠ 0. Then However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when g(a) = 0. In the former case we say that the limit does not exist. In the latter case we can write g(x) = (x – a) k g1 (x), where k is the maximum of powers of (x – a) in g(x) Similarly, h(x) = (x – a) l h1 (x) as h (a) = 0. Now, if k > l, we have lim lim lim lim lim x a x a l x a x a x a ( ) ( ) lim lim lim lim x a x a x a x a g x g x g a f x h x h x h a → → → → = = = g x x a g x f x h x x a h x → → → → → ( ) ( ) ( ) ( ) ( ) − = = − Reprint 2025-26 ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) k 1 230 MATHEMATICS If k < l, the limit is not defined. Example 1 Find the limits: (i) 3 2 1 lim 1 x x x → − + (ii) ( ) 3 lim 1 x x x → + Solution The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have Example 2 Find the limits: (iii) 2 10 1 lim 1 ... x x x x →− + + + + . (iii) 2 10 1 lim 1 ... x x x x →− + + + + ( ) ( ) ( ) 2 10 = + − + − + + − 1 1 1 ... 1 (ii) ( ) ( ) ( ) 3 lim 1 3 3 1 3 4 12 x x x → + = + = = (i) 1 lim x→ [x 3 – x 2 + 1] = 13 – 12 + 1 = 1 (i) 2 1 lim x 100 x → x + + (ii) 1 = ( )( ) ( ) lim 0. 0 lim x a → x a g x g a x a − = = → h x h a 1 1 k l ( ) − ( ) ( ) 1 1 4 4 lim x 4 x x x → x − + − = − + + = 1 1 1... 1 1. 2 2 3 2 Solution All the functions under consideration are rational functions. Hence, we first evaluate these functions at the prescribed points. If this is of the form 0 0 , we try to rewrite the function cancelling the factors which are causing the limit to be of the form 0 0 . (iii) 2 (v) 2 3 2 1 2 1 lim x 3 2 x → x x x x x − − − − + . 3 2 2 4 lim x 4 4 x → x x x − − + (iv) 3 2 Reprint 2025-26 2 lim x 5 6 x x → x x − − + 2 2 (ii) Evaluating the function at 2, it is of the form 0 0 . Hence 3 2 Hence 2 (iii) Evaluating the function at 2, we get it of the form 0 0 . (i) We have 2 2 3 2 2 4 lim x 4 4 x → x x x − − + = ( )( ) 4 4 lim x 4 x x x → x − + − = ( ) ( )( ) 2 2 2 1 1 1 2 lim x 100 1 100 101 x → x 1 + + = = + + = 2 2 2 ( ) 0 0 2 2 4 − = = + . = ( ) = ( ) 2 lim x 2 2 x x ( ) 2 2 lim as 2 x 2 x x x → x 2 2 lim 2 x ( ) ( ) 2 2 2 2 4 lim x 2 2 2 2 0 x → x x − + − x x → + − → x x + + = = − − 2 ( )2 2 x x − ≠ + − LIMITS AND DERIVATIVES 231 which is not defined. (iv) Evaluating the function at 2, we get it of the form 0 0 . Hence 3 2 2 lim x 5 6 x x → x x − 2 2 − + = ( ) = ( ) ( )2 2 Reprint 2025-26 2 lim x 2 3 x x 2 4 lim 4 x 3 2 3 1 x → x = = = − − − − . → x x − − − 2 2 ( ) ( ) 2 232 MATHEMATICS Evaluating the function at 1, we get it of the form 0 0 . Hence 2 (v) First, we rewrite the function as a rational function. 2 3 2 2 1 3 2 x x x x x x − − − − + = ( ) ( ) 2 2 1 1 3 2 x x x x x x 2 3 2 1 2 1 lim x 3 2 x → x x x x x − − − − + = ( )( ) 2 = ( ) ( ) ( ) 2 1 1 1 2 x x x x x x − − − − − = ( ) ( ) 2 4 4 1 1 2 x x x x x − + − − − = ( ) ( ) 2 4 3 1 2 x x x x x − + − − − − − − + = ( )( ) 4 3 lim x 1 2 x x → x x x − + − − ( ) ( ) 1 3 1 lim x 1 2 x x → x x x − − 1 − − We remark that we could cancel the term (x – 1) in the above evaluation because x ≠ 1 . Evaluation of an important limit which will be used in the sequel is given as a theorem below. Theorem 2 For any positive integer n, Remark The expression in the above theorem for the limit is true even if n is any rational number and a is positive. 1 lim n n n x a x a na x a − → Reprint 2025-26 − = − . = ( ) 1 3 lim x 2 x → x x − − = ( ) 1 3 1 1 2 − − = 2. Proof Dividing (x n – a n ) by (x – a), we see that Thus, lim lim n n Example 3 Evaluate: Solution (i) We have x n – a n = (x–a) (x n–1 + x n–2 a + x n–3 a 2 + ... + x an–2 + a n–1) x a x a x a → x a → − = − (x n–1 + x n–2 a + x n–3 a 2 + ... + x an–2 + a n–1) (i) 15 10 1 1 lim x 1 x → x = a n – l + a an–2 +. . . + a n–2 (a) +an–l = a n–1 + a n – 1 +...+a n–1 + a n–1 (n terms) = n 1 na − − − (ii) 0 1 1 lim x 10 1 1 lim x 1 x → x 15 − − = 15 10 = 15 10 = 15 (1)14 ÷ 10(1)9 (by the theorem above) 1 1 lim x 1 1 x x → x x − − ÷ − − 1 1 1 1 lim lim x 1 x 1 x x → x → x − − ÷ − − 1 → x + − LIMITS AND DERIVATIVES 233 x Then 0 1 1 lim x (ii) Put y = 1 + x, so that y →1 as x → 0. → x + − = 1 1 lim y –1 y → y x = 15 ÷ 10 3 2 = = = 1 1 2 1 (1) 2 Reprint 2025-26 1 1 lim y 1 y → y − − − (by the remark above) = 1 2 1 1 2 2 − 234 MATHEMATICS 12.4 Limits of Trigonometric Functions The following facts (stated as theorems) about functions in general come in handy in calculating limits of some trigonometric functions. Theorem 3 Let f and g be two real valued functions with the same domain such that f (x) ≤ g( x) for all x in the domain of definition, For some a, if both lim x a → f(x) and Theorem 4 (Sandwich Theorem) Let f, g and h be real functions such that f(x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if lim x a → f(x) = l = lim x a → h(x), then lim x a → g(x) = l. This is illustrated in Fig 12.9. lim x a → g(x) exist, then lim x a → f(x) ≤ lim x a → g(x). This is illustrated in Fig 12.8. Fig 12.8 inequality relating trigonometric functions. Given below is a beautiful geometric proof of the following important sin cos 1 x x x < < for π 0 2 < < x (*) Reprint 2025-26 Fig 12.9 Proof We know that sin (– x) = – sin x and cos( – x) = cos x. Hence, it is sufficient to prove the inequality for π 0 2 < 24. Find ( ) 1 lim x f x → , where ( ) 2 25. Evaluate ( ) 0 lim x f x → , where ( ) ( ) π sin π lim x π π tan 2 lim π 2 x → x − − 16. 0 cos lim x π → x − x x 1, 1 x x f x x x − ≤ = − − > x x f x x x ≠ = = 2 | | , 0 1, 1 0, 0 → − x 17. 0 cos 2 1 lim x cos 1 x → x − − x 26. Find ( ) 0 lim x f x → , where ( ) , 0 | | 0, 0 27. Find ( ) 5 lim x f x → , where f x x ( ) = − | | 5 28. Suppose ( ) and if 1 lim x→ f (x) = f (1) what are possible values of a and b? a bx x f x x b ax x + < = = − > , 1 4, 1 , 1 x x f x x x ≠ = = Reprint 2025-26 29. Let a1 , a2 , . . ., an be fixed real numbers and define a function 30. If ( ) 31. If the function f(x) satisfies ( ) 2 1 32. If ( ) 12.5 Derivatives What is 1 lim x a → f (x) ? For some a ≠ a1 , a2 , ..., an , compute limx a → f (x). For what value (s) of a does lim x a → f (x) exists? and ( ) 1 lim x f x → exist? f x x a x a x a ( ) = − − − ( 1 2 ) ( )...( n ) . mx n x f x nx m x x x f x x + < = = − > + < = + ≤ ≤ + > nx m x 1, 0 0, 0 x x 3 2 1, 0 , 0 , 0 1 , 1 . 2 lim π x 1 f x → x . For what integers m and n does both ( ) 0 lim x f x → − = − , evaluate ( ) 1 lim x f x → . LIMITS AND DERIVATIVES 239 We have seen in the Section 13.2, that by knowing the position of a body at various time intervals it is possible to find the rate at which the position of the body is changing. It is of very general interest to know a certain parameter at various instants of time and try to finding the rate at which it is changing. There are several real life situations where such a process needs to be carried out. For instance, people maintaining a reservoir need to know when will a reservoir overflow knowing the depth of the water at several instances of time, Rocket Scientists need to compute the precise velocity with which the satellite needs to be shot out from the rocket knowing the height of the rocket at various times. Financial institutions need to predict the changes in the value of a particular stock knowing its present value. In these, and many such cases it is desirable to know how a particular parameter is changing with respect to some other parameter. The heart of the matter is derivative of a function at a given point in its domain of definition. Reprint 2025-26 240 MATHEMATICS Definition 1 Suppose f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by provided this limit exists. Derivative of f (x) at a is denoted by f′(a). Observe that f′(a) quantifies the change in f(x) at a with respect to x. Example 5 Find the derivative at x = 2 of the function f(x) = 3x. Solution We have The derivative of the function 3x at x = 2 is 3. Example 6 Find the derivative of the function f(x) = 2x 2 + 3x – 5 at x = –1. Also prove that f ′ (0) + 3f ′ ( –1) = 0. Solution We first find the derivatives of f(x) at x = –1 and at x = 0. We have f ' 1 (− ) = ( ) ( ) f′ (2) = ( ) ( ) 1 1 lim h → h − + − − = 0 0 0 6 3 6 3 lim lim lim3 3 h h h h h → h h → → + − = = = . 0 2 2 lim h f h f → h + − = ( ) ( ) 0 3 2 3 2 lim h 0 lim h 0 → h + − f h f f a h f a ( ) ( ) → h + − h and f ' 0( ) = ( ) ( ) = ( ) ( ) ( ) ( ) 2 2 = ( ) ( ) 2 = ( ) ( ) ( ) ( ) 2 2 2 1 3 1 5 2 1 3 1 5 lim h 0 0 2 lim lim 2 1 2 0 1 1 h h h h h → h → − = − = − = − 2 0 3 0 5 2 0 3 0 5 lim h 0 0 lim h → h → h → h + − 0 0 0 − + + − + − − − + − − + + + − − + − f h f Reprint 2025-26 h h h h Clearly f f ' 0 3 ' 1 0 ( ) + − = ( ) Remark At this stage note that evaluating derivative at a point involves effective use of various rules, limits are subjected to. The following illustrates this. Example 7 Find the derivative of sin x at x = 0. Solution Let f(x) = sin x. Then Example 8 Find the derivative of f(x) = 3 at x = 0 and at x = 3. Solution Since the derivative measures the change in function, intuitively it is clear that the derivative of the constant function must be zero at every point. This is indeed, supported by the following computation. Similarly f ' 3( ) = ( ) ( ) f ' 0( ) = ( ) ( ) f ′(0) = ( ) ( ) = ( ) ( ) = ( ) ( ) 2 0 0 lim h 0 sin 0 sin 0 lim h 0 0 0 0 0 3 3 0 lim lim lim 0 h h h f h f 0 0 3 3 3 3 lim lim 0 h h f h f → h + − → h + − = 0 sin lim 1 h 0 0 2 3 lim lim 2 3 2 0 3 3 h h h h h → h → + = + = + = → h → h h → + − − = = = . → h → h + − − = = . 0 f h f h → h = LIMITS AND DERIVATIVES 241 h We now present a geometric interpretation of derivative of a function at a point. Let y = f(x) be a function and let P = (a, f(a)) and Q = (a + h, f(a + h) be two points close to each other on the graph of this function. The Fig 12.11 is now self explanatory. Reprint 2025-26 Fig 12.11 242 MATHEMATICS We know that ( ) ( ) ( ) 0 lim h precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, as h tends to 0, the point Q tends to P and we have curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence exists at every point, it defines a new function called the derivative of f . Formally, we define derivative of a function as follows. Definition 2 Suppose f is a real valued function, the function defined by wherever the limit exists is defined to be the derivative of f at x and is denoted by f′(x). This definition of derivative is also called the first principle of derivative. Thus ( ) ( ) ( ) 0 ' lim h From the triangle PQR, it is clear that the ratio whose limit we are taking is This is equivalent to the fact that the chord PQ tends to the tangent at P of the For a given function f we can find the derivative at every point. If the derivative 0 Q P QR lim lim h PR f a h f a 0 lim h f a ′( ) = tan ψ . f x h f x f x → h + − = → h → + − = → h + − f a h f a f a → h + − ′ = f x h f x ( ) ( ) ( ) ( ) are different notations for derivative of a function. Sometimes f′(x) is denoted by ( ) ( ) d f x dx or if y = f(x), it is denoted by dy dx . This is referred to as derivative of f(x) or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a is also denoted by ( ) or a a d df f x dx dx or even x a df dx = . Example 9 Find the derivative of f(x) = 10x. Solution Since f′ ( x) = ( ) ( ) Clearly the domain of definition of f′ (x) is wherever the above limit exists. There 0 lim h → h + − f x h f x Reprint 2025-26 Example 10 Find the derivative of f(x) = x 2 . Solution We have, f ′(x) = ( ) ( ) Example 11 Find the derivative of the constant function f (x) = a for a fixed real number a. Solution We have, f ′(x) = ( ) ( ) Example 12 Find the derivative of f(x) = 1 x = ( ) ( ) 2 2 = 0 0 0 lim lim 0 h h a a → h h → − = = as h ≠ 0 0 lim h = ( ) ( ) = 0 10 lim h → h + − = ( ) 0 lim 2 2 h h x x → + = 0 10 10 lim h → h + − → h = ( ) 0 lim 10 10 h→ = . x h x 0 lim h 0 lim h → h + − h → h + − x h x f x h f x f x h f x LIMITS AND DERIVATIVES 243 Solution We have f ′(x) = ( ) ( ) = 0 = ( ) ( ) 0 1 lim h 1 1 – ( ) lim h = ( ) 0 1 lim h → h + → h x x h − + + → h x x h − + = ( ) 0 1 lim h→ x x h − + = 2 1 x h x 0 lim h x x h → h + − f x h f x h Reprint 2025-26 x − 244 MATHEMATICS 12.5.1 Algebra of derivative of functions Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules for derivatives to follow closely that of limits. We collect these in the following theorem. Theorem 5 Let f and g be two functions such that their derivatives are defined in a common domain. Then (iv) Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero). (iii) Derivative of product of two functions is given by the following product rule. (ii) Derivative of difference of two functions is difference of the derivatives of the functions. (i) Derivative of sum of two functions is sum of the derivatives of the functions. ( ) ( ) ( ) ( ) d d d f x g x f x g x dx dx dx + = + . ( ) ( ) ( ) ( ) d d d f x g x f x g x dx dx dx − = − . ( ) ( ) . ( ) . ( ) ( ) . ( ) d d d f x g x f x g x f x g x dx dx dx = + d d f x g x f x g x d f x dx dx dx g x g x − = ( )2 ( ) . ( ) ( ) ( ) ( ) ( ) ( ) The proofs of these follow essentially from the analogous theorem for limits. We will not prove these here. As in the case of limits this theorem tells us how to compute derivatives of special types of functions. The last two statements in the theorem may be restated in the following fashion which aids in recalling them easily: product rule. Similarly, the quotient rule is Let u f x = ( ) and v = g (x). Then This is referred to a Leibnitz rule for differentiating product of functions or the ( ) uv ′ = u v uv ′ ′ + Reprint 2025-26 function 1. This is because f x ′( ) = ( ) ( ) We use this and the above theorem to compute the derivative of f(x) = 10x = x + .... + x (ten terms). By (i) of the above theorem = 1 ... 1 + + (ten terms) = 10. We note that this limit may be evaluated using product rule too. Write f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also derivative of v(x) = x equals 1. Thus by the product rule we have Now, let us tackle derivatives of some standard functions. It is easy to see that the derivative of the function f(x) = x is the constant f x ′( ) = ( ) ( ) 10x uv u v uv x 0. 10.1 10 ′ ′ = = + = + = ′ ′ ′ = 2 u v uv v ′ ′ − u v df x( ) dx = d dx ( x x + + ... ) (ten terms) = . . . d d x x dx dx + + (ten terms) = 0 lim1 1 h→ = . 0 lim h → h + − = 0 lim h f x h f x LIMITS AND DERIVATIVES 245 → h + − x h x f(x) = x 2 = x .x and hence Theorem 6 Derivative of f(x) = x n is nxn – 1 for any positive integer n. Proof By definition of the derivative function, we have On similar lines the derivative of f(x) = x 2 may be evaluated. We have = 1. .1 2 x x x + = . More generally, we have the following theorem. ( ) ( ) ( ) 0 ' lim h f x h f x f x → h + − = = ( ) df dx = ( ) . . . ( ) ( ) d d d x x x x x x dx dx dx = + Reprint 2025-26 0 lim n n h → h + − . x h x 246 MATHEMATICS hence (x + h) n – x n = h(nxn – 1 +... + h n – 1). Thus follows. The result is true for n = 1, which has been proved earlier. We have Remark The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here). Binomial theorem tells that (x + h) n = ( ) ( ) ( ) 1 C C ... C 0 1 n n n n n n n x x h h − + + + and Alternatively, we may also prove this by induction on n and the product rule as ( ) d n x dx = ( ) 1 . d n x x dx − df x( ) dx = ( ) = ( ) 1 1 = ( ) 1 1 0 lim ... n n h nx h − − → + + = n 1 nx − . = ( ) ( ) ( ) 1 1 . . d n d n x x x x dx dx − − + (by product rule) = (( ) ) 1 2 1. . 1 n n x x n x − − + − (by induction hypothesis) = ( ) 1 1 1 1 n n n x n x nx − − − + − = . 0 lim n n h .... lim n n h → h + − → 0 h nx h x h x h − − + + 12.5.2 Derivative of polynomials and trigonometric functions We start with the following theorem which tells us the derivative of a polynomial function. Theorem 7 Let f(x) = 1 1 1 0 .... n n n n a x a x a x a − + + + + − be a polynomial function, where ai s are all real numbers and an ≠ 0. Then, the derivative function is given by Example 13 Compute the derivative of 6x 100 – x 55 + x. Solution A direct application of the above theorem tells that the derivative of the above function is 99 54 600 55 1 x x − + . Proof of this theorem is just putting together part (i) of Theorem 5 and Theorem 6. ( ) 1 2 1 ( ) 1 ... n x n n df x na x n a x dx − − = + − + + − 2 1 2a x a + . Reprint 2025-26 Example 14 Find the derivative of f(x) = 1 + x + x 2 + x 3 +... + x 50 at x = 1. Solution A direct application of the above Theorem 6 tells that the derivative of the above function is 1 + 2x + 3x 2 + . . . + 50x 49 . At x = 1 the value of this function equals 1 + 2(1) + 3(1)2 + .. . + 50(1)49 = 1 + 2 + 3 + . . . + 50 = (50 51 )( ) Example 15 Find the derivative of f(x) = x 1 x + Solution Clearly this function is defined everywhere except at x = 0. We use the quotient rule with u = x + 1 and v = x. Hence u′ = 1 and v′ = 1. Therefore Example 16 Compute the derivative of sin x. Solution Let f(x) = sin x. Then df x d x d u ( ) 1 dx dx x dx v + = = ( ) ( ) 2 2 2 u v uv 1 1 1 x x 1 v x x ′ ′ − − + = = = − df x( ) dx = ( ) ( ) ( ) ( ) = 0 0 0 sin sin lim lim h h f x h f x x h x 2 2cos sin 2 2 lim h → h → h + − + − = → h + (using formula for sin A – sin B) x h h LIMITS AND DERIVATIVES 247 2 = 1275. Example 17 Compute the derivative of tan x. Solution Let f(x) = tan x. Then df x( ) dx = ( ) ( ) ( ) ( ) = 0 0 sin 2 lim cos .lim cos .1 cos 2 2 h h = ( ) 0 0 tan tan lim lim h h f x h f x x h x ( ) 0 1 sin sin lim h cos cos x h x h h x x x → → h + = = . → h → h + − + − = → h x h x + − + Reprint 2025-26 248 MATHEMATICS Example 18 Compute the derivative of f(x) = sin2 x. Solution We use the Leibnitz product rule to evaluate this. ( ) ( ) sin sin df x d x x dx dx = ( ) sin sin sin sin x x x x ( ) ′ ′ = + = (cos sin sin cos x x x x ) + ( ) = = 2sin cos sin 2 x x x . = ( ) ( ) ( ) 0 sin cos cos sin lim h cos cos x h x x h x = ( ) = ( ) 0 0 sin 1 lim .lim h h cos cos h → h x h x → + = 2 2 1 1. sec cos x x = . ( ) 0 sin lim h cos cos x h x → h x h x + − → h x h x + − + + + (using formula for sin (A + B)) 1. Find the derivative of x 2 – 2 at x = 10. 2. Find the derivative of x at x = 1. 3. Find the derivative of 99x at x = l00. 4. Find the derivative of the following functions from first principle. (iii) 2 1 x (iv) 1 1 x x + − 5. For the function (i) 3 x − 27 (ii) ( x x − − 1 2 )( ) ( ) 100 99 2 . . . 1 100 99 2 x x x f x = + + + + +x . EXERCISE 12.2 Reprint 2025-26 Prove that f f ′ ′ (1 100 0 ) = ( ). 10. Find the derivative of cos x from first principle. 11. Find the derivative of the following functions: 6. Find the derivative of 1 2 2 1 . . . n n n n n x ax a x a x a − − − + + + + + for some fixed real number a. 7. For some constants a and b, find the derivative of 8. Find the derivative of n n x a x a − − for some constant a. 9. Find the derivative of (i) ( x a x b − − ) ( ) (ii) ( ) 2 2 ax b + (iii) x a x b − − (i) sin cos x x (ii) sec x (iii) 5sec 4cos x x + (iv) cosec x (v) 3cot 5cosec x x + (iii) ( ) 3 x x 5 3 − + (iv) ( ) 5 9 x x 3 6 − − (v) ( ) 4 5 x x 3 4 − − − (vi) 2 2 1 3 1 x x x − + − (i) 3 2 4 x − (ii) ( ) ( ) 3 5 3 1 1 x x x + − − LIMITS AND DERIVATIVES 249 Example 19 Find the derivative of f from the first principle, where f is given by Solution (i) Note that function is not defined at x = 2. But, we have (vi) 5sin 6cos 7 x x − + (vii) 2tan 7sec x x − (i) f (x) = 2 3 2 x x x h x f x h f x x h x f x → h → h + + + − + − + − − ′ = = ( ) ( ) ( ) + − (ii) f (x) = 1 x x + 0 0 2 3 2 3 2 2 lim lim h h Miscellaneous Examples Reprint 2025-26 ( ) 250 MATHEMATICS Again, note that the function f ′ is also not defined at x = 2. (ii) The function is not defined at x = 0. But, we have f x ′( ) = ( ) ( ) = 0 1 1 1 lim h h → h x h x + − + = ( ) ( ) 0 0 1 1 1 lim lim 1 h h x x h h h → h x x h h x x h → − − + = − + + = ( )( ) ( )( ) = ( )( ) ( ) ( )( ) ( ) = ( ) ( ) ( )2 0 lim lim h h 1 1 lim 1 1 h→ x x h x − = − + ( ) ( ) 0 2 2 3 2 2 3 2 lim h 2 2 x h x x x h ( )( ) 0 2 3 2 2 2 2 3 2 2 3 lim h 2 2 x x h x x x h x –7 7 lim 2 2 2 h x x h x → = − − + − − → h → h → h x x h + + − − + + − − + − → h x x h + − + − − + − − + 0 0 x h x f x h f x x h x + + − + + − + = − + − 1 1 Again, note that the function f ′ is not defined at x = 0. Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin cos x x + (ii) x x sin Solution (i) we have f x '( ) = f x h f x ( ) ( ) h + − = ( ) 2 0 = ( ) ( ) = 0 sin cos cos sin cos cos sin sin sin cos lim h 0 sin cos sin cos lim h → h + + + − − → h + + − − − x h x h x h x h x x x h x h x x Reprint 2025-26 (ii) f x '( ) = ( ) ( ) ( ) ( ) = ( ) 0 0 = x x x cos sin + Example 21 Compute derivative of (i) f(x) = sin 2x (ii) g(x) = cot x Solution (i) Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus = ( ) ( ) ( ) = ( ) ( ) 0 0 sin cos 1 lim cos sin limsin h h h h x x x → h → h − − + ( ) = cos sin x x − df x( ) dx = ( ) 2sin cos 2 sin cos ( ) d d x x x x dx dx = sin cos 1 sin lim lim cos h h 0 sin cos sin sin cos 1 cos cos 1 lim h x x h h x x h h → → − + ( ) 0 lim sin cos sin cos h x h h x → + + → h − + − + − = ( )( ) = ( ) ( ) sin sin lim lim h h f x h f x x h x h x x sin cos sin cos sin lim h sin cos 1 cos sin sin cos sin cos lim h → h → h + − + + − = → h + + − → h − + + + h x x x h x h 0 0 0 0 x x h x x h h x h h x x h x h h x x x LIMITS AND DERIVATIVES 251 h x → h − + cos 1 lim cos h 0 (ii) By definition, g(x) = cos cot sin x x x = . We use the quotient rule on this function wherever it is defined. cos (cot ) sin dg d d x x dx dx dx x = = 2 sin cos sin cos ( ) x x x x ( ) ′ ′ = + = 2 cos cos sin sin ( x x x x ) + −( ) ( ) 2 2 = − 2 cos sin x x Reprint 2025-26 252 MATHEMATICS Alternatively, this may be computed by noting that 1 cot tan x x = . Here, we use the fact that the derivative of tan x is sec2 x which we saw in Example 17 and also that the derivative of the constant function is 0. dg dx = 1 (cot ) tan d d x dx dx x = = 2 (cos ) (sin ) (cos )(sin ) (sin ) x x x x x ′ − ′ = 2 ( sin )(sin ) (cos )(cos ) (sin ) x x x x x − − = 2 2 2 2 sin cos cosec sin x x x x + − = − = 2 (1) (tan ) (1)(tan ) = 2 = 2 2 2 sec cosec tan 2 (0)(tan ) (sec ) (tan ) x x x − x x x − = − x ′ − ′ (tan ) x x Example 22 Find the derivative of Solution (i) Let 5 cos ( ) sin x x h x x − = . We use the quotient rule on this function wherever it is defined. (i) 5 cos sin x x x − (ii) cos tan x x x + 2 ( cos ) sin ( cos )(sin ) ( ) (sin ) x x x x x x h x x − − − ′ ′ ′ = 5 5 Reprint 2025-26 (ii) We use quotient rule on the function cos tan x x x + wherever it is defined. p, q, r and s are fixed non-zero constants and m and n are integers): 1. Find the derivative of the following functions from first principle: 2. (x + a) 3. (px + q) r s x + 4. ( ) ( )2 ax b cx d + + (i) −x (ii) 1 ( ) x − − (iii) sin (x + 1) (iv) cos (x – π 8 ) Find the derivative of the following functions (it is to be understood that a, b, c, d, h x ′( ) = 2 ( cos ) tan ( cos )(tan ) (tan ) x x x x x x x + − + ′ ′ Miscellaneous Exercise on Chapter 12 = 4 5 = 5 4 = 2 2 (1 sin ) tan ( cos )sec (tan ) x x x x x x − − + 2 (5 sin )sin ( cos )cos sin x x x x x x x + − − 2 cos 5 sin 1 (sin ) x x x x x − + + LIMITS AND DERIVATIVES 253 14. sin (x + a) 15. cosec x cot x 16. cos 1 sin x + x 11. 4 2 x − 12. ( )n ax b + 13. ( ) ( ) n m ax b cx d + + 5. ax b cx d + + 6. 8. 2 ax b px qx r + + + 9. 2 px qx r ax b + + + 10. 4 2 cos a b x x x − + 1 1 1 1 + − 7. 2 1 ax bx c + + x x Reprint 2025-26 254 MATHEMATICS 17. sin cos sin cos x x x x + − 18. sec 1 sec 1 x x − + 19. sinn x 20. sin cos a b x c d x + + 21. sin( ) cos x a x + 22. 4 x x x (5sin 3cos ) − 23. ( ) 2 x x +1 cos 24. ( )( ) 2 ax x p q x + + sin cos 25. ( x x x x + − cos tan ) ( ) 26. 4 5sin 3 7cos x x x x + + 27. 28. 1 tan x + x 29. ( x x x x + − sec tan ) ( ) 30. sinn x x Summary ÆThe expected value of the function as dictated by the points to the left of a point defines the left hand limit of the function at that point. Similarly the right hand limit. ÆLimit of a function at a point is the common value of the left and right hand limits, if they coincide. ÆFor a function f and a real number a, lim x a → f(x) and f (a) may not be same (In fact, one may be defined and not the other one). ÆFor functions f and g the following holds: x 2 cos 4 sin π x = ÆFollowing are some of the standard limits lim ( ) ( ) lim ( ) lim ( ) [ ] x a x a x a f x g x f x g x → → → ± = ± lim ( ). ( ) lim ( ).lim ( ) [ ] x a x a x a f x g x f x g x → → → = lim ( ) ( ) lim ( ) lim ( ) x a x a x a f x f x g x g x → → → 1 lim n n n x a x a na x a − → − = − Reprint 2025-26 ÆThe derivative of a function f at a is defined by ÆDerivative of a function f at any point x is defined by ÆFor functions u and v the following holds: 2 u u v uv v v ′ ′ ′ − = provided all are defined. ÆFollowing are some of the standard derivatives. 0 sin lim 1 x 0 1 cos lim 0 x ( ) u v u v ± = ± ′ ′ ′ ( ) uv u v uv ′ ′ ′ = + 1 ( ) d n n x nx dx − = f a h f a f a → h + − ′ = 0 ( ) ( ) ( ) ( ) lim h df x f x h f x f x dx → h + − ′ = = → x = → x − = ( ) ( ) ( ) lim h x 0 x LIMITS AND DERIVATIVES 255 In the history of mathematics two names are prominent to share the credit for inventing calculus, Issac Newton (1642 – 1727) and G.W. Leibnitz (1646 – 1717). Both of them independently invented calculus around the seventeenth century. After the advent of calculus many mathematicians contributed for further development of calculus. The rigorous concept is mainly attributed to the great (sin ) cos d x x dx = (cos ) sin d x x dx = − Historical Note Reprint 2025-26 256 MATHEMATICS mathematicians, A.L. Cauchy, J.L.Lagrange and Karl Weierstrass. Cauchy gave the foundation of calculus as we have now generally accepted in our textbooks. Cauchy used D’Alembert’s limit concept to define the derivative of a function. Starting with definition of a limit, Cauchy gave examples such as the limit of Before 1900, it was thought that calculus is quite difficult to teach. So calculus became beyond the reach of youngsters. But just in 1900, John Perry and others in England started propagating the view that essential ideas and methods of calculus were simple and could be taught even in schools. F.L. Griffin, pioneered the teaching of calculus to first year students. This was regarded as one of the most daring act in those days. Today not only the mathematics but many other subjects such as Physics, Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus. i → 0, the “function derive’e, y′ for f ′ (x)”. sinα α for α = 0. He wrote ( ) ( ) , y f x i f x x i ∆ + − = ∆ and called the limit for — v — Reprint 2025-26" class_11,13,Statistics,ncert_books/class_11/kemh1dd/kemh113.pdf,"13.1 Introduction We know that statistics deals with data collected for specific purposes. We can make decisions about the data by analysing and interpreting it. In earlier classes, we have studied methods of representing data graphically and in tabular form. This representation reveals certain salient features or characteristics of the data. We have also studied the methods of finding a representative value for the given data. This value is called the measure of central tendency. Recall mean (arithmetic mean), median and mode are three measures of central tendency. A measure of central tendency gives us a rough idea where data points are centred. But, in order to make better interpretation from the data, we should also have an idea how the data are scattered or how much they are bunched around a measure of central tendency. Consider now the runs scored by two batsmen in their last ten matches as follows: v“Statistics may be rightly called the science of averages and their estimates.” – A.L.BOWLEY & A.L. BODDINGTON v STATISTICS Chapter 13 Karl Pearson (1857-1936) Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52 Clearly, the mean and median of the data are Batsman A Batsman B Mean 53 53 Median 53 53 Recall that, we calculate the mean of a data (denoted by x ) by dividing the sum of the observations by the number of observations, i.e., Reprint 2025-26 258 MATHEMATICS order and applying the following rule. B are same i.e., 53. Can we say that the performance of two players is same? Clearly No, because the variability in the scores of batsman A is from 0 (minimum) to 117 (maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60. Let us now plot the above scores as dots on a number line. We find the following diagrams: For batsman A For batsman B th 1 2 n + observations. Also, the median is obtained by first arranging the data in ascending or descending If the number of observations is odd, then the median is th 1 2 n + observation. If the number of observations is even, then median is the mean of th We find that the mean and median of the runs scored by both the batsmen A and 1 1 n i i x x n = = ∑ Fig 13.1 2 n and are clustering around the measure of central tendency (mean and median), while those corresponding to batsman A are scattered or more spread out. Thus, the measures of central tendency are not sufficient to give complete information about a given data. Variability is another factor which is required to be studied under statistics. Like ‘measures of central tendency’ we want to have a single number to describe variability. This single number is called a ‘measure of dispersion’. In this Chapter, we shall learn some of the important measures of dispersion and their methods of calculation for ungrouped and grouped data. We can see that the dots corresponding to batsman B are close to each other and Reprint 2025-26 Fig 13.2 13.2 Measures of Dispersion The dispersion or scatter in a data is measured on the basis of the observations and the types of the measure of central tendency, used there. There are following measures of dispersion: (i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation. In this Chapter, we shall study all of these measures of dispersion except the quartile deviation. 13.3 Range Recall that, in the example of runs scored by two batsmen A and B, we had some idea of variability in the scores on the basis of minimum and maximum runs in each series. To obtain a single number for this, we find the difference of maximum and minimum values of each series. This difference is called the ‘Range’ of the data. In case of batsman A, Range = 117 – 0 = 117 and for batsman B, Range = 60 – 46 = 14. Clearly, Range of A > Range of B. Therefore, the scores are scattered or dispersed in case of A while for B these are close to each other. Thus, Range of a series = Maximum value – Minimum value. The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency. For this purpose, we need some other measure of variability. Clearly, such measure must depend upon the difference (or deviation) of the values from the central tendency. The important measures of dispersion, which depend upon the deviations of the observations from a central tendency are mean deviation and standard deviation. Let us discuss them in detail. 13.4 Mean Deviation STATISTICS 259 Recall that the deviation of an observation x from a fixed value ‘a’ is the difference x – a. In order to find the dispersion of values of x from a central value ‘a’ , we find the deviations about a. An absolute measure of dispersion is the mean of these deviations. To find the mean, we must obtain the sum of the deviations. But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations. Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations from mean ( x ) is zero. Also Mean of deviations Sum of deviations 0 0 Number of observations n = = = as the measure of dispersion is concerned. Thus, finding the mean of deviations about mean is not of any use for us, as far Reprint 2025-26 260 MATHEMATICS of each value from a central tendency or a fixed number ‘a’. Recall, that the absolute value of the difference of two numbers gives the distance between the numbers when represented on a number line. Thus, to find the measure of dispersion from a fixed number ‘a’ we may take the mean of the absolute values of the deviations from the central value. This mean is called the ‘mean deviation’. Thus mean deviation about a central value ‘a’ is the mean of the absolute values of the deviations of the observations from ‘a’. The mean deviation from ‘a’ is denoted as M.D. (a). Therefore, Remark Mean deviation may be obtained from any measure of central tendency. However, mean deviation from mean and median are commonly used in statistical studies. Let us now learn how to calculate mean deviation about mean and mean deviation about median for various types of data 13.4.1 Mean deviation for ungrouped data Let n observations be x1 , x2 , x3 , ...., xn . The following steps are involved in the calculation of mean deviation about mean or median: Step 1 Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘a’. Step 2 Find the deviation of each x i from a, i.e., x 1 – a, x 2 – a, x 3 – a,. . . , x n – a Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is Remember that, in finding a suitable measure of dispersion, we require the distance M.D.(a) = Sum of absolute values of deviations from ' ' Number of observations a . Step 4 Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e., Thus M.D. ( x ) = 1 1 n i i x x n = ∑ − , where x = Mean and M.D. (M) = 1 1 M n i i x n = ∑ − , where M = Median there, i.e., axaxaxax1 2 3 ....,,,, n −−−− 1 M.D.( ) n i i x a a n = − = ∑ Reprint 2025-26 Example 1 Find the mean deviation about the mean for the following data: Solution We proceed step-wise and get the following: Step 1 Mean of the given data is Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are 6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9, or –3, –2, 1, 3, 4, –5, –1, 3 Step 3 The absolute values of the deviations, i.e., i x x − are Step 4 The required mean deviation about the mean is ANote In this Chapter, we shall use the symbol M to denote median unless stated otherwise.Let us now illustrate the steps of the above method in following examples. 3, 2, 1, 3, 4, 5, 1, 3 M.D. ( x ) = 6 7 10 12 13 4 8 12 72 9 8 8 x + + + + + + + = = = = 3 2 1 3 4 5 1 3 22 2 75 8 8 . + + + + + + + = = 6, 7, 10, 12, 13, 4, 8, 12 1 8 i i x x = ∑ − 8 STATISTICS 261 Example 2 Find the mean deviation about the mean for the following data : Solution We have to first find the mean ( x ) of the given data ANote Instead of carrying out the steps every time, we can carry on calculation, step-wise without referring to steps. 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5 1 1 20 i i x x = = ∑ = 20 200 = 10 20 Reprint 2025-26 262 MATHEMATICS The respective absolute values of the deviations from mean, i.e., xxi − are 2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5 Example 3 Find the mean deviation about the median for the following data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Solution Here the number of observations is 11 which is odd. Arranging the data into ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 Now Median = th 11 1 2 + or 6th observation = 9 The absolute values of the respective deviations from the median, i.e., xi − M are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12 Therefore 11 and ( ) 11 Therefore 20 and M.D. ( x ) = 124 20 = 6.2 1 124 i i x x = ∑ − = 1 1 1 M.D. M M 58 5.27 11 11 i i x = = − = × = ∑ 1 M 58 i i x = ∑ − = 13.4.2 Mean deviation for grouped data We know that data can be grouped into two ways : (a) Discrete frequency distribution Let the given data consist of n distinct values x 1 , x 2 , ..., x n occurring with frequencies f 1 , f 2 , ..., f n respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution: x : x1 x2 x3 ... xn (a) Discrete frequency distribution, (b) Continuous frequency distribution. Let us discuss the method of finding mean deviation for both types of the data. f : f 1 f2 f 3 ... f n Reprint 2025-26 (i) Mean deviation about mean First of all we find the mean x of the given data by using the formula where ∑= n frequencies f i and ∑ = = n Then, we find the deviations of observations xi from the mean x and take their absolute values, i.e., xxi − for all i =1, 2,..., n. After this, find the mean of the absolute values of the deviations, which is the required mean deviation about the mean. Thus (ii) Mean deviation about median To find mean deviation about median, we find the median of the given discrete frequency distribution. For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify i ii fx 1 denotes the sum of the products of observations xi with their respective i i f 1 N is the sum of the frequencies. M.D. ( ) x f x x f f = = = = ∑ ∑ ∑ , f x x x f = i i n i n i i i i i n 1 − = = 1 ∑ i i i n i i n 1 ∑ 1 N = 1 1 = xxf i n i ∑ i − N =1 1 STATISTICS 263 the observation whose cumulative frequency is equal to or just greater than N N is the sum of frequencies. This value of the observation lies in the middle of the data, therefore, it is the required median. After finding median, we obtain the mean of the absolute values of the deviations from median.Thus, Example 4 Find mean deviation about the mean for the following data : xi 2 5 6 8 10 12 f i 2 8 10 7 8 5 1 1 M.D.(M) M N i i i f x = = ∑ − Reprint 2025-26 n 2 , where 264 MATHEMATICS Solution Let us make a Table 13.1 of the given data and append other columns after calculations. Table 13.1 Therefore 6 and 6 Example 5 Find the mean deviation about the median for the following data: xi f i f i xi xxi − f i xxi − 10 8 80 2.5 20 12 5 60 4.5 22.5 2 2 4 5.5 11 5 8 40 2.5 20 6 10 60 1.5 15 8 7 56 0.5 3.5 N 40 6 1 1 1 300 7.5 N 40 i i i x f x = = = × = ∑ 1 1 1 M. D. ( ) 92 2.3 N 40 i i i x f x x = = ∑ − = × = 1 ∑ == i= i f , 300 6 40 300 92 1 ∑ = i= ii xf , 92 6 1 ∑ =− = xxf i i i Solution The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get (Table 13.2). Now, N=30 which is even. f i 3 4 5 2 4 5 4 3 c.f. 3 7 12 14 18 23 27 30 x i 3 6 9 12 13 15 21 22 xi 3 6 9 12 13 15 21 22 f i 3 4 5 2 4 5 4 3 Reprint 2025-26 Table 13.2 lie in the cumulative frequency 18, for which the corresponding observation is 13. Now, absolute values of the deviations from median, i.e., xi − M are shown in Table 13.3. Table 13.3 We have 8 8 Therefore 8 (b) Continuous frequency distribution A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies. For example, marks obtained by 100 students are presented in a continuous frequency distribution as follows : th th 15 observation 16 observation 13 13 Therefore, Median M 13 2 2 + + = = = Median is the mean of the 15th and 16th observations. Both of these observations f i xi − M 30 28 20 2 0 10 32 27 xi − M 10 7 4 1 0 2 8 9 f i 3 4 5 2 4 5 4 3 1 1 i 30 and M 149 i i i i f f x = = ∑ ∑ = − = 1 1 M. D. (M) M N i i i f x = = ∑ − = 1 149 30 × = 4.97. STATISTICS 265 (i) Mean deviation about mean While calculating the mean of a continuous frequency distribution, we had made the assumption that the frequency in each class is centred at its mid-point. Here also, we write the mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation. Let us take the following example. Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 Number of Students 12 18 27 20 17 6 Reprint 2025-26 266 MATHEMATICS Example 6 Find the mean deviation about the mean for the following data. Solution We make the following Table 13.4 from the given data : Here 7 7 7 Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Number of students 2 3 8 14 8 3 2 Marks Number of Mid-points f i x i xxi − f i xxi − obtained students 10-20 2 15 30 30 60 20-30 3 25 75 20 60 30-40 8 35 280 10 80 40-50 14 45 630 0 0 50-60 8 55 440 10 80 60-70 3 65 195 20 60 70-80 2 75 150 30 60 1 1 1 N 40, 1800, i i i i i 400 i i i f f x f x x = = = = = = ∑ ∑ ∑ − = f i xi 40 1800 400 Table 13.4 Therefore 7 and ( ) 7 Shortcut method for calculating mean deviation about mean We can avoid the tedious calculations of computing x by following step-deviation method. Recall that in this method, we take an assumed mean which is in the middle or just close to it in the data. Then deviations of the observations (or mid-points of classes) are taken from the 1 1 1800 45 N 40 i i i x f x = = = = ∑ 1 1 1 M.D. 400 10 N 40 i i i x f x x = = − = × = ∑ Reprint 2025-26 assumed mean. This is nothing but the shifting of origin from zero to the assumed mean on the number line, as shown in Fig 13.3 If there is a common factor of all the deviations, we divide them by this common factor to further simplify the deviations. These are known as step-deviations. The process of taking step-deviations is the change of scale on the number line as shown in Fig 13.4 Let us take the data of Example 6 and find the mean deviation by using stepdeviation method. Fig 13.3 Fig 13.4 STATISTICS 267 The deviations and step-deviations reduce the size of the observations, so that the computations viz. multiplication, etc., become simpler. Let, the new variable be denoted by h ax d i i − = , where ‘a’ is the assumed mean and h is the common factor. Then, the mean x by step-deviation method is given by n f d i i i x a h ∑ = = + × Reprint 2025-26 1 N 268 MATHEMATICS Take the assumed mean a = 45 and h = 10, and form the following Table 13.5. Therefore Number of students Marks obtained f i x i 10-20 2 15 – 3 – 6 30 60 20-30 3 25 – 2 – 6 20 60 30-40 8 35 – 1 – 8 10 80 40-50 14 45 0 0 0 0 50-60 8 55 1 8 10 80 60-70 3 65 2 6 20 60 70-80 2 75 3 6 30 60 40 0 400 Mid-points 45 10 i i x d − = i i f d xxi − f i xxi − = 0 45 10 45 40 + × = f d i i i x a h ∑ = = + × Table 13.5 7 1 N and 7 (ii) Mean deviation about median The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations. Let us recall the process of finding median for a continuous frequency distribution. The data is first arranged in ascending order. Then, the median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula ANote The step deviation method is applied to compute x . Rest of the procedure is same. 1 1 400 M D ( ) 10 N 40 i i i x f x x = . . = ∑ − = = Reprint 2025-26 where median class is the class interval whose cumulative frequency is just greater than or equal to N 2 , N is the sum of frequencies, l, f, h andC are,respectively the lower limit , the frequency, the width of the median class and C the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values of the deviations of mid-point x i of each class from the median i.e., xi − M are obtained. Then 1 M.D. (M) M N 1 n f x i i i = − ∑ = The process is illustrated in the following example: Example 7 Calculate the mean deviation about median for the following data : Solution Form the following Table 13.6 from the given data : Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 6 7 15 16 4 2 N C Median 2 l h f − = + × Table 13.6 STATISTICS 269 Class Frequency Cumulative Mid-points x Med. i − f i x Med. i − f i (c.f.) xi 0-10 6 6 5 23 138 10-20 7 13 15 13 91 20-30 15 28 25 3 45 30-40 16 44 35 7 112 40-50 4 48 45 17 68 50-60 2 50 55 27 54 50 508 frequency Reprint 2025-26 270 MATHEMATICS The class interval containing th N 2 or 25th item is 20-30. Therefore, 20–30 is the median class. We know that Here l = 20, C = 13, f = 15, h = 10 and N = 50 Therefore, Median 25 13 20 10 15 − = + × = 20 + 8 = 28 Thus, Mean deviation about median is given by Find the mean deviation about the mean for the data in Exercises 1 and 2. 1. 4, 7, 8, 9, 10, 12, 13, 17 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Find the mean deviation about the median for the data in Exercises 3 and 4. 3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 M.D. (M) = 6 Median = 1 1 M N i i i f x = ∑ − = 1 508 50 × = 10.16 N C 2 l h f − + × EXERCISE 13.1 Find the mean deviation about the mean for the data in Exercises 5 and 6. 5. xi 5 10 15 20 25 Find the mean deviation about the median for the data in Exercises 7 and 8. f i 7 4 6 3 5 6. xi 10 30 50 70 90 7. xi 5 7 9 10 12 15 f i 8 6 2 2 2 6 8. xi 15 21 27 30 35 f i 4 24 28 16 8 f i 3 5 6 7 8 Reprint 2025-26 Find the mean deviation about the mean for the data in Exercises 9 and 10. 9. Income per 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 day in ` Number 4 8 9 10 7 5 4 3 of persons 10. Height 95-105 105-115 115-125 125-135 135-145 145-155 in cms Number of 9 13 26 30 12 10 boys 12. Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 (in years) Number 5 6 12 14 26 12 16 9 [Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval] 13.4.3 Limitations of mean deviation In a series, where the degree of variability is very high, the median is not a representative central tendency. Thus, the mean deviation about median calculated for such series can not be fully relied. The sum of the deviations from the mean (minus signs ignored) is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion. 11. Find the mean deviation about median for the following data : Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of 6 8 14 16 4 2 Girls STATISTICS 271 13.5 Variance and Standard Deviation Recall that while calculating mean deviation about mean or median, the absolute values of the deviations were taken. The absolute values were taken to give meaning to the mean deviation, otherwise the deviations may cancel among themselves. Another way to overcome this difficulty which arose due to the signs of deviations, is to take squares of all the deviations. Obviously all these squares of deviations are Reprint 2025-26 272 MATHEMATICS non-negative. Let x1 , x2 , x3 , ..., xn be n observations and x be their mean. Then If this sum is zero, then each xx )( i − has to be zero. This implies that there is no dispersion at all as all observations are equal to the mean x . close to the mean x and therefore, there is a lower degree of dispersion. On the contrary, if this sum is large, there is a higher degree of dispersion of the observations from the mean x . Can we thus say that the sum ∑= − n of the degree of dispersion or scatter? Let us take the set A of six observations 5, 15, 25, 35, 45, 55. The mean of the observations is x = 30. The sum of squares of deviations from x for this set is = 625 + 225 + 25 + 25 + 225 + 625 = 1750 Let us now take another set B of 31 observations 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45. The mean of these observations is y = 30 Note that both the sets A and B of observations have a mean of 30. Now, the sum of squares of deviations of observations for set B from the mean y is given by If ∑= − n ∑ = − 6 2 )( i i xx = (5–30)2 + (15–30)2 + (25–30)2 + (35–30)2 + (45–30)2 +(55–30)2 i i xx 1 1 2 )( is small , this indicates that the observations x1 , x2 , x3 ,...,xn are 2 2 2 2 2 1 1 ( ) ( ) ....... ( ) ( ) n n i i x x x x x x x x = − + − + + − = − . i i xx 1 2 )( is a reasonable indicator (Because sum of squares of first n natural numbers = ( 1) (2 1) 6 n n n + + . Here n = 15) ∑= − 31 2 )( i i yy = (15–30)2 +(16–30)2 + (17–30)2 + ...+ (44–30)2 +(45–30)2 1 = (–15)2 +(–14)2 + ...+ (–1)2 + 02 + 12 + 22 + 32 + ...+ 142 + 152 = 2 [152 + 142 + ... + 12 ] = 15 (15 1) (30 1) 2 6 × + + × = 5 × 16 × 31 = 2480 Reprint 2025-26 will tend to say that the set A of six observations has a lesser dispersion about the mean than the set B of 31 observations, even though the observations in set A are more scattered from the mean (the range of deviations being from –25 to 25) than in the set B (where the range of deviations is from –15 to 15). This is also clear from the following diagrams. For the set A, we have For the set B, we have Thus, we can say that the sum of squares of deviations from the mean is not a proper measure of dispersion. To overcome this difficulty we take the mean of the squares of the deviations, i.e., we take ∑= − n If ∑= − n i i xx 1 2 )( is simply our measure of dispersion or scatter about mean, we i i xx n 1 2 )( 1 . In case of the set A, we have Fig 13.5 Fig 13.6 STATISTICS 273 1 Mean 6 = × 1750 = 291.67 and in case of the set B, it is 1 31× 2480 = 80. This indicates that the scatter or dispersion is more in set A than the scatter or dispersion in set B, which confirms with the geometrical representation of the two sets. of dispersion. This number, i.e., mean of the squares of the deviations from mean is called the variance and is denoted by 2 σ (read as sigma square). Therefore, the variance of n observations x1 , x2 ,..., xn is given by Thus, we can take ∑ − 2 )( 1 xx n i as a quantity which leads to a proper measure Reprint 2025-26 274 MATHEMATICS 13.5.1 Standard Deviation In the calculation of variance, we find that the units of individual observations xi and the unit of their mean x are different from that of variance, since variance involves the sum of squares of (x i – x ). For this reason, the proper measure of dispersion about the mean of a set of observations is expressed as positive square-root of the variance and is called standard deviation. Therefore, the standard deviation, usually denoted by σ , is given by hence, standard deviation of ungrouped data. Example 8 Find the variance of the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 Solution From the given data we can form the following Table 13.7. The mean is calculated by step-deviation method taking 14 as assumed mean. The number of observations is n = 10 Table 13.7 Let us take the following example to illustrate the calculation of variance and xi 14 2 i i x d − = (xi – x ) i i xx n 1 2 )( 1 σ ... (1) ∑= = − n i i xx n 1 2 2 )( 1 σ ∑ = −= n Deviations from mean (xi – x ) 10 –2 –5 25 12 –1 –3 9 14 0 –1 1 16 1 1 1 18 2 3 9 20 3 5 25 22 4 7 49 24 5 9 81 5 330 6 –4 –9 81 8 –3 –7 49 Reprint 2025-26 Therefore Mean x = assumed mean + h n d n and Variance ( 2 σ ) = 10 2 Thus Standard deviation (σ ) = 33 5 74 = . 13.5.2 Standard deviation of a discrete frequency distribution Let the given discrete frequency distribution be x : x1 , x2 , x3 ,. . . , xn f : f 1 , f2 , f 3 ,. . . , f n In this case standard deviation ( ) 2 1 1 ( ) N where 1 N n i i f = =∑ . Example 9 Find the variance and standard deviation for the following data: Let us take up following example. x i 4 8 11 17 20 24 32 f i 3 5 9 5 4 3 1 i i i σ f x x = = − ∑ ... (2) 1 1 ) i i ( x x n = ∑ − = 1 330 10 × = 33 n i i × ∑ =1 = 5 14 2 15 10 + × = STATISTICS 275 Solution Presenting the data in tabular form (Table 13.8), we get 4 3 12 –10 100 300 8 5 40 –6 36 180 11 9 99 –3 9 81 17 5 85 3 9 45 20 4 80 6 36 144 24 3 72 10 100 300 32 1 32 18 324 324 30 420 1374 xi f i f i xi xi – x 2 xx )( i − f i 2 xx )( i − Reprint 2025-26 Table 13.8 276 MATHEMATICS Therefore Hence variance 2 ( ) σ = 7 2 and Standard deviation σ = 8.45)( = 6.77 13.5.3 Standard deviation of a continuous frequency distribution The given continuous frequency distribution can be represented as a discrete frequency distribution by replacing each class by its mid-point. Then, the standard deviation is calculated by the technique adopted in the case of a discrete frequency distribution. If there is a frequency distribution of n classes each class defined by its mid-point xi with frequency f i , the standard deviation will be obtained by the formula i i i σ f x x = = − ∑ , N = 30, ( ) 7 7 2 1 1 420 14 N 30 i i i f x x = = = × = ∑ 1 1 ( ) N 7 n 1 1 420, 1374 i i i i i i f x f x x = = ∑ ∑ = − = = 1 30 × 1374 = 45.8 1 1 ( ) N i i i f x x = ∑ − 2 where x is the mean of the distribution and 1 N n i i f = =∑ . Another formula for standard deviation We know that Variance 2 ( ) σ = 2 1 1 ( ) N = 2 2 = 2 2 1 1 1 1 2 N 1 1 1 1 2 N i i i f x x = ∑ − = 2 2 1 1 ( 2 ) N i i i i i i i i f x x f x f x = = = + − ∑ ∑ ∑ i i i i i i i i f x x f x x f = = = + − ∑ ∑ ∑ n n n n n n n Reprint 2025-26 i i i i f x x x x = ∑ + − n or 2 σ = Thus, standard deviation ( ) 2 2 Example 10 Calculate the mean, variance and standard deviation for the following distribution : Solution From the given data, we construct the following Table 13.9. Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2 Class Frequency Mid-point f i xi (xi – x ) 2 f i (xi – x ) 2 = 2 2 1 1 N 2 . N N = + − n i i i f x x x x 1 1 1 Here or N N = 2 2 2 1 2 1 N 2 =1 2 2 1 1 =1 1 1 N N N N ∑ ∑ ∑ ∑ n i i n n i i i i i i i i i i f x f x f x f x − = i i i f x x x = ∑ + − 2 2 1 1 N n − = − i i i i i i σ f x f x = = − ∑ ∑ ... (3) n 1 =1 1 N N 2 Table 13.9 n n i i i f x x = = ∑ − i i i i i i x f x x f x = = = = ∑ ∑ n n n STATISTICS 277 2 30-40 3 35 105 729 2187 40-50 7 45 315 289 2023 50-60 12 55 660 49 588 60-70 15 65 975 9 135 70-80 8 75 600 169 1352 80-90 3 85 255 529 1587 90-100 2 95 190 1089 2178 (f i ) (xi ) 50 3100 10050 Reprint 2025-26 278 MATHEMATICS Thus 7 and Standard deviation (σ ) = = 201 14 18 . Example 11 Find the standard deviation for the following data : Solution Let us form the following Table 13.10: Variance ( ) 2 σ = 7 2 1 1 3100 Mean 62 N 50 i i i x f x = = ∑ = = x i 3 8 13 18 23 f i 7 10 15 10 6 13 15 195 169 2535 18 10 180 324 3240 = 1 10050 50 × = 201 xi f i f i xi xi 2 f i xi 2 3 7 21 9 63 8 10 80 64 640 1 1 ( ) N i i i f x x = ∑ − Table 13.10 Now, by formula (3), we have σ = ( )2 1 2 N N i i i i ∑ ∑ f x f x − 23 6 138 529 3174 = 1 2 48 9652 (614) 48 × − = 1 463296 376996 48 − 48 614 9652 Reprint 2025-26 Therefore, Standard deviation (σ ) = 6.12 13.5.4. Shortcut method to find variance and standard deviation Sometimes the values of xi in a discrete distribution or the mid points xi of different classes in a continuous distribution are large and so the calculation of mean and variance becomes tedious and time consuming. By using step-deviation method, it is possible to simplify the procedure. width of class-intervals). Let the step-deviations or the new values be yi . i.e. i A i x y h − = or xi = A + hyi ... (1) We know that 1 N Replacing x i from (1) in (2), we get Let the assumed mean be ‘A’ and the scale be reduced to h 1 times (h being the x = 1 A ) i i i f ( hy = ∑ + n N i i i f x x = = ∑ ... (2) = 1 293 77 48 × . = 6.12 n STATISTICS 279 Thus x = A + h y ... (3) Now Variance of the variable x, 2 2 = 1 1 1 A N = N 1 A N N = 2 1 1 (A A ) N i i i f y . h = + ∑ 1 because N n i i f = = ∑ i i i i i f h f y = = + ∑ ∑ = 1 1 1 A N i n n i i i i f h f y = = + ∑ ∑ i i i f hy h y = ∑ + − − (Using (1) and (3)) n n n n Reprint 2025-26 x i i i σ f ( x x = = − ∑ 1 1 ) N n 280 MATHEMATICS i.e. 2 σ x = 2 2 h σ y or σ x = hσ y ... (4) From (3) and (4), we have Examples 12 Calculate mean, variance and standard deviation for the following distribution. Solution Let the assumed mean A = 65. Here h = 10 We obtain the following Table 13.11 from the given data : Let us solve Example 11 by the short-cut method and using formula (5) Classes 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2 σ x = = 2 2 = 2 2 1 1 ( ) N i i i h f y y = ∑ − = h 2 × variance of the variable yi 1 1 N N 1 ( ) N i i i i i i h f y f y = = − ∑ ∑ ... (5) i i i f h y y = ∑ − n n n n 2 2 Table 13.11 f i x i 30-40 3 35 – 3 9 – 9 27 40-50 7 45 – 2 4 – 14 28 50-60 12 55 – 1 1 – 12 12 60-70 15 65 0 0 0 0 70-80 8 75 1 1 8 8 80-90 3 85 2 4 6 12 9 0-100 2 95 3 9 6 18 Class Frequency Mid-point yi = 65 10 i x − yi 2 f i yi f i yi 2 N=50 – 15 105 Reprint 2025-26 Therefore x = 15 A 65 10 62 50 50 i i f y + × = − × = ∑ h Variance 2 σ = ( ) 2 2 2 N 2 N i i h f y f y i i ∑ ∑ − and standard deviation (σ ) = 201 = 14.18 Find the mean and variance for each of the data in Exercies 1 to 5. 1. 6, 7, 10, 12, 13, 4, 8, 12 2. First n natural numbers 3. First 10 multiples of 3 4. x i 6 10 14 18 24 28 30 f i 2 4 7 12 8 4 3 = 1 [5250 225] 201 25 − = = ( )2 10 2 50 105 (–15) 2 (50) × − EXERCISE 13.2 STATISTICS 281 5. xi 92 93 97 98 102 104 109 6. Find the mean and standard deviation using short-cut method. Find the mean and variance for the following frequency distributions in Exercises 7 and 8. 7. Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2 xi 60 61 62 63 64 65 66 67 68 f i 2 1 12 29 25 12 10 4 5 f i 3 2 3 2 6 3 3 Reprint 2025-26 282 MATHEMATICS 8. Classes 0-10 10-20 20-30 30-40 40-50 9. Find the mean, variance and standard deviation using short-cut method 10. The diameters of circles (in mm) drawn in a design are given below: Calculate the standard deviation and mean diameter of the circles. [ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.] Example 13 The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations. Solution Let the observations be x 1 , x 2 , ..., x 20 and x be their mean. Given that variance = 5 and n = 20. We know that Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25 Frequencies 5 8 15 16 6 Height 70-75 75-80 80-85 85-90 90-95 95-100 100-105105-110 110-115 in cms No. of 3 4 7 7 15 9 6 6 3 children Miscellaneous Examples or 20 2 ... (1) If each observation is multiplied by 2, and the new resulting observations are yi , then Variance ( ) 2 20 2 yi = 2xi i.e., xi = i y 2 1 1 ( ) i i x x = ∑ − = 100 1 1 ( ) i i x x n σ = = − ∑ , i.e., 20 2 Reprint 2025-26 1 1 5 ( ) 20 i i x x = = − ∑ Therefore 20 20 i.e. y = 2 x or x = y 2 1 Substituting the values of xi and x in (1), we get Thus the variance of new observations = 1 2 400 20 2 5 20 × = = × Example14 The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations. Solution Let the other two observations be x and y. Therefore, the series is 1, 2, 6, x, y. Now Mean x = 4.4 = 1 2 6 5 + + + +x y or 22 = 9 + x + y Therefore x + y = 13 ... (1) ANote The reader may note that if each observation is multiplied by a constant k, the variance of the resulting observations becomes k 2 times the original variance. 1 1 100 2 2 i i y y = − = ∑ , i.e., ∑= =− 20 2 20 1 1 1 1 1 2 20 i i i i y y x n = = = = ∑ ∑ = 20 2 400)( i i yy 1 1 1 2 20 i i . x = ∑ STATISTICS 283 Also variance = 8.24 = 2 5 i.e. 8.24 = ( ) ( ) ( ) ( ) 1 2 2 2 2 2 2 3 4 2 4 1 6 2 4 4 ( ) 2 4 4 5 . . . x y . x y . + + + + − × + + × or 41.20 = 11.56 + 5.76 + 2.56 + x 2 + y 2 –8.8 × 13 + 38.72 Therefore x 2 + y 2 = 97 ... (2) But from (1), we have x 2 + y 2 + 2xy = 169 ... (3) From (2) and (3), we have 2xy = 72 ... (4) Subtracting (4) from (2), we get 1 )( 1 xx n i ∑ i = − Reprint 2025-26 284 MATHEMATICS or x – y = ± 5 ... (5) So, from (1) and (5), we get x = 9, y = 4 when x – y = 5 or x = 4, y = 9 when x – y = – 5 Thus, the remaining observations are 4 and 9. Example 15 If each of the observation x1 , x2 , ...,xn is increased by ‘a’, where a is a negative or positive number, show that the variance remains unchanged. Solution Let x be the mean of x1 , x2 , ...,xn . Then the variance is given by If ‘a is added to each observation, the new observations will be yi = xi + a ... (1) Let the mean of the new observations be y . Then i.e. y = x + a ... (2) Thus, the variance of the new observations 2 σ 2 = 2 1 1 ( ) n i i y y n = ∑ − = 2 ( ) 1 y = 1 1 1 1 ( ) n n i i i i y x a n n = = ∑ ∑ = + x 2 + y 2 – 2xy = 97 – 72 i.e. (x – y) 2 = 25 = 1 1 1 n n i i i x a n = = + ∑ ∑ = ax n na x n n 2 σ1 = 2 1 1 ( ) n i i x x n = ∑ − i ∑ i +=+ =1 1 n Thus, the variance of the new observations is same as that of the original observations. ANote We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance. Example 16 The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation? = 2 1 1 ( ) n i i x x n = ∑ − = 2 σ1 1 axax n i ∑ i −−+ = [Using (1) and (2)] Reprint 2025-26 Solution Given that number of observations (n) = 100 We know that ∑ = = n i.e. 100 i.e. Incorrect sum of observations = 4000 Thus the correct sum of observations = Incorrect sum – 50 + 40 Hence Correct mean = correct sum 3990 100 100 = = 39.9 Also Standard deviation σ = i.e. 5.1 = 2 2 1 1 Incorrect (40) 100 i i x n x 1 1 Incorrect mean ( x ) = 40, Incorrect standard deviation (σ) = 5.1 1 1 40 100 i i x = = ∑ or 100 i i x = × ∑ − = ( )2 1 1 2 xx n n = 4000 – 50 + 40 = 3990 2 2 2 1 1 1 1 n n i i i i x x n = = n − ∑ ∑ i ∑ i − = n 1 i i x = ∑ = 4000 STATISTICS 285 or 26.01 = 2 1 1 Incorrect 100 Therefore Incorrect 2 1 n i i x = ∑ = 100 (26.01 + 1600) = 162601 Now Correct 2 Therefore Correct standard deviation 1 n i i x = ∑ = Incorrect ∑ = n i i x = × ∑ – 1600 = 162601 – 2500 + 1600 = 161701 Reprint 2025-26 n i i x 1 2 – (50)2 + (40)2 286 MATHEMATICS 1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. 2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. 3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. 4. Given that x is the mean and σ 2 is the variance of n observations x1 , x2 , ...,xn . Prove that the mean and variance of the observations ax1 , ax2 , ax3 , ...., axn are a x and a 2 σ 2 , respectively, (a ≠ 0). 5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12. 6. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted. = 2 Correct 2 (Correct mean) i x = 161701 2 (39 9) 100 − . = 1617 01 1592 01 . . − = 25 = 5 Miscellaneous Exercise On Chapter 13 n − ∑ Summary ÆMeasures of dispersion Range, Quartile deviation, mean deviation, variance, standard deviation are measures of dispersion. Range = Maximum Value – Minimum Value ÆMean deviation for ungrouped data M M.D. ( ) M.D. (M) i i x – x x – x , n n = ∑ = ∑ Reprint 2025-26 ÆMean deviation for grouped data M.D. N M.D. M M N ( ) x , ( ) , where N f x x f x f i i i i = = = i ∑ ∑ ∑ ÆVariance and standard deviation for ungrouped data 2 1 2 ( ) i x – x n σ = ∑ , 1 2 ( – ) i x x n σ = ∑ ÆVariance and standard deviation of a discrete frequency distribution ( ) ( ) 2 1 1 2 2 , N N i i i i σ = ∑ f x x − = σ ∑ f x x − ÆVariance and standard deviation of a continuous frequency distribution ( ) ( ) 2 2 2 1 1 2 , N N N i i i i i i σ = ∑ f x x − = σ ∑ ∑ f x f x − ÆShortcut method to find variance and standard deviation. where i A i x y h − = ( ) 2 2 2 2 2 N N i i i i h σ f y f y = − ∑ ∑ , ( )2 2 N N i i i i h σ = ∑ ∑ f y f y − , Historical Note STATISTICS 287 ‘Statistics’ is derived from the Latin word ‘status’ which means a political state. This suggests that statistics is as old as human civilisation. In the year 3050 B.C., perhaps the first census was held in Egypt. In India also, about 2000 years ago, we had an efficient system of collecting administrative statistics, particularly, during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of collecting data related to births and deaths is mentioned in Kautilya’s Arthshastra (around 300 B.C.) A detailed account of administrative surveys conducted during Akbar’s regime is given in Ain-I-Akbari written by Abul Fazl. Captain John Graunt of London (1620-1674) is known as father of vital statistics due to his studies on statistics of births and deaths. Jacob Bernoulli (1654-1705) stated the Law of Large numbers in his book “Ars Conjectandi’, published in 1713. Reprint 2025-26 288 MATHEMATICS The theoretical development of statistics came during the mid seventeenth century and continued after that with the introduction of theory of games and chance (i.e., probability). Francis Galton (1822-1921), an Englishman, pioneered the use of statistical methods, in the field of Biometry. Karl Pearson (1857-1936) contributed a lot to the development of statistical studies with his discovery of Chi square test and foundation of statistical laboratory in England (1911). Sir Ronald A. Fisher (1890-1962), known as the Father of modern statistics, applied it to various diversified fields such as Genetics, Biometry, Education, Agriculture, etc. — v — Reprint 2025-26" class_11,14,Probability,ncert_books/class_11/kemh1dd/kemh114.pdf,"14.1 Event We have studied about random experiment and sample space associated with an experiment. The sample space serves as an universal set for all questions concerned with the experiment. Consider the experiment of tossing a coin two times. An associated sample space is S = {HH, HT, TH, TT}. Now suppose that we are interested in those outcomes which correspond to the occurrence of exactly one head. We find that HT and TH are the only elements of S corresponding to the occurrence of this happening (event). These two elements form the set E = { HT, TH} We know that the set E is a subset of the sample space S . Similarly, we find the following correspondence between events and subsets of S. Description of events Corresponding subset of ‘S’ Number of tails is exactly 2 A = {TT} Number of tails is atleast one B = {HT, TH, TT} Number of heads is atmost one C = {HT, TH, TT} Second toss is not head D = { HT, TT} Number of tails is atmost two S = {HH, HT, TH, TT} Number of tails is more than two φ vWhere a mathematical reasoning can be had, it is as great a folly to make use of any other, as to grope for a thing in the dark, when you have a candle in your hand. – JOHN ARBUTHNOT v PROBABILITY Chapter 14 an event and an event is associated with a subset of sample space. In the light of this we define an event as follows. Definition Any subset E of a sample space S is called an event. The above discussion suggests that a subset of sample space is associated with Reprint 2025-26 290 MATHEMATICS 14.1.1 Occurrence of an event Consider the experiment of throwing a die. Let E denotes the event “ a number less than 4 appears”. If actually ‘1’ had appeared on the die then we say that event E has occurred. As a matter of fact if outcomes are 2 or 3, we say that event E has occurred Thus, the event E of a sample space S is said to have occurred if the outcome ω of the experiment is such that ω ∈ E. If the outcome ω is such that ω ∉ E, we say that the event E has not occurred. 14.1.2 Types of events Events can be classified into various types on the basis of the elements they have. 1. Impossible and Sure Events The empty set φ and the sample space S describe events. In fact φ is called an impossible event and S, i.e., the whole sample space is called the sure event. To understand these let us consider the experiment of rolling a die. The associated sample space is write the subset associated with the event E? Clearly no outcome satisfies the condition given in the event, i.e., no element of the sample space ensures the occurrence of the event E. Thus, we say that the empty set only correspond to the event E. In other words we can say that it is impossible to have a multiple of 7 on the upper face of the die. Thus, the event E = φ is an impossible event. Now let us take up another event F “the number turns up is odd or even”. Clearly F = {1, 2, 3, 4, 5, 6,} = S, i.e., all outcomes of the experiment ensure the occurrence of the event F. Thus, the event F = S is a sure event. Let E be the event “ the number appears on the die is a multiple of 7”. Can you S = {1, 2, 3, 4, 5, 6} 2. Simple Event If an event E has only one sample point of a sample space, it is called a simple (or elementary) event. In a sample space containing n distinct elements, there are exactly n simple events. For example in the experiment of tossing two coins, a sample space is There are four simple events corresponding to this sample space. These are E1 = {HH}, E2 ={HT}, E3 = { TH} and E4 ={TT}. S={HH, HT, TH, TT} Reprint 2025-26 3. Compound Event If an event has more than one sample point, it is called a Compound event. For example, in the experiment of “tossing a coin thrice” the events E: ‘Exactly one head appeared’ F: ‘Atleast one head appeared’ G: ‘Atmost one head appeared’ etc. are all compound events. The subsets of S associated with these events are E={HTT,THT,TTH} F={HTT,THT, TTH, HHT, HTH, THH, HHH} G= {TTT, THT, HTT, TTH} Each of the above subsets contain more than one sample point, hence they are all compound events. 14.1.3 Algebra of events In the Chapter on Sets, we have studied about different ways of combining two or more sets, viz, union, intersection, difference, complement of a set etc. Like-wise we can combine two or more events by using the analogous set notations. Let A, B, C be events associated with an experiment whose sample space is S. 1. Complementary Event For every event A, there corresponds another event A′ called the complementary event to A. It is also called the event ‘not A’. For example, take the experiment ‘of tossing three coins’. An associated sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Let A={HTH, HHT, THH} be the event ‘only one tail appears’ Clearly for the outcome HTT, the event A has not occurred. But we may say that the event ‘not A’ has occurred. Thus, with every outcome which is not in A, we say that ‘not A’ occurs. Thus the complementary event ‘not A’ to the event A is PROBABILITY 291 or A′ = {ω : ω ∈ S and ω ∉A} = S – A. 2. The Event ‘A or B’ Recall that union of two sets A and B denoted by A ∪ B contains all those elements which are either in A or in B or in both. When the sets A and B are two events associated with a sample space, then ‘A ∪ B’ is the event ‘either A or B or both’. This event ‘A∪ B’ is also called ‘A or B’. Therefore Event ‘A or B’ = A ∪ B = {ω : ω ∈ A or ω ∈ B} A′ = {HHH, HTT, THT, TTH, TTT} Reprint 2025-26 292 MATHEMATICS 3. The Event ‘A and B’ We know that intersection of two sets A ∩ B is the set of those elements which are common to both A and B. i.e., which belong to both ‘A and B’. If A and B are two events, then the set A ∩ B denotes the event ‘A and B’. Thus, A ∩ B = {ω : ω ∈ A and ω ∈ B} For example, in the experiment of ‘throwing a die twice’ Let A be the event ‘score on the first throw is six’ and B is the event ‘sum of two scores is atleast 11’ then A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)} so A ∩ B = {(6,5), (6,6)} Note that the set A ∩ B = {(6,5), (6,6)} may represent the event ‘the score on the first throw is six and the sum of the scores is atleast 11’. 4. The Event ‘A but not B’ We know that A–B is the set of all those elements which are in A but not in B. Therefore, the set A–B may denote the event ‘A but not B’.We know that A – B = A ∩ B´ Example 1 Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events (i) Aor B (ii) A and B (iii) A but not B (iv) ‘not A’. Solution Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5} Obviously (i) ‘A or B’ = A ∪ B = {1, 2, 3, 5} (ii) ‘A and B’ = A ∩ B = {3,5} (iv) ‘not A’ = A′ = {1,4,6} 14.1.4 Mutually exclusive events In the experiment of rolling a die, a sample space is S = {1, 2, 3, 4, 5, 6}. Consider events, A ‘an odd number appears’ and B ‘an even number appears’ Clearly the event A excludes the event B and vice versa. In other words, there is no outcome which ensures the occurrence of events A and B simultaneously. Here Clearly A ∩ B = φ, i.e., A and B are disjoint sets. In general, two events A and B are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets A and B are disjoint. A = {1, 3, 5} and B = {2, 4, 6} (iii) ‘A but not B’ = A – B = {2} Reprint 2025-26 appears’ and event B ‘a number less than 4 appears’ Obviously A = {1, 3, 5} and B = {1, 2, 3} Now 3 ∈ A as well as 3 ∈ B Therefore, A and B are not mutually exclusive events. Remark Simple events of a sample space are always mutually exclusive. 14.1.5 Exhaustive events Consider the experiment of throwing a die. We have S = {1, 2, 3, 4, 5, 6}. Let us define the following events A: ‘a number less than 4 appears’, B: ‘a number greater than 2 but less than 5 appears’ and C: ‘a number greater than 4 appears’. Then A = {1, 2, 3}, B = {3,4} and C = {5, 6}. We observe that A ∪ B ∪ C = {1, 2, 3} ∪ {3, 4} ∪ {5, 6} = S. Such events A, B and C are called exhaustive events. In general, if E1 , E2 , ..., En are n events of a sample space S and if then E1 , E2 , ...., En are called exhaustive events.In other words, events E1 , E2 , ..., En are said to be exhaustive if atleast one of them necessarily occurs whenever the experiment is performed. Further, if Ei ∩ Ej = φ for i ≠ j i.e., events Ei and Ej are pairwise disjoint and Again in the experiment of rolling a die, consider the events A ‘an odd number 1 2 3 1 E E E E E S n n i i ... = ∪ ∪ ∪ ∪ = ∪ = PROBABILITY 293 events. We now consider some examples. Example 2 Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with this experiment A: ‘the sum is even’. B: ‘the sum is a multiple of 3’. Which pairs of these events are mutually exclusive? SE 1 =∪ = i n i , then events E1 , E2 , ..., En are called mutually exclusive and exhaustive C: ‘the sum is less than 4’. D: ‘the sum is greater than 11’. Reprint 2025-26 294 MATHEMATICS Solution There are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}. Then A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)} C = {(1, 1), (2, 1), (1, 2)} and D = {(6, 6)} We find that A ∩ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)} ≠ φ Therefore, A and B are not mutually exclusive events. Similarly A ∩ C ≠ φ, A ∩ D ≠ φ, B ∩ C ≠ φ and B ∩ D ≠ φ. Thus, the pairs of events, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive events. Also C ∩ D = φ and so C and D are mutually exclusive events. Example 3 A coin is tossed three times, consider the following events. A: ‘No head appears’, B: ‘Exactly one head appears’ and C: ‘Atleast two heads appear’. Do they form a set of mutually exclusive and exhaustive events? Solution The sample space of the experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} Now A ∪ B ∪ C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S Therefore, A, B and C are exhaustive events. Also, A ∩ B = φ, A ∩ C = φ and B ∩ C = φ Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive. Hence, A, B and C form a set of mutually exclusive and exhaustive events. 1. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive? 2. A die is thrown. Describe the following events: (i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3 (iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3 Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′ EXERCISE 14.1 Reprint 2025-26 3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive? 4. Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”. Which events are (i) mutually exclusive? (ii) simple? (iii) Compound? 5. Three coins are tossed. Describe (i) Two events which are mutually exclusive. (ii) Three events which are mutually exclusive and exhaustive. (iii) Two events, which are not mutually exclusive. (iv) Two events which are mutually exclusive but not exhaustive. (v) Three events which are mutually exclusive but not exhaustive. 6. Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤ 5. Describe the events (i) A′ (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) A ∩ B′ ∩ C′ PROBABILITY 295 14.2 Axiomatic Approach to Probability In earlier sections, we have considered random experiments, sample space and events associated with these experiments. In our day to day life we use many words about the chances of occurrence of events. Probability theory attempts to quantify these chances of occurrence or non occurrence of events. 7. Refer to question 6 above, state true or false: (give reason for your answer) (i) A and B are mutually exclusive (ii) A and B are mutually exclusive and exhaustive (iii) A = B′ (iv) A and C are mutually exclusive (v) A and B′ are mutually exclusive. (vi) A′, B′, C are mutually exclusive and exhaustive. Reprint 2025-26 296 MATHEMATICS event associated with an experiment having known the number of total outcomes. Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities. Let S be the sample space of a random experiment. The probability P is a real valued function whose domain is the power set of S and range is the interval [0,1] satisfying the following axioms It follows from (iii) that P(φ) = 0. To prove this, we take F = φ and note that E and φ are disjoint events. Therefore, from axiom (iii), we get P (E ∪ φ) = P (E) + P (φ) or P(E) = P(E) + P (φ) i.e. P (φ) = 0. Let S be a sample space containing outcomes 1 2 , ,..., ω ω ωn , i.e., S = {ω1 , ω2 , ..., ωn } It follows from the axiomatic definition of probability that (i) 0 ≤ P (ωi ) ≤ 1 for each ωi ∈ S (ii) P (ω1 ) + P (ω2 ) + ... + P (ωn ) = 1 (iii) For any event A, P(A) = ∑ P(ωi ), ωi ∈ A. ANote It may be noted that the singleton {ωi } is called elementary event and for notational convenience, we write P(ωi ) for P({ωi }). of the outcomes H and T. In earlier classes, we have studied some methods of assigning probability to an For example, in ‘a coin tossing’ experiment we can assign the number 1 2 to each (i) For any event E, P (E) ≥ 0 (ii) P(S) = 1 (iii) If E and F are mutually exclusive events, then P(E ∪ F) = P(E) + P(F). i.e. P(H) = 1 2 and P(T) = 1 2 (1) Clearly this assignment satisfies both the conditions i.e., each number is neither less than zero nor greater than 1 and Therefore, in this case we can say that probability of H = 2 1 , and probability of T = 2 1 If we take P(H) = 4 1 and P(T) = 4 3 ... (2) P(H) + P(T) = 2 1 + 2 1 = 1 Reprint 2025-26 H and T. In fact, we can assign the numbers p and (1 – p) to both the outcomes such that 0 ≤ p ≤ 1 and P(H) + P(T) = p + (1 – p) = 1 This assignment, too, satisfies both conditions of the axiomatic approach of probability. Hence, we can say that there are many ways (rather infinite) to assign probabilities to outcomes of an experiment. We now consider some examples. Example 4 Let a sample space be S = {ω1 , ω2 ,..., ω6 }.Which of the following assignments of probabilities to each outcome are valid? Outcomes ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 Does this assignment satisfy the conditions of axiomatic approach? Yes, in this case, probability of H = 1 We find that both the assignments (1) and (2) are valid for probability of (a) 6 1 6 1 6 1 6 1 6 1 6 1 (b) 1 0 0 0 0 0 (c) 8 1 3 2 3 1 3 1 4 1 − 3 1 − (d) 12 1 12 1 6 1 6 1 6 1 2 3 (e) 0.1 0.2 0.3 0.4 0.5 0.6 4 and probability of T = 4 3 . PROBABILITY 297 Solution (a) Condition (i): Each of the number p(ωi ) is positive and less than one. Condition (ii): Sum of probabilities Therefore, the assignment is valid (b) Condition (i): Each of the number p(ωi ) is either 0 or 1. Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1 Therefore, the assignment is valid (c) Condition (i) Two of the probabilities p(ω5 ) and p(ω6 ) are negative, the assignment is not valid (d) Since p(ω6 ) = 3 2 > 1, the assignment is not valid = 1 6 1 6 1 6 1 6 1 6 1 6 1 =+++++ Reprint 2025-26 298 MATHEMATICS (e) Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1, the assignment is not valid. 14.2.1 Probability of an event Let S be a sample space associated with the experiment ‘examining three consecutive pens produced by a machine and classified as Good (non-defective) and bad (defective)’. We may get 0, 1, 2 or 3 defective pens as result of this examination. A sample space associated with this experiment is S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, where B stands for a defective or bad pen and G for a non – defective or good pen. Let the probabilities assigned to the outcomes be as follows Sample point: BBB BBG BGB GBB BGG GBG GGB GGG Probability: 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 Let event A: there is exactly one defective pen and event B: there are atleast two defective pens. Hence A = {BGG, GBG, GGB} and B = {BBG, BGB, GBB, BBB} Now P(A) = ∑ ∀ ∈ P(ω ), ω A i i and P(B) = ∑ ∀ ∈ P(ω ), ω B i i = P(BGG) + P(GBG) + P(GGB) = 8 3 8 1 8 1 8 1 =++ = P(BBG) + P(BGB) + P(GBB) + P(BBB) = 2 1 8 4 8 1 8 1 8 1 8 1 ==+++ The sample space of this experiment is S = {HH, HT, TH, TT} Let the following probabilities be assigned to the outcomes us find the probability of the event E: ‘Both the tosses yield the same result’. Here E = {HH, TT} Now P(E) = Σ P(wi ), for all wi ∈ E Let us consider another experiment of ‘tossing a coin “twice” Clearly this assignment satisfies the conditions of axiomatic approach. Now, let P(HH) = 4 1 , P(HT) = 7 1 , P(TH) = 7 2 , P(TT) = 28 9 Reprint 2025-26 For the event F: ‘exactly two heads’, we have F = {HH} and P(F) = P(HH) = 1 4 14.2.2 Probabilities of equally likely outcomes Let a sample space of an experiment be S = {ω1 , ω2 ,..., ωn }. Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same. i.e. P(ωi ) = p, for all ωi ∈ S where 0 ≤ p ≤ 1 Since 1 P(ω ) 1 n i i= ∑ = i.e., p + p + ... + p (n times) = 1 or np = 1 i.e., p = 1 n Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If each out come is equally likely, then it follows that 14.2.3 Probability of the event ‘A or B’ Let us now find the probability of event ‘A or B’, i.e., P (A ∪ B) Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated with ‘tossing of a coin thrice’ = P(HH) + P(TT) = 7 4 28 9 4 1 =+ P(E) m n = = Number of outcomes favourable to E Total possible outcomes PROBABILITY 299 Clearly A ∪ B = {HHT, HTH, THH, HHH} Now P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH) If all the outcomes are equally likely, then Also P(A) = P(HHT) + P(HTH) + P(THH) = 3 8 ( ) 1 1 1 1 4 1 P A B 8 8 8 8 8 2 ∪ = + + + = = Reprint 2025-26 300 MATHEMATICS and P(B) = P(HTH) + P(THH) + P(HHH) = 3 8 Therefore P(A) + P(B) = 3 3 6 8 8 8 + = It is clear that P(A ∪ B) ≠ P(A) + P(B) The points HTH and THH are common to both A and B. In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A∪ B) we have to subtract the probabilities of the sample points in A ∩ B from P(A) + P(B) i.e. P(A B) ∪ = P(A) P(B) P(ω ) ω A B i i + − ∑ ∀ ∈ ∩ , = P A P B P A B ( ) ( ) ( ) + − ∩ Thus we observe that, ∪ = + − ∩ )BA(P)B(P)A(P)BA(P In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have Since A B = (A–B) (A B) (B–A) ∪ ∪ ∩ ∪ , we have P(A ∪ B) = [ P(ω ) ω (A–B) P(ω ) ω A B + i ] [ i i ] ∑ ∀ ∈ + ∑ ∀ ∈ ∩ i [∑ ∀ ∈ P(ω ) ω B – A i i ] (because A–B, A ∩ B and B – A are mutually exclusive) ... (1) Also P(A) P(B) ( + = ∑ ∀ ∈ ∑ ∀ ∈ [ p ω ) A + (ω ) ω B i i ω ] [ p i i ] P A B ( ∪ = ∑ ∀ ∈ ∪ ) p , (ω ω A B i i ) . Hence P(A B) P (A) +P(B) – P(A B) ∪ = ∩ . Alternatively, it can also be proved as follows: A ∪ B = A ∪ (B – A), where A and B – A are mutually exclusive, and B = (A ∩ B) ∪ (B – A), where A ∩ B and B – A are mutually exclusive. Using Axiom (iii) of probability, we get = [∑ ∀ ∈ ∪ ∩ P(ω ) ω (A–B) (A B) + i i ] [∑ ∀ ∈ ∪ ∩ P(ω ) ω (B – A) (A B) i i ] = [∑ ∀ ∈ ∑ ∀ ∈ ∩ P(ω ) ω (A – B) + P(ω ) ω (A B) i i ] [ i i ] + [∑ ∀ ∈ P(ω ) ω (B–A) i i ] + [∑ ∀ ∈ ∩ P(ω ) ω (A B) i i ] = P(A B) P( ∪ + ∑ ∀ ∈ ∩ [ ω ) ω A B i i ] [using (1)] = P(A B)+ P(A B) ∪ ∩ . Reprint 2025-26 and P(B) = P ( A ∩ B) + P (B – A) ... (3) Subtracting (3) from (2) gives P (A ∪ B) – P(B) = P(A) – P (A ∩ B) or P(A ∪ B) = P(A) + P (B) – P (A ∩ B) The above result can further be verified by observing the Venn Diagram (Fig 14.1) If A and B are disjoint sets, i.e., they are mutually exclusive events, then A ∩ B = φ Therefore P(A B) = P ( ) = 0 ∩ φ Thus, for mutually exclusive events A and B, we have ∪ = + )B(P)A(P)BA(P , which is Axiom (iii) of probability. 14.2.4 Probability of event ‘not A’ Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10} If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability P (A ∪B) = P (A) + P (B – A) ... (2) Fig 14.1 PROBABILITY 301 of each outcome is 10 1 Now P(A) = P(2) + P(4) + P(6) + P(8) Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10} Now P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10) = 1 1 1 1 4 2 10 10 10 10 10 5 + + + = = Reprint 2025-26 302 MATHEMATICS Thus, P(A′) = 3 5 = )A(P1 5 2 1 −=− Also, we know that A′ and A are mutually exclusive and exhaustive events i.e., A ∩ A′ = φ and A ∪ A′ = S or P(A ∪ A′) = P(S) Now P(A) + P(A′) = 1, by using axioms (ii) and (iii). or P( A′) = P(not A) = 1 – P(A) We now consider some examples and exercises having equally likely outcomes unless stated otherwise. Example 5 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be (i) a diamond (ii) not an ace (iii) a black card (i.e., a club or, a spade) (iv) not a diamond (v) not a black card. Solution When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. (i) Let A be the event 'the card drawn is a diamond' Clearly the number of elements in set A is 13. Therefore, P(A) = 13 1 52 4 = = 6 3 10 5 = i.e. probability of a diamond card = 1 4 (ii) We assume that the event ‘Card drawn is an ace’ is B Therefore ‘Card drawn is not an ace’ should be B′. We know that P(B′) = 1 – P(B) = 13 12 13 1 1 52 4 1 =−=− (iii) Let C denote the event ‘card drawn is black card’ Therefore, number of elements in the set C = 26 i.e. P(C) = 2 1 52 26 = Reprint 2025-26 Thus, probability of a black card = 2 1 . (iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’, so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’ Now P(not A) = 1 – P(A) = 4 3 4 1 1 =− (v) The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’. We know that P(not C) = 1 – P(C) = 2 1 2 1 1 =− Therefore, probability of not a black card = 2 1 Example 6 A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue, (v) either red or blue. Solution There are 9 discs in all so the total number of possible outcomes is 9. Let the events A, B, C be defined as A: ‘the disc drawn is red’ B: ‘the disc drawn is yellow’ C: ‘the disc drawn is blue’. PROBABILITY 303 Hence P(A) = 9 4 Therefore, P(B) = 9 2 (iii) The number of blue discs = 3, i.e., n(C) = 3 Therefore, P(C) = 3 1 9 3 = (iv) Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 – P(C) (ii) The number of yellow discs = 2, i.e., n (B) = 2 (i) The number of red discs = 4, i.e., n (A) = 4 Reprint 2025-26 304 MATHEMATICS Therefore P(not C) = 3 2 3 1 1 =− (v) The event ‘either red or blue’ may be described by the set ‘A or C’ Since, A and C are mutually exclusive events, we have Example 7Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that (a) Both Anil and Ashima will not qualify the examination. Solution Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02. Then (a) The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E´ ∩ F´. Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e., Ashima will not qualify the examination. (b) Atleast one of them will not qualify the examination and (c) Only one of them will qualify the examination. P(A or C) = P (A ∪ C) = P(A) + P(C) = 9 7 3 1 9 4 =+ Also E´ ∩ F´ = (E ∪ F)´ (by Demorgan's Law) Now P(E ∪ F) = P(E) + P(F) – P(E ∩ F) or P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13 Therefore P(E´ ∩ F´) = P(E ∪ F)´ = 1 – P(E ∪ F) = 1 – 0.13 = 0.87 (b) P (atleast one of them will not qualify) = 1 – P(both of them will qualify) = 1 – 0.02 = 0.98 (c) The event only one of them will qualify the examination is same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima Reprint 2025-26 will qualify) i.e., E ∩ F´ or E´ ∩ F, where E ∩ F´ and E´ ∩ F are mutually exclusive. Therefore, P(only one of them will qualify) = P(E ∩ F´ or E´ ∩ F) Example 8 A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men? Solution The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in 4C2 ways. (a) No men in the committee of two means there will be two women in the committee. Therefore ( ) 2 2 4 2 C 1 2 1 1 P no man C 4 3 6 × × = = = × (b) One man in the committee means that there is one woman. One man out of 2 can be selected in 2C1 ways and one woman out of 2 can be selected in 2C1 ways. Together they can be selected in 2 2 C C 1 1 × ways. Therefore ( ) 2 2 1 1 4 2 C C 2 2 2 P One man C 2 3 3 × × = = = × = P(E ∩ F´) + P(E´ ∩ F) = P (E) – P(E ∩ F) + P(F) – P (E ∩ F) = 0.05 – 0.02 + 0.10 – 0.02 = 0.11 Out of two women, two can be selected in 2C 1 2 = way. PROBABILITY 305 (c) Two men can be selected in 2C2 way. Hence ( ) 2 2 4 4 2 2 C 1 1 P Two men C C 6 = = = 1. Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = {ω ω ω ω ω ω ω 1 2 3 4 5 6 7 , , , , , , } EXERCISE 14.2 Reprint 2025-26 306 MATHEMATICS Assignment ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 ω 7 2. A coin is tossed twice, what is the probability that atleast one tail occurs? 3. A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear. 4. A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card. 5. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12 6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? 7. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. 8. Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) atleast 2 heads (iv) atmost 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) atmost two tails (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) 7 1 7 1 7 1 7 1 7 1 7 1 7 1 (c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (d) – 0.1 0.2 0.3 0.4 – 0.2 0.1 0.3 (e) 14 1 14 2 14 3 14 4 14 5 14 6 14 15 10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) a consonant 9. If 11 2 is the probability of an event, what is the probability of the event ‘not A’. Reprint 2025-26 11. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.] 12. Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 13. Fill in the blanks in following table: P(A) P(B) P(A ∩ B) P(A ∪ B) 14. Given P(A) = 5 3 and P(B) = 5 1 . Find P(A or B), if A and B are mutually exclusive 15. If E and F are events such that P(E) = 4 1 , P(F) = 2 1 and P(E and F) = 8 1 , find (i) P(E or F), (ii) P(not E and not F). 16. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive. 17. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B) 18. In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology. 19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both? 20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination? events. (iii) 0.5 0.35 . . . 0.7 (ii) 0.35 . . . 0.25 0.6 (i) 1 3 1 5 15 . . . 1 PROBABILITY 307 Reprint 2025-26 308 MATHEMATICS 21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that Example 9 On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) A before B? (ii) A before B and B before C? (iii) A first and B last? (iv) A either first or second? (v) A just before B? Solution The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24. Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC. S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB BACD, BADC, BDAC, BDCA, BCAD, BCDA CABD, CADB, CBDA, CBAD, CDAB, CDBA DABC, DACB, DBCA, DBAC, DCAB, DCBA} Miscellaneous Examples (i) Let the event ‘she visits A before B’ be denoted by E Therefore,E = {ABCD, CABD, DABC, ABDC, CADB, DACB ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} Thus ( ) ( ) (ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F. Here F = {ABCD, DABC, ABDC, ADBC} Therefore, ( ) ( ) Students are advised to find the probability in case of (iii), (iv) and (v). ( ) E 12 1 P E S 24 2 n ( ) F 4 1 P F S 24 6 n n = = = n = = = Reprint 2025-26 Example 10 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings. Solution Total number of possible hands = 52C7 (i) Number of hands with 4 Kings = 4 48 C C 4 3 × (other 3 cards must be chosen from the rest 48 cards) Hence P (a hand will have 4 Kings) = 4 48 4 3 52 7 C C 1 C 7735 × = (ii) Number of hands with 3 Kings and 4 non-King cards = 4 48 C C 3 4 × Therefore P (3 Kings) = 4 48 3 4 52 7 C C 9 C 1547 × = (iii) P(atleast 3 King) = P(3 Kings or 4 Kings) = P(3 Kings) + P(4 Kings) Example 11 If A, B, C are three events associated with a random experiment, prove that P A B C ( ∪ ∪ ) = P A P B +P C P A B P A C ( ) + ( ) ( ) − ∩ − ∩ ( ) ( ) – P ( B ∩ C) + P ( A ∩ B ∩ C) = 9 1 46 1547 7735 7735 + = PROBABILITY 309 Solution Consider E = B ∪ C so that P (A ∪ B ∪ C ) = P (A ∪ E ) = P A P E P ( ) + − ( ) ( ) A E ∩ ... (1) Now Also A E A B C ∩ = ∩ ∪ ( ) = (A B A C ∩ ∪ ∩ ) ( ) [using distribution property of intersection of sets over the union]. Thus P E P B C ( ) = ∪ ( ) = + − ∩ P B P C P B C ( ) ( ) ( ) ... (2) P A E P A B P A C ( ∩ = ∩ + ∩ ) ( ) ( ) – P A B A C ( ∩ ∩ ∩ ) ( ) Reprint 2025-26 310 MATHEMATICS Example 12 In a relay race there are five teams A, B, C, D and E. Solution If we consider the sample space consisting of all finishing orders in the first three places, we will have 5 P3 , i.e., ( ) 5! 5 3 ! − = 5 × 4 × 3 = 60 sample points, each with a probability of 1 60 . (a) A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., ABC. Thus P(A, B and C finish first, second and third respectively) = 1 60 . (b) A, B and C are the first three finishers. There will be 3! arrangements for A, B and C. Therefore, the sample points corresponding to this event will be 3! in number. Using (2) and (3) in (1), we get (a) What is the probability that A, B and C finish first, second and third, respectively. (b) What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely) P A B C P A P B P C P B C [ ∪ ∪ = + + − ∩ ] ( ) ( ) ( ) ( ) – P A B P A C P A B C ( ∩ − ∩ + ∩ ∩ ) ( ) ( ) = P A B P A C ( ∩ + ∩ ) ( ) – P A B C [ ∩ ∩ ] ... (3) So P (A, B and C are first three to finish) 3! 6 1 60 60 10 = = = 1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? 2. 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade? Miscellaneous Exercise on Chapter 14 Reprint 2025-26 3. A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine 4. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets. 5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that 6. Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. 7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. 8. From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: (i) P(2) (ii) P(1 or 3) (iii) P(not 3) (a) you both enter the same section? (b) you both enter the different sections? Find (i) P(A ∪ B) (ii) P(A´ ∩ B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´) S. No. Name Sex Age in years 1. Harish M 30 2. Rohan M 33 PROBABILITY 311 10. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase? 9. If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed? A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? 3. Sheetal F 46 4. Alis F 28 5. Salim M 41 Reprint 2025-26 312 MATHEMATICS In this Chapter, we studied about the axiomatic approach of probability. The main features of this Chapter are as follows: ÆEvent: A subset of the sample space ÆImpossible event : The empty set ÆSure event: The whole sample space ÆComplementary event or ‘not event’ : The set A′ or S – A ÆEvent A or B: The set A ∪ B ÆEvent A and B: The set A ∩ B ÆEvent A and not B: The set A – B ÆMutually exclusive event: A and B are mutually exclusive if A ∩ B = φ ÆExhaustive and mutually exclusive events: Events E1 , E2 ,..., En are mutually exclusive and exhaustive if E1 ∪ E2 ∪ ...∪ En = S and Ei ∩ Ej = φ V i ≠ j ÆProbability: Number P (ωi ) associated with sample point ω i such that of the outcome ωi . ÆEqually likely outcomes: All outcomes with equal probability ÆProbability of an event: For a finite sample space with equally likely outcomes (iii) P(A) = ∑P(ωi) for all ωi ∈A. The number P (ωi ) is called probability (i) 0 ≤ P (ωi ) ≤ 1 (ii) ∑P(ωi) for all ωi ∈ S = 1 Summary the set A, n(S) = number of elements in the set S. ÆIf A and B are any two events, then P(A or B) = P(A) + P(B) – P(A and B) equivalently, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ÆIf A and B are mutually exclusive, then P(A or B) = P(A) + P(B) ÆIf A is any event, then P(not A) = 1 – P(A) Probability of an event (A) P(A) (S) n n = , where n(A) = number of elements in Reprint 2025-26 Probability theory like many other branches of mathematics, evolved out of practical consideration. It had its origin in the 16th century when an Italian physician and mathematician Jerome Cardan (1501–1576) wrote the first book on the subject “Book on Games of Chance” (Biber de Ludo Aleae). It was published in 1663 after his death. In 1654, a gambler Chevalier de Metre approached the well known French Philosopher and Mathematician Blaise Pascal (1623–1662) for certain dice problem. Pascal became interested in these problems and discussed with famous French Mathematician Pierre de Fermat (1601–1665). Both Pascal and Fermat solved the problem independently. Besides, Pascal and Fermat, outstanding contributions to probability theory were also made by Christian Huygenes (1629– 1665), a Dutchman, J. Bernoulli (1654–1705), De Moivre (1667–1754), a Frenchman Pierre Laplace (1749–1827), the Russian P.L Chebyshev (1821– 1897), A. A Markov (1856–1922) and A. N Kolmogorove (1903–1987). Kolmogorov is credited with the axiomatic theory of probability. His book ‘Foundations of Probability’ published in 1933, introduces probability as a set function and is considered a classic. Historical Note — v — PROBABILITY 313 Reprint 2025-26" class_12,1,Relations and Functions,ncert_books/class_12/lemh1dd/lemh101.pdf,"1.1 Introduction Recall that the notion of relations and functions, domain, co-domain and range have been introduced in Class XI along with different types of specific real valued functions and their graphs. The concept of the term ‘relation’ in mathematics has been drawn from the meaning of relation in English language, according to which two objects or quantities are related if there is a recognisable connection or link between the two objects or quantities. Let A be the set of students of Class XII of a school and B be the set of students of Class XI of the same school. Then some of the examples of relations from A to B are (i) {(a, b) ∈ A × B: a is brother of b}, (ii) {(a, b) ∈ A × B: a is sister of b}, (iii) {(a, b) ∈ A × B: age of a is greater than age of b}, (iv) {(a, b) ∈ A × B: total marks obtained by a in the final examination is less than the total marks obtained by b in the final examination}, (v) {(a, b) ∈ A × B: a lives in the same locality as b}. However, abstracting from this, we define mathematically a relation R from A to B as an arbitrary subset of A × B. If (a, b) ∈ R, we say that a is related to b under the relation R and we write as a R b. In general, (a, b) ∈ R, we do not bother whether there is a recognisable connection or link between a and b. As seen in Class XI, functions are special kind of relations. In this chapter, we will study different types of relations and functions, composition of functions, invertible functions and binary operations. vThere is no permanent place in the world for ugly mathematics ... . It may be very hard to define mathematical beauty but that is just as true of beauty of any kind, we may not know quite what we mean by a beautiful poem, but that does not prevent us from recognising one when we read it. — G. H. HARDY v RELATIONS AND FUNCTIONS Chapter 1 Reprint 2025-26 Lejeune Dirichlet (1805-1859) 1.2 Types of Relations In this section, we would like to study different types of relations. We know that a relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two extreme relations. For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by R = {(a, b): a – b = 10}. This is the empty set, as no pair (a, b) satisfies the condition a – b = 10. Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole set A × A, as all pairs (a, b) in A × A satisfy | a – b | ≥ 0. These two extreme examples lead us to the following definitions. Definition 1 A relation R in a set A is called empty relation, if no element of A is related to any element of A, i.e., R = φ ⊂ A × A. Definition 2 A relation R in a set A is called universal relation, if each element of A is related to every element of A, i.e., R = A × A. relations. Example 1 Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R′ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation. Solution Since the school is boys school, no student of the school can be sister of any student of the school. Hence, R = φ, showing that R is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be less than 3 meters. This shows that R′ = A × A is the universal relation. Remark In Class XI, we have seen two ways of representing a relation, namely raster method and set builder method. However, a relation R in the set {1, 2, 3, 4} defined by R = {(a, b) : b = a + 1} is also expressed as a R b if and only if b = a + 1 by many authors. We may also use this notation, as and when convenient. If (a, b) ∈ R, we say that a is related to b and we denote it as a R b. 2 MATHEMATICS Both the empty relation and the universal relation are some times called trivial is an equivalence relation. To study equivalence relation, we first consider three types of relations, namely reflexive, symmetric and transitive. Definition 3 A relation R in a set A is called (i) reflexive, if (a, a) ∈ R, for every a ∈ A, (ii) symmetric, if (a1 , a2 ) ∈ R implies that (a2 , a1 ) ∈ R, for all a1 , a2 ∈ A. (iii) transitive, if (a1 , a2 ) ∈ R and (a2 , a3 ) ∈ R implies that (a1 , a3 )∈ R, for all a1 , a2 , a3 ∈ A. One of the most important relation, which plays a significant role in Mathematics, Reprint 2025-26 Definition 4 A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive. Example 2 Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1 , T2 ) : T1 is congruent to T2 }. Show that R is an equivalence relation. Solution R is reflexive, since every triangle is congruent to itself. Further, (T1 , T2 ) ∈ R ⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 ⇒ (T2 , T1 ) ∈ R. Hence, R is symmetric. Moreover, (T1 , T2 ), (T2 , T3 ) ∈ R ⇒ T1 is congruent to T2 and T2 is congruent to T3 ⇒ T1 is congruent to T3 ⇒ (T1 , T3 ) ∈ R. Therefore, R is an equivalence relation. Example 3 Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive. Solution R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example 4 Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive. R is not transitive. Indeed, if L1 is perpendicular to L2 and RELATIONS AND FUNCTIONS 3 Fig 1.1 Solution R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R. Example 5 Show that the relation R in the set Z of integers given by is an equivalence relation. Solution R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b – c) is even (Why?). So, (a – c) is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z. R = {(a, b) : 2 divides a – b} Reprint 2025-26 etc., lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc., do not lie in R. Similarly, all odd integers are related to one and no even integer is related to one. Therefore, the set E of all even integers and the set O of all odd integers are subsets of Z satisfying following conditions: (i) All elements of E are related to each other and all elements of O are related to each other. (ii) No element of E is related to any element of O and vice-versa. (iii) E and O are disjoint and Z = E ∪ O. The subset E is called the equivalence class containing zero and is denoted by [0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that [0] ≠ [1], [0] = [2r] and [1] = [2r + 1], r ∈ Z. Infact, what we have seen above is true for an arbitrary equivalence relation R in a set X. Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai called partitions or subdivisions of X satisfying: (i) all elements of Ai are related to each other, for all i. (iii) ∪ Aj = X and Ai ∩ Aj = φ, i ≠ j. The subsets Ai are called equivalence classes. The interesting part of the situation is that we can go reverse also. For example, consider a subdivision of the set Z given by three mutually disjoint subsets A1 , A2 and A3 whose union is Z with A1 = {x ∈ Z : x is a multiple of 3} = {..., – 6, – 3, 0, 3, 6, ...} (ii) no element of Ai is related to any element of Aj , i ≠ j. 4 MATHEMATICS In Example 5, note that all even integers are related to zero, as (0, ± 2), (0, ± 4) A2 = {x ∈ Z : x – 1 is a multiple of 3} = {..., – 5, – 2, 1, 4, 7, ...} arguments similar to those used in Example 5, we can show that R is an equivalence relation. Also, A1 coincides with the set of all integers in Z which are related to zero, A2 coincides with the set of all integers which are related to 1 and A3 coincides with the set of all integers in Z which are related to 2. Thus, A1 = [0], A2 = [1] and A3 = [2]. In fact, A1 = [3r], A2 = [3r + 1] and A3 = [3r + 2], for all r ∈ Z. Example 6 Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}. Define a relation R in Z given by R = {(a, b) : 3 divides a – b}. Following the A3 = {x ∈ Z : x – 2 is a multiple of 3} = {..., – 4, – 1, 2, 5, 8, ...} Reprint 2025-26 Solution Given any element a in A, both a and a must be either odd or even, so that (a, a) ∈ R. Further, (a, b) ∈ R ⇒ both a and b must be either odd or even ⇒ (b, a) ∈ R. Similarly, (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be either even or odd simultaneously ⇒ (a, c) ∈ R. Hence, R is an equivalence relation. Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements of this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related to each other, as all of them are even. Also, no element of the subset {1, 3, 5, 7} can be related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements of {2, 4, 6} are even. 1. Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y) : x and y work at the same place} (b) R = {(x, y) : x and y live in the same locality} (c) R = {(x, y) : x is exactly 7 cm taller than y} (d) R = {(x, y) : x is wife of y} (e) R = {(x, y) : x is father of y} EXERCISE 1.1 RELATIONS AND FUNCTIONS 5 2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b 2} is neither reflexive nor symmetric nor transitive. 3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. 4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric. 5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b 3} is reflexive, symmetric or transitive. Reprint 2025-26 10. Give an example of a relation. Which is 11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre. 6 MATHEMATICS 6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive. 7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation. 8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by 9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. is an equivalence relation. Find the set of all elements related to 1 in each case. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive. (iii) Reflexive and symmetric but not transitive. (i) R = {(a, b) : |a – b| is a multiple of 4} (ii) R = {(a, b) : a = b} (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. 12. Show that the relation R defined in the set A of all triangles as R = {(T1 , T2 ) : T1 is similar to T2 }, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1 , T2 and T3 are related? 13. Show that the relation R defined in the set A of all polygons as R = {(P1 , P2 ) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5? 14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is parallel to L2 }. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4. Reprint 2025-26 1.3 Types of Functions The notion of a function along with some special functions like identity function, constant function, polynomial function, rational function, modulus function, signum function etc. along with their graphs have been given in Class XI. studied. As the concept of function is of paramount importance in mathematics and among other disciplines as well, we would like to extend our study about function from where we finished earlier. In this section, we would like to study different types of functions. Consider the functions f 1 , f 2 , f 3 and f 4 given by the following diagrams. In Fig 1.2, we observe that the images of distinct elements of X1 under the function f 1 are distinct, but the image of two distinct elements 1 and 2 of X1 under f 2 is same, namely b. Further, there are some elements like e and f in X2 which are not images of any element of X1 under f 1 , while all elements of X3 are images of some elements of X1 under f 3 . The above observations lead to the following definitions: 15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. 16. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R Addition, subtraction, multiplication and division of two functions have also been (A) R is reflexive and symmetric but not transitive. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation. (B) R is reflexive and transitive but not symmetric. RELATIONS AND FUNCTIONS 7 Definition 5 A function f : X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct, i.e., for every x1 , x2 ∈ X, f(x1 ) = f(x2 ) implies x1 = x2 . Otherwise, f is called many-one. The function f 1 and f 4 in Fig 1.2 (i) and (iv) are one-one and the function f 2 and f 3 in Fig 1.2 (ii) and (iii) are many-one. Definition 6 A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f(x) = y. The function f 3 and f 4 in Fig 1.2 (iii), (iv) are onto and the function f 1 in Fig 1.2 (i) is not onto as elements e, f in X2 are not the image of any element in X1 under f 1 . Reprint 2025-26 Remark f : X → Y is onto if and only if Range of f = Y. Definition 7 A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto. Example 7 Let A be the set of all 50 students of Class X in a school. Let f : A → N be function defined by f(x) = roll number of the student x. Show that f is one-one but not onto. 8 MATHEMATICS The function f 4 in Fig 1.2 (iv) is one-one and onto. Fig 1.2 (i) to (iv) Solution No two different students of the class can have same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51 in N is not roll number of any student of the class, so that 51 can not be image of any element of X under f. Hence, f is not onto. Example 8 Show that the function f : N → N, given by f(x) = 2x, is one-one but not onto. Solution The function f is one-one, for f(x1 ) = f(x2 ) ⇒ 2x1 = 2x2 ⇒ x1 = x2 . Further, f is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1. Reprint 2025-26 Example 9 Prove that the function f : R → R, given by f(x) = 2x, is one-one and onto. Solution f is one-one, as f(x1 ) = f(x2 ) ⇒ 2x1 = 2x2 ⇒ x1 = x2 . Also, given any real number y in R, there exists 2 y in R such that f( 2 y ) = 2 . ( 2 y ) = y. Hence, f is onto. Example 10 Show that the function f : N→ N, given by f (1) = f(2) = 1 and f(x) = x – 1, for every x > 2, is onto but not one-one. Solution f is not one-one, as f(1) = f(2) = 1. But f is onto, as given any y ∈ N, y ≠ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y. Also for 1 ∈ N, we have f(1) = 1. Fig 1.3 RELATIONS AND FUNCTIONS 9 Example 11 Show that the function f : R → R, defined as f(x) = x 2 , is neither one-one nor onto. Solution Since f(– 1) = 1 = f(1), f is not oneone. Also, the element – 2 in the co-domain R is not image of any element x in the domain R (Why?). Therefore f is not onto. Example 12 Show that f : N → N, given by is both one-one and onto. Fig 1.4 1,if is odd, ( ) 1,if is even x x f x x x + = − Reprint 2025-26 Solution Suppose f(x1 ) = f(x2 ). Note that if x1 is odd and x2 is even, then we will have x1 + 1 = x2 – 1, i.e., x2 – x1 = 2 which is impossible. Similarly, the possibility of x1 being even and x2 being odd can also be ruled out, using the similar argument. Therefore, both x1 and x2 must be either odd or even. Suppose both x1 and x2 are odd. Then f(x1 ) = f(x2 ) ⇒ x1 + 1 = x2 + 1 ⇒ x1 = x2 . Similarly, if both x1 and x2 are even, then also f(x1 ) = f(x2 ) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2 . Thus, f is one-one. Also, any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number 2r in the co-domain N is the image of 2r – 1 in the domain N. Thus, f is onto. Example 13 Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one. Solution Suppose f is not one-one. Then there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same. Also, the image of 3 under f can be only one element. Therefore, the range set can have at the most two elements of the co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one. Example 14 Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto. Solution Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto. Remark The results mentioned in Examples 13 and 14 are also true for an arbitrary finite set X, i.e., a one-one function f : X → X is necessarily onto and an onto map f : X → X is necessarily one-one, for every finite set X. In contrast to this, Examples 8 and 10 show that for an infinite set, this may not be true. In fact, this is a characteristic difference between a finite and an infinite set. 10 MATHEMATICS EXERCISE 1.2 1. Show that the function f : R∗ → R∗ defined by f(x) = 1 x is one-one and onto, 2. Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f(x) = x 2 3. Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗ ? (iv) f : N → N given by f(x) = x 3 (iii) f : R → R given by f(x) = x 2 (v) f : Z → Z given by f(x) = x 3 (ii) f : Z → Z given by f(x) = x 2 Reprint 2025-26 4. Show that the Modulus Function f : R → R, given by f(x) = | x |, is neither oneone nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative. 5. Show that the Signum Function f : R → R, given by 6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one. 7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. 8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function. 9. Let f : N → N be defined by f (n) = is neither one-one nor onto. (ii) f : R → R defined by f(x) = 1 + x 2 (i) f : R → R defined by f(x) = 3 – 4x f x x x ( ) , , = 1 0 0 0 1 0 , if + if if n n n n 2 x 1 2 > = < , if is odd , if is even for all n ∈ N. RELATIONS AND FUNCTIONS 11 10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by 12. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. 11. Let f : R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (A) f is one-one onto (B) f is many-one onto State whether the function f is bijective. Justify your answer. f(x) = 2 3 x x − − . Is f one-one and onto? Justify your answer. (C) f is one-one but not onto (D) f is neither one-one nor onto. (C) f is one-one but not onto (D) f is neither one-one nor onto. Reprint 2025-26 1.4 Composition of Functions and Invertible Function Definition 8 Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by Example 15 Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof. Solution We have gof(2) = g (f(2)) = g (3) = 7, gof (3) = g (f(3)) = g (4) = 7, gof(4) = g (f(4)) = g (5) = 11 and gof(5) = g (5) = 11. Example 16 Find gof and fog, if f : R → R and g : R → R are given by f(x) = cos x and g (x) = 3x 2 . Show that gof ≠ fog. Solution We have gof(x) = g (f(x)) = g (cos x) = 3 (cos x) 2 = 3 cos2 x. Similarly, fog(x) = f(g (x)) = f(3x 2 ) = cos (3x 2 ). Note that 3cos2 x ≠ cos 3x 2 , for x = 0. Hence, gof ≠ fog. Definition 9 A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = IX and fog = IY . The function g is called the inverse of f and is denoted by f –1 . 12 MATHEMATICS gof (x) = g(f (x)), ∀ x ∈ A. Fig 1.5 Thus, if f is invertible, then f must be one-one and onto and conversely, if f is one-one and onto, then f must be invertible. This fact significantly helps for proving a function f to be invertible by showing that f is one-one and onto, specially when the actual inverse of f is not to be determined. Example 17 Let f : N → Y be a function defined as f(x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. Solution Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3, for some x in the domain N. This shows that ( 3) 4 y x − = . Define g : Y → N by Reprint 2025-26 fog (y) = f(g (y)) = f ( 3) 4 ( 3) 3 4 4 y y − − = + = y – 3 + 3 = y. This shows that gof = IN and fog = IY , which implies that f is invertible and g is the inverse of f. Example 18 If R1 and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation. Solution Since R1 and R2 are equivalence relations, (a, a) ∈ R1 , and (a, a) ∈ R2 ∀ a ∈A. This implies that (a, a) ∈ R1 ∩ R2 , ∀ a, showing R1 ∩ R2 is reflexive. Further, (a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1 and (b, a) ∈ R2 ⇒ (b, a) ∈ R1 ∩ R2 , hence, R1 ∩ R2 is symmetric. Similarly, (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2 ⇒ (a, c) ∈ R1 and (a, c) ∈ R2 ⇒ (a, c) ∈ R1 ∩ R2 . This shows that R1 ∩ R2 is transitive. Thus, R1 ∩ R2 is an equivalence relation. Example 19 Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y) R (u, v) if and only if xv = yu. Show that R is an equivalence relation. Solution Clearly, (x, y) R (x, y), ∀ (x, y) ∈ A, since xy = yx. This shows that R is reflexive. Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y). This shows that R is symmetric. Similarly, (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and ( 3) ( ) 4 y g y − = . Now, gof(x) = g (f(x)) = g (4x + 3) = (4 3 3) 4 x x + − = and Miscellaneous Examples RELATIONS AND FUNCTIONS 13 ub = va ⇒ a a xv yu u u = ⇒ b a xv yu v u = ⇒ xb = ya and hence (x, y) R (a, b). Thus, R is transitive. Thus, R is an equivalence relation. Example 20 Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R1 = R2 . Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3. Therefore, (x, y) ∈ R1 ⇒ x – y is a multiple of 3 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ (x, y) ∈ R2 . Hence, R1 ⊂ R2 . Similarly, {x, y} ∈ R2 ⇒ {x, y} Reprint 2025-26 ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible by 3 ⇒ {x, y} ∈ R1 . This shows that R2 ⊂ R1 . Hence, R1 = R2 . Example 21 Let f : X → Y be a function. Define a relation R in X given by R = {(a, b): f(a) = f(b)}. Examine whether R is an equivalence relation or not. Solution For every a ∈ X, (a, a) ∈ R, since f(a) = f(a), showing that R is reflexive. Similarly, (a, b) ∈ R ⇒ f(a) = f(b) ⇒ f(b) = f(a) ⇒ (b, a) ∈ R. Therefore, R is symmetric. Further, (a, b) ∈ R and (b, c) ∈ R ⇒ f(a) = f(b) and f(b) = f(c) ⇒ f(a) = f(c) ⇒ (a, c) ∈ R, which implies that R is transitive. Hence, R is an equivalence relation. Example 22 Find the number of all one-one functions from set A = {1, 2, 3} to itself. Solution One-one function from {1, 2, 3} to itself is simply a permutation on three symbols 1, 2, 3. Therefore, total number of one-one maps from {1, 2, 3} to itself is same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6. Example 23 Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three. Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Now, if we add the pair (2, 1) to R1 to get R2 , then the relation R2 will be reflexive, transitive but not symmetric. Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation. However, we can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R1 at a time, as by doing so, we will be forced to add the remaining pair in order to maintain transitivity and in the process, the relation will become symmetric also which is not required. Thus, the total number of desired relations is three. 14 MATHEMATICS Example 24 Show that the number of equivalence relation in the set {1, 2, 3} containing (1, 2) and (2, 1) is two. Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1). If we add any one, say (2, 3) to R1 , then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1). Thus, the only equivalence relation bigger than R1 is the universal relation. This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two. Example 25 Consider the identity function I N : N → N defined as I N (x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as (IN + IN ) (x) = IN (x) + IN (x) = x + x = 2x is not onto. Reprint 2025-26 Solution Clearly IN is onto. But IN + IN is not onto, as we can find an element 3 in the co-domain N such that there does not exist any x in the domain N with (IN + IN ) (x) = 2x = 3. Example 26 Consider a function f : 0, 2 π → R given by f(x) = sin x and g : 0, 2 π → R given by g(x) = cos x. Show that f and g are one-one, but f + g is not one-one. Solution Since for any two distinct elements x1 and x2 in 0, 2 π , sin x1 ≠ sin x2 and cos x1 ≠ cos x2 , both f and g must be one-one. But (f + g) (0) = sin 0 + cos 0 = 1 and (f + g) 2 π = sin cos 1 2 2 π π + = . Therefore, f + g is not one-one. 1. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by ( ) 1 | | x f x x = + , 2. Show that the function f : R → R given by f(x) = x 3 is injective. 3. Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer. 4. Find the number of all onto functions from the set {1, 2, 3,......, n} to itself. 5. Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined x ∈ R is one one and onto function. Miscellaneous Exercise on Chapter 1 RELATIONS AND FUNCTIONS 15 by f(x) = x 2 – x, x ∈ A and 1 ( ) 2 1, 2 g x x = − − x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀ a ∈ A, are called equal functions). Reprint 2025-26 In this chapter, we studied different types of relations and equivalence relation, composition of functions, invertible functions and binary operations. The main features of this chapter are as follows: Æ Empty relation is the relation R in X given by R = φ ⊂ X × X. Æ Universal relation is the relation R in X given by R = X × X. Æ Reflexive relation R in X is a relation with (a, a) ∈ R ∀ a ∈ X. Æ Symmetric relation R in X is a relation satisfying (a, b) ∈ R implies (b, a) ∈ R. Æ Transitive relation R in X is a relation satisfying (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R. Æ Equivalence relation R in X is a relation which is reflexive, symmetric and transitive. Æ Equivalence class [a] containing a ∈ X for an equivalence relation R in X is the subset of X containing all elements b related to a. Æ A function f : X → Y is one-one (or injective) if 16 MATHEMATICS 6. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4 f(x1 ) = f(x2 ) ⇒ x1 = x2 ∀ x1 , x2 ∈ X. Æ A function f : X → Y is onto (or surjective) if given any y ∈ Y, ∃ x ∈ X such that f(x) = y. Æ A function f : X → Y is one-one and onto (or bijective), if f is both one-one and onto. Æ Given a finite set X, a function f : X → X is one-one (respectively onto) if and only if f is onto (respectively one-one). This is the characteristic property of a finite set. This is not true for infinite set Summary Reprint 2025-26 Historical Note The concept of function has evolved over a long period of time starting from R. Descartes (1596-1650), who used the word ‘function’ in his manuscript “Geometrie” in 1637 to mean some positive integral power x n of a variable x while studying geometrical curves like hyperbola, parabola and ellipse. James Gregory (1636-1675) in his work “ Vera Circuli et Hyperbolae Quadratura” (1667) considered function as a quantity obtained from other quantities by successive use of algebraic operations or by any other operations. Later G. W. Leibnitz (1646-1716) in his manuscript “Methodus tangentium inversa, seu de functionibus” written in 1673 used the word ‘function’ to mean a quantity varying from point to point on a curve such as the coordinates of a point on the curve, the slope of the curve, the tangent and the normal to the curve at a point. However, in his manuscript “Historia” (1714), Leibnitz used the word ‘function’ to mean quantities that depend on a variable. He was the first to use the phrase ‘function of x’. John Bernoulli (1667-1748) used the notation φx for the first time in 1718 to indicate a function of x. But the general adoption of symbols like f, F, φ, ψ ... to represent functions was made by Leonhard Euler (1707-1783) in 1734 in the first part of his manuscript “Analysis Infinitorium”. Later on, Joeph Louis Lagrange (1736-1813) published his manuscripts “Theorie des functions analytiques” in 1793, where he discussed about analytic function and used the notion f (x), F(x), φ(x) etc. for different function of x. Subsequently, Lejeunne Dirichlet (1805-1859) gave the definition of function which was being used till the set theoretic definition of function presently used, was given after set theory was developed by Georg Cantor (1845-1918). The set theoretic definition of function known to us presently is simply an abstraction of the definition given by Dirichlet in a rigorous manner. RELATIONS AND FUNCTIONS 17 Reprint 2025-26 —v—" class_12,2,Inverse Trigonometric Functions,ncert_books/class_12/lemh1dd/lemh102.pdf,"18 MATHEMATICS 2.1 Introduction In Chapter 1, we have studied that the inverse of a function f, denoted by f –1, exists if f is one-one and onto. There are many functions which are not one-one, onto or both and hence we can not talk of their inverses. In Class XI, we studied that trigonometric functions are not one-one and onto over their natural domains and ranges and hence their inverses do not exist. In this chapter, we shall study about the restrictions on domains and ranges of trigonometric functions which ensure the existence of their inverses and observe their behaviour through graphical representations. Besides, some elementary properties will also be discussed. The inverse trigonometric functions play an important role in calculus for they serve to define many integrals. The concepts of inverse trigonometric functions is also used in science and engineering. INVERSE TRIGONOMETRIC FUNCTIONS vMathematics, in general, is fundamentally the science of self-evident things. — FELIX KLEIN v Chapter 2 Aryabhata (476-550 A. D.) 2.2 Basic Concepts In Class XI, we have studied trigonometric functions, which are defined as follows: sine function, i.e., sine : R → [– 1, 1] cosine function, i.e., cos : R → [– 1, 1] tangent function, i.e., tan : R – { x : x = (2n + 1) 2 π , n ∈ Z} → R cotangent function, i.e., cot : R – { x : x = nπ, n ∈ Z} → R secant function, i.e., sec : R – { x : x = (2n + 1) 2 π , n ∈ Z} → R – (– 1, 1) cosecant function, i.e., cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1) Reprint 2025-26 onto, then we can define a unique function g : Y→X such that g(y) = x, where x ∈ X and y = f (x), y ∈ Y. Here, the domain of g = range of f and the range of g = domain of f. The function g is called the inverse of f and is denoted by f –1. Further, g is also one-one and onto and inverse of g is f. Thus, g –1 = (f –1) –1 = f. We also have (f –1 o f ) (x) = f –1 (f (x)) = f –1(y) = x and (f o f –1) (y) = f (f –1(y)) = f(x) = y Since the domain of sine function is the set of all real numbers and range is the closed interval [–1, 1]. If we restrict its domain to , 2 2 −π π , then it becomes one-one and onto with range [– 1, 1]. Actually, sine function restricted to any of the intervals therefore, define the inverse of sine function in each of these intervals. We denote the inverse of sine function by sin–1 (arc sine function). Thus, sin–1 is a function whose domain is [– 1, 1] and range could be any of the intervals 3 , 2 2 − π −π , , 2 2 −π π or function sin–1. The branch with range , 2 2 −π π is called the principal value branch, whereas other intervals as range give different branches of sin–1 . When we refer to the function sin–1, we take it as the function whose domain is [–1, 1] and range is − 3 , 2 2 π π , and so on. Corresponding to each such interval, we get a branch of the 3 2 2 π π , , , 2 2 −π π , 3 , 2 2 π π etc., is one-one and its range is [–1, 1]. We can, We have also learnt in Chapter 1 that if f : X→Y such that f(x) = y is one-one and INVERSE TRIGONOMETRIC FUNCTIONS 19 if – 1 ≤ x ≤ 1 and sin–1 (sin x) = x if 2 2 x π π − ≤ ≤ . In other words, if y = sin–1 x, then sin y = x. Remarks (i) We know from Chapter 1, that if y = f(x) is an invertible function, then x = f –1 (y). Thus, the graph of sin–1 function can be obtained from the graph of original function by interchanging x and y axes, i.e., if (a, b) is a point on the graph of sine function, then (b, a) becomes the corresponding point on the graph of inverse , 2 2 −π π . We write sin–1 : [–1, 1] → , 2 2 −π π From the definition of the inverse functions, it follows that sin (sin–1 x) = x Reprint 2025-26 20 MATHEMATICS of sine function. Thus, the graph of the function y = sin–1 x can be obtained from the graph of y = sin x by interchanging x and y axes. The graphs of y = sin x and y = sin–1 x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of y = sin–1 x represent the principal value branch. (ii) It can be shown that the graph of an inverse function can be obtained from the corresponding graph of original function as a mirror image (i.e., reflection) along the line y = x. This can be visualised by looking the graphs of y = sin x and y = sin–1 x as given in the same axes (Fig 2.1 (iii)). Fig 2.1 (i) real numbers and range is the set [–1, 1]. If we restrict the domain of cosine function to [0, π], then it becomes one-one and onto with range [–1, 1]. Actually, cosine function Like sine function, the cosine function is a function whose domain is the set of all Fig 2.1 (ii) Fig 2.1 (iii) Reprint 2025-26 restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with range as [–1, 1]. We can, therefore, define the inverse of cosine function in each of these intervals. We denote the inverse of the cosine function by cos–1 (arc cosine function). Thus, cos–1 is a function whose domain is [–1, 1] and range could be any of the intervals [–π, 0], [0, π], [π, 2π] etc. Corresponding to each such interval, we get a branch of the function cos–1. The branch with range [0, π] is called the principal value branch of the function cos–1 . We write The graph of the function given by y = cos–1 x can be drawn in the same way as discussed about the graph of y = sin–1 x. The graphs of y = cos x and y = cos–1 x are given in Fig 2.2 (i) and (ii). cos–1 : [–1, 1] → [0, π]. Fig 2.2 (i) INVERSE TRIGONOMETRIC FUNCTIONS 21 Fig 2.2 (ii) x ≠ nπ, n ∈ Z} and the range is the set {y : y ∈ R, y ≥ 1 or y ≤ –1} i.e., the set R – (–1, 1). It means that y = cosec x assumes all real values except –1 < y < 1 and is not defined for integral multiple of π. If we restrict the domain of cosec function to cosec function restricted to any of the intervals 3 , { } 2 2 − π −π − −π , , 2 2 −π π – {0}, 3 , { } 2 2 π π − π etc., is bijective and its range is the set of all real numbers R – (–1, 1). , 2 2 π π − – {0}, then it is one to one and onto with its range as the set R – (– 1, 1). Actually, Let us now discuss cosec–1x and sec–1x as follows: Since, cosec x = 1 sin x , the domain of the cosec function is the set {x : x ∈ R and Reprint 2025-26 22 MATHEMATICS Thus cosec–1 can be defined as a function whose domain is R – (–1, 1) and range could be any of the intervals − − − − 3 2 2 π π , { }π , − − π π 2 2 0, { } , 3 , { } 2 2 π π − π etc. The function corresponding to the range , {0} 2 2 −π π − is called the principal value branch of cosec–1 . We thus have principal branch as cosec–1 : R – (–1, 1) → , {0} 2 2 −π π − The graphs of y = cosec x and y = cosec–1 x are given in Fig 2.3 (i), (ii). n ∈ Z} and range is the set R – (–1, 1). It means that sec (secant function) assumes all real values except –1 < y < 1 and is not defined for odd multiples of 2 π . If we restrict the domain of secant function to [0, π] – { 2 π }, then it is one-one and onto with Also, since sec x = 1 cos x , the domain of y = sec x is the set R – {x : x = (2n + 1) 2 π , Fig 2.3 (i) Fig 2.3 (ii) Reprint 2025-26 its range as the set R – (–1, 1). Actually, secant function restricted to any of the intervals [–π, 0] – { 2 −π }, [0, ] – 2 π π , [π, 2π] – { 3 2 π } etc., is bijective and its range is R – {–1, 1}. Thus sec–1 can be defined as a function whose domain is R– (–1, 1) and range could be any of the intervals [– π, 0] – { 2 −π }, [0, π] – { 2 π }, [π, 2π] – { 3 2 π } etc. Corresponding to each of these intervals, we get different branches of the function sec–1 . The branch with range [0, π] – { 2 π } is called the principal value branch of the function sec–1 . We thus have The graphs of the functions y = sec x and y = sec-1 x are given in Fig 2.4 (i), (ii). sec–1 : R – (–1,1) → [0, π] – { 2 π } INVERSE TRIGONOMETRIC FUNCTIONS 23 {x : x ∈ R and x ≠ (2n +1) 2 π , n ∈ Z} and the range is R. It means that tan function is not defined for odd multiples of 2 π . If we restrict the domain of tangent function to Finally, we now discuss tan–1 and cot–1 We know that the domain of the tan function (tangent function) is the set Fig 2.4 (i) Fig 2.4 (ii) Reprint 2025-26 24 MATHEMATICS restricted to any of the intervals 3 , 2 2 − π −π , , 2 2 −π π , 3 , 2 2 π π etc., is bijective and its range is R. Thus tan–1 can be defined as a function whose domain is R and range could be any of the intervals 3 , 2 2 − π −π , , 2 2 −π π , 3 , 2 2 π π and so on. These intervals give different branches of the function tan–1. The branch with range , 2 2 −π π is called the principal value branch of the function tan–1 . We thus have , 2 2 −π π , then it is one-one and onto with its range as R. Actually, tangent function tan–1 : R → , 2 2 −π π The graphs of the function y = tan x and y = tan–1x are given in Fig 2.5 (i), (ii). {x : x ∈ R and x ≠ nπ, n ∈ Z} and range is R. It means that cotangent function is not defined for integral multiples of π. If we restrict the domain of cotangent function to (0, π), then it is bijective with and its range as R. In fact, cotangent function restricted to any of the intervals (–π, 0), (0, π), (π, 2π) etc., is bijective and its range is R. Thus cot –1 can be defined as a function whose domain is the R and range as any of the We know that domain of the cot function (cotangent function) is the set Fig 2.5 (i) Fig 2.5 (ii) Reprint 2025-26 Fig 2.6 (i) Fig 2.6 (ii) The following table gives the inverse trigonometric function (principal value branches) along with their domains and ranges. intervals (–π, 0), (0, π), (π, 2π) etc. These intervals give different branches of the function cot –1. The function with range (0, π) is called the principal value branch of the function cot –1 . We thus have cot–1 : R → (0, π) The graphs of y = cot x and y = cot–1x are given in Fig 2.6 (i), (ii). INVERSE TRIGONOMETRIC FUNCTIONS 25 sin–1 : [–1, 1] → , 2 2 π π − cos–1 : [–1, 1] → [0, π] cosec–1 : R – (–1,1) → , 2 2 π π − – {0} sec–1 : R – (–1, 1) → [0, π] – { } 2 π tan–1 : R → , 2 2 −π π cot–1 : R → (0, π) Reprint 2025-26 26 MATHEMATICS Example 1 Find the principal value of sin–1 1 2 . Solution Let sin–1 1 Example 2 Find the principal value of cot–1 1 Solution Let cot–1 1 sin 4 π = 1 ANote We now consider some examples: We know that the range of the principal value branch of sin–1 is − π π 2 2 , and 3 − = y. Then, 1 cot cot 3 3 y − π = = − = cot 3 π π − = 2 cot 3 π We know that the range of principal value branch of cot–1 is (0, π) and 1. sin–1x should not be confused with (sin x) –1. In fact (sin x) –1 = 1 sin x and similarly for other trigonometric functions. 2. Whenever no branch of an inverse trigonometric functions is mentioned, we mean the principal value branch of that function. 3. The value of an inverse trigonometric functions which lies in the range of principal branch is called the principal value of that inverse trigonometric functions. 2 . Therefore, principal value of sin–1 1 2 = y. Then, sin y = 1 3 − 2 . 2 is 4 π cot 2 3 π = 1 Find the principal values of the following: 1. sin–1 1 2 − 2. cos–1 3 2 3. cosec–1 (2) 4. tan–1 ( 3) − 5. cos–1 1 2 − 6. tan–1 (–1) 3 − . Hence, principal value of cot–1 1 3 − is 2 3 π EXERCISE 2.1 Reprint 2025-26 10. cosec–1 ( − 2 ) Find the values of the following: 2.3 Properties of Inverse Trigonometric Functions In this section, we shall prove some important properties of inverse trigonometric functions. It may be mentioned here that these results are valid within the principal value branches of the corresponding inverse trigonometric functions and wherever they are defined. Some results may not be valid for all values of the domains of inverse trigonometric functions. In fact, they will be valid only for some values of x for which inverse trigonometric functions are defined. We will not go into the details of these values of x in the domain as this discussion goes beyond the scope of this textbook. 11. tan–1(1) + cos–1 1 2 − + sin–1 1 2 − 12. cos–1 1 2 + 2 sin–1 1 2 13. If sin–1 x = y, then 14. tan–1 ( ) 1 3 sec 2 − − − is equal to 7. sec–1 2 (A) 0 ≤ y ≤ π (B) 2 2 y π π − ≤ ≤ (A) π (B) 3 π − (C) 3 π (D) 2 3 π (C) 0 < y < π (D) 2 2 y π π − < < 3 8. cot–1 ( 3) 9. cos–1 1 INVERSE TRIGONOMETRIC FUNCTIONS 27 2 − is equivalent to functions. Let us recall that if y = sin–1x, then x = sin y and if x = sin y, then y = sin–1x. This For suitable values of domain similar results follow for remaining trigonometric sin (sin–1 x) = x, x ∈ [– 1, 1] and sin–1 (sin x) = x, x ∈ , 2 2 −π π Reprint 2025-26 28 MATHEMATICS Example 3 Show that Solution Example 4 Express 1 cos tan 1 sin x x − − , 3 2 2 − π π < 1 in the simplest form. Solution Let x = sec θ, then 2 x −1 = 2 sec 1 tan θ − = θ Prove the following: Write the following functions in the simplest form: 1. 3sin–1 x = sin–1 (3x – 4x 3 ), 1 1 – , 2 2 x ∈ 2. 3cos–1 x = cos–1 (4x 3 – 3x), 1 , 1 2 x ∈ 3. 2 1 1 1 tan x x − + − , x ≠ 0 4. 1 1 cos tan 1 cos x x − − + , 0 < x < π 5. 1 cos sin tan cos sin x x x x − − + , 4 −π < x < 3 4 π 6. 1 2 2 tan x Therefore, –1 2 1 cot x −1 = cot–1 (cot θ) = θ = sec–1 x, which is the simplest form. EXERCISE 2.2 INVERSE TRIGONOMETRIC FUNCTIONS 29 Find the values of each of the following: 7. 2 3 1 3 2 3 tan 3 a x x a ax − − − , a > 0; 3 3 − < < a a x 8. –1 –1 1 tan 2cos 2sin 2 9. 2 –1 –1 2 2 1 2 1 tan sin cos 2 1 1 x y x y − + + + , | x | < 1, y > 0 and xy < 1 a x − − , | x | < a Reprint 2025-26 30 MATHEMATICS Find the values of each of the expressions in Exercises 16 to 18. Example 6 Find the value of 1 3 sin (sin ) 5 − π 10. –1 2 sin sin 3 π 11. –1 3 tan tan 4 π 12. –1 –1 3 3 tan sin cot 5 2 + 13. 1 7 cos cos is equal to 6 − π 14. 1 1 sin sin ( ) 3 2 π − − − is equal to 15. 1 1 tan 3 cot ( 3) − − − − is equal to (A) 7 6 π (B) 5 6 π (C) 3 π (D) 6 π (A) 1 2 (B) 1 3 (C) 1 4 (D) 1 (A) π (B) 2 π − (C) 0 (D) 2 3 Miscellaneous Examples Solution We know that 1 sin (sin ) x x − = . Therefore, 1 3 3 sin (sin ) 5 5 − π π = But 3 , 5 2 2 π π π ∉ − , which is the principal branch of sin–1 x However 3 3 2 sin ( ) sin( ) sin 5 5 5 π π π = π − = and 2 , 5 2 2 π π π ∈ − Therefore 1 3 1 2 2 sin (sin ) sin (sin ) 5 5 5 − π − π π = = Reprint 2025-26 Find the value of the following: Prove that Prove that 10. –1 1 1 1 –1 tan cos 1 1 4 2 x x x x x + − − π = − + + − , 1 1 2 − ≤ ≤x [Hint: Put x = cos 2θ] 1. –1 13 cos cos 6 π 2. –1 7 tan tan 6 π 3. –1 –1 3 24 2sin tan 5 7 = 4. –1 8 3 77 –1 –1 sin sin tan 17 5 36 + = 5. –1 –1 4 12 33 –1 cos cos cos 5 13 65 + = 6. –1 12 3 56 –1 –1 cos sin sin 13 5 65 + = 7. –1 63 5 3 –1 –1 tan sin cos 16 13 5 = + 8. –1 1 1 –1 tan cos 2 1 9. –1 1 sin 1 sin cot 1 sin 1 sin 2 x x x x x + + − = + − − , 0, 4 x π ∈ x x x − = + , x ∈ [0, 1] Miscellaneous Exercise on Chapter 2 INVERSE TRIGONOMETRIC FUNCTIONS 31 Solve the following equations: 13. sin (tan–1 x), | x | < 1 is equal to 14. sin–1 (1 – x) – 2 sin–1 x = 2 π , then x is equal to 11. 2tan–1 (cos x) = tan–1 (2 cosec x) 12. –1 1 1 –1 tan tan ,( 0) 1 2 x x x x − = > + (A) 2 1 x (A) 0, 1 2 (B) 1, 1 2 (C) 0 (D) 1 2 − x (B) 2 1 1− x (C) 2 1 Reprint 2025-26 1+ x (D) 2 1 x + x 32 MATHEMATICS Summary Æ The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table: Æ sin–1x should not be confused with (sin x) –1. In fact (sin x) –1 = 1 sin x and similarly for other trigonometric functions. Æ The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions. For suitable values of domain, we have Æ y = sin–1 x ⇒ x = sin y Æ x = sin y ⇒ y = sin–1 x Æ sin (sin–1 x) = x Æ sin–1 (sin x) = x Functions Domain Range (Principal Value Branches) y = sin–1 x [–1, 1] , 2 2 −π π y = cos–1 x [–1, 1] [0, π] y = cosec–1 x R – (–1,1) , 2 2 −π π – {0} y = sec–1 x R – (–1, 1) [0, π] – { } 2 π y = tan–1 x R , 2 2 π π − y = cot–1 x R (0, π) Reprint 2025-26 The study of trigonometry was first started in India. The ancient Indian Mathematicians, Aryabhata (476A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. All this knowledge went from India to Arabia and then from there to Europe. The Greeks had also started the study of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately adopted throughout the world. In India, the predecessor of the modern trigonometric functions, known as the sine of an angle, and the introduction of the sine function represents one of the main contribution of the siddhantas (Sanskrit astronomical works) to mathematics. Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa contains a proof for the expansion of sin (A + B). Exact expression for sines or cosines of 18°, 36°, 54°, 72°, etc., were given by Bhaskara II. The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were suggested by the astronomer Sir John F.W. Hersehel (1813) The name of Thales (about 600 B.C.) is invariably associated with height and distance problems. He is credited with the determination of the height of a great pyramid in Egypt by measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known height, and comparing the ratios: H S h s = = tan (sun’s altitude) Historical Note INVERSE TRIGONOMETRIC FUNCTIONS 33 Thales is also said to have calculated the distance of a ship at sea through the proportionality of sides of similar triangles. Problems on height and distance using the similarity property are also found in ancient Indian works. Reprint 2025-26 —v—" class_12,3,Matrices,ncert_books/class_12/lemh1dd/lemh103.pdf,"34 MATHEMATICS 3.1 Introduction The knowledge of matrices is necessary in various branches of mathematics. Matrices are one of the most powerful tools in mathematics. This mathematical tool simplifies our work to a great extent when compared with other straight forward methods. The evolution of concept of matrices is the result of an attempt to obtain compact and simple methods of solving system of linear equations. Matrices are not only used as a representation of the coefficients in system of linear equations, but utility of matrices far exceeds that use. Matrix notation and operations are used in electronic spreadsheet programs for personal computer, which in turn is used in different areas of business and science like budgeting, sales projection, cost estimation, analysing the results of an experiment etc. Also, many physical operations such as magnification, rotation and reflection through a plane can be represented mathematically by matrices. Matrices are also used in cryptography. This mathematical tool is not only used in certain branches of sciences, but also in genetics, economics, sociology, modern psychology and industrial management. vThe essence of Mathematics lies in its freedom. — CANTOR v MATRICES Chapter 3 In this chapter, we shall find it interesting to become acquainted with the fundamentals of matrix and matrix algebra. 3.2 Matrix Suppose we wish to express the information that Radha has 15 notebooks. We may express it as [15] with the understanding that the number inside [ ] is the number of notebooks that Radha has. Now, if we have to express that Radha has 15 notebooks and 6 pens. We may express it as [15 6] with the understanding that first number inside [ ] is the number of notebooks while the other one is the number of pens possessed by Radha. Let us now suppose that we wish to express the information of possession Reprint 2025-26 of notebooks and pens by Radha and her two friends Fauzia and Simran which is as follows: Radha has 15 notebooks and 6 pens, Fauzia has 10 notebooks and 2 pens, Simran has 13 notebooks and 5 pens. Now this could be arranged in the tabular form as follows: Notebooks Pens Radha 15 6 Fauzia 10 2 Simran 13 5 and this can be expressed as or Radha Fauzia Simran Notebooks 15 10 13 MATRICES 35 note books possessed by Radha, Fauzia and Simran, respectively and the entries in the second column represent the number of pens possessed by Radha, Fauzia and Simran, Pens 6 2 5 which can be expressed as: In the first arrangement the entries in the first column represent the number of Reprint 2025-26 36 MATHEMATICS respectively. Similarly, in the second arrangement, the entries in the first row represent the number of notebooks possessed by Radha, Fauzia and Simran, respectively. The entries in the second row represent the number of pens possessed by Radha, Fauzia and Simran, respectively. An arrangement or display of the above kind is called a matrix. Formally, we define matrix as: Definition 1 A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix. In the above examples, the horizontal lines of elements are said to constitute, rows of the matrix and the vertical lines of elements are said to constitute, columns of the matrix. Thus A has 3 rows and 2 columns, B has 3 rows and 3 columns while C has 2 rows and 3 columns. 3.2.1 Order of a matrix A matrix having m rows and n columns is called a matrix of order m × n or simply m × n matrix (read as an m by n matrix). So referring to the above examples of matrices, we have A as 3 × 2 matrix, B as 3 × 3 matrix and C as 2 × 3 matrix. We observe that A has 3 × 2 = 6 elements, B and C have 9 and 6 elements, respectively. We denote matrices by capital letters. The following are some examples of matrices: – 2 5 A 0 5 = , 3 6 1 2 3 2 B 3.5 –1 2 i + − = , 3 1 3 C cos sin 2 tan x x x x x + = + 5 3 5 7 or A = [aij] m × n , 1≤ i ≤ m, 1≤ j ≤ n i, j ∈ N Thus the i th row consists of the elements ai1 , ai2 , ai3 ,..., ain, while the j th column consists of the elements a1j , a2j , a3j ,..., amj , In general aij, is an element lying in the i th row and j th column. We can also call it as the (i, j) th element of A. The number of elements in an m × n matrix will be equal to mn. In general, an m × n matrix has the following rectangular array: Reprint 2025-26 figure in the form of a matrix. For example, consider a quadrilateral ABCD with vertices A (1, 0), B (3, 2), C (1, 3), D (–1, 2). y (or [x, y]). For example point P(0, 1) as a matrix representation may be given as ANote In this chapter x We can also represent any point (x, y) in a plane by a matrix (column or row) as Observe that in this way we can also express the vertices of a closed rectilinear Now, quadrilateral ABCD in the matrix form, can be represented as 1. We shall follow the notation, namely A = [aij] m × n to indicate that A is a matrix 2. We shall consider only those matrices whose elements are real numbers or functions taking real values. of order m × n. A B C D 1 3 1 1 X 0 2 3 2 × − = or 0 P 1 = or [0 1]. 2 4 B 3 2 Y C 1 3 D 1 2 × = − A 1 0 4 2 MATRICES 37 a plane. Now, let us consider some examples. Example 1 Consider the following information regarding the number of men and women workers in three factories I, II and III in the third row and second column represent? Thus, matrices can be used as representation of vertices of geometrical figures in Represent the above information in the form of a 3 × 2 matrix. What does the entry I 30 25 II 25 31 III 27 26 Men workers Women workers Reprint 2025-26 38 MATHEMATICS Solution The information is represented in the form of a 3 × 2 matrix as follows: The entry in the third row and second column represents the number of women workers in factory III. Example 2 If a matrix has 8 elements, what are the possible orders it can have? Solution We know that if a matrix is of order m × n, it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8. Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4) Hence, possible orders are 1 × 8, 8 ×1, 4 × 2, 2 × 4 Example 3 Construct a 3 × 2 matrix whose elements are given by 1 | 3 | 2 ij a i j = − . Solution In general a 3 × 2 matrix is given by 11 12 Now 1 | 3 | 2 ij a i j = − , i = 1, 2, 3 and j = 1, 2. 30 25 A 25 31 = 27 26 31 32 A = . a a a a a a 21 22 Therefore 11 1 |1 3 1| 1 2 a = − × = 12 1 5 |1 3 2 | 2 2 a = − × = Hence the required matrix is given by 21 1 1 | 2 3 1| 2 2 a = − × = 22 1 | 2 3 2 | 2 2 a = − × = 31 1 | 3 3 1| 0 2 a = − × = 32 1 3 | 3 3 2 | 2 2 a = − × = Reprint 2025-26 5 1 2 1 A 2 2 3 0 2 = . 3.3 Types of Matrices In this section, we shall discuss different types of matrices. (iii) Square matrix In general, A = [aij] m × 1 is a column matrix of order m × 1. (ii) Row matrix (i) Column matrix A matrix is said to be a column matrix if it has only one column. For example, A matrix is said to be a row matrix if it has only one row. For example, 1 4 1 B 5 2 3 2 × = − is a row matrix. In general, B = [bij] 1 × n is a row matrix of order 1 × n. A matrix in which the number of rows are equal to the number of columns, is said to be a square matrix. Thus an m × n matrix is said to be a square matrix if m = n and is known as a square matrix of order ‘n’. A 1 1/ 2 − = − is a square matrix of order 3. = − is a column matrix of order 4 × 1. 3 1 0 0 3 MATRICES 39 ANote If A = [aij] is a square matrix of order n, then elements (entries) a11, a22, ..., ann are said to constitute the diagonal, of the matrix A. Thus, if Then the elements of the diagonal of A are 1, 4, 6. For example In general, A = [aij] m × m is a square matrix of order m. 3 A 3 2 1 2 4 3 1 Reprint 2025-26 A 2 4 1 3 5 6 − = − . 1 3 1 40 MATHEMATICS (iv) Diagonal matrix (vi) Identity matrix A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix. In other words, the square matrix A = [aij] n × n is an of order 1, 2, 3, respectively. (v) Scalar matrix A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [bij] n × n is said to be a scalar matrix if bij = 0, when i ≠ j bij = k, when i = j, for some constant k. For example A square matrix B = [bij] m × m is said to be a diagonal matrix if all its non diagonal elements are zero, that is a matrix B = [bij] m × m is said to be a diagonal matrix if bij = 0, when i ≠ j. For example, A = [4], 1 0 B 0 2 − = , A = [3], 1 0 B 0 1 − = − , are scalar matrices of order 1, 2 and 3, respectively. C 0 3 0 = 0 0 3 C 0 2 0 0 0 3 3 0 0 − = , are diagonal matrices 1.1 0 0 identity matrix, if 1 if 0 if ij i j a i j = = ≠ . We denote the identity matrix of order n by I n . When order is clear from the context, we simply write it as I. For example [1], 1 0 respectively. Observe that a scalar matrix is an identity matrix when k = 1. But every identity matrix is clearly a scalar matrix. 0 1 , are identity matrices of order 1, 2 and 3, 1 0 0 0 1 0 0 0 1 Reprint 2025-26 3.3.1 Equality of matrices Definition 2 Two matrices A = [aij] and B = [bij] are said to be equal if (i) they are of the same order not equal matrices. Symbolically, if two matrices A and B are equal, we write A = B. Example 4 If (vii) Zero matrix (ii) each element of A is equal to the corresponding element of B, that is aij = bij for all i and j. For example, 2 3 2 3 and 0 1 0 1 are equal matrices but 3 2 2 3 and 0 1 0 1 are If A matrix is said to be zero matrix or null matrix if all its elements are zero. For example, [0], 0 0 0 0 0 , [0, 0] are all zero matrices. We denote zero matrix by O. Its order will be clear from the context. − = , then x = – 1.5, y = 0, z = 2, a = 6 , b = 3, c = 2 x y z a b c + + − − − − = − − + − − + − x z y y a c b b 3 2 2 6 1.5 0 3 4 2 7 0 6 3 2 6 1 0 6 3 2 2 3 21 0 2 4 21 0 0 0 , 0 0 0 MATRICES 41 Find the values of a, b, c, x, y and z. Solution As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get x + 3 = 0, z + 4 = 6, 2y – 7 = 3y – 2 a – 1 = – 3, 0 = 2c + 2 b – 3 = 2b + 4, Simplifying, we get a = – 2, b = – 7, c = – 1, x = – 3, y = –5, z = 2 Example 5 Find the values of a, b, c, and d from the following equation: 2 2 4 3 5 4 3 11 24 a b a b c d c d + − − = − + Reprint 2025-26 42 MATHEMATICS Solution By equality of two matrices, equating the corresponding elements, we get 1. In the matrix (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23. 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements? 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? 4. Construct a 2 × 2 matrix, A = [aij], whose elements are given by: 2a + b = 4 5c – d = 11 a – 2b = – 3 4c + 3d = 24 Solving these equations, we get a = 1, b = 2, c = 3 and d = 4 (i) 2 ( ) 2 ij i j a + = (ii) ij i a j = (iii) 2 ( 2 ) 2 ij i j a + = 2 5 19 7 5 A 35 2 12 2 − = − − , write: 3 1 5 17 EXERCISE 3.1 5. Construct a 3 × 4 matrix, whose elements are given by: 6. Find the values of x, y and z from the following equations: 7. Find the value of a, b, c and d from the equation: (i) 1 | 3 | 2 ij a i j = − + (ii) 2 ij a i j = − (i) 4 3 x = (ii) 2 6 2 5 5 8 x y z xy + = + (iii) 5 1 5 y z 2 1 5 2 3 0 13 a b a c a b c d − + − = − + Reprint 2025-26 + + + = + x y z x z y z 9 5 7 3.4 Operations on Matrices In this section, we shall introduce certain operations on matrices, namely, addition of matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices. 3.4.1 Addition of matrices Suppose Fatima has two factories at places A and B. Each factory produces sport shoes for boys and girls in three different price categories labelled 1, 2 and 3. The quantities produced by each factory are represented as matrices given below: 10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512 8. A = [aij] m × n\ is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these 9. Which of the given values of x and y make the following pair of matrices equal 3 7 5 1 2 3 x y x + + − , 0 2 8 4 y − (A) 1 , 7 3 x y − = = (B) Not possible to find (C) y = 7, 2 3 x − = (D) 1 2 , 3 3 x y − − = = MATRICES 43 category. Then the total production Suppose Fatima wants to know the total production of sport shoes in each price In category 1 : for boys (80 + 90), for girls (60 + 50) In category 2 : for boys (75 + 70), for girls (65 + 55) In category 3 : for boys (90 + 75), for girls (85 + 75) This can be represented in the matrix form as Reprint 2025-26 + + + + + + . 80 90 60 50 75 70 65 55 90 75 85 75 44 MATHEMATICS This new matrix is the sum of the above two matrices. We observe that the sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. Furthermore, the two matrices have to be of the same order. 2×3 matrix. Then, we define 11 11 12 12 13 13 21 21 22 22 23 23 A + B a b a b a b a b a b a b + + + = + + + . In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n. Then, the sum of the two matrices A and B is defined as a matrix C = [cij] m × n , where cij = aij + bij, for all possible values of i and j. Example 6 Given 3 1 1 A 2 3 0 − = and 2 5 1 B 1 2 3 2 Since A, B are of the same order 2 × 3. Therefore, addition of A and B is defined and is given by Thus, if 11 12 13 21 22 23 A a a a 2 3 1 5 1 1 2 3 1 5 0 A + B 1 1 2 2 3 3 0 0 6 2 2 + + − + + = = − + + a a a = is a 2 × 3 matrix and 11 12 13 21 22 23 B b b b = − , find A + B b b b = is another 3.4.2 Multiplication of a matrix by a scalar Now suppose that Fatima has doubled the production at a factory A in all categories (refer to 3.4.1). ANote 1. We emphasise that if A and B are not of the same order, then A + B is not 2. We may observe that addition of matrices is an example of binary operation on the set of matrices of the same order. defined. For example if 2 3 A 1 0 = , 1 2 3 B , 1 0 1 = then A + B is not defined. Reprint 2025-26 the new matrix is obtained by multiplying each element of the previous matrix by 2. A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar k. In other words, kA = k [aij] m × n = [k (aij)] m × n , that is, (i, j) th element of kA is kaij for all possible values of i and j. Previously quantities (in standard units) produced by factory A were Revised quantities produced by factory A are as given below: This can be represented in the matrix form as In general, we may define multiplication of a matrix by a scalar as follows: if 1 2 80 2 60 2 2 75 2 65 3 2 90 2 85 × × × × × × Boys Girls . We observe that 160 120 150 130 180 170 MATRICES 45 –A = (– 1) A. For example, if A = Negative of a matrix The negative of a matrix is denoted by – A. We define 3A = − , then 3 5 7 3 3 5 21 9 − = − 3 1 1.5 5 7 3 2 0 5 3 1 1.5 9 3 4.5 2 0 5 6 0 15 Reprint 2025-26 46 MATHEMATICS – A = (– 1) 3 1 3 1 A ( 1) 5 5 x x − − = − = − − Difference of matrices If A = [aij], B = [bij] are two matrices of the same order, say m × n, then difference A – B is defined as a matrix D = [dij], where dij = aij – bij, for all value of i and j. In other words, D = A – B = A + (–1) B, that is sum of the matrix A and the matrix – B. Example 7 If 1 2 3 3 1 3 A and B 2 3 1 1 0 2 − = = − , then find 2A – B. Solution We have 3.4.3 Properties of matrix addition The addition of matrices satisfy the following properties: For example, let A = 3 1 5 x − , then – A is given by 2A – B = 2 1 2 3 2 3 1 = 2 4 6 3 1 3 4 6 2 1 0 2 − − + − = 2 3 4 1 6 3 1 5 3 4 1 6 0 2 2 5 6 0 − + − − = + + − 3 1 3 1 0 2 − − − (i) Commutative Law If A = [aij], B = [bij] are matrices of the same order, say m × n, then A + B = B + A. Now A + B = [aij] + [bij] = [aij + bij] = [bij + aij] (addition of numbers is commutative) = ([bij] + [aij]) = B + A (ii) Associative Law For any three matrices A = [aij], B = [bij], C = [cij] of the same order, say m × n, (A + B) + C = A + (B + C). Now (A + B) + C = ([aij] + [bij]) + [c ij] = [aij + bij] + [cij] = [(aij + bij) + cij] = [aij + (bij + c ij)] (Why?) = [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C) Reprint 2025-26 3.4.4 Properties of scalar multiplication of a matrix If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are scalars, then Example 8 If 2A + 3X = 5B. (iv) The existence of additive inverse Let A = [aij] m × n be any matrix, then we (iii) Existence of additive identity Let A = [aij] be an m × n matrix and O be an m × n zero matrix, then A + O = O + A = A. In other words, O is the additive identity for matrix addition. (iii) ( k + l) A = (k + l) [aij] (ii) k (A + B) = k ([aij] + [bij]) = k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)] = [k aij] + [k bij] = k [aij] + k [bij] = kA + kB (i) k(A +B) = k A + kB, (ii) (k + l)A = k A + l A have another matrix as – A = [– aij] m × n such that A + (– A) = (– A) + A= O. So – A is the additive inverse of A or negative of A. A 4 2 and B 4 2 3 6 5 1 = [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A − = − = − , then find the matrix X, such that 8 0 2 2 MATRICES 47 Solution We have 2A + 3X = 5B or 2A + 3X – 2A = 5B – 2A or 2A – 2A + 3X = 5B – 2A (Matrix addition is commutative) or O + 3X = 5B – 2A (– 2A is the additive inverse of 2A) or 3X = 5B – 2A (O is the additive identity) or X = 1 3 (5B – 2A) or 2 2 8 0 1 X 5 4 2 2 4 2 3 5 1 3 6 − = − − − = Reprint 2025-26 10 10 16 0 1 20 10 8 4 3 25 5 6 12 − − + − − − − 48 MATHEMATICS Example 9 Find X and Y, if 5 2 X Y 0 9 + = and 3 6 X Y 0 1 − = − . Solution We have ( ) ( ) 5 2 3 6 X Y X Y 0 9 0 1 + + − = + − . or (X + X) + (Y – Y) = 8 8 or X = 1 8 8 4 4 2 0 8 0 4 = Also (X + Y) – (X – Y) = 5 2 3 6 or (X – X) + (Y + Y) = 5 3 2 6 0 9 1 − − + ⇒ 2 4 2Y 0 10 − = = 10 16 10 0 1 20 8 10 4 3 25 6 5 12 − − + − + − − − = 0 8 ⇒ 8 8 2X 0 8 = 0 9 0 1 − − 6 10 1 12 14 3 31 7 − − − − = − − − − 10 2 3 14 4 3 31 7 3 3 or Y = 1 2 4 1 2 2 0 10 0 5 − − = Example 10 Find the values of x and y from the following equation: Solution We have 5 3 4 2 7 3 1 2 x y − + − = 7 6 5 3 4 2 7 3 1 2 x y − + − = 7 6 15 14 ⇒ 2 10 3 4 7 6 14 2 6 1 2 15 14 x y − + = − Reprint 2025-26 15 14 or 2 3 10 4 14 1 2 6 2 x y + − + − + = 7 6 15 14 ⇒ 2 3 6 7 6 15 2 4 15 14 x y + = − or 2x + 3 = 7 and 2y – 4 = 14 (Why?) or 2x = 7 – 3 and 2y = 18 or x = 4 2 and y = 18 2 i.e. x = 2 and y = 9. Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. MATRICES 49 Solution (i) Combined sales in September and October for each farmer in each variety is given by (i) Find the combined sales in September and October for each farmer in each variety. (ii) Find the decrease in sales from September to October. (iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October. Reprint 2025-26 50 MATHEMATICS sale of each variety of rice, respectively, and Grucharan Singh receives profit of `400, ` 200 and ` 200 in the sale of each variety of rice, respectively. 3.4.5 Multiplication of matrices Suppose Meera and Nadeem are two friends. Meera wants to buy 2 pens and 5 story books, while Nadeem needs 8 pens and 10 story books. They both go to a shop to enquire about the rates which are quoted as follows: (iii) 2% of B = 2 B 100 × = 0.02 × B (ii) Change in sales from September to October is given by Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the = 0.02 = Pen – ` 5 each, story book – ` 50 each. How much money does each need to spend? Clearly, Meera needs ` (5 × 2 + 50 × 5) that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540. In terms of matrix representation, we can write the above information as follows: Suppose that they enquire about the rates from another shop, quoted as follows: pen – ` 4 each, story book – ` 40 each. Now, the money required by Meera and Nadeem to make purchases will be respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432 Requirements Prices per piece (in Rupees) Money needed (in Rupees) 8 10 2 5 50 Reprint 2025-26 5 8 5 10 50 540 × + × = × + × 5 2 5 50 260 matrices as follows: Requirements Prices per piece (in Rupees) Money needed (in Rupees) multiplication of two matrices A and B, the number of columns in A should be equal to the number of rows in B. Furthermore for getting the elements of the product matrix, we take rows of A and columns of B, multiply them element-wise and take the sum. Formally, we define multiplication of matrices as follows: The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Let A = [aij] be an m × n matrix and B = [bjk] be an n × p matrix. Then the product of the matrices A and B is the matrix C of order m × p. To get the (i, k) th element cik of the matrix C, we take the i th row of A and k th column of B, multiply them elementwise and take the sum of all these products. In other words, if A = [aij] m × n , B = [bjk] n × p , then the i th row of A is [ai1 ai2 ... ain] and the k th column of Requirements Prices per piece (in Rupees) Money needed (in Rupees) Again, the above information can be represented as follows: Now, the information in both the cases can be combined and expressed in terms of The above is an example of multiplication of matrices. We observe that, for 8 10 2 5 8 10 2 5 5 4 50 40 40 4 = 260 208 540 432 5 2 5 50 4 2 40 5 8 5 10 5 0 8 4 10 4 0 × + × × + × × + × × + × 8 4 10 4 0 432 × + × = × + × 4 2 40 5 208 MATRICES 51 B is The matrix C = [cik] m × p is the product of A and B. For example, if 1 1 2 C 0 3 4 − = and , then cik = ai1 b1k + ai2 b2k + ai3 b3k + ... + ain bnk = 1 n ij jk j a b = ∑ . b b b 2 . . . k 1 nk k Reprint 2025-26 2 7 D 1 1 = − − , then the product CD is defined 5 4 52 MATHEMATICS and is given by entry is the sum of the products across some row of C with the corresponding entries down some column of D. These four computations are Thus 13 2 CD 17 13 − = − 2 7 1 1 2 CD 1 1 0 3 4 5 4 − = − − . This is a 2 × 2 matrix in which each Example 12 Find AB, if 6 9 2 6 0 A and B 2 3 7 9 8 = = . Solution The matrix A has 2 columns which is equal to the number of rows of B. Hence AB is defined. Now 6(2) 9(7) 6(6) 9(9) 6(0) 9(8) AB 2(2) 3(7) 2(6) 3(9) 2(0) 3(8) + + + = + + + = 12 63 36 81 0 72 4 21 12 27 0 24 + + + + + + = 75 117 72 Reprint 2025-26 25 39 24 Remark If AB is defined, then BA need not be defined. In the above example, AB is defined but BA is not defined because B has 3 column while A has only 2 (and not 3) rows. If A, B are, respectively m × n, k × l matrices, then both AB and BA are defined if and only if n = k and l = m. In particular, if both A and B are square matrices of the same order, then both AB and BA are defined. Non-commutativity of multiplication of matrices Now, we shall see by an example that even if AB and BA are both defined, it is not necessary that AB = BA. Example 13 If AB ≠ BA. Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix. Hence AB and BA are both defined and are matrices of order 2 × 2 and 3 × 3, respectively. Note that and 2 3 1 2 3 AB 4 5 4 2 5 2 1 2 3 2 12 4 6 6 15 1 2 3 BA 4 5 4 20 8 10 12 25 4 2 5 2 1 2 4 4 2 6 5 − = − = 2 8 6 3 10 3 0 4 − − + + − = = − − + + − − − + + 2 3 1 2 3 A and B 4 5 4 2 5 2 1 − = = − , then find AB, BA. Show that 8 8 10 12 10 5 10 3 − + − + − = − + + − + + − = − − − MATRICES 53 10 2 21 16 2 37 2 2 11 Clearly AB ≠ BA In the above example both AB and BA are of different order and so AB ≠ BA. But one may think that perhaps AB and BA could be the same if they were of the same order. But it is not so, here we give an example to show that even if AB and BA are of same order they may not be same. Example 14 If 1 0 A 0 1 = − and 0 1 B 1 0 = , then 0 1 AB 1 0 = − . and 0 1 BA 1 0 − = . Clearly AB ≠ BA. Thus matrix multiplication is not commutative. Reprint 2025-26 54 MATHEMATICS Zero matrix as the product of two non zero matrices not be true for matrices, we will observe this through an example. Example 15 Find AB, if 0 1 A 0 2 − = and 3 5 B 0 0 = . Solution We have 0 1 3 5 0 0 AB 0 2 0 0 0 0 − = = . the matrices is a zero matrix. 3.4.6 Properties of multiplication of matrices The multiplication of matrices possesses the following properties, which we state without proof. ANote This does not mean that AB ≠ BA for every pair of matrices A, B for which AB and BA, are defined. For instance, If 1 0 3 0 A , B 0 2 0 4 = = , then AB = BA = 3 0 0 8 Observe that multiplication of diagonal matrices of same order will be commutative. 1. The associative law For any three matrices A, B and C. We have We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. This need Thus, if the product of two matrices is a zero matrix, it is not necessary that one of (AB) C = A (BC), whenever both sides of the equality are defined. Example 16 If A(BC), (AB)C and show that (AB)C = A(BC). 2. The distributive law For three matrices A, B and C. 3. The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that IA = AI = A. Now, we shall verify these properties by examples. (ii) (A+B) C = AC + BC, whenever both sides of equality are defined. (i) A (B+C) = AB + AC 1 1 1 1 3 1 2 3 4 A 2 0 3 , B 0 2 and C 2 0 2 1 3 1 2 1 4 − − = = = − − − Reprint 2025-26 , find Solution We have Now BC = (AB) (C) = = + + − − + − = − = − + − + − − + − + + − − + − − − − + + − − + − = + + − + − − − + − + − − + − − − − − 35 2 39 22 31 2 27 11 7 2 3 1 4 0 4 2 7 2 11 8 4 4 4 7 2 1 2 2 4 0 6 2 8 1 1 2 3 4 1 18 1 36 2 0 3 36 4 18 2 0 2 1 1 15 1 30 2 0 3 30 4 15 1 3 1 6 2 0 3 6 4 3 1 2 3 4 0 2 0 4 0 0 0 4 0 2 2 0 2 1 1 4 1 8 2 0 3 8 4 4 AB 2 0 3 0 2 2 0 3 6 0 12 1 18 − + + + − = = + − + + = − − − + − − + 1 1 1 1 3 1 0 1 3 2 4 2 1 3 1 2 1 4 3 0 2 9 2 8 1 15 MATRICES 55 Therefore A(BC) = = = − − − − − − − + − + + − − + − + − + + + − − + − − + + − + + − − + − − − + − − − − . Clearly, (AB) C = A (BC) 7 4 7 2 0 2 3 4 11 1 2 8 14 0 21 4 0 6 6 0 33 2 0 24 21 4 14 6 0 4 9 4 22 3 2 16 1 1 1 7 2 3 1 2 0 3 4 0 4 2 3 1 2 7 2 11 8 4 4 4 7 35 2 39 22 31 2 27 11 Reprint 2025-26 56 MATHEMATICS Example 17 If Solution Now, So (A + B) C = Further AC = and BC = Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC 0 7 8 A +B 5 0 10 A 6 0 8 , B 1 0 2 , C 2 7 8 0 1 2 0 3 = − = = − − = − − − + − − = − + + = − + + − + − − = − + + = − + + − + − = + + = − + − + = − 0 6 7 0 1 1 2 9 1 10 12 8 20 0 1 1 2 0 2 3 1 1 0 2 2 2 0 6 8 1 2 0 3 2 4 0 2 0 6 7 2 0 12 21 9 6 0 8 2 12 0 24 12 7 8 0 3 14 16 0 30 0 7 8 2 0 14 24 10 5 0 10 2 10 0 30 20 8 6 0 3 16 12 0 28 8 6 0 So AC + BC = Clearly, (A + B) C = AC + BC Example 18 If Solution We have 2 A 3 2 1 4 2 1 = − , then show that A3 – 23A – 40 I = O A A.A 3 2 1 3 2 1 1 12 8 1 2 3 = = − − = 30 2 28 1 2 3 1 2 3 19 4 8 4 2 1 4 2 1 14 6 15 Reprint 2025-26 So A3 = A A2 = Now Example 19 In a legislative assembly election, a political group hired a public relations firm to promote its candidate in three ways: telephone, house calls, and letters. The cost per contact (in paise) is given in matrix A as A3 – 23A – 40I = = = = − − − − − − − + − − + − − − − − − − − + − + − + − + − − + − + − + − − = 63 46 69 1 2 3 1 0 0 69 6 23 – 23 3 2 1 – 40 0 1 0 92 46 63 4 2 1 0 0 1 69 6 23 69 46 23 0 40 0 92 46 63 92 46 23 0 0 40 63 23 40 46 46 0 69 69 0 69 69 0 6 46 40 23 23 0 0 0 0 0 0 0 O 0 0 0 63 46 69 23 46 69 40 0 0 92 92 0 46 46 0 63 23 40 − = − 1 2 3 19 4 8 63 46 69 3 2 1 1 12 8 69 6 23 4 2 1 14 6 15 92 46 63 MATRICES 57 cities X and Y. Telephone Housecall Letter 1000 500 5000 X B 3000 1000 10,000 Y → = → . Find the total amount spent by the group in the two The number of contacts of each type made in two cities X and Y is given by A = Cost per contact Reprint 2025-26 40 Telephone 100 Housecall 50 Letter 58 MATHEMATICS Solution We have = 340,000 X 720,000 Y → → So the total amount spent by the group in the two cities is 340,000 paise and 720,000 paise, i.e., `3400 and ` 7200, respectively. 1. Let 2 4 1 3 2 5 A , B , C 3 2 2 5 3 4 − = = = − Find each of the following: (i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA 2. Compute the following: (iii) (i) a b a b b a b a + − (ii) 2 2 2 2 − − + (iv) 2 2 2 2 8 5 16 8 0 5 2 8 5 3 2 4 1 4 6 12 7 6 BA = 40,000 50,000 250,000 X 120,000 +100,000 +500,000 Y + + → → EXERCISE 3.2 a c a b ac ab + + + + + − − x x x x + 2 2 2 2 cos sin sin cos 2 2 2 2 2 2 2 2 a b b c ab bc sin cos cos sin x x x x 3. Compute the indicated products. (iv) (vi) (i) a b a b b a b a − − (ii) − − − − 2 3 4 1 3 5 3 4 5 0 2 4 4 5 6 3 0 5 2 3 3 1 3 1 0 1 0 2 3 1 Reprint 2025-26 [2 3 4] (iii) 1 2 1 2 3 1 2 3 (v) 2 3 2 3 1 − − − 2 1 1 0 1 3 2 1 2 1 1 1 4. If 5. If 6. Simplify cos sin sin cos cos + sin sin cos cos sin θ θ θ − θ θ θ − θ θ θ θ 7. Find X and Y, if 8. Find X, if Y = 3 2 (A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C. (ii) 2 3 2 2 2X + 3Y and 3X 2Y 4 0 1 5 − = + = − (i) 7 0 3 0 X + Y and X – Y 2 5 0 3 = = 2 5 2 3 1 1 3 3 5 5 1 2 4 1 2 4 A and B 3 3 3 5 5 5 7 2 7 6 2 2 3 3 5 5 5 A 5 0 2 , B 4 2 5 and C 0 3 2 1 1 1 2 0 3 1 2 3 = = − − = = = , then compute 3A – 5B. − − , then compute 1 2 3 3 1 2 4 1 2 1 4 and 2X + Y = 1 0 3 2 − MATRICES 59 10. Solve the equation for x, y, z and t, if 1 1 3 5 2 3 3 0 2 4 6 x z y t − + = 12. Given 6 4 3 1 2 3 x y x x y z w w z w + = + − + , find the values of x, y, z and w. 11. If 2 1 10 9. Find x and y, if 1 3 0 5 6 2 0 1 2 1 8 y 3 1 5 x y − + = , find the values of x and y. x + = Reprint 2025-26 60 MATHEMATICS 13. If 14. Show that 15. Find A2 – 5A + 6I, if 16. If 17. If 3 2 1 0 A and I= 4 2 0 1 − = − , find k so that A2 = kA – 2I (ii) (i) 5 1 2 1 2 1 5 1 6 7 3 4 3 4 6 7 − − ≠ F ( ) sin cos 0 0 0 1 A 0 2 1 2 0 3 = , prove that A3 – 6A2 + 7A + 2I = 0 x x x − = , show that F(x) F(y) = F(x + y). − − − ≠ − 1 2 3 1 1 0 1 1 0 1 2 3 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 2 3 4 2 3 4 1 1 0 1 0 2 cos sin 0 x x 2 0 1 A 2 1 3 = − 1 1 0 18. If 0 tan 2 A tan 0 2 I + A = (I – A) cos sin sin cos α − α α α 19. A trust fund has ` 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) `1800 (b) `2000 α − = α and I is the identity matrix of order 2, show that Reprint 2025-26 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are `80, `60 and ` 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22. 21. The restriction on n, k and p so that PY + WY will be defined are: 3.5. Transpose of a Matrix In this section, we shall learn about transpose of a matrix and special types of matrices such as symmetric and skew symmetric matrices. Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A′ or (AT ). In other words, if A = [aij] m × n , then A′ = [aji] n × m . For example, if (A) k = 3, p = n (B) k is arbitrary, p = 2 (C) p is arbitrary, k = 3 (D) k = 2, p = 3 22. If n = p, then the order of the matrix 7X – 5Z is: (A) p × 2 (B) 2 × n (C) n × 3 (D) p × n 3 5 3 3 0 A 3 1 , then A 1 5 1 0 1 5 5 × × = ′ = − − 3 2 2 3 MATRICES 61 3.5.1 Properties of transpose of the matrices We now state the following properties of transpose of matrices without proof. These may be verified by taking suitable examples. For any matrices A and B of suitable orders, we have (i) (A′)′ = A, (ii) (kA)′ = kA′ (where k is any constant) (iii) (A + B)′ = A′ + B′ (iv) (A B)′ = B′ A′ Example 20 If 3 3 2 2 1 2 A and B 4 2 0 1 2 4 − = = , verify that (i) (A′)′ = A, (ii) (A + B)′ = A′ + B′, (iii) (kB)′ = kB′, where k is any constant. Reprint 2025-26 62 MATHEMATICS Solution Thus (A′)′ = A (ii) We have (i) We have Therefore (A + B)′ = Now A′ = So A′ + B′ = A = 3 3 2 , 4 2 0 B = 2 1 2 5 3 1 4 A B 1 2 4 5 4 4 − − ⇒ + = A = ( ) 3 4 3 3 2 3 3 2 A 3 2 A A 4 2 0 4 2 0 2 0 ′ ⇒ = ⇒ = ′ ′ = − ′ = − − 3 4 2 1 3 2 , B 1 2 , 2 0 2 4 5 5 3 1 4 5 5 3 1 4 4 4 Thus (A + B)′ = A′ + B′ (iii) We have Then (kB)′ = Thus (kB)′ = kB′ kB = k 2 1 2 2 2 Reprint 2025-26 − = − = ′ 2 2 1 2 1 2 B 2 4 2 4 k k k − − = 1 2 4 2 4 k k k k k k k k k k k 4 4 Example 21 If [ ] 2 Solution We have then AB = [ ] 2 4 1 3 6 Now A′ = [–2 4 5] , Clearly (AB)′ = B′A′ B′A′ = [ ] 1 2 4 5 3 2 4 5 6 12 15 (AB) A 4 , B 1 3 6 5 A = [ ] 2 4 , B 1 3 6 − = = − − = − − − − − = − = ′ − − − 5 5 6 12 24 30 B 3 6 ′ = − = 1 , verify that (AB)′ = B′A′. − − − − 4 12 24 5 15 30 2 6 12 MATRICES 63 3.6 Symmetric and Skew Symmetric Matrices Definition 4 A square matrix A = [aij] is said to be symmetric if A′ = A, that is, [aij] = [aji] for all possible values of i and j. Definition 5 A square matrix A = [aij] is said to be skew symmetric matrix if A′ = – A, that is aji = – aij for all possible values of i and j. Now, if we put i = j, we have aii = – aii. Therefore 2aii = 0 or aii = 0 for all i’s. This means that all the diagonal elements of a skew symmetric matrix are zero. For example A 2 1.5 1 3 1 1 = − − − is a symmetric matrix as A′ = A 3 2 3 Reprint 2025-26 64 MATHEMATICS Now, we are going to prove some results of symmetric and skew-symmetric matrices. Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric matrix and A – A′ is a skew symmetric matrix. Proof Let B = A + A′, then Therefore B = A + A′ is a symmetric matrix Now let C = A – A′ Therefore C = A – A′ is a skew symmetric matrix. For example, the matrix 0 B 0 B′ = (A + A′)′ C′ = (A – A′)′ = A′ – (A′)′ (Why?) = − − − is a skew symmetric matrix as B′= –B = A′ + (A′)′ (as (A + B)′ = A′ + B′) = A′ + A (as (A′)′ = A) = A + A′ (as A + B = B + A) = B = A′ – A (Why?) = – (A – A′) = – C e f e g f g 0 Theorem 2 Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Proof Let A be a square matrix, then we can write a skew symmetric matrix. Since for any matrix A, (kA)′ = kA′, it follows that 1 (A A ) 2 + ′ From the Theorem 1, we know that (A + A′) is a symmetric matrix and (A – A′) is 1 1 A (A A ) (A A ) 2 2 = + + − ′ ′ Reprint 2025-26 is symmetric matrix and 1 (A A ) 2 − ′ is skew symmetric matrix. Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Example 22 Express the matrix skew symmetric matrix. Solution Here Let P = Now P′ = B′ = − − − − − − − − − − = P 4 3 3 1 1 (B + B ) 3 6 2 2 2 3 2 6 3 3 2 2 2 3 3 1 2 3 1 3 2 2 1 1 2 3 2 4 4 3 − − ′ = − − − = B 1 3 4 − − = − − − as the sum of a symmetric and a 1 2 3 2 2 4 − − − − − , 3 3 2 2 2 3 3 1 2 3 1 3 2 MATRICES 65 Thus P = 1 (B + B ) 2 ′ is a symmetric matrix. Also, let Q = 1 5 0 2 2 0 1 5 1 1 1 (B – B ) 1 0 6 0 3 2 2 2 5 6 0 5 3 0 2 − − − − ′ = = − − Reprint 2025-26 66 MATHEMATICS Then Q′ = Thus Q = 1 (B – B ) 2 ′ is a skew symmetric matrix. Now Thus, B is represented as the sum of a symmetric and a skew symmetric matrix. 1. Find the transpose of each of the following matrices: (i) − 3 3 1 5 2 0 2 2 2 2 2 2 4 3 1 P + Q 3 1 0 3 1 3 4 B 2 2 1 2 3 3 5 1 3 3 0 2 2 5 1 2 1 − − − − − − − = + = − = − − − − − − − = − − 1 5 0 2 3 1 0 3 Q 2 5 3 0 2 (ii) 1 1 EXERCISE 3.3 2 3 − (iii) − − 1 5 6 3 5 6 2. If 3. If (i) (A + B)′ = A′ + B′, (ii) (A – B)′ = A′ – B′ (i) (A + B)′ = A′ + B′ (ii) (A – B)′ = A′ – B′ 1 2 3 4 1 5 A 5 7 9 and B 1 2 0 3 4 1 2 1 A 1 2 and B 1 2 3 0 1 − ′ = − = , then verify that − − − = = − , then verify that 2 1 1 1 3 1 Reprint 2025-26 2 3 1 4. If 2 3 1 0 A and B 1 2 1 2 − − ′ = = , then find (A + 2B)′ 5. For the matrices A and B, verify that (AB)′ = B′A′, where 6. If (i) cos sin A sin cos α α = − α α , then verify that A′ A = I 7. (i) Show that the matrix 8. For the matrix 1 5 A 6 7 = , verify that (ii) If sin cos A cos sin α α = − α α , then verify that A′ A = I (ii) Show that the matrix (i) [ ] 1 A 4 , B 1 2 1 3 = − = − 1 1 5 A 1 2 1 0 1 1 A 1 0 1 − = − is a symmetric matrix. − = − − is a skew symmetric matrix. 5 1 3 1 1 0 (ii) [ ] 0 A 1 , B 1 5 7 2 = = MATRICES 67 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix: 9. Find ( ) 1 A A 2 + ′ and ( ) 1 A A 2 − ′ , when (ii) (A – A′) is a skew symmetric matrix (i) (A + A′) is a symmetric matrix Reprint 2025-26 A 0 = − − − 0 a c b c a b 0 68 MATHEMATICS Choose the correct answer in the Exercises 11 and 12. 11. If A, B are symmetric matrices of same order, then AB – BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix 3.7 Invertible Matrices Definition 6 If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by A– 1. In that case A is said to be invertible. 12. If cos sin A , sin cos α − α = α α and A + A′ = I, then the value of α is (A) 6 π (B) 3 π (C) π (D) 3 2 π (iii) (i) 3 5 1 1 − (ii) − − − − − 3 3 1 2 2 1 4 5 2 (iv) 1 5 1 2 − − − − − 6 2 2 2 3 1 2 1 3 For example, let A = 2 3 Now AB = 2 3 2 3 Also BA = 1 0 I 0 1 = . Thus B is the inverse of A, in other words B = A– 1 and A is inverse of B, i.e., A = B–1 Reprint 2025-26 = 4 3 6 6 1 0 I 2 2 3 4 0 1 − − + = = − − + 1 2 and B = 2 3 1 2 − − be two matrices. 1 2 1 2 − − Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique. Proof Let A = [aij] be a square matrix of order m. If possible, let B and C be two inverses of A. We shall show that B = C. Since B is the inverse of A AB = BA = I ... (1) Since C is also the inverse of A AC = CA = I ... (2) Thus B = BI = B (AC) = (BA) C = IC = C Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1 . Proof From the definition of inverse of a matrix, we have or A–1 (AB) (AB)–1 = A–1I (Pre multiplying both sides by A–1) or (A–1A) B (AB)–1 = A–1 (Since A–1 I = A–1) or IB (AB)–1 = A–1 or B (AB)–1 = A–1 ANote 1. A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order. 2. If B is the inverse of A, then A is also the inverse of B. (AB) (AB)–1 = 1 MATRICES 69 1. Matrices A and B will be inverse of each other only if (A) AB = BA (B) AB = BA = 0 (C) AB = 0, BA = I (D) AB = BA = I or B–1 B (AB)–1 = B–1 A–1 or I (AB)–1 = B–1 A–1 Hence (AB)–1 = B–1 A–1 EXERCISE 3.4 Reprint 2025-26 70 MATHEMATICS Example 23 If cos sin A sin cos θ θ = − θ θ , then prove that cos sin A sin cos n n n n n θ θ = − θ θ , n ∈ N. Solution We shall prove the result by using principle of mathematical induction. We have P(n) : If cos sin A sin cos θ θ = − θ θ , then cos sin A sin cos n n n n n θ θ = − θ θ , n ∈ N P(1) : cos sin A sin cos θ θ = − θ θ , so 1 cos sin A sin cos θ θ = − θ θ Therefore, the result is true for n = 1. Let the result be true for n = k. So Now, we prove that the result holds for n = k +1 Now Ak + 1 = cos sin cos sin A A sin cos sin cos k k k k k θ θ θ θ ⋅ = − θ θ − θ θ P(k) : cos sin A sin cos θ θ = − θ θ , then cos sin A sin cos k k k k k θ θ = − θ θ = cos cos – sin sin cos sin sin cos k k k k θ θ θ θ θ θ + θ θ − θ θ + θ θ − θ θ + θ θ sin cos cos sin sin sin cos cos k k k k Miscellaneous Examples we have cos sin A sin cos n n n n n θ θ = − θ θ , holds for all natural numbers. Example 24 If A and B are symmetric matrices of the same order, then show that AB is symmetric if and only if A and B commute, that is AB = BA. Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B. Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, = cos( ) sin ( ) cos( 1) sin ( 1) sin ( ) cos( ) sin ( 1) cos( 1) k k k k k k k k θ + θ θ + θ + θ + θ = − θ + θ θ + θ − + θ + θ Reprint 2025-26 Let AB be symmetric, then (AB)′ = AB But (AB)′ = B′A′= BA (Why?) Therefore BA = AB Conversely, if AB = BA, then we shall show that AB is symmetric. Now (AB)′ = B′A′ = B A (as A and B are symmetric) = AB Hence AB is symmetric. Example 25 Let 2 1 5 2 2 5 A , B , C 3 4 7 4 3 8 − = = = . Find a matrix D such that CD – AB = O. Solution Since A, B, C are all square matrices of order 2, and CD – AB is well defined, D must be a square matrix of order 2. Let D = a b c d . Then CD – AB = 0 gives or 2 5 2 5 3 0 2 5 2 1 5 2 3 8 3 4 7 4 a b c d − − = O a c b d + + − + + = 0 0 0 0 3 8 3 8 43 22 a c b d MATRICES 71 or 2 5 3 2 5 3 8 43 3 8 22 a c b d a c b d + − + + − + − = 0 0 0 0 By equality of matrices, we get 2a + 5c – 3 = 0 ... (1) 3a + 8c – 43 = 0 ... (2) 2b + 5d = 0 ... (3) and 3b + 8d – 22 = 0 ... (4) Solving (1) and (2), we get a = –191, c = 77. Solving (3) and (4), we get b = – 110, d = 44. Reprint 2025-26 72 MATHEMATICS Therefore D = 191 110 77 44 a b c d − − = 1. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix. 2. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric. 3. Find the values of x, y, z if the matrix 4. For what values of x : [ ] 1 2 0 0 1 2 1 2 0 1 2 1 0 2 x 5. If 3 1 A 1 2 = − , show that A2 – 5A + 7I = 0. A′A = I. Miscellaneous Exercise on Chapter 3 = O? A = − − satisfy the equation 0 2 x y z x y z y z 6. Find x, if [ ] 1 0 2 7. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below: Market Products x − − = 5 1 0 2 1 4 O 2 0 3 1 I 10,000 2,000 18,000 II 6,000 20,000 8,000 Reprint 2025-26 x Choose the correct answer in the following questions: 10. If the matrix A is both symmetric and skew symmetric, then (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these 11. If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to (A) A (B) I – A (C) I (D) 3A 8. Find the matrix X so that 1 2 3 7 8 9 X 4 5 6 2 4 6 − − − = 9. If A = is such that A² = I, then Summary Æ A matrix is an ordered rectangular array of numbers or functions. Æ A matrix having m rows and n columns is called a matrix of order m × n. Æ [aij] m × 1 is a column matrix. Æ [aij] 1 × n is a row matrix. Æ An m × n matrix is a square matrix if m = n. Æ A = [aij] m × m is a diagonal matrix if aij = 0, when i ≠ j. Æ A = [aij] n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some constant), when i = j. Æ A = [aij] n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j. Æ A zero matrix has all its elements as zero. Æ A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all possible values of i and j. (a) If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit. (A) 1 + α² + βγ = 0 (B) 1 – α² + βγ = 0 (C) 1 – α² – βγ = 0 (D) 1 + α² – βγ = 0 γ α− α β MATRICES 73 Reprint 2025-26 74 MATHEMATICS Æ kA = k[aij] m × n = [k(aij)] m × n Æ – A = (–1)A Æ A – B = A + (–1) B Æ A + B = B + A Æ (A + B) + C = A + (B + C), where A, B and C are of same order. Æ k(A + B) = kA + kB, where A and B are of same order, k is constant. Æ (k + l) A = kA + lA, where k and l are constant. Æ If A = [aij] m × n and B = [bjk] n × p , then AB = C = [cik] m × p , where = Æ (i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC Æ If A = [aij] m × n , then A′ or AT = [aji] n × m Æ (i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′ Æ A is a symmetric matrix if A′ = A. Æ A is a skew symmetric matrix if A′ = –A. Æ Any square matrix can be represented as the sum of a symmetric and a skew symmetric matrix. Æ If A and B are two square matrices such that AB = BA = I, then B is the inverse matrix of A and is denoted by A–1 and A is the inverse of B. Æ Inverse of a square matrix, if it exists, is unique. —v— ik ij jk j c a b ∑ =1 n Reprint 2025-26 NOTES MATRICES 75 Reprint 2025-26" class_12,4,Determinants,ncert_books/class_12/lemh1dd/lemh104.pdf,"76 MATHEMATICS v All Mathematical truths are relative and conditional. — C.P. STEINMETZ v 4.1 Introduction In the previous chapter, we have studied about matrices and algebra of matrices. We have also learnt that a system of algebraic equations can be expressed in the form of matrices. This means, a system of linear equations like can be represented as 1 1 1 2 2 2 a b c x a b c y = . Now, this system of equations has a unique solution or not, is determined by the number a1 b2 – a2 b1 . (Recall that if equations has a unique solution). The number a1 b2 – a2 b1 2 2 a b a b ≠ or, a1 b2 – a2 b1 ≠ 0, then the system of linear 1 1 DETERMINANTS a1 x + b1 y = c 1 a2 x + b2 y = c 2 Chapter 4 P.S. Laplace (1749-1827) which determines uniqueness of solution is associated with the matrix 1 1 2 2 A a b a b = and is called the determinant of A or det A. Determinants have wide applications in Engineering, Science, Economics, Social Science, etc. In this chapter, we shall study determinants up to order three only with real entries. Also, we will study various properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle, adjoint and inverse of a square matrix, consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using inverse of a matrix. 4.2 Determinant To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where aij = (i, j) th element of A. Reprint 2025-26 This may be thought of as a function which associates each square matrix with a unique number (real or complex). If M is the set of square matrices, K is the set of numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and k ∈ K, then f(A) is called the determinant of A. It is also denoted by |A| or det A or ∆. Remarks (i) For matrix A, |A| is read as determinant of A and not modulus of A. (ii) Only square matrices have determinants. 4.2.1 Determinant of a matrix of order one Let A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a 4.2.2 Determinant of a matrix of order two Let A = 11 12 21 22 a a a a be a matrix of order 2 × 2, then the determinant of A is defined as: Example 1 Evaluate 2 4 If A = a b c d , then determinant of A is written as |A| = a b c d = det (A) det (A) = |A| = ∆ = = a11a22 – a21a12 –1 2 . DETERMINANTS 77 Solution We have 2 4 Example 2 Evaluate 1 Solution We have 4.2.3 Determinant of a matrix of order 3 × 3 Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order x x + = x (x) – (x + 1) (x – 1) = x 2 – (x 2 – 1) = x 2 – x 2 + 1 = 1 – 1 x x 1 –1 2 = 2(2) – 4(–1) = 4 + 4 = 8. x x + – 1 x x Reprint 2025-26 78 MATHEMATICS 3 corresponding to each of three rows (R1 , R2 and R3 ) and three columns (C1 , C2 and C3 ) giving the same value as shown below. Consider the determinant of square matrix A = [aij] 3 × 3 i.e., | A | = 21 22 23 Expansion along first Row (R1 ) Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a 11] and with the second order determinant obtained by deleting the elements of first row (R1 ) and first column (C1 ) of | A | as a11 lies in R1 and C1 , i.e., (–1)1 + 1 a11 22 23 Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a 12] and the second order determinant obtained by deleting elements of first row (R1 ) and 2nd column (C2 ) of | A | as a12 lies in R1 and C2 , i.e., (–1)1 + 2 a12 21 23 Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a 13] and the second order determinant obtained by deleting elements of first row (R1 ) and third column (C3 ) of | A | as a13 lies in R1 and C3 , 32 33 a a a a 31 33 a a a a 11 12 13 a a a 31 32 33 a a a a a a i.e., (–1)1 + 3 a13 21 22 Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by or |A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23) + a13 (a21 a32 – a31 a22) det A = |A| = (–1)1 + 1 a11 22 23 1 2 21 23 12 32 33 31 33 (–1) a a a a a a a a a + + + 1 3 21 22 13 31 32 (–1) a a a a a + 31 32 a a a a Reprint 2025-26 = a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32 – a13 a31 a22 ... (1) ANote We shall apply all four steps together. Expansion along second row (R2 ) Expanding along R2 ,we get = – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13) – a23 (a11 a32 – a31 a12) | A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32 + a23 a31 a12 = a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 – a13 a31 a22 ... (2) Expansion along first Column (C1 ) | A | = 2 1 12 13 2 2 11 13 21 22 32 33 31 33 (–1) (–1) a a a a a a a a a a + + + 2 3 11 12 23 31 32 (–1) a a a a a + + | A | = a a a a a a 21 22 23 a a a 11 12 13 31 32 33 DETERMINANTS 79 By expanding along C1 , we get | A | = 1 1 22 23 2 1 12 13 11 21 32 33 32 33 (–1) ( 1) a a a a a a a a a a + + + − = a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22) + 3 1 12 13 31 22 23 (–1) a a a a a + | A | = 12 13 Reprint 2025-26 a a a 11 21 31 a a a a a a 22 23 32 33 80 MATHEMATICS | A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23 – a31 a13 a22 = a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 – a13 a31 a22 ... (3) Clearly, values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along R3 , C2 and C3 are equal to the value of |A| obtained in (1), (2) or (3). Hence, expanding a determinant along any row or column gives same value. Remarks (i) For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros. (ii) While expanding, instead of multiplying by (–1)i + j , we can multiply by +1 or –1 according as (i + j) is even or odd. Observe that, |A| = 4(– 2) = 22 |B| or |A| = 2n |B|, where n = 2 is the order of square matrices A and B. | B |, where n = 1, 2, 3 (iii) Let A = 2 2 In general, if A = kB where A and B are square matrices of order n, then | A| = k n 4 0 and B = 1 1 2 0 . Then, it is easy to verify that A = 2B. Also |A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2. 1 2 4 –1 3 0 Example 3 Evaluate the determinant ∆ = Solution Note that in the third column, two entries are zero. So expanding along third column (C3 ), we get Example 4 Evaluate ∆ = –sin 0 sin cos –sin 0 0 sin – cos α β α α α β ∆ = –1 3 1 2 1 2 4 – 0 0 4 1 4 1 –1 3 + Reprint 2025-26 = 4 (–1 – 12) – 0 + 0 = – 52 4 1 0 . . Solution Expanding along R1 , we get Example 5 Find values of x for which 3 3 2 Solution We have 3 3 2 1 4 1 x x = i.e. 3 – x 2 = 3 – 8 i.e. x 2 = 8 Hence x = ± 2 2 Evaluate the determinants in Exercises 1 and 2. 1. 2 4 –5 –1 2. (i) cos – sin ∆ = 0 sin – sin sin – sin 0 0 – sin – cos – sin 0 cos 0 cos – sin β α β α α α β α α β sin cos θ θ θ θ (ii) = 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0) = sin α sin β cos α – cos α sin α sin β = 0 EXERCISE 4.1 x = . 1 4 1 x 2 – 1 – 1 1 1 x x x x x + + + DETERMINANTS 81 3. If A = 1 2 4 2 , then show that | 2A | = 4 | A | 4. If A = 5. Evaluate the determinants (i) 3 –1 –2 0 0 –1 3 –5 0 , then show that | 3 A | = 27 | A | 1 0 1 0 1 2 0 0 4 Reprint 2025-26 (ii) 3 – 4 5 1 1 –2 2 3 1 82 MATHEMATICS 4.3 Area of a Triangle In earlier classes, we have studied that the area of a triangle whose vertices are (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ), is given by the expression 1 2 [x1 (y2 –y3 ) + x2 (y3 –y1 ) + x3 (y1 –y2 )]. Now this expression can be written in the form of a determinant as 6. If A = 7. Find values of x, if 8. If 2 6 2 18 18 6 x x = , then x is equal to (A) 6 (B) ± 6 (C) – 6 (D) 0 (iii) (i) 2 4 2 4 , find | A | –1 0 –3 –2 3 0 (iv) 5 1 6 x 0 1 2 1 1 –2 2 1 –3 5 4 –9 x = (ii) 2 3 3 ∆ = 1 1 1 1 1 2 1 x y x y 2 2 2 –1 –2 0 2 –1 3 –5 0 4 5 2 5 x x = ... (1) Remarks (i) Since area is a positive quantity, we always take the absolute value of the determinant in (1). (ii) If area is given, use both positive and negative values of the determinant for calculation. (iii) The area of the triangle formed by three collinear points is zero. Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1). Solution The area of triangle is given by ∆ = 3 8 1 1 4 2 1 2 5 1 1 – Reprint 2025-26 x y 3 3 Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units. Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So This gives ( ) 1 3 2 y – x = 0 or y = 3x, which is the equation of required line AB. Also, since the area of the triangle ABD is 3 sq. units, we have This gives, 3 3 2 − k = ± , i.e., k = ∓ 2. = ( ) ( ) ( ) 1 3 2 –1 – 8 – 4 – 5 1 – 4 –10 2 + = ( ) 1 61 3 72 14 2 2 + = – 0 0 1 1 1 3 1 2 x y 1 = 0 1 3 1 1 0 0 1 2 k 0 1 = ± 3 EXERCISE 4.2 DETERMINANTS 83 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8) 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear. 3. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k) 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants. 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is (A) 12 (B) –2 (C) –12, –2 (D) 12, –2 Reprint 2025-26 84 MATHEMATICS 4.4 Minors and Cofactors In this section, we will learn to write the expansion of a determinant in compact form using minors and cofactors. Definition 1 Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of an element aij is denoted by Mij. Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of order n – 1. Example 8 Find the minor of element 6 in the determinant Solution Since 6 lies in the second row and third column, its minor M23 is given by Definition 2 Cofactor of an element aij , denoted by Aij is defined by Aij = (–1)i + j Mij , where Mij is minor of aij . Example 9 Find minors and cofactors of all the elements of the determinant 1 –2 Solution Minor of the element aij is Mij Here a11 = 1. So M11 = Minor of a11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2 M22 = Minor of the element a22 = 1 Now, cofactor of aij is Aij. So A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3 A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4 A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2 A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 M23 = 1 2 7 8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆). 7 8 9 ∆ = 1 2 3 4 5 6 4 3 Reprint 2025-26 Example 10 Find minors and cofactors of the elements a11, a21 in the determinant Solution By definition of minors and cofactors, we have Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 Remark Expanding the determinant ∆, in Example 21, along R1 , we have = a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij = sum of product of elements of R1 with their corresponding cofactors Similarly, ∆ can be calculated by other five ways of expansion that is along R2 , R3 , C1 , C2 and C3 . Hence ∆ = sum of the product of elements of any row (or column) with their corresponding cofactors. Minor of a11 = M11 = 22 23 Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32 Minor of a21 = M21 = 12 13 ∆ = (–1)1+1 a11 22 23 32 33 a a a a + (–1)1+2 a12 21 23 32 33 a a a a = a22 a33– a23 a32 32 33 a a a a = a12 a33 – a13 a32 ∆ = a a a a a a a a a 11 12 13 21 22 23 31 32 33 31 33 a a a a + (–1)1+3 a13 21 22 DETERMINANTS 85 31 32 a a a a ANote If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. For example, Similarly, we can try for other rows and columns. ∆ = a11 A21 + a12 A22 + a13 A23 = a11 (–1)1+1 12 13 32 33 a a a a + a12 (–1)1+2 11 13 31 33 a a a a + a13 (–1)1+3 11 12 31 32 a a = 11 12 13 a a a a a a a a a 11 12 13 31 32 33 = 0 (since R1 and R2 are identical) Reprint 2025-26 a a 86 MATHEMATICS Example 11 Find minors and cofactors of the elements of the determinant Solution We have M11 = 0 4 6 0 4 1 5 7 2 3 5 – M12 = 6 4 1 7– = – 42 – 4 = – 46; A12 = (–1)1+2 (– 46) = 46 M13 = 6 0 1 5 = 30 – 0 = 30; A13 = (–1)1+3 (30) = 30 M21 = 3 5 M22 = 2 5 M23 = 2 3 1 5 – = 10 + 3 = 13; A23 = (–1)2+3 (13) = –13 – and verify that a11 A31 + a12 A32 + a13 A33= 0 5 7 – 1 7– = –14 – 5 = –19; A22 = (–1)2+2 (–19) = –19 – = 21 – 25 = – 4; A21 = (–1)2+1 (– 4) = 4 5 7– = 0 –20 = –20; A11 = (–1)1+1 (–20) = –20 and M33 = 2 3 6 0 – = 0 + 18 = 18; A33 = (–1)3+3 (18) = 18 Now a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18 So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 M31 = 3 5 M32 = 2 5 6 4 = 8 – 30 = –22; A32 = (–1)3+2 (–22) = 22 0 4 – = –12 – 0 = –12; A31 = (–1)3+1 (–12) = –12 Reprint 2025-26 Write Minors and Cofactors of the elements of following determinants: (A) a11 A31+ a12 A32 + a13 A33 (B) a11 A11+ a12 A21 + a13 A31 (C) a21 A11 + a22 A12 + a23 A13 (D) a11 A11 + a21 A21 + a31 A31 4.5 Adjoint and Inverse of a Matrix In the previous chapter, we have studied inverse of a matrix. In this section, we shall discuss the condition for existence of inverse of a matrix. To find inverse of a matrix A, i.e., A–1 we shall first define adjoint of a matrix. 1. (i) 2 4 0 3 – (ii) a c b d 2. (i) 3. Using Cofactors of elements of second row, evaluate ∆ = 4. Using Cofactors of elements of third column, evaluate ∆ = 1 5. If ∆ = 11 12 13 1 0 0 0 1 0 0 0 1 a a a a a a a a a 21 22 23 31 32 33 (ii) and Aij is Cofactors of aij , then value of ∆ is given by 1 0 4 3 5 1 0 1 2 – EXERCISE 4.3 DETERMINANTS 87 5 3 8 2 0 1 1 2 3 . 1 1 x yz y zx z xy . 4.5.1 Adjoint of a matrix Definition 3 The adjoint of a square matrix A = [aij] n × n is defined as the transpose of the matrix [Aij] n × n , where Aij is the cofactor of the element aij . Adjoint of the matrix A is denoted by adj A. Let 11 12 13 31 32 33 A = a a a a a a a a a 21 22 23 Reprint 2025-26 88 MATHEMATICS Then 11 12 13 Example 12 2 3 Find A for A = 1 4 adj Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2 Hence adj A = 11 21 12 22 A A 4 –3 = A A –1 2 Remark For a square matrix of order 2, given by a a The adj A can also be obtained by interchanging a11 and a22 and by changing signs of a12 and a21, i.e., We state the following theorem without proof. A =Transposeof A A A A A A adj A A A 31 32 33 21 22 23 A = 11 12 21 22 a a = A A A A A A A A A 11 21 31 12 22 32 13 23 33 Theorem 1 If A be any given square matrix of order n, then where I is the identity matrix of order n Verification Let A = Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have , then adj A = 11 21 31 a a a a a a a a a 11 12 13 21 22 23 31 32 33 A(adj A) = (adj A) A = A I , Reprint 2025-26 A A A A A A A A A 12 22 32 13 23 33 Similarly, we can show (adj A) A = A I Hence A (adj A) = (adj A) A = A I Definition 4 A square matrix A is said to be singular if A = 0. For example, the determinant of matrix A = 1 2 4 8 Hence A is a singular matrix. Definition 5 A square matrix A is said to be non-singular if A ≠ 0 Let A = 1 2 3 4 . Then A = 1 2 3 4 = 4 – 6 = – 2 ≠ 0. Hence A is a nonsingular matrix We state the following theorems without proof. Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants, that is, AB = A B , where A and B are square matrices of the same order A (adj A) = = A A 0 0 0 A 0 0 0 A is zero = A I 1 0 0 0 1 0 0 0 1 DETERMINANTS 89 Remark We know that (adj A) A = A I = Writing determinants of matrices on both sides, we have ( A) A adj = Reprint 2025-26 A 0 0 0 A 0 0 0 A 0 A 0 0 0 0 A A A 0 0 , ≠ 90 MATHEMATICS i.e. |(adj A)| |A| = 3 1 0 0 A 0 1 0 0 0 1 i.e. |(adj A)| |A| = |A| 3 (1) i.e. |(adj A)| = | A | 2 In general, if A is a square matrix of order n, then |adj(A)| = |A| n – 1 . Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. Proof Let A be invertible matrix of order n and I be the identity matrix of order n. Then, there exists a square matrix B of order n such that AB = BA = I Now AB = I. So AB = I or A B = 1 (since I 1, AB A B ) = = This gives A ≠ 0. Hence A is nonsingular. Conversely, let A be nonsingular. Then A ≠ 0 Now A (adj A) = (adj A) A = A I (Theorem 1) or A 1 1 A A A I | A | | A | adj adj = = or AB = BA = I, where B = 1 A | A | adj Thus A is invertible and A–1 = 1 A | A | adj (Why?) Example 13 If A = Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0 Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0, A33 = 1 Therefore adj A = 1 3 3 1 4 3 1 3 4 , then verify that A adj A = |A| I. Also find A–1 . Reprint 2025-26 − − − − 7 3 3 1 1 0 1 0 1 Now A (adj A) = Also A–1 1 A A = a d j = Example 14 If A = 2 3 1 2 and B 1 4 1 3 − = − − , then verify that (AB)–1 = B–1A–1 . Solution We have AB = 2 3 1 2 1 5 1 4 1 3 5 14 − − = − − − Since, AB = –11 ≠ 0, (AB)–1 exists and is given by = = − − − − − − − + + − + + − − − + + − + + − − − + + − + + = (1) 7 3 3 1 1 1 0 1 1 0 1 1 3 3 7 3 3 1 4 3 1 1 0 1 3 4 1 0 1 1 0 0 7 3 3 3 3 0 3 0 3 7 4 3 3 4 0 3 0 3 7 3 4 3 3 0 3 0 4 0 1 0 0 0 1 − − − − = = A . I − − − − 1 0 0 0 1 0 0 0 1 7 3 3 1 1 0 1 0 1 DETERMINANTS 91 Further, A = –11 ≠ 0 and B = 1 ≠ 0. Therefore, A–1 and B–1 both exist and are given by Therefore B A− − = − Hence (AB)–1 = B–1 A–1 (AB)–1 = 1 1 14 5 (AB) AB 11 5 1 adj − − =− − − 1 1 1 11 A–1 = − − − 3 2 1 1 1 − 11 − − − Reprint 2025-26 − 4 3 1 2 4 3 1 2 = − − − − − = 3 2 1 1 1 ,B 1 14 5 11 5 1 = 1 11 14 5 5 1 1 14 5 11 5 1 = 92 MATHEMATICS Example 15 Show that the matrix A = 2 3 1 2 where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A–1 . Solution We have 2 2 3 2 3 7 12 A A.A 1 2 1 2 4 7 = = = Hence 2 7 12 8 12 1 0 A 4A I 4 7 4 8 0 1 − + = − + Now A2 – 4A + I = O Therefore A A – 4A = – I or A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0) or A (A A–1) – 4I = – A–1 or AI – 4I = – A–1 or A–1 = 4I – A = 4 0 2 3 2 3 Hence 1 2 3 A 1 2 − − = − Find adjoint of each of the matrices in Exercises 1 and 2. EXERCISE 4.4 0 4 1 2 1 2 − − = − satisfies the equation A2 – 4A + I = O, 0 0 O 0 0 = = Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4 Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 1. 1 2 3 4 3. 2 3 5. 2 2 − − 4 6 4 3 − 2. 6. − − 4. − − 1 1 2 3 0 2 1 0 3 − 1 1 2 2 3 5 2 0 1 1 5 3 2 7. Reprint 2025-26 − 1 2 3 0 2 4 0 0 5 12. Let A = 3 7 13. If A = 3 1 14. For the matrix A = 3 2 15. For the matrix A = 16. If A = 11. 8. Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1 . α α α − α 1 0 0 3 3 0 5 2 1− 1 0 0 0 cos sin 0 sin cos −1 2 − − − 2 1 1 1 2 1 2 5 9. − , show that A2 – 5A + 7I = O. Hence find A–1 . and B = 6 8 1 1 1 1 1 1 2 3 2 1 3 − − , find the numbers a and b such that A2 + aA + bI = O. − 7 9 2 1 3 4 1 0 7 2 1 − . Verify that (AB)–1 = B–1 A–1 . 10. 1 1 2 0 2 3 3 2 4 − − − DETERMINANTS 93 4.6 Applications of Determinants and Matrices In this section, we shall discuss application of determinants and matrices for solving the system of linear equations in two or three variables and for checking the consistency of the system of linear equations. Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to (A) | A | (B) | A | 2 (C) | A | 3 (D) 3|A| 18. If A is an invertible matrix of order 2, then det (A–1) is equal to (A) det (A) (B) 1 det (A) (C) 1 (D) 0 1 1 2 Reprint 2025-26 94 MATHEMATICS Consistent system A system of equations is said to be consistent if its solution (one or more) exists. Inconsistent system A system of equations is said to be inconsistent if its solution does not exist. ANote In this chapter, we restrict ourselves to the system of linear equations having unique solutions only. 4.6.1 Solution of system of linear equations using inverse of a matrix Let us express the system of linear equations as matrix equations and solve them using inverse of the coefficient matrix. Consider the system of equations Let A = 1 1 1 1 = = Then, the system of equations can be written as, AX = B, i.e., = 1 a b c x a b c y a b c z 3 3 3 3 , X and B a b c x d a b c y d a b c z d a1 x + b1 y + c 1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c 3 z = d3 1 1 1 2 2 2 3 3 3 2 2 2 2 d d d 2 3 Case I If A is a nonsingular matrix, then its inverse exists. Now AX = B or A–1 (AX) = A–1 B (premultiplying by A–1) or (A–1A) X = A–1 B (by associative property) or I X = A–1 B or X = A–1 B This matrix equation provides unique solution for the given system of equations as inverse of a matrix is unique. This method of solving system of equations is known as Matrix Method. Case II If A is a singular matrix, then |A| = 0. In this case, we calculate (adj A) B. If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the system of equations is called inconsistent. Reprint 2025-26 If (adj A) B = O, then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. Example 16 Solve the system of equations Solution The system of equations can be written in the form AX = B, where Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution. Note that A–1 = − − − Therefore X = A–1B = – 1 11 i.e. x y = − − Hence x = 3, y = – 1 Example 17 Solve the following system of equations by matrix method. 3x – 2y + 3z = 8 2x + y – z = 1 4x – 3y + 2z = 4 2x + 5y = 1 3x + 2y = 7 A = 2 5 1 ,X and B 3 2 7 x y = = 1 11 1 11 11 2 5 33 3 2 = − 7 − − 3 2 5 1 3 2 1 DETERMINANTS 95 Solution The system of equations can be written in the form AX = B, where We see that Hence, A is nonsingular and so its inverse exists. Now A11 = –1, A12 = – 8, A13 = –10 A21 = –5, A22 = – 6, A23 = 1 A31 = –1, A32 = 9, A33 = 7 3 2 3 8 A 2 1 1 , X and B 1 A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0 − = − = = − 4 3 2 4 x y z Reprint 2025-26 96 MATHEMATICS Therefore A–1 = So X = –1 i.e. Hence x = 1, y = 2 and z = 3. Example 18 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method. Solution Let first, second and third numbers be denoted by x, y and z, respectively. Then, according to given conditions, we have x + y + z = 6 y + 3z = 11 x + z = 2y or x – 2y + z = 0 This system can be written as A X = B, where = x y z − − − − − − − 1 5 1 8 1 A B = 8 6 9 1 17 10 1 7 4 − − − = − 1 5 1 1 8 6 9 17 10 1 7 17 1 1 34 2 17 51 3 − − − − − − − Here A 1 1 6 – (0 – 3) 0 –1 9 0 = + + = ≠ ( ) ( ) . Now we find adj A Hence adj A = A11 = 1 (1 + 6) = 7, A12 = – (0 – 3) = 3, A13 = – 1 A21 = – (1 + 2) = – 3, A22 = 0, A23 = – (– 2 – 1) = 3 A31 = (3 – 1) = 2, A32 = – (3 – 0) = – 3, A33 = (1 – 0) = 1 A = Reprint 2025-26 –1 3 1 1 1 1 0 1 3 1 2 1 7 –3 2 3 0 –3 , X = x y z and B = 11 0 6 Thus A –1 = 1 A adj (A) = Since X = A–1 B or Thus x = 1, y = 2, z = 3 Examine the consistency of the system of equations in Exercises 1 to 6. 1. x + 2y = 2 2. 2x – y = 5 3. x + 3y = 5 2x + 3y = 3 x + y = 4 2x + 6y = 8 4. x + y + z = 1 5. 3x–y – 2z = 2 6. 5x – y + 4z = 5 2x + 3y + 2z = 2 2y – z = –1 2x + 3y + 5z = 2 ax + ay + 2az = 4 3x – 5y = 3 5x – 2y + 6z = –1 Solve system of linear equations, using matrix method, in Exercises 7 to 14. 7. 5x + 2y = 4 8. 2x – y = –2 9. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10. 5x + 2y = 3 11. 2x + y + z = 1 12. x – y + z = 4 EXERCISE 4.5 x y z X = = 1 9 7 3 2 6 1 3 0 3 11 9 1 3 1 0 – – − + + + − + + = 1 9 42 33 0 18 0 0 6 33 0 – 7 3 2 1 3 0 3 9 1 3 1 – – – DETERMINANTS 97 18 27 9 = 1 2 3 3y – 5z = 9 x + y + z = 2 13. 2x + 3y +3 z = 5 14. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12 3x + 2y = 5 x – 2y – z = 3 2 2x + y – 3z = 0 Reprint 2025-26 98 MATHEMATICS Example 19 Use product Solution Consider the product 15. If A = 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ` 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ` 70. Find cost of each item per kg by matrix method. , find A–1. Using A–1 solve the system of equations 3 2 – 4 1 1 –2 2 –3 5 x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 1 1 2 0 2 3 3 2 4 Miscellaneous Examples 1 1 2 2 0 1 0 2 3 9 2 3 3 2 4 6 1 2 – – – – – – 9 2 3 6 1 2 2 0 1 to solve the system of equations Hence Now, given system of equations can be written, in matrix form, as follows = − − + − + + − + − + − − + − − + − + + − = – = 1 1 2 0 2 3 3 2 4 1 –1 2 0 2 –3 3 –2 4 0 18 18 0 4 3 0 6 6 6 18 24 0 4 4 3 6 8 2 9 12 0 2 2 1 3 4 – 1 – – = x y z Reprint 2025-26 – – 9 2 3 6 1 2 2 0 1 1 1 2 – 1 0 0 0 1 0 0 0 1 or Hence x = 0, y = 5 and z = 3 1. Prove that the determinant 2. Evaluate cos cos cos sin – sin –sin cos 0 3. If A–1 = ( ) 1 3 1 1 1 2 2 = – – – – – sin cos sin sin cos 15 6 5 and B 1 3 0 , find AB 5 2 2 0 2 1 β β α β α β α α β α β α Miscellaneous Exercises on Chapter 4 – – – x sin cos – sin – 1 y z cos 1 x x = = θ θ − − − − = − + + + − = + − 1 1 1 2 1 0 2 3 1 3 2 4 2 2 0 2 0 9 2 6 5 6 1 4 3 θ θ . x is independent of θ. 9 2 3 6 1 2 DETERMINANTS 99 2 0 1 1 1 2 4. Let A = 5. Evaluate 6. Evaluate 1 1 (i) [adj A]–1 = adj (A–1) (ii) (A–1) –1 = A 1 x y x y 1 2 1 2 3 1 1 1 5 + + x y x y y x y x x y x y y x x+ y + . Verify that + Reprint 2025-26 100 MATHEMATICS Using properties of determinants in Exercises 11 to 15, prove that: 7. Solve the system of equations Choose the correct answer in Exercise 17 to 19. 8. If x, y, z are nonzero real numbers, then the inverse of matrix (A) 2 3 10 + + = 4 x y z 4 6 5 – + =1 x y z 6 9 20 + = – 2 x y z x 0 0 1 0 0 0 0 0 0 − 1 x y 0 0 − 1 z − 1 (B) xyz y 1 0 0 1 0 1 0 0 0 1 xyz x 0 0 0 0 − 1 0 0 − 1 z 0 0 A 0 0 − 1 = is 0 0 x y z 9. Let A = (A) Det(A) = 0 (B) Det(A) ∈ (2, ∞) (C) (C) Det(A) ∈ (2, 4) (D) Det(A) ∈ [2, 4] y xyz z θ − θ θ − − θ , where 0 ≤ θ ≤ 2π. Then 0 0 sin 1 sin 1 sin 1 1 sin 1 Reprint 2025-26 (D) Summary Æ Determinant of a matrix A = [a11] 1×1 is given by | a11 | = a11 Æ Determinant of a matrix A = Æ Determinant of a matrix A = For any square matrix A, the |A| satisfy following properties. Æ Area of a triangle with vertices (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by 1 1 1 2 2 2 2 2 2 2 2 2 1 1 1 3 3 3 3 3 3 3 3 3 A a b c b c a c a b a b c a b c b c a c a b a b c = = − + 21 22 A a a a a = = a11 a22 – a12 a21 11 12 x y ∆ = a b c a b c a b c a a a a 11 12 1 1 1 2 2 2 3 3 3 21 22 is given by 1 1 1 2 1 x y x y 1 1 2 2 3 3 is given by (expanding along R1 ) DETERMINANTS 101 Æ Minor of an element aij of the determinant of matrix A is the determinant obtained by deleting i th row and j th column and denoted by Mij. Æ Cofactor of aij of given by Aij = (– 1)i+ j Mij Æ Value of determinant of a matrix A is obtained by sum of product of elements of a row (or a column) with corresponding cofactors. For example, A = a11 A11 + a12 A12 + a13 A13. Æ If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column), then their sum is zero. For example, a11 A21 + a12 A22 + a13 A23 = 0 Reprint 2025-26 102 MATHEMATICS Æ If 11 12 13 cofactor of aij Æ A (adj A) = (adj A) A = |A| I, where A is square matrix of order n. Æ A square matrix A is said to be singular or non-singular according as |A| = 0 or |A| ≠ 0. Æ If AB = BA = I, where B is square matrix, then B is called inverse of A. Also A–1 = B or B–1 = A and hence (A–1) –1 = A. Æ A square matrix A has inverse if and only if A is non-singular. Æ –1 1 A ( A) A = adj Æ If a1 x + b1 y + c1 z = d1 a2 x + b2 y + c 2 z = d2 a3 x + b3 y + c 3 z = d3 , then these equations can be written as A X = B, where Æ Unique solution of equation AX = B is given by X = A–1 B, where A 0 ≠ . Æ A system of equation is consistent or inconsistent according as its solution exists or not. Æ For a square matrix A in matrix equation AX = B (i) |A| ≠ 0, there exists unique solution (ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution 3 3 3 3 A ,X = and B= a b c x d a b c y d = 31 32 33 A , a a a a a a = a b c z d 1 1 1 1 2 2 2 2 a a a 21 22 23 then 11 21 31 A A A A A A A adj = , where Aij is A A A 12 22 32 13 23 33 (iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent. Reprint 2025-26 The Chinese method of representing the coefficients of the unknowns of several linear equations by using rods on a calculating board naturally led to the discovery of simple method of elimination. The arrangement of rods was precisely that of the numbers in a determinant. The Chinese, therefore, early developed the idea of subtracting columns and rows as in simplification of a determinant Mikami, China, pp 30, 93. Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of determinants and of their expansion. But he used this device only in eliminating a quantity from two equations and not directly in the solution of a set of simultaneous linear equations. T. Hayashi, “The Fakudoi and Determinants in Japanese Mathematics,” in the proc. of the Tokyo Math. Soc., V. Vendermonde was the first to recognise determinants as independent functions. He may be called the formal founder. Laplace (1772), gave general method of expanding a determinant in terms of its complementary minors. In 1773 Lagrange treated determinants of the second and third orders and used them for purpose other than the solution of equations. In 1801, Gauss used determinants in his theory of numbers. The next great contributor was Jacques - Philippe - Marie Binet, (1812) who stated the theorem relating to the product of two matrices of m-columns and nrows, which for the special case of m = n reduces to the multiplication theorem. Also on the same day, Cauchy (1812) presented one on the same subject. He used the word ‘determinant’ in its present sense. He gave the proof of multiplication theorem more satisfactory than Binet’s. The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after this the word determinant received its final acceptance. Historical Note DETERMINANTS 103 Reprint 2025-26" class_12,5,Continuity and Differentiability,ncert_books/class_12/lemh1dd/lemh105.pdf,"5.1 Introduction This chapter is essentially a continuation of our study of differentiation of functions in Class XI. We had learnt to differentiate certain functions like polynomial functions and trigonometric functions. In this chapter, we introduce the very important concepts of continuity, differentiability and relations between them. We will also learn differentiation of inverse trigonometric functions. Further, we introduce a new class of functions called exponential and logarithmic functions. These functions lead to powerful techniques of differentiation. We illustrate certain geometrically obvious conditions through differential calculus. In the process, we will learn some fundamental theorems in this area. 5.2 Continuity 104 MATHEMATICS vThe whole of science is nothing more than a refinement of everyday thinking.” — ALBERT EINSTEIN v CONTINUITY AND DIFFERENTIABILITY Chapter 5 Sir Issac Newton (1642-1727) We start the section with two informal examples to get a feel of continuity. Consider the function 1, if 0 ( ) 2, if 0 x f x x ≤ = > This function is of course defined at every point of the real line. Graph of this function is given in the Fig 5.1. One can deduce from the graph that the value of the function at nearby points on x-axis remain close to each other except at x = 0. At the points near and to the left of 0, i.e., at points like – 0.1, – 0.01, – 0.001, the value of the function is 1. At the points near and to the right of 0, i.e., at points like 0.1, 0.01, Reprint 2025-26 Fig 5.1 0.001, the value of the function is 2. Using the language of left and right hand limits, we may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2). In particular the left and right hand limits do not coincide. We also observe that the value of the function at x = 0 concides with the left hand limit. Note that when we try to draw the graph, we cannot draw it in one stroke, i.e., without lifting pen from the plane of the paper, we can not draw the graph of this function. In fact, we need to lift the pen when we come to 0 from left. This is one instance of function being not continuous at x = 0. Now, consider the function defined as are both equal to 1. But the value of the function at x = 0 equals 2 which does not coincide with the common value of the left and right hand limits. Again, we note that we cannot draw the graph of the function without lifting the pen. This is yet another instance of a function being not continuous at x = 0. Naively, we may say that a function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper. This function is also defined at every point. Left and the right hand limits at x = 0 Mathematically, it may be phrased precisely as follows: f x x x ( ) , , = ≠ = 1 0 2 0 if if CONTINUITY AND DIFFERENTIABILITY 105 Fig 5.2 Definition 1 Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if at x = c exist and equal to each other, then f is said to be continuous at x = c. Recall that if the right hand and left hand limits at x = c coincide, then we say that the common value is the limit of the function at x = c. Hence we may also rephrase the definition of continuity as follows: a function is continuous at x = c if the function is defined at x = c and if the value of the function at x = c equals the limit of the function at x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a point of discontinuity of f. More elaborately, if the left hand limit, right hand limit and the value of the function lim ( ) ( ) x c f x f c → = Reprint 2025-26 Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1. Solution First note that the function is defined at the given point x = 1 and its value is 5. Then find the limit of the function at x = 1. Clearly Thus 1 lim ( ) 5 (1) x f x f → = = Hence, f is continuous at x = 1. Example 2 Examine whether the function f given by f(x) = x 2 is continuous at x = 0. Solution First note that the function is defined at the given point x = 0 and its value is 0. Then find the limit of the function at x = 0. Clearly Thus 0 lim ( ) 0 (0) x f x f → = = Hence, f is continuous at x = 0. Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0. Solution By definition f(x) = , if 0 , if 0 x x x x − < ≥ Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is 106 MATHEMATICS 1 1 lim ( ) lim (2 3) 2(1) 3 5 x x f x x → → = + = + = 0 0 lim ( ) lim 0 0 x x f x x → → = = = 2 2 Similarly, the right hand limit of f at 0 is x = 0. Hence, f is continuous at x = 0. Example 4 Show that the function f given by is not continuous at x = 0. Thus, the left hand limit, right hand limit and the value of the function coincide at 0 0 lim ( ) lim (– ) 0 x x f x x → − → − = = 0 0 lim ( ) lim 0 x x f x x → + → + = = f(x) = 3 3, if 0 1, if 0 x x x + ≠ = Reprint 2025-26 Solution The function is defined at x = 0 and its value at x = 0 is 1. When x ≠ 0, the function is given by a polynomial. Hence, at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function. Example 5 Check the points where the constant function f(x) = k is continuous. Solution The function is defined at all real numbers and by definition, its value at any real number equals k. Let c be any real number. Then Since f(c) = k = lim x c → f(x) for any real number c, the function f is continuous at every real number. Example 6 Prove that the identity function on real numbers given by f(x) = x is continuous at every real number. Solution The function is clearly defined at every point and f(c) = c for every real number c. Also, Thus, lim x c → f(x) = c = f(c) and hence the function is continuous at every real number. extension of this definition to discuss continuity of a function. Definition 2 A real function f is said to be continuous if it is continuous at every point in the domain of f. This definition requires a bit of elaboration. Suppose f is a function defined on a closed interval [a, b], then for f to be continuous, it needs to be continuous at every point in [a, b] including the end points a and b. Continuity of f at a means Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous Having defined continuity of a function at a given point, now we make a natural 0 lim ( ) x f x → = 3 3 0 lim ( 3) 0 3 3 x x → + = + = lim ( ) x c f x → = lim x c k k → = lim ( ) x c f x → = lim x c x c → = CONTINUITY AND DIFFERENTIABILITY 107 and continuity of f at b means of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of f is a singleton, f is a continuous function. Observe that lim ( ) x a f x → − and lim ( ) x b f x → + do not make sense. As a consequence – lim ( ) x b f x → = f(b) lim ( ) x a f x → + = f (a) Reprint 2025-26 Example 7 Is the function defined by f(x) = | x |, a continuous function? Solution We may rewrite f as By Example 3, we know that f is continuous at x = 0. Let c be a real number such that c < 0. Then f(c) = – c. Also Since lim ( ) ( ) x c f x f c → = , f is continuous at all negative real numbers. Now, let c be a real number such that c > 0. Then f(c) = c. Also is continuous at all points. Example 8 Discuss the continuity of the function f given by f (x) = x 3 + x 2 – 1. Solution Clearly f is defined at every real number c and its value at c is c 3 + c 2 – 1. We also know that 108 MATHEMATICS Since lim ( ) ( ) x c f x f c → = , f is continuous at all positive real numbers. Hence, f lim ( ) x c f x → = lim ( ) – x c x c → − = (Why?) lim ( ) x c f x → = lim x c x c → = (Why?) lim ( ) x c f x → = 3 2 3 2 lim ( 1) 1 x c x x c c → + − = + − f (x) = , if 0 x x − < ≥ , if 0 x x f is a continuous function. Example 9 Discuss the continuity of the function f defined by f (x) = 1 x , x ≠ 0. Solution Fix any non zero real number c, we have at every point in the domain of f. Thus f is a continuous function. Thus lim ( ) ( ) x c f x f c → = , and hence f is continuous at every real number. This means Also, since for c ≠ 0, 1 f c( ) c = , we have lim ( ) ( ) x c f x f c → = and hence, f is continuous 1 1 lim ( ) lim x c x c f x → → x c = = Reprint 2025-26 the function f (x) = 1 x near x = 0. To carry out this analysis we follow the usual trick of finding the value of the function at real numbers close to 0. Essentially we are trying to find the right hand limit of f at 0. We tabulate this in the following (Table 5.1). higher. This may be rephrased as: the value of f (x) may be made larger than any given number by choosing a positive real number very close to 0. In symbols, we write (to be read as: the right hand limit of f (x) at 0 is plus infinity). We wish to emphasise that + ∞ is NOT a real number and hence the right hand limit of f at 0 does not exist (as a real number). Similarly, the left hand limit of f at 0 may be found. The following table is self explanatory. Table 5.2 From the Table 5.2, we deduce that the value of f(x) may be made smaller than any given number by choosing a negative real number very close to 0. In symbols, we write We take this opportunity to explain the concept of infinity. This we do by analysing We observe that as x gets closer to 0 from the right, the value of f (x) shoots up f (x) – 1 – 3.333... – 5 – 10 – 102 – 103 – 10n f (x) 1 3.333... 5 10 100 = 102 1000 = 103 10n x – 1 – 0.3 – 0.2 – 10–1 – 10–2 – 10–3 – 10–n x 1 0.3 0.2 0.1 = 10–1 0.01 = 10–2 0.001 = 10–3 10–n 0 lim ( ) x f x → + = + ∞ Table 5.1 CONTINUITY AND DIFFERENTIABILITY 109 (to be read as: the left hand limit of f(x) at 0 is minus infinity). Again, we wish to emphasise that – ∞ is NOT a real number and hence the left hand limit of f at 0 does not exist (as a real number). The graph of the reciprocal function given in Fig 5.3 is a geometric representation of the above mentioned facts. Fig 5.3 0 lim ( ) x f x → − = − ∞ Reprint 2025-26 Example 10 Discuss the continuity of the function f defined by Solution The function f is defined at all points of the real line. Case 1 If c < 1, then f(c) = c + 2. Therefore, lim ( ) lim( 2) 2 x c x c f x x c → → = + = + Thus, f is continuous at all real numbers lessthan 1. Case 2 If c > 1, then f(c) = c – 2. Therefore, Thus, f is continuous at all points x > 1. Case 3 If c = 1, then the left hand limit of f at x = 1 is The right hand limit of f at x = 1 is Since the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f. The graph of the function is given in Fig 5.4. Example 11 Find all the points of discontinuity of the function f defined by + < = − > Solution As in the previous example we find that f is continuous at all real numbers x ≠ 1. The left hand limit of f at x = 1 is 110 MATHEMATICS – – 1 1 lim ( ) lim ( 2) 1 2 3 x x f x x → → = + = + = lim ( ) lim x c x c f x → → = (x – 2) = c – 2 = f (c) 1 1 lim ( ) lim ( 2) 1 2 1 x x f x x → + → + = − = − = − x x x 2, if 1 0, if 1 f (x) = 2, if 1 2, if 1 x x x x + ≤ − > Fig 5.4 The right hand limit of f at x = 1 is Since, the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f. The graph of the function is given in the Fig 5.5. – 1 1 lim ( ) lim ( 2) 1 2 3 x x f x x → − → = + = + = 1 1 lim ( ) lim ( 2) 1 2 1 x x f x x → + → + = − = − = − f (x) = x x 2, if 1 Reprint 2025-26 Fig 5.5 Example 12 Discuss the continuity of the function defined by Solution Observe that the function is defined at all real numbers except at 0. Domain of definition of this function is Case 1 If c ∈ D1 , then lim ( ) lim x c x c f x → → = (x + 2) = c + 2 = f (c) and hence f is continuous in D1 . Case 2 If c ∈ D2 , then lim ( ) lim x c x c f x → → = (– x + 2) = – c + 2 = f (c) and hence f is continuous in D2 . Since f is continuous at all points in the domain of f, we deduce that f is continuous. Graph of this function is given in the Fig 5.6. Note that to graph this function we need to lift the pen from the plane of the paper, but we need to do that only for those points where the function is not defined. Example 13 Discuss the continuity of the function f given by D1 ∪ D2 where D1 = {x ∈ R : x < 0} and D2 = {x ∈ R : x > 0} f(x) = 2 , if 0 x x ≥ < , if 0 x x f(x) = 2, if 0 x x + < − + > CONTINUITY AND DIFFERENTIABILITY 111 2, if 0 x x Fig 5.6 Solution Clearly the function is defined at every real number. Graph of the function is given in Fig 5.7. By inspection, it seems prudent to partition the domain of definition of f into three disjoint subsets of the real line. Let D1 = {x ∈ R : x < 0}, D2 = {0} and D3 = {x ∈ R : x > 0} Case 1 At any point in D1 , we have f(x) = x 2 and it is easy to see that it is continuous there (see Example 2). Case 2 At any point in D3 , we have f(x) = x and it is easy to see that it is continuous there (see Example 6). Reprint 2025-26 Fig 5.7 Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f(0) = 0. The left hand limit of f at 0 is The right hand limit of f at 0 is continuous at every point in its domain and hence, f is a continuous function. Example 14 Show that every polynomial function is continuous. Solution Recall that a function p is a polynomial function if it is defined by p(x) = a0 + a1 x + ... + an x n for some natural number n, an ≠ 0 and ai ∈ R. Clearly this function is defined for every real number. For a fixed real number c, we have every real number and hence p is a continuous function. Example 15 Find all the points of discontinuity of the greatest integer function defined by f(x) = [x], where [x] denotes the greatest integer less than or equal to x. Solution First observe that f is defined for all real numbers. Graph of the function is given in Fig 5.8. From the graph it looks like that f is discontinuous at every integral point. Below we explore, if this is true. 112 MATHEMATICS Thus 0 lim ( ) 0 x f x → = = f(0) and hence f is continuous at 0. This means that f is By definition, p is continuous at c. Since c is any real number, p is continuous at – 2 2 0 0 lim ( ) lim 0 0 x x f x x → → − = = = 0 0 lim ( ) lim 0 x x f x x → + → + = = lim ( ) ( ) x c p x p c → = Reprint 2025-26 Fig 5.8 Case 1 Let c be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to c the value of the function is equal to [c]; i.e., numbers not equal to integers. Case 2 Let c be an integer. Then we can find a sufficiently small real number r > 0 such that [c – r] = c – 1 whereas [c + r] = c. This, in terms of limits mean that lim x c → − f(x) = c – 1, lim x c → + f(x) = c Since these limits cannot be equal to each other for any c, the function is discontinuous at every integral point. 5.2.1 Algebra of continuous functions In the previous class, after having understood the concept of limits, we learnt some algebra of limits. Analogously, now we will study some algebra of continuous functions. Since continuity of a function at a point is entirely dictated by the limit of the function at that point, it is reasonable to expect results analogous to the case of limits. Theorem 1 Suppose f and g be two real functions continuous at a real number c. Then (1) f + g is continuous at x = c. lim ( ) lim [ ] [ ] x c x c f x x c → → = = . Also f(c) = [c] and hence the function is continuous at all real (2) f – g is continuous at x = c. (3) f . g is continuous at x = c. (4) f g is continuous at x = c, (provided g (c) ≠ 0). CONTINUITY AND DIFFERENTIABILITY 113 Proof We are investigating continuity of (f + g) at x = c. Clearly it is defined at x = c. We have Hence, f + g is continuous at x = c. Proofs for the remaining parts are similar and left as an exercise to the reader. lim( )( ) x c f g x → + = lim [ ( ) ( )] x c f x g x → + (by definition of f + g) = lim ( ) lim ( ) x c x c f x g x → → + (by the theorem on limits) = f(c) + g(c) (as f and g are continuous) = (f + g) (c) (by definition of f + g) Reprint 2025-26 Remarks (i) As a special case of (3) above, if f is a constant function, i.e., f(x) = λ for some real number λ, then the function (λ . g) defined by (λ . g) (x) = λ . g(x) is also continuous. In particular if λ = – 1, the continuity of f implies continuity of – f. (ii) As a special case of (4) above, if f is the constant function f(x) = λ, then the The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. The following examples illustrate this: Example 16 Prove that every rational function is continuous. Solution Recall that every rational function f is given by where p and q are polynomial functions. The domain of f is all real numbers except points at which q is zero. Since polynomial functions are continuous (Example 14), f is continuous by (4) of Theorem 1. Example 17 Discuss the continuity of sine function. Solution To see this we use the following facts 114 MATHEMATICS function g λ defined by ( ) ( ) x g g x λ λ = is also continuous wherever g(x) ≠ 0. In particular, the continuity of g implies continuity of 1 g . ( ) ( ) , ( ) 0 ( ) p x f x q x q x = ≠ 0 lim sin 0 x x → = We have not proved it, but is intuitively clear from the graph of sin x near 0. Now, observe that f(x) = sin x is defined for every real number. Let c be a real number. Put x = c + h. If x → c we know that h → 0. Therefore = sin c + 0 = sin c = f(c) Thus lim x c → f(x) = f(c) and hence f is a continuous function. lim ( ) x c f x → = lim sin x c x → Reprint 2025-26 = 0 lim sin( ) h c h → + = 0 lim [sin cos cos sin ] h c h c h → + = 0 0 lim [sin cos ] lim [cos sin ] h h c h c h → → + Remark A similar proof may be given for the continuity of cosine function. Example 18 Prove that the function defined by f(x) = tan x is a continuous function. Solution The function f(x) = tan x = sin cos x x . This is defined for all real numbers such that cos x ≠ 0, i.e., x ≠ (2n +1) 2 π . We have just proved that both sine and cosine functions are continuous. Thus tan x being a quotient of two continuous functions is continuous wherever it is defined. composition of functions. Recall that if f and g are two real functions, then (f o g) (x) = f(g (x)) is defined whenever the range of g is a subset of domain of f. The following theorem (stated without proof) captures the continuity of composite functions. Theorem 2 Suppose f and g are real valued functions such that (f o g) is defined at c. If g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c. The following examples illustrate this theorem. Example 19 Show that the function defined by f(x) = sin (x 2 ) is a continuous function. Solution Observe that the function is defined for every real number. The function f may be thought of as a composition g o h of the two functions g and h, where g (x) = sin x and h (x) = x 2 . Since both g and h are continuous functions, by Theorem 2, it can be deduced that f is a continuous function. An interesting fact is the behaviour of continuous functions with respect to CONTINUITY AND DIFFERENTIABILITY 115 Example 20 Show that the function f defined by where x is any real number, is a continuous function. Solution Define g by g (x) = 1 – x + | x| and h by h (x) = | x| for all real x. Then of a polynomial function and the modulus function is continuous. But then f being a composite of two continuous functions is continuous. In Example 7, we have seen that h is a continuous function. Hence g being a sum (h o g) (x) = h (g (x)) f(x) = |1 – x + | x | |, Reprint 2025-26 = h (1– x + | x |) = | 1– x + | x | | = f(x) , if 1 ( ) 5, if > 1 x x f x x ≤ = continuous at x = 0? At x = 1? At x = 2? Find all points of discontinuity of f, where f is defined by 116 MATHEMATICS 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. 2. Examine the continuity of the function f(x) = 2x 2 – 1 at x = 3. 3. Examine the following functions for continuity. 4. Prove that the function f(x) = x n is continuous at x = n, where n is a positive integer. 5. Is the function f defined by 6. 2 3, if 2 ( ) 2 3, if > 2 x x f x x x + ≤ = − 7. (a) f(x) = x – 5 (b) f(x) = 1 x − 5 , x ≠ 5 (c) f(x) = 2 25 5 x x − + , x ≠ –5 (d) f(x) = | x – 5 | x x f x x x | | , if 0 ( ) 0, if 0 ≠ = = 9. , if 0 ( ) | | 1, if 0 EXERCISE 5.1 f x x x x x x x f x x x ( ) 2 , if 3 < 3 6 2, if 3 + ≤ − = − − < + ≥ < = − ≥ | | 3, if 3 x x 10. 2 1, if 1 ( ) 1, if 1 x x f x x x + ≥ = + < 11. 12. 10 1, if 1 ( ) , if 1 x x f x x x − ≤ = > 13. Is the function defined by 8. 5, if 1 ( ) 5, if 1 x x f x x x + ≤ = − > a continuous function? 2 Reprint 2025-26 3, if 2 ( ) 1, if 2 x x f x x x − ≤ = + > 3 2 Discuss the continuity of the function f, where f is defined by 14. 16. − ≤ − = − < ≤ > 17. Find the relationship between a and b so that the function f defined by 1, if 3 ( ) 3, if 3 ax x f x bx x + ≤ = + > is continuous at x = 3. 18. For what value of λ is the function defined by 2 ( 2 ), if 0 ( ) 4 1, if 0 x x x f x x x λ − ≤ = + > continuous at x = 0? What about continuity at x = 1? 19. Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x. 20. Is the function defined by f(x) = x 2 – sin x + 5 continuous at x = π? 21. Discuss the continuity of the following functions: (a) f(x) = sin x + cos x (b) f(x) = sin x – cos x (c) f(x) = sin x . cos x 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions. 23. Find all points of discontinuity of f, where x f x x f x x x x 3, if 0 1 ( ) 4, if 1 3 ( ) 2 , if 1 1 2, if 1 ≤ ≤ = < < ≤ ≤ 15. 5, if 3 10 2, if 1 x x CONTINUITY AND DIFFERENTIABILITY 117 x x f x x 2 , if 0 ( ) 0, if 0 1 < = ≤ ≤ 4 , if > 1 x x < = + ≥ 24. Determine if f defined by ≠ = = is a continuous function? x x f x x x x x f x x x x 2 1 sin , if 0 ( ) 0, if 0 sin , if 0 ( ) 1, if 0 Reprint 2025-26 sin cos , if 0 ( ) 1, if 0 x x x f x x − ≠ = − = Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29. 118 MATHEMATICS 25. Examine the continuity of f, where f is defined by 26. 27. 2 , if 2 ( ) 3, if 2 kx x f x x ≤ = > at x = 2 28. 1, if ( ) cos , if kx x f x x x + ≤ π = > π at x = π 29. 1, if 5 ( ) 3 5, if 5 kx x f x x x + ≤ = − > at x = 5 30. Find the values of a and b such that the function defined by ≤ = + < < ≥ is a continuous function. 31. Show that the function defined by f(x) = cos (x 2 ) is a continuous function. 32. Show that the function defined by f(x) = | cos x | is a continuous function. 33. Examine that sin | x | is a continuous function. 34. Find all the points of discontinuity of f defined by f(x) = | x | – | x + 1 |. k x x x f x x cos , if 2 2 ( ) 3, if 2 π ≠ π − = π = x f x ax b x 5, if 2 ( ) , if 2 10 at x = 2 π 21, if 10 x 5.3. Differentiability Recall the following facts from previous class. We had defined the derivative of a real function as follows: Suppose f is a real function and c is a point in its domain. The derivative of f at c is defined by ( ) ( ) lim h → h + − 0 Reprint 2025-26 f c h f c provided this limit exists. Derivative of f at c is denoted by f ′(c) or ( ( )) | c d f x dx . The function defined by wherever the limit exists is defined to be the derivative of f. The derivative of f is denoted by f ′(x) or ( ( )) d f x dx or if y = f(x) by dy dx or y′. The process of finding derivative of a function is called differentiation. We also use the phrase differentiate f(x) with respect to x to mean find f ′(x). The following rules were established as a part of algebra of derivatives: (1) (u ± v)′ = u′ ± v′ (2) (uv)′ = u′v + uv′ (Leibnitz or product rule) (3) 2 u u v uv v v ′ ′ − ′ = , wherever v ≠ 0 (Quotient rule). The following table gives a list of derivatives of certain standard functions: Table 5.3 f (x) x n sin x cos x tan x f ′(x) nxn – 1 cos x – sin x sec2 x f x h f x f x → h + − ′ = ( ) ( ) ( ) lim h 0 CONTINUITY AND DIFFERENTIABILITY 119 Now the natural question is; what if it doesn’t? The question is quite pertinent and so is its answer. If 0 In other words, we say that a function f is differentiable at a point c in its domain if both to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. As in case of continuity, at the end points a and b, we take the right hand limit and left hand limit, which are nothing but left hand derivative and right hand derivative of the function at a and b respectively. Similarly, a function is said to be differentiable in an interval (a, b) if it is differentiable at every point of (a, b). ( ) ( ) lim h → h + − and 0 ( ) ( ) lim h – 0 Whenever we defined derivative, we had put a caution provided the limit exists. f c h f c ( ) ( ) lim h → h + − does not exist, we say that f is not differentiable at c. f c h f c f c h f c h → + + − are finite and equal. A function is said Reprint 2025-26 Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that point. Proof Since f is differentiable at c, we have But for x ≠ c, we have Therefore lim [ ( ) ( )] x c f x f c → − = ( ) ( ) lim . ( ) x c or lim [ ( )] lim [ ( )] x c x c f x f c → → − = ( ) ( ) lim . lim [( )] x c x c f x f c x c → → x c − − − = f ′(c) . 0 = 0 or lim ( ) x c f x → = f (c) Hence f is continuous at x = c. Corollary 1 Every differentiable function is continuous. We remark that the converse of the above statement is not true. Indeed we have seen that the function defined by f(x) = | x | is a continuous function. Consider the left hand limit 120 MATHEMATICS f(x) – f(c) = ( ) ( ) . ( ) f x f c x c x c − − − (0 ) (0) lim 1 h → h h + − − = = − ( ) ( ) lim ( ) x c f x f c f c → x c − = ′ − f h f h f x f c x c → x c − − − does not exist and hence f is not differentiable at 0. Thus f is not a differentiable function. 5.3.1 Derivatives of composite functions To study derivative of composite functions, we start with an illustrative example. Say, we want to find the derivative of f, where The right hand limit Since the above left and right hand limits at 0 are not equal, 0 (0 ) (0) lim h (0 ) (0) lim 1 h – 0 f h f h h h → + + − = = 0 f (x) = (2x + 1)3 Reprint 2025-26 → h + − f h f a polynomial function as illustrated below. Now, observe that f(x) = (h o g) (x) where g(x) = 2x + 1 and h(x) = x 3 . Put t = g(x) = 2x + 1. Then f(x) = h(t) = t 3 . Thus the derivative of, say, (2x + 1)100 . We may formalise this observation in the following theorem called the chain rule. Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two functions u and v ; i.e., f = v o u. Suppose t = u(x) and if both dt dx and dv dt exist, we have One way is to expand (2x + 1)3 using binomial theorem and find the derivative as The advantage with such observation is that it simplifies the calculation in finding We skip the proof of this theorem. Chain rule may be extended as follows. Suppose df dx = 6 (2x + 1)2 = 3(2x + 1)2 . 2 = 3t 2 . 2 = dh dt dt dx ⋅ ( ) d f x dx = 3 (2 1) d x dx + df dv dt dx dt dx = ⋅ = 3 2 (8 12 6 1) d x x x dx + + + = 24x 2 + 24x + 6 = 6 (2x + 1)2 CONTINUITY AND DIFFERENTIABILITY 121 f is a real valued function which is a composite of three functions u, v and w; i.e., f = (w o u) o v. If t = v (x) and s = u (t), then provided all the derivatives in the statement exist. Reader is invited to formulate chain rule for composite of more functions. Example 21 Find the derivative of the function given by f(x) = sin (x 2 ). Solution Observe that the given function is a composite of two functions. Indeed, if t = u(x) = x 2 and v(t) = sin t, then f(x) = (v o u) (x) = v(u(x)) = v(x 2 ) = sin x 2 df d w u dt dw ds dt ( o ) dx dt dx ds dt dx = ⋅ = ⋅ ⋅ Reprint 2025-26 Put t = u(x) = x 2 . Observe that cos dv t dt = and 2 dt x dx = exist. Hence, by chain rule It is normal practice to express the final result only in terms of x. Thus Differentiate the functions with respect to x in Exercises 1 to 8. 122 MATHEMATICS f(x) = | x – 1|, x ∈ R is not differentiable at x = 1. 10. Prove that the greatest integer function defined by 1. sin (x 2 + 5) 2. cos (sin x) 3. sin (ax + b) 4. sec (tan ( x )) 5. sin ( ) cos ( ) ax b cx d + + 6. cos x 3 . sin2 (x 5 ) 7. ( ) 2 2 cot x 8. cos( x ) 9. Prove that the function f given by df dx = cos 2 dv dt t x dt dx ⋅ = ⋅ df dx = 2 cos 2 2 cos t x x x ⋅ = EXERCISE 5.2 5.3.2 Derivatives of implicit functions Until now we have been differentiating various functions given in the form y = f(x). But it is not necessary that functions are always expressed in this form. For example, consider one of the following relationships between x and y: x – y – π = 0 x + sin xy – y = 0 In the first case, we can solve for y and rewrite the relationship as y = x – π. In the second case, it does not seem that there is an easy way to solve for y. Nevertheless, there is no doubt about the dependence of y on x in either of the cases. When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f(x), we say that y is given as an explicit function of x. In the latter case it f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2. Reprint 2025-26 is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to differentiate implicit functions. Example 22 Find dy dx if x – y = π. Solution One way is to solve for y and rewrite the above as y = x – π But then dy dx = 1 Alternatively, directly differentiating the relationship w.r.t., x, we have everywhere w.r.t., x. Thus which implies that dy dx = 1 dx dx = Example 23 Find dy dx , if y + sin y = cos x. Recall that d dx π means to differentiate the constant function taking value π ( ) ( ) d d x y dx dx − = 0 ( ) d x y dx − = d dx π CONTINUITY AND DIFFERENTIABILITY 123 Solution We differentiate the relationship directly with respect to x, i.e., (sin ) dy d y dx dx + = (cos ) d x dx which implies using chain rule This gives dy dx = sin 1 cos x y − + where y ≠ (2n + 1) π cos dy dy y dx dx + ⋅ = – sin x Reprint 2025-26 5.3.3 Derivatives of inverse trigonometric functions We remark that inverse trigonometric functions are continuous functions, but we will not prove this. Now we use chain rule to find derivatives of these functions. Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists. Solution Let y = sin–1 x. Then, x = sin y. Differentiating both sides w.r.t. x, we get which implies that dy dx = 1 1 1 cos y cos(sin ) x − = i.e., x ∈ (– 1, 1). To make this result a bit more attractive, we carry out the following manipulation. Recall that for x ∈ (– 1, 1), sin (sin–1 x) = x and hence Also, since y ∈ , 2 2 π π − , cos y is positive and hence cos y = 2 1− x Thus, for x ∈ (– 1, 1), 124 MATHEMATICS Observe that this is defined only for cos y ≠ 0, i.e., sin–1 x ≠ , 2 2 π π − , i.e., x ≠ – 1, 1, cos2 y = 1 – (sin y) 2 = 1 – (sin (sin–1 x))2 = 1 – x 2 1 = cos y dy dx Domain off (-1, 1) (-1, 1) R f(x) sin–1 x cos-1 x tan-1x f 1 (x) 2 1 1− x 2 1 1 x − − 2 1 1+ x 2 1 1 cos 1 dy dx y x = = − Reprint 2025-26 Find dy dx in the following: 10. y = tan–1 3 11. 2 1 2 1 cos 0 1 , 1 12. 2 1 2 1 sin 0 1 , 1 13. 1 2 2 cos , 1 1 1 14. ( ) 1 2 1 1 sin 2 1 , 2 2 y x x x − = − − < < 15. 1 2 1 1 sec , 0 2 1 2 y x x − = < < − 1. 2x + 3y = sin x 2. 2x + 3y = sin y 3. ax + by2 = cos y 4. xy + y2 = tan x + y 5. x 2 + xy + y 2 = 100 6. x 3 + x2 y + xy2 + y 3 = 81 7. sin2 y + cos xy = κ 8. sin2 x + cos2 y = 1 9. y = sin–1 2 2 1 x y x x − − = < < + x y x x − − = < < + x y x x − = − < < + 2 3 , 1 3 x x x − − 3 3 − < 1), the increment in the value of y = f n (x) increases as n increases for n = 1, 2, 3, 4. It is conceivable that such a statement is true for all positive values of n, The Fig 5.9 gives a sketch of y = f 1 (x) = x, y = f 2 (x) = x 2 , y = f 3 (x) = x 3 and y = f 4 (x) Reprint 2025-26 where f n (x) = x n . Essentially, this means that the graph of y = f n (x) leans more towards the y-axis as n increases. For example, consider f 10(x) = x 10 and f 15(x) = x 15. If x increases from 1 to 2, f 10 increases from 1 to 210 whereas f 15 increases from 1 to 215. Thus, for the same increment in x, f 15 grow faster than f 10. Upshot of the above discussion is that the growth of polynomial functions is dependent on the degree of the polynomial function – higher the degree, greater is the growth. The next natural question is: Is there a function which grows faster than any polynomial function. The answer is in affirmative and an example of such a function is For example, we can prove that 10x grows faster than f 100 (x) = x 100. For large values of x like x = 103 , note that f 100 (x) = (103 ) 100 = 10300 whereas f(103 ) = 3 10 10 = 101000 . Clearly f(x) is much greater than f 100 (x). It is not difficult to prove that for all x > 103 , f(x) > f 100 (x). But we will not attempt to give a proof of this here. Similarly, by choosing large values of x, one can verify that f(x) grows faster than f n (x) for any positive integer n. 126 MATHEMATICS Our claim is that this function f grows faster than f n (x) = x n for any positive integer n. y = f(x) = 10x . Fig 5.9 Definition 3 The exponential function with positive base b > 1 is the function The graph of y = 10x is given in the Fig 5.9. It is advised that the reader plots this graph for particular values of b like 2, 3 and 4. Following are some of the salient features of the exponential functions: (1) Domain of the exponential function is R, the set of all real numbers. (2) Range of the exponential function is the set of all positive real numbers. (3) The point (0, 1) is always on the graph of the exponential function (this is a restatement of the fact that b 0 = 1 for any real b > 1). (4) Exponential function is ever increasing; i.e., as we move from left to right, the graph rises above. Reprint 2025-26 y = f(x) = b x (5) For very large negative values of x, the exponential function is very close to 0. In other words, in the second quadrant, the graph approaches x-axis (but never meets it). Exponential function with base 10 is called the common exponential function. In the Appendix A.1.4 of Class XI, it was observed that the sum of the series is a number between 2 and 3 and is denoted by e. Using this e as the base we obtain an extremely important exponential function y = e x . This is called natural exponential function. It would be interesting to know if the inverse of the exponential function exists and has nice interpretation. This search motivates the following definition. Definition 4 Let b > 1 be a real number. Then we say logarithm of a to base b is x if b x = a. Logarithm of a to base b is denoted by logb a. Thus logb a = x if b x = a. Let us work with a few explicit examples to get a feel for this. We know 23 = 8. In terms of logarithms, we may rewrite this as log2 8 = 3. Similarly, 104 = 10000 is equivalent to saying log10 10000 = 4. Also, 625 = 54 = 252 is equivalent to saying log5 625 = 4 or log25 625 = 2. On a slightly more mature note, fixing a base b > 1, we may look at logarithm as a function from positive real numbers to all real numbers. This function, called the logarithmic function, is defined by logb : R+ → R x → logb x = y if b y = x As before if the base b = 10, we say it is common logarithms and if b = e, then we say it is natural logarithms. Often natural logarithm is denoted by ln. In this chapter, log x denotes the logarithm function to base e, i.e., ln x will be written as simply log x. The Fig 5.10 gives the plots of logarithm function to base 2, e and 10. 1 1 1 ... 1! 2! + + + CONTINUITY AND DIFFERENTIABILITY 127 Some of the important observations about the logarithm function to any base b > 1 are listed below: Fig 5.10 Reprint 2025-26 (6) Fig 5.11 gives the plot of y = e x and y = ln x. It is of interest to observe that the two curves are the mirror images of each other reflected in the line y = x. Two properties of ‘log’ functions are proved below: 128 MATHEMATICS (1) We cannot make a meaningful definition of logarithm of non-positive numbers and hence the domain of log function is R+ . (2) The range of log function is the set of all real numbers. (3) The point (1, 0) is always on the graph of the log function. (4) The log function is ever increasing, i.e., as we move from left to right the graph rises above. (5) For x very near to zero, the value of log x can be made lesser than any given real number. In other words in the fourth quadrant the graph approaches y-axis (but never meets it). (1) There is a standard change of base rule to obtain loga p in terms of logb p. Let loga p = α, logb p = β and logb a = γ. This means a α = p, b β = p and b γ = a. Substituting the third equation in the first one, we have (b γ ) α = b γα = p Using this in the second equation, we get b β = p = b γα Fig 5.11 (2) Another interesting property of the log function is its effect on products. Let logb pq = α. Then b α = pq. If logb p = β and logb q = γ, then b β = p and b γ = q. But then b α = pq = bβb γ = b β + γ which implies β = αγ or α = β γ . But then which implies α = β + γ, i.e., logb pq = logb p + logb q loga p = log log b Reprint 2025-26 b p a Example 25 Is it true that x = e log x for all real x? Solution First, observe that the domain of log function is set of all positive real numbers. So the above equation is not true for non-positive real numbers. Now, let y = e log x . If y > 0, we may take logarithm which gives us log y = log (e log x ) = log x . log e = log x. Thus y = x. Hence x = e log x is true only for positive values of x. calculus is that it doesn’t change during the process of differentiation. This is captured in the following theorem whose proof we skip. Theorem 5* (1) The derivative of e x w.r.t., x is e x ; i.e., d dx (e x ) = e x . One of the striking properties of the natural exponential function in differential A particularly interesting and important consequence of this is when p = q. In this case the above may be rewritten as An easy generalisation of this (left as an exercise!) is for any positive integer n. In fact this is true for any real number n, but we will not attempt to prove this. On the similar lines the reader is invited to verify logb p2 = logb p + logb p = 2 log p logb pn = n log p logb x y = logb x – logb y CONTINUITY AND DIFFERENTIABILITY 129 Example 26 Differentiate the following w.r.t. x: Solution (i) Let y = e – x . Using chain rule, we have * Please see supplementary material on Page 222. (2) The derivative of log x w.r.t., x is 1 x ; i.e., d dx (log x) = 1 x . (ii) Let y = sin (log x). Using chain rule, we have (i) e –x (ii) sin (log x), x > 0 (iii) cos–1 (e x ) (iv) e cos x dy dx = x d e dx − ⋅ (– x) = – e – x dy dx = cos (log ) cos (log ) (log ) d x x x dx x ⋅ = Reprint 2025-26 Differentiate the following w.r.t. x: 5.5. Logarithmic Differentiation In this section, we will learn to differentiate certain special class of functions given in the form y = f(x) = [u(x)]v (x) By taking logarithm (to base e) the above may be rewritten as 130 MATHEMATICS (iv) Let y = e cos x . Using chain rule, we have 10. cos (log x + e x ), x > 0 (iii) Let y = cos–1 (e x ). Using chain rule, we have 1. sin x e x 2. 1 sin x e − 3. 3 x e 4. sin (tan–1 e –x ) 5. log (cos e x ) 6. 2 5 ... x x x e e e + + + 7. , 0 x e x > 8. log (log x), x > 1 9. cos , 0 log x x x > dy dx = 2 2 1 ( ) 1 ( ) 1 dy dx = cos cos ( sin ) (sin ) x x e x x e ⋅ − = − EXERCISE 5.4 x x x x d e e dx e e − − ⋅ = − − log y = v(x) log [u(x)] Using chain rule we may differentiate this to get which implies that [ ] ( ) ( ) ( ) log ( ) ( ) dy v x y u x v x u x dx u x = ⋅ ′ + ′ ⋅ The main point to be noted in this method is that f(x) and u(x) must always be positive as otherwise their logarithms are not defined. This process of differentiation is known as logarithms differentiation and is illustrated by the following examples: 1 1 ( ) ( ) dy v x y dx u x ⋅ = ⋅ . u′(x) + v′(x) . log [u(x)] Reprint 2025-26 Example 27 Differentiate 2 Solution Let 2 Taking logarithm on both sides, we have Now, differentiating both sides w.r.t. x, we get or dy dx = 2 2 1 2 6 4 2 ( 3) 4 3 4 5 y x x x x x x + + − − + + + Example 28 Differentiate a x w.r.t. x, where a is a positive constant. Solution Let y = a x . Then log y = x log a Differentiating both sides w.r.t. x, we have 1 dy y dx ⋅ = 2 2 1 1 2 6 4 2 ( 3) 4 3 4 5 x x x x x x + + − − + + + 2 ( 3) ( 4) (3 4 5) x x y x x − + = + + log y = 1 2 [log (x – 3) + log (x 2 + 4) – log (3x 2 + 4x + 5)] = 2 2 2 2 1 ( 3)( 4) 1 2 6 4 2 ( 3) 3 4 5 4 3 4 5 x x x x x x x x x x − + + + − + + − + + + 2 ( 3) ( 4) 3 4 5 x x x x − + + + w.r.t. x. CONTINUITY AND DIFFERENTIABILITY 131 or dy dx = y log a Thus ( ) d x a dx = a x log a Alternatively ( ) d x a dx = log log ( ) ( log ) d d x a x a e e x a dx dx = 1 dy y dx = log a Reprint 2025-26 = e x log a . log a = a x log a. Example 29 Differentiate x sin x , x > 0 w.r.t. x. Solution Let y = x sin x . Taking logarithm on both sides, we have Therefore 1 . dy y dx = sin (log ) log (sin ) d d x x x x dx dx + or 1 dy y dx = 1 (sin ) log cos x x x x + or dy dx = sin cos log x y x x x + Example 30 Find dy dx , if y x + x y + x x = a b . Solution Given that y x + x y + x x = a b . Putting u = y x , v = x y and w = x x , we get u + v + w = a b Therefore 0 du dv dw dx dx dx + + = ... (1) 132 MATHEMATICS log y = sin x log x = sin sin cos log x x x x x x + = sin 1 sin sin cos log x x x x x x x − ⋅ + ⋅ Now, u = y x . Taking logarithm on both sides, we have log u = x log y Differentiating both sides w.r.t. x, we have So du dx = log log x dy x dy x u y y y y dx y dx + = + ... (2) Also v = x y 1 du u dx ⋅ = (log ) log ( ) d d x y y x dx dx + Reprint 2025-26 = 1 log 1 dy x y y dx ⋅ + ⋅ Taking logarithm on both sides, we have log v = y log x Differentiating both sides w.r.t. x, we have So dv dx = log y dy v x x dx + = log y y dy x x x dx + ... (3) Again w = x x Taking logarithm on both sides, we have log w = x log x. Differentiating both sides w.r.t. x, we have 1 dw w dx ⋅ = (log ) log ( ) d d x x x x dx dx + ⋅ 1 dv v dx ⋅ = (log ) log d dy y x x dx dx + = 1 log dy y x x dx ⋅ + ⋅ = 1 x x log 1 x ⋅ + ⋅ CONTINUITY AND DIFFERENTIABILITY 133 i.e. dw dx = w (1 + log x) = x x (1 + log x) ... (4) From (1), (2), (3), (4), we have or (x . yx – 1 + x y . log x) dy dx = – x x (1 + log x) – y . x y–1 – y x log y Therefore dy dx = log log x x dy y y dy y y x x y dx x dx + + + + x x (1 + log x) = 0 Reprint 2025-26 − − + + + + 1 [ log . (1 log )] . log x y x x y y y y x x x x y x x − 1 Differentiate the functions given in Exercises 1 to 11 w.r.t. x. Find dy dx of the functions given in Exercises 12 to 15. 134 MATHEMATICS 12. x y + y x = 1 13. y x = x y 14. (cos x) y = (cos y) x 15. xy = e (x – y) 16. Find the derivative of the function given by f(x) = (1 + x) (1 + x 2 ) (1 + x 4 ) (1 + x 8 ) and hence find f ′(1). 17. Differentiate (x 2 – 5x + 8) (x 3 + 7x + 9) in three ways mentioned below: (i) by using product rule (ii) by expanding the product to obtain a single polynomial. (iii) by logarithmic differentiation. Do they all give the same answer? 18. If u, v and w are functions of x, then show that 11. (x cos x) x + 1 ( sin ) x x x 1. cos x . cos 2x . cos 3x 2. ( 1) ( 2) ( 3) ( 4) ( 5) x x x x x − − − − − 3. (log x) cos x 4. x x – 2sin x 5. (x + 3)2 . (x + 4)3 . (x + 5)4 6. 1 1 1 x x x x x 7. (log x) x + x log x 8. (sin x) x + sin–1 x 9. x sin x + (sin x) cos x 10. 2 cos 2 1 1 x x x x x + + − EXERCISE 5.5 + + + 5.6 Derivatives of Functions in Parametric Forms Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between d dx (u. v. w) = du dx v. w + u . dv dx . w + u . v dw dx in two ways - first by repeated application of product rule, second by logarithmic differentiation. Reprint 2025-26 them is expressed via a third variable. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the form x = f(t), y = g (t) is said to be parametric form with t as a parameter. or dy dx = whenever 0 dy dt dx dx dt dt Thus dy dx = ( ) as ( ) and ( ) ( ) g t dy dx g t f t f t dt dt ′ = ′ = ′ ′ [provided f ′(t) ≠ 0] Example 31 Find dy dx , if x = a cos θ, y = a sin θ. Solution Given that x = a cos θ, y = a sin θ Therefore dx dθ = – a sin θ, dy dθ = a cos θ In order to find derivative of function in such form, we have by chain rule. dy dt = dy dx dx dt ⋅ dy d a dx a d θ θ = = − θ − θ θ ≠ CONTINUITY AND DIFFERENTIABILITY 135 Hence dy dx = cos cot sin Example 32 Find dy dx , if x = at2 , y = 2at. Solution Given that x = at2 , y = 2at So dx dt = 2at and dy dt = 2a Therefore dy dx = 2 1 2 Reprint 2025-26 dy dt a dx at t dt = = Example 33 Find dy dx , if x = a (θ + sin θ), y = a (1 – cos θ). Solution We have dx dθ = a(1 + cos θ), dy dθ = a (sin θ) Therefore dy dx = sin tan (1 cos ) 2 Example 34 Find 2 2 2 3 3 3 , if dy x y a dx + = . Solution Let x = a cos3 θ, y = a sin3 θ. Then 136 MATHEMATICS ANote It may be noted here that dy dx is expressed in terms of parameter only without directly involving the main variables x and y. 2 2 3 3 x y + = 2 2 3 3 3 3 ( cos ) ( sin ) a a θ + θ = 2 2 3 2 2 3 a (cos (sin ) θ + θ = a dy d a dx a d θ θ θ = = + θ θ Hence, x = a cos3θ, y = a sin3θ is parametric equation of 2 2 2 3 3 3 x y a + = Now dx dθ = – 3a cos2 θ sin θ and dy dθ = 3a sin2 θ cos θ Therefore dy dx = Reprint 2025-26 dy d a y dx a x d θ θ θ = = − θ = − − θ θ θ 2 3 2 3 sin cos tan 3 cos sin If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy dx . 5.7 Second Order Derivative Let y = f(x). Then 10. x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) 11. If 1 1 sin cos , , show that t t dy y x a y a dx x − − = = = − 1. x = 2at2 , y = at4 2. x = a cos θ, y = b cos θ 3. x = sin t, y = cos 2t 4. x = 4t, y = 4 t 5. x = cos θ – cos 2θ, y = sin θ – sin 2θ 6. x = a (θ – sin θ), y = a (1 + cos θ) 7. x = 3 sin cos 2 t t , 3 cos cos 2 t y t = 8. cos log tan 2 t x a t = + y = a sin t 9. x = a sec θ, y = b tan θ dy dx = f ′(x) ... (1) If f ′(x) is differentiable, we may differentiate (1) again w.r.t. x. Then, the left hand EXERCISE 5.6 CONTINUITY AND DIFFERENTIABILITY 137 side becomes d dy dx dx which is called the second order derivative of y w.r.t. x and is denoted by 2 denoted by D2 y or y″ or y 2 if y = f(x). We remark that higher order derivatives may be defined similarly. 2 d y dx . The second order derivative of f(x) is denoted by f ″(x). It is also Reprint 2025-26 Example 35 Find 2 Solution Given that y = x 3 + tan x. Then Therefore 2 Example 36 If y = A sin x + B cos x, then prove that 2 Solution We have and 2 Hence 2 Example 37 If y = 3e 2x + 2e 3x , prove that 2 138 MATHEMATICS 2 d y dx , if y = x 3 + tan x. 2 d y dx + y = 0 2 d y dx = ( ) 2 2 3 sec d x x dx + 2 d y dx = d dx (A cos x – B sin x) = – A sin x – B cos x = – y dy dx = 3x 2 + sec2 x dy dx = A cos x – B sin x = 6x + 2 sec x . sec x tan x = 6x + 2 sec2 x tan x 2 5 6 0 d y dy y dx dx − + = . 2 0 d y y dx + = . Solution Given that y = 3e 2x + 2e 3x . Then Therefore 2 Hence 2 2 5 d y dy dx dx − + 6y = 6 (2e 2x + 3e 3x ) 2 d y dx = 12e 2x + 18e 3x = 6 (2e 2x + 3e 3x ) dy dx = 6e 2x + 6e 3x = 6 (e 2x + e 3x ) Reprint 2025-26 – 30 (e 2x + e 3x ) + 6 (3e 2x + 2e 3x ) = 0 Example 38 If y = sin–1 x, show that (1 – x 2 ) 2 Solution We have y = sin–1 x. Then or 2 (1 ) 1 dy x dx − = So 2 (1 ) . 0 d dy x dx dx − = or ( ) 2 2 2 2 (1 ) (1 ) 0 d y dy d x x dx dx dx − ⋅ + ⋅ − = or Hence 2 2 2 (1 ) 0 d y dy x x dx dx − − = Alternatively, Given that y = sin–1 x, we have 1 y x = − , i.e., ( ) 2 2 1 1 1 − = x y 1 2 1 dy dx = 2 1 2 2 2 2 2 (1 ) 0 2 1 d y dy x x dx dx x − ⋅ − ⋅ = − (1 ) − x CONTINUITY AND DIFFERENTIABILITY 139 2 0 d y dy x dx dx − = . So 2 2 1 2 1 (1 ) . 2 (0 2 ) 0 − + − = x y y y x Hence (1 – x 2 ) y2 – xy1 = 0 Find the second order derivatives of the functions given in Exercises 1 to 10. 7. e 6x cos 3x 8. tan–1 x 9. log (log x) 10. sin (log x) 11. If y = 5 cos x – 3 sin x, prove that 2 1. x 2 + 3x + 2 2. x 20 3. x . cos x 4. log x 5. x 3 log x 6. e x sin 5x EXERCISE 5.7 Reprint 2025-26 2 0 d y y dx + = Example 39 Differentiate w.r.t. x, the following function: Solution 140 MATHEMATICS 12. If y = cos–1 x, Find 2 13. If y = 3 cos (log x) + 4 sin (log x), show that x 2 y2 + xy1 + y = 0 14. If y = Ae mx + Be nx, show that 2 15. If y = 500e 7x + 600e –7x , show that 2 16. If e y (x + 1) = 1, show that 2 2 17. If y = (tan–1 x) 2 , show that (x 2 + 1)2 y 2 + 2x (x 2 + 1) y 1 = 2 (i) 2 1 3 2 2 4 x x + + + (ii) log7 (log x) (i) Let y = 2 1 3 2 2 4 x x + + + = 1 1 2 2 2 (3 2) (2 4) x x − + + + 2 d y dx in terms of y alone. Miscellaneous Examples 2 d y dy dx dx = 2 ( ) 0 d y dy m n mny dx dx − + + = 2 49 d y y dx = Note that this function is defined at all real numbers 2 3 x > − . Therefore This is defined for all real numbers 2 3 x > − . dy dx = = 1 2 3 2 3 1 2 2 4 4 1 2 2 3 2 ( ) ( ) ( ) x + ⋅ − x x + ⋅ − − = ( ) 3 2 2 3 2 2 3 2 2 4 1 1 1 1 2 2 2 2 1 1 (3 2) (3 2) (2 4) (2 4) 2 2 d d x x x x dx dx − − − + ⋅ + + − + ⋅ + x x x − + + Reprint 2025-26 Example 40 Differentiate the following w.r.t. x. Solution (i) Let f (x) = cos –1 (sin x). Observe that this function is defined for all real numbers. We may rewrite this function as f(x) = cos –1 (sin x) (ii) Let y = log7 (log x) = log (log ) log 7 x (by change of base formula). (i) cos –1 (sin x) (ii) 1 sin tan 1 cos x x − + (iii) 1 1 2 sin 1 4 x The function is defined for all real numbers x > 1. Therefore dy dx = 1 (log (log )) log 7 d x dx = 1 1 (log ) log 7 log = 1 x x log 7 log = cos cos − − = 2 x π − 1 2 π x CONTINUITY AND DIFFERENTIABILITY 141 d x x dx ⋅ x + − + (ii) Let f(x) = tan –1 sin 1 cos x x + . Observe that this function is defined for all real Thus f ′(x) = – 1. numbers, where cos x ≠ – 1; i.e., at all odd multiplies of π. We may rewrite this function as f(x) = 1 sin tan 1 cos x x − + = 1 2 2 sin cos 2 2 tan 2cos 2 Reprint 2025-26 x − x x 142 MATHEMATICS (iii) Let f(x) = sin –1 1 2 1 4 x Observe that we could cancel cos 2 x in both numerator and denominator as it is not equal to zero. Thus f ′(x) = 1 . 2 x such that 1 2 1 1 1 4 x we need to find all x such that 1 2 1 1 4 x may rewrite this as 2 ≤ 1 2 x + 2x which is true for all x. Hence the function is defined at every real number. By putting 2x = tan θ, this function may be rewritten as x + − ≤ ≤ + . Since the quantity in the middle is always positive, x + + . To find the domain of this function we need to find all = 1 tan tan 2 2 − x x = f(x) = 1 1 2 sin 1 4 x = sin− ⋅ x + ≤ + , i.e., all x such that 2x + 1 ≤ 1 + 4x . We x + − + 1 2 2 2 + ( ) 1 2 x x Thus f ′(x) = ( ) 2 1 2 (2 ) 1 2 Reprint 2025-26 = 1 2 2tan sin 1 tan − θ + θ = sin –1 [sin 2θ] = 2θ = 2 tan – 1 (2x ) = 2 (2 )log 2 1 4 x x ⋅ + = 1 2 log 2 d dx ⋅ ⋅ + 1 4 x x + + x x Example 41 Find f ′(x) if f(x) = (sin x) sin x for all 0 < x < π. Solution The function y = (sin x) sin x is defined for all positive real numbers. Taking logarithms, we have log y = log (sin x) sin x = sin x log (sin x) Then 1 dy y dx = d dx (sin x log (sin x)) Thus dy dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x) sin x cos x Example 42 For a positive constant a find dy dx , where 1 1 , and a t t y a x t t + = = + Solution Observe that both y and x are defined for all real t ≠ 0. Clearly dy dt = ( ) 1 t t d a dt + = 1 1 log t t d a t a dt t + + ⋅ = 1 2 1 1 log t t a a t + − = cos x log (sin x) + sin x . 1 (sin ) sin = cos x log (sin x) + cos x = (1 + log (sin x)) cos x CONTINUITY AND DIFFERENTIABILITY 143 d x x dx ⋅ Similarly dx dt = dx dt ≠ 0 only if t ≠ ± 1. Thus for t ≠ ± 1, = 1 1 1 1 a d a t t t dt t − + ⋅ + 2 1 1 1 a a t t t − + ⋅ − Reprint 2025-26 − + Example 43 Differentiate sin2 x w.r.t. e cos x . Solution Let u (x) = sin2 x and v (x) = e cos x . We want to find / / du du dx dv dv dx = . Clearly Thus du dv = cos cos 2sin cos 2cos sin x x x x x x e e = − − Differentiate w.r.t. x the function in Exercises 1 to 11. 1. (3x 2 – 9x + 5)9 2. sin3 x + cos6 x 144 MATHEMATICS 3. (5x) 3 cos 2x 4. sin–1 (x x ), 0 ≤ x ≤ 1 du dx = 2 sin x cos x and dv dx = e cos x (– sin x) = – (sin x) e cos x Miscellaneous Exercise on Chapter 5 dy dy dt dx dx dt = = = a t t t a t t a a a + ⋅ − a t a t t + t t + 1 1 1 log 1 1 1 − − 1 1 1 a 1 2 log 2 10. x x + x a + a x + a a , for some fixed a > 0 and x > 0 5. 6. 1 1 sin 1 sin cot 1 sin 1 sin x x x x − + + − + − − , 0 < x < 2 π 7. (log x) log x , x > 1 8. cos (a cos x + b sin x), for some constant a and b. 9. (sin x – cos x) (sin x – cos x) , 3 4 4 x π π < < 1 cos 2 2 7 x − + , – 2 < x < 2 x Reprint 2025-26 12. Find dy dx , if y = 12 (1 – cos t), x = 10 (t – sin t), 2 2 t π π − < < 13. Find dy dx , if y = sin–1 x + sin–1 2 1− x , 0 < x < 1 14. If x y y x 1 1 0 + + + = , for , – 1 < x < 1, prove that 1 dy dx x = − + 15. If (x – a) 2 + (y – b) 2 = c 2 , for some c > 0, prove that 16. If cos y = x cos (a + y), with cos a ≠ ± 1, prove that 2 cos ( ) sin dy a y dx a + = . 17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find 2 18. If f(x) = | x | 3 , show that f ″(x) exists for all real x and find it. 11. ( ) 2 2 3 3 x x x x − + − , for x > 3 is a constant independent of a and b. + 1 dy dx d y dx 2 ( )2 1 CONTINUITY AND DIFFERENTIABILITY 145 2 3 2 2 2 d y dx . 19. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines. 20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer. 21. If 22. If y = 1 a x cos e − , – 1 ≤ x ≤ 1, show that ( ) 2 2 2 2 1 0 d y dy x x a y dx dx − − − = . f x g x h x ( ) ( ) ( ) y l m n a b c = , prove that Reprint 2025-26 f x g x h x ( ) ( ) ( ) dy l m n dx a b c = ′ ′ ′ 146 MATHEMATICS Summary Æ A real valued function is continuous at a point in its domain if the limit of the function at that point equals the value of the function at that point. A function is continuous if it is continuous on the whole of its domain. Æ Sum, difference, product and quotient of continuous functions are continuous. i.e., if f and g are continuous functions, then Æ Every differentiable function is continuous, but the converse is not true. Æ Chain rule is rule to differentiate composites of functions. If f = v o u, t = u (x) Æ Following are some of the standard derivatives (in appropriate domains): (f ± g) (x) = f(x) ± g(x) is continuous. (f . g) (x) = f(x) . g(x) is continuous. and if both dt dx and dv dt exist then ( ) ( ) ( ) f f x x g g x = (wherever g(x) ≠ 0) is continuous. ( ) 1 2 1 sin 1 d x dx x − = − ( ) 1 2 1 cos 1 d x dx x − = − − df dv dt dx dt dx = ⋅ Æ Logarithmic differentiation is a powerful technique to differentiate functions of the form f(x) = [u (x)]v (x) . Here both f(x) and u(x) need to be positive for this technique to make sense. ( ) 1 2 1 tan 1 d x dx x − = + ( ) d x x e e dx = ( ) 1 log d x dx x = Reprint 2025-26 —v—" class_12,6,Application of Derivatives,ncert_books/class_12/lemh1dd/lemh106.pdf,"6.1 Introduction In Chapter 5, we have learnt how to find derivative of composite functions, inverse trigonometric functions, implicit functions, exponential functions and logarithmic functions. In this chapter, we will study applications of the derivative in various disciplines, e.g., in engineering, science, social science, and many other fields. For instance, we will learn how the derivative can be used (i) to determine rate of change of quantities, (ii) to find the equations of tangent and normal to a curve at a point, (iii) to find turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs. We will also use derivative to find intervals on which a function is increasing or decreasing. Finally, we use the derivative to find approximate value of certain quantities. 6.2 Rate of Change of Quantities Recall that by the derivative ds dt , we mean the rate of change of distance s with respect to the time t. In a similar fashion, whenever one quantity y varies with another v With the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature.” — WHITEHEAD v APPLICATION OF DERIVATIVES Chapter 6 quantity x, satisfying some rule y f x = ( ), then dy dx (or f′(x)) represents the rate of change of y with respect to x and dy dx x x = 0 (or f′(x0 )) represents the rate of change of y with respect to x at 0 x x = . Further, if two variables x and y are varying with respect to another variable t, i.e., if x f t = ( ) and y g t = ( ), then by Chain Rule dy dx = dy dx dt dt , if 0 dx dt ≠ Reprint 2025-26 change of y and that of x both with respect to t. Example 1 Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm. Solution The area A of a circle with radius r is given by A = πr 2 . Therefore, the rate of change of the area A with respect to its radius r is given by A 2 ( ) 2 d d r r dr dr = π = π . When r = 5 cm, A 10 d dr = π . Thus, the area of the circle is changing at the rate of 10π cm2 /s. Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of an edge is 10 centimetres ? Solution Let x be the length of a side, V be the volume and S be the surface area of the cube. Then, V = x 3 and S = 6x 2 , where x is a function of time t. Now dV dt = 9cm3 /s (Given) Therefore 9 = V 3 3 ( ) ( ) d d d dx x x dt dt dx dt = = ⋅ (By Chain Rule) 148 MATHEMATICS Thus, the rate of change of y with respect to x can be calculated using the rate of Let us consider some examples. or dx dt = 2 3 x ... (1) Now dS dt = 2 2 (6 ) (6 ) d d dx x x dt dx dt = ⋅ (By Chain Rule) Hence, when x = 10 cm, 2 3.6 cm /s dS dt = Reprint 2025-26 = 2 3 dx x dt ⋅ = 2 3 36 12x x x ⋅ = (Using (1)) Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing? Solution The area A of a circle with radius r is given by A = πr 2 . Therefore, the rate of change of area A with respect to time t is It is given that dr dt = 4cm/s Therefore, when r = 10 cm, dA dt = 2π (10) (4) = 80π Thus, the enclosed area is increasing at the rate of 80π cm2 /s, when r = 10 cm. Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2cm/minute. When x =10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Solution Since the length x is decreasing and the width y is increasing with respect to time, we have ANote dy dx is positive if y increases as x increases and is negative if y decreases as x increases. dA dt = 2 2 ( ) ( ) d d dr r r dt dr dt π = π ⋅ = 2π r dr dt (By Chain Rule) 3 cm/min dx dt = − and 2 cm/min dy dt = APPLICATION OF DERIVATIVES 149 (a) The perimeter P of a rectangle is given by (b) The area A of the rectangle is given by A = x . y Therefore dP dt = 2 2 3 2 2 dx dt Therefore dA dt = dx dy y x dt dt ⋅ + ⋅ P = 2 (x + y) Reprint 2025-26 = – 3(6) + 10(2) (as x = 10 cm and y = 6 cm) = 2 cm2 /min dy dt + = − + = − ( ) cm/min Example 5 The total cost C(x) in Rupees, associated with the production of x units of an item is given by C(x) = 0.005 x 3 – 0.02 x 2 + 30x + 5000 Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. Solution Since marginal cost is the rate of change of total cost with respect to the output, we have Marginal cost (MC) = 2 0.005(3 ) 0.02(2 ) 30 dC x x dx = − + When x = 3, MC = 2 0.015(3 ) 0.04(3) 30 − + Hence, the required marginal cost is ` 30.02 (nearly). Example 6 The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x 2 + 36x + 5. Find the marginal revenue, when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant. Solution Since marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have Marginal Revenue (MR) = R 6 36 d x dx = + When x = 5, MR = 6(5) + 36 = 66 Hence, the required marginal revenue is ` 66. 150 MATHEMATICS = 0.135 – 0.12 + 30 = 30.015 1. Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm 2. The volume of a cube is increasing at the rate of 8 cm3 /s. How fast is the surface area increasing when the length of an edge is 12 cm? 3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. 4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? 5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? EXERCISE 6.1 Reprint 2025-26 Find the rate of change of its volume with respect to x. 14. Sand is pouring from a pipe at the rate of 12 cm3 /s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm? 15. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x 3 – 0.003x 2 + 15x + 4000. Find the marginal cost when 17 units are produced. 16. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 13x 2 + 26x + 15. Find the marginal revenue when x = 7. Choose the correct answer for questions 17 and 18. 17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π (B) 12π (C) 8π (D) 11π 6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? 7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle. 8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. 9. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm. 10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ? 11. A particle moves along the curve 6y = x 3 +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. 12. The radius of an air bubble is increasing at the rate of 1 2 cm/s. At what rate is the 13. A balloon, which always remains spherical, has a variable diameter 3 (2 1) 2 x + . volume of the bubble increasing when the radius is 1 cm? APPLICATION OF DERIVATIVES 151 Reprint 2025-26 6.3 Increasing and Decreasing Functions In this section, we will use differentiation to find out whether a function is increasing or decreasing or none. Consider the function f given by f (x) = x 2 , x Î R. The graph of this function is a parabola as given in Fig 6.1. as we move from left to right, the height of the graph decreases 152 MATHEMATICS 18. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x 2 + 36x + 5. The marginal revenue, when x = 15 is (A) 116 (B) 96 (C) 90 (D) 126 First consider the graph (Fig 6.1) to the right of the origin. Observe that as we Values left to origin –2 4 3 2 9 4 –1 1 1 2 1 4 0 0 x f (x) = x 2 Fig 6.1 as we move from left to right, the height of the graph increases Values right to origin 1 1 3 2 2 4 0 0 1 2 x f (x) = x 2 1 4 9 4 move from left to right along the graph, the height of the graph continuously increases. For this reason, the function is said to be increasing for the real numbers x > 0. from left to right along the graph, the height of the graph continuously decreases. Consequently, the function is said to be decreasing for the real numbers x < 0. increasing or decreasing on an interval. Definition 1 Let I be an interval contained in the domain of a real valued function f. Then f is said to be (i) increasing on I if x1 < x2 in I Þ f(x1 ) < f(x2 ) for all x1 , x2 Î I. (ii) decreasing on I, if x 1 < x 2 in I Þ f(x 1 ) > f(x 2 ) for all x 1 , x 2 Î I. (iii) constant on I, if f(x) = c for all x Î I, where c is a constant. Now consider the graph to the left of the origin and observe here that as we move We shall now give the following analytical definitions for a function which is Reprint 2025-26 Definition 2 Let x 0 be a point in the domain of definition of a real valued function f. Then f is said to be increasing, decreasing at x0 if there exists an open interval I containing x 0 such that f is increasing, decreasing, respectively, in I. Let us clarify this definition for the case of increasing function. Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R. Solution Let x1 and x2 be any two numbers in R. Then (iv) strictly increasing on I if x1 < x2 in I Þ f (x1 ) < f(x2 ) for all x1 , x2 Î I. (v) strictly decreasing on I if x1 < x2 in I Þ f(x1 ) > f(x2 ) for all x1 , x2 Î I. For graphical representation of such functions see Fig 6.2. Strictly Increasing function We shall now define when a function is increasing or decreasing at a point. (i) Neither Increasing nor Decreasing function (iii) Strictly Decreasing function Fig 6.2 (ii) APPLICATION OF DERIVATIVES 153 We shall now give the first derivative test for increasing and decreasing functions. The proof of this test requires the Mean Value Theorem studied in Chapter 5. Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval (a,b). Then (a) f is increasing in [a,b] if f ¢(x) > 0 for each x Î (a, b) (b) f is decreasing in [a,b] if f ¢(x) < 0 for each x Î (a, b) (c) f is a constant function in [a,b] if f ¢(x) = 0 for each x Î (a, b) Thus, by Definition 1, it follows that f is strictly increasing on R. x1 < x2 Þ 7x1 < 7x2 Þ 7x1 – 3 < 7x2 – 3 Þ f(x1 ) < f(x2 ) Reprint 2025-26 Proof (a) Let x1 , x2 Î [a, b] be such that x1 < x2 . Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c between x1 and x2 such that i.e. f(x2 ) – f(x1 ) > 0 (as f ¢(c) > 0 (given)) i.e. f(x2 ) > f(x1 ) Thus, we have 1 2 1 2 1 2 x x f x f x x x a b ( ) ( ), for all , [ , ] Hence, f is an increasing function in [a,b]. The proofs of part (b) and (c) are similar. It is left as an exercise to the reader. Remarks There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval excluding the end points and f is continuous in the interval, then f is increasing. Similarly, if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the interval, then f is decreasing. Example 8 Show that the function f given by f(x) = x 3 – 3x 2 + 4x, x Î R is increasing on R. Solution Note that 154 MATHEMATICS f(x2 ) – f(x1 ) = f ¢(c) (x2 – x1 ) = 3(x – 1)2 + 1 > 0, in every interval of R Therefore, the function f is increasing on R. Example 9 Prove that the function given by f(x) = cos x is (a) decreasing in (0, p) (b) increasing in (p, 2p), and (c) neither increasing nor decreasing in (0, 2p). f ¢(x) = 3x 2 – 6x + 4 Reprint 2025-26 = 3(x 2 – 2x + 1) + 1 Solution Note that f ′(x) = – sin x Example 10 Find the intervals in which the function f given by f(x) = x 2 – 4x + 6 is Solution We have f (x) = x 2 – 4x + 6 or f ′(x) = 2x – 4 Therefore, f ′(x) = 0 gives x = 2. Now the point x = 2 divides the real line into two disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6.3). In the interval (– ∞, 2), f ′(x) = 2x – 4 < 0. Therefore, f is decreasing in this interval. Also, in the interval (2, ) ∞ , f x ′( ) 0 > and so the function f is increasing in this interval. Example 11 Find the intervals in which the function f given by f (x) = 4x 3 – 6x 2 – 72x + 30 is (a) increasing (b) decreasing. (a) Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in (0, π). (b) Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in (π, 2π). (c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π). (a) increasing (b) decreasing APPLICATION OF DERIVATIVES 155 Fig 6.3 Therefore, f ′(x) = 0 gives x = – 2, 3. The points x = – 2 and x = 3 divides the real line into three disjoint intervals, namely, (– ∞, – 2), (– 2, 3) and (3, ∞). Solution We have f(x) = 4x 3 – 6x 2 – 72x + 30 or f ′(x) = 12x 2 – 12x – 72 = 12(x 2 – x – 6) = 12(x – 3) (x + 2) Reprint 2025-26 Fig 6.4 In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3), f ′(x) is negative. Consequently, the function f is increasing in the intervals (– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3). However, f is neither increasing nor decreasing in R. Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈ 0 2 , π is (a) increasing (b) decreasing. Solution We have f(x) = sin 3x or f ′(x) = 3cos 3x implies 3 3 0, 2 x π ∈ ). So 6 x π = and 2 π . The point 6 x π = divides the interval 0 2 , π into two disjoint intervals 0, 6 π and π π 6 2 , . 156 MATHEMATICS Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives 3 3 , 2 2 x π π = (as x ∈ 0 2 , π (– ∞, – 2) (–) (–) > 0 f is increasing Interval Sign of f ′(x) Nature of function f (– 2, 3) (–) (+) < 0 f is decreasing (3, ∞) (+) (+) > 0 f is increasing all , 6 2 x π π ∈ as 3 3 6 2 2 2 x x π π π π < < ⇒ < < . Now, f x ′( ) 0 > for all 0, 6 x π ∈ as 0 0 3 6 2 x x π π ≤ < ⇒ ≤ < and f x ′( ) 0 < for Therefore, f is increasing in 0, 6 π and decreasing in , 6 2 π π . Reprint 2025-26 Fig 6.5 f is increasing on 0 6 , π and decreasing on π π 6 2 , . Example 13 Find the intervals in which the function f given by f(x) = sin x + cos x, 0 ≤ x ≤ 2π is increasing or decreasing. Solution We have namely, 0, 4 π , π π 4 Also, the given function is continuous at x = 0 and 6 x π = . Therefore, by Theorem 1, or f ′(x) = cos x – sin x Now f x ′( ) 0 = gives sin x = cos x which gives that 4 x π = , 5 4 π as 0 2 ≤ ≤ π x The points 4 x π = and 5 4 x π = divide the interval [0, 2π] into three disjoint intervals, Note that 5 ( ) 0 if 0, ,2 4 4 f x x π π ′ > ∈ ∪ π 5 4 , and 5 ,2 4 π π . f(x) = sin x + cos x, APPLICATION OF DERIVATIVES 157 Fig 6.6 or f is increasing in the intervals 0, 4 π and 5 ,2 4 π π Also ′ < ∈ f x x ( ) 0 , 4 or f is decreasing in π π 4 5 4 if π π 5 4 , Reprint 2025-26 158 MATHEMATICS 1. Show that the function given by f (x) = 3x + 17 is increasing on R. 2. Show that the function given by f (x) = e 2x is increasing on R. 3. Show that the function given by f (x) = sin x is 4. Find the intervals in which the function f given by f(x) = 2x 2 – 3x is (a) increasing (b) decreasing 5. Find the intervals in which the function f given by f(x) = 2x 3 – 3x 2 – 36x + 7 is (a) increasing (b) decreasing 6. Find the intervals in which the following functions are strictly increasing or decreasing: (a) x 2 + 2x – 5 (b) 10 – 6x – 2x 2 (a) increasing in 0, 2 π (b) decreasing in , 2 π π (c) neither increasing nor decreasing in (0, π) 5 ,2 4 π π > 0 f is increasing 5 4 , < 0 f is decreasing Interval Sign of f x ′( ) Nature of function 0, 4 π > 0 f is increasing π π 4 EXERCISE 6.2 7. Show that 2 log(1 ) 2 x y x x = + − + , x > – 1, is an increasing function of x throughout its domain. 8. Find the values of x for which y = [x(x – 2)]2 is an increasing function. 9. Prove that 4sin (2 cos ) y θ = − θ + θ is an increasing function of θ in 0 2 , π . (c) –2x 3 – 9x 2 – 12x + 1 (d) 6 – 9x – x 2 (e) (x + 1)3 (x – 3)3 Reprint 2025-26 10. Prove that the logarithmic function is increasing on (0, ∞). 12. Which of the following functions are decreasing on 0, 2 π ? 14. For what values of a the function f given by f(x) = x 2 + ax + 1 is increasing on [1, 2]? 15. Let I be any interval disjoint from [–1, 1]. Prove that the function f given by 16. Prove that the function f given by f(x) = log sin x is increasing on 0 2 , π and 17. Prove that the function f given by f (x) = log |cos x| is decreasing on 0, 2 π and 11. Prove that the function f given by f(x) = x 2 – x + 1 is neither strictly increasing (A) cos x (B) cos 2x (C) cos 3x (D) tan x 13. On which of the following intervals is the function f given by f(x) = x 100 + sin x –1 decreasing ? (A) (0,1) (B) , 2 π π (C) 0, 2 π (D) None of these nor decreasing on (– 1, 1). decreasing on π π 2 , . 1 f x x ( ) x = + is increasing on I. APPLICATION OF DERIVATIVES 159 6.4 Maxima and Minima In this section, we will use the concept of derivatives to calculate the maximum or minimum values of various functions. In fact, we will find the ‘turning points’ of the graph of a function and thus find points at which the graph reaches its highest (or 18. Prove that the function given by f (x) = x 3 – 3x 2 + 3x – 100 is increasing in R. 19. The interval in which y = x 2 e –x is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2) increasing on 3 , 2 2 π π . Reprint 2025-26 lowest) locally. The knowledge of such points is very useful in sketching the graph of a given function. Further, we will also find the absolute maximum and absolute minimum of a function that are necessary for the solution of many applied problems. Let us consider the following problems that arise in day to day life. (i) The profit from a grove of orange trees is given by P(x) = ax + bx2 , where a,b are constants and x is the number of orange trees per acre. How many trees per acre will maximise the profit? (ii) A ball, thrown into the air from a building 60 metres high, travels along a path and h(x) is the height of the ball . What is the maximum height the ball will reach? (iii) An Apache helicopter of enemy is flying along the path given by the curve f (x) = x 2 + 7. A soldier, placed at the point (1, 2), wants to shoot the helicopter when it is nearest to him. What is the nearest distance? In each of the above problem, there is something common, i.e., we wish to find out the maximum or minimum values of the given functions. In order to tackle such problems, we first formally define maximum or minimum values of a function, points of local maxima and minima and test for determining such points. Definition 3 Let f be a function defined on an interval I. Then (a) f is said to have a maximum value in I, if there exists a point c in I such that f c f x ( ) ( ) > , for all x ∈ I. The number f(c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I. (b) f is said to have a minimum value in I, if there exists a point c in I such that f (c) < f (x), for all x ∈ I. The number f (c), in this case, is called the minimum value of f in I and the point c, in this case, is called a point of minimum value of f in I. (c) f is said to have an extreme value in I if there exists a point c in I such that f (c) is either a maximum value or a minimum value of f in I. The number f (c), in this case, is called an extreme value of f in I and the point c is called an extreme point. 160 MATHEMATICS given by 2 ( ) 60 60 x h x x = + − , where x is the horizontal distance from the building Remark In Fig 6.7(a), (b) and (c), we have exhibited that graphs of certain particular functions help us to find maximum value and minimum value at a point. Infact, through graphs, we can even find maximum/minimum value of a function at a point at which it is not even differentiable (Example 15). Reprint 2025-26 Example 14 Find the maximum and the minimum values, if any, of the function f given by f(x) = x 2 , x ∈ R. Solution From the graph of the given function (Fig 6.8), we have f(x) = 0 if x = 0. Also f(x) ≥ 0, for all x ∈ R. Therefore, the minimum value of f is 0 and the point of minimum value of f is x = 0. Further, it may be observed from the graph of the function that f has no maximum value and hence no point of maximum value of f in R. Example 15 Find the maximum and minimum values of f , if any, of the function given by f(x) = |x |, x ∈ R. ANote If we restrict the domain of f to [– 2, 1] only, then f will have maximum value(– 2)2 = 4 at x = – 2. Fig 6.7 APPLICATION OF DERIVATIVES 161 Fig 6.8 Solution From the graph of the given function (Fig 6.9) , note that f (x) ≥ 0, for all x ∈ R and f(x) = 0 if x = 0. Therefore, the function f has a minimum value 0 and the point of minimum value of f is x = 0. Also, the graph clearly shows that f has no maximum value in R and hence no point of maximum value in R. ANote (i) If we restrict the domain of f to [– 2, 1] only, then f will have maximum value | – 2| = 2. Reprint 2025-26 Fig 6.9 Example 16 Find the maximum and the minimum values, if any, of the function given by f (x) = x, x ∈ (0, 1). Solution The given function is an increasing (strictly) function in the given interval (0, 1). From the graph (Fig 6.10) of the function f , it seems that, it should have the minimum value at a point closest to 0 on its right and the maximum value at a point closest to 1 on its left. Are such points available? Of course, not. It is not possible to locate such points. Infact, if a point x0 is closest to 0, then we find 0 0 2 x < x for all 0 x ∈(0,1). Also, if x1 is closest to 1, then 1 1 1 2 x x + > for all 1 x ∈(0,1) . Therefore, the given function has neither the maximum value nor the minimum value in the interval (0,1). Remark The reader may observe that in Example 16, if we include the points 0 and 1 in the domain of f , i.e., if we extend the domain of f to [0,1], then the function f has minimum value 0 at x = 0 and maximum value 1 at x = 1. Infact, we have the following results (The proof of these results are beyond the scope of the present text) Every monotonic function assumes its maximum/minimum value at the end points of the domain of definition of the function. A more general result is Every continuous function on a closed interval has a maximum and a minimum value. ANote By a monotonic function f in an interval I, we mean that f is either increasing in I or decreasing in I. 162 MATHEMATICS (ii) One may note that the function f in Example 27 is not differentiable at x = 0. Fig 6.10 Maximum and minimum values of a function defined on a closed interval will be discussed later in this section. Let us now examine the graph of a function as shown in Fig 6.11. Observe that at points A, B, C and D on the graph, the function changes its nature from decreasing to increasing or vice-versa. These points may be called turning points of the given function. Further, observe that at turning points, the graph has either a little hill or a little valley. Roughly speaking, the function has minimum value in some neighbourhood (interval) of each of the points A and C which are at the bottom of their respective Reprint 2025-26 valleys. Similarly, the function has maximum value in some neighbourhood of points B and D which are at the top of their respective hills. For this reason, the points A and C may be regarded as points of local minimum value (or relative minimum value) and points B and D may be regarded as points of local maximum value (or relative maximum value) for the function. The local maximum value and local minimum value of the function are referred to as local maxima and local minima, respectively, of the function. We now formally give the following definition Definition 4 Let f be a real valued function and let c be an interior point in the domain of f. Then (a) c is called a point of local maxima if there is an h > 0 such that f (c) ≥ f (x), for all x in (c – h, c + h), x ≠ c The value f(c) is called the local maximum value of f. (b) c is called a point of local minima if there is an h > 0 such that f (c) ≤ f (x), for all x in (c – h, c + h) The value f(c) is called the local minimum value of f . Geometrically, the above definition states that if x = c is a point of local maxima of f, then the graph of f around c will be as shown in Fig 6.12(a). Note that the function f is increasing (i.e., f ′(x) > 0) in the interval (c – h, c) and decreasing (i.e., f ′(x) < 0) in the interval (c, c + h). This suggests that f ′(c) must be zero. Fig 6.11 APPLICATION OF DERIVATIVES 163 Reprint 2025-26 Fig 6.12 shown in Fig 6.14(b). Here f is decreasing (i.e., f ′(x) < 0) in the interval (c – h, c) and increasing (i.e., f ′(x) > 0) in the interval (c, c + h). This again suggest that f ′(c) must be zero. The above discussion lead us to the following theorem (without proof). Theorem 2 Let f be a function defined on an open interval I. Suppose c ∈ I be any point. If f has a local maxima or a local minima at x = c, then either f ′(c) = 0 or f is not differentiable at c. Remark The converse of above theorem need not be true, that is, a point at which the derivative vanishes need not be a point of local maxima or local minima. For example, if f (x) = x 3 , then f ′(x) = 3x 2 and so f ′(0) = 0. But 0 is neither a point of local maxima nor a point of local minima (Fig 6.13). ANote A point c in the domain of a function f at which either f ′(c) = 0 or f is not differentiable is called a critical point off. Note that if f is continuous at c and f ′(c) = 0, then there exists an h > 0 such that f is differentiable in the interval (c – h, c + h). local minima using only the first order derivatives. 164 MATHEMATICS Similarly, if c is a point of local minima of f , then the graph of f around c will be as We shall now give a working rule for finding points of local maxima or points of Fig 6.13 Theorem 3 (First Derivative Test) Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then (i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima. (ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima. (iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection (Fig 6.13). Reprint 2025-26 Example 17 Find all points of local maxima and local minima of the function f given by f(x) = x 3 – 3x + 3. Solution We have f(x) = x 3 – 3x + 3 or f ′(x) = 3x 2 – 3 = 3(x – 1) (x + 1) or f ′(x) = 0 at x = 1 and x = – 1 Thus, x = ± 1 are the only critical points which could possibly be the points of local maxima and/or local minima of f . Let us first examine the point x = 1. Note that for values close to 1 and to the right of 1, f ′(x) > 0 and for values close to 1 and to the left of 1, f ′(x) < 0. Therefore, by first derivative test, x = 1 is a point of local minima and local minimum value is f (1) = 1. In the case of x = –1, note that f ′(x) > 0, for values close to and to the left of –1 and f ′(x) < 0, for values close to and to the right of – 1. Therefore, by first derivative test, x = – 1 is a point of local maxima and local maximum value is f(–1) = 5. ANote If c is a point of local maxima of f , then f(c) is a local maximum value of f. Similarly, if c is a point of local minima of f , then f(c) is a local minimum value of f. Figures 6.13 and 6.14, geometrically explain Theorem 3. Fig 6.14 APPLICATION OF DERIVATIVES 165 Close to 1 to the right (say 1.1 etc.) >0 to the left (say 0.9 etc.) <0 Close to –1 to the right (say 0.9 etc.) 0 to the left (say 1.1 etc.) 0 − < − > Values of x Sign of f ′(x) = 3(x – 1) (x + 1) Reprint 2025-26 Example 18 Find all the points of local maxima and local minima of the function f given by Solution We have f (x) = 2x 3 – 6x 2 + 6x + 5 or f ′(x) = 6x 2 – 12x + 6 = 6 (x – 1)2 or f ′(x) = 0 at x = 1 Thus, x = 1 is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f. Observe that f ′(x) ≥ 0, for all x ∈ R and in particular f ′(x) > 0, for values close to 1 and to the left and to the right of 1. Therefore, by first derivative test, the point x = 1 is neither a point of local maxima nor a point of local minima. Hence x = 1 is a point of inflexion. Remark One may note that since f ′(x), in Example 30, never changes its sign on R, graph of f has no turning points and hence no point of local maxima or local minima. given function. This test is often easier to apply than the first derivative test. Theorem 4 (Second Derivative Test) Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then (i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0 The value f (c) is local maximum value of f . In this case, f (c) is local minimum value of f . (iii) The test fails if f ′(c) = 0 and f ″(c) = 0. In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion. ANote As f is twice differentiable at c, we mean second order derivative of f exists at c. 166 MATHEMATICS (ii) x = c is a point of local minima if f c ′( ) 0 = and f ″(c) > 0 We shall now give another test to examine local maxima and local minima of a f(x) = 2x 3 – 6x 2 + 6x +5. Example 19 Find local minimum value of the function f given by f (x) = 3 + |x |, x ∈ R. Solution Note that the given function is not differentiable at x = 0. So, second derivative test fails. Let us try first derivative test. Note that 0 is a critical point of f . Now to the left of 0, f(x) = 3 – x and so f ′(x) = – 1 < 0. Also to Reprint 2025-26 Fig 6.15 the right of 0, f(x) = 3 + x and so f ′(x) = 1 > 0. Therefore, by first derivative test, x = 0 is a point of local minima of f and local minimum value of f is f (0) = 3. Example 20 Find local maximum and local minimum values of the function f given by Solution We have f (x) = 3x 4 + 4x 3 – 12x 2 + 12 or f ′(x) = 12x 3 + 12x 2 – 24x = 12x (x – 1) (x + 2) or f ′(x) = 0 at x = 0, x = 1 and x = – 2. Now f ″(x) = 36x 2 + 24x – 24 = 12 (3x 2 + 2x – 2) or maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively. Example 21 Find all the points of local maxima and local minima of the function f given by Solution We have f(x) = 2x 3 – 6x 2 + 6x +5 Therefore, by second derivative test, x = 0 is a point of local maxima and local f (x) = 3x 4 + 4x 3 – 12x 2 + 12 f(x) = 2x 3 – 6x 2 + 6x +5. f f f ′′ = − < ′′ = > ′′ − = > ( ) ( ) ( ) 0 24 0 1 36 0 2 72 0 APPLICATION OF DERIVATIVES 167 or 2 2 ( ) 6 12 6 6( 1) fails in this case. So, we shall go back to the first derivative test. We have already seen (Example 18) that, using first derivative test, x =1 is neither a point of local maxima nor a point of local minima and so it is a point of inflexion. Example 22 Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. Solution Let one of the numbers be x. Then the other number is (15 – x). Let S(x) denote the sum of the squares of these numbers. Then Now f ′(x) = 0 gives x =1. Also f ″(1) = 0. Therefore, the second derivative test f x x ′ = − + = − ′′ = − ( ) 12( 1) f x x x x Reprint 2025-26 or S ( ) 4 30 test, 15 2 x = is the point of local minima of S. Hence the sum of squares of numbers is minimum when the numbers are 15 2 and 15 15 15 2 2 − = . Remark Proceeding as in Example 34 one may prove that the two positive numbers, whose sum is k and the sum of whose squares is minimum, are and 2 2 k k . Example 23 Find the shortest distance of the point (0, c) from the parabola y = x 2 , where 1 2 ≤ c ≤ 5. Solution Let (h, k) be any point on the parabola y = x 2 . Let D be the required distance between (h, k) and (0, c). Then 168 MATHEMATICS Now S′(x) = 0 gives 15 2 x = . Also 15 S 4 0 2 ′′ = > . Therefore, by second derivative 2 2 2 2 D ( 0) ( ) ( ) = − + − = + − h k c h k c ... (1) S(x) = x 2 + (15 – x) 2 = 2x 2 – 30x + 225 x ′ = − ′′ = S ( ) 4 x x or D′(k) = 2 1 2( ) Now D′(k) = 0 gives 2 1 2 c k − = 2 1 2 c k − > , then D ( ) 0 ′ > k . So, by first derivative test, D (k) is minimum at 2 1 2 c k − = . Since (h, k) lies on the parabola y = x 2 , we have k = h 2 . So (1) gives Observe that when 2 1 2 c k − < , then 2( ) 1 0 k c − + < , i.e., D ( ) 0 ′ < k . Also when D ≡ D(k) = 2 k k c + − ( ) Reprint 2025-26 2 ( ) k c k k c + − + − Hence, the required shortest distance is given by Example 24 Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m, BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP2 + RQ2 is minimum. Solution Let R be a point on AB such that AR = x m. Then RB = (20 – x) m (as AB = 20 m). From Fig 6.16, we have RP2 = AR2 + AP2 and RQ2 = RB2 + BQ2 Therefore RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2 = x 2 + (16)2 + (20 – x) 2 + (22)2 = 2x 2 – 40x + 1140 Let S ≡ S(x) = RP2 + RQ2 = 2x 2 – 40x + 1140. Therefore S′(x) = 4x – 40. Now S′(x) = 0 gives x = 10. Also S″(x) = 4 > 0, for all x and so S″(10) > 0. Therefore, by second derivative test, x = 10 is the point of local minima of S. Thus, the distance of R from A on AB is AR = x =10 m. ANote The reader may note that in Example 35, we have used first derivative test instead of the second derivative test as the former is easy and short. 2 2 1 2 1 2 1 4 1 D 2 2 2 2 c c c c c − − − − = + − = APPLICATION OF DERIVATIVES 169 Fig 6.16 Example 25 If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum. Solution The required trapezium is as given in Fig 6.17. Draw perpendiculars DP and Reprint 2025-26 Fig 6.17 CQ on AB. Let AP = x cm. Note that ∆APD ~ ∆BQC. Therefore, QB = x cm. Also, by Pythagoras theorem, DP = QC = 2 100 − x . Let A be the area of the trapezium. Then or A′(x) = ( ) 2 2 ( 2 ) ( 10) 100 2 100 x x x x − + + − − Now A′(x) = 0 gives 2x 2 + 10x – 100 = 0, i.e., x = 5 and x = –10. Since x represents distance, it can not be negative. So, x = 5. Now 170 MATHEMATICS A″(x) = = 3 3 2 2 2 300 1000 ( 2 ) 100 ( 4 10) ( 2 10 100) 2 100 100 x x − − − − − − − − − + − − A ≡ A(x) = 1 2 (sum of parallel sides) (height) x x x x x x x 2 2 2 2 = ( ) 1 2 (2 10 10) 100 2 x + + − x = ( ) 2 ( 10) 100 x + − x = 2 (on simplification) x − − + 2 2 10 100 100 x x − or A″(5) = 3 Thus, area of trapezium is maximum at x = 5 and the area is given by Example 26 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Solution Let OC = r be the radius of the cone and OA = h be its height. Let a cylinder with radius OE = x inscribed in the given cone (Fig 6.18). The height QE of the cylinder is given by A(5) = 2 2 (5 10) 100 (5) 15 75 75 3 cm + − = = 3 2 2 2(5) 300(5) 1000 2250 30 0 75 75 75 (100 (5) ) (100 ) − − − − = = < − − x Reprint 2025-26 or QE h = r x r − or QE = h r x ( ) r − Let S be the curved surface area of the given cylinder. Then or point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. 6.4.1 Maximum and Minimum Values of a Function in a Closed Interval Let us consider a function f given by Now S′(x) = 0 gives 2 r x = . Since S″(x) < 0 for all x, S 0 2 r ′′ < . So 2 r x = is a QE OA = EC OC (since ∆QEC ~ ∆AOC) S ≡ S(x) = 2 ( ) xh r x r π − = 2 2 ( ) h rx x r π − π ′ = − − π ′′ = 2 S ( ) ( 2 ) 4 S ( ) h x r x r h x r APPLICATION OF DERIVATIVES 171 Fig 6.18 nor has a minimum value. Further, we may note that the function even has neither a local maximum value nor a local minimum value. not have a local maximum (minimum) values but it certainly does have maximum value 3 = f(1) and minimum value 2 = f(0). The maximum value 3 of f at x = 1 is called absolute maximum value (global maximum or greatest value) of f on the interval [0, 1]. Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum value (global minimum or least value) of f on [0, 1]. interval [a, d]. Observe that the function f has a local minima at x = b and local Observe that the function is continuous on (0, 1) and neither has a maximum value However, if we extend the domain of f to the closed interval [0, 1], then f still may Consider the graph given in Fig 6.19 of a continuous function defined on a closed f(x) = x + 2, x ∈ (0, 1) Reprint 2025-26 minimum value is f(b). The function also has a local maxima at x = c and local maximum value is f (c). Also from the graph, it is evident that f has absolute maximum value f (a) and absolute minimum value f (d). Further note that the absolute maximum (minimum) value of f is different from local maximum (minimum) value of f. We will now state two results (without proof) regarding absolute maximum and absolute minimum values of a function on a closed interval I. Theorem 5 Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I. Theorem 6 Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then (i) f ′(c) = 0 if f attains its absolute maximum value at c. (ii) f ′(c) = 0 if f attains its absolute minimum value at c. In view of the above results, we have the following working rule for finding absolute maximum and/or absolute minimum values of a function in a given closed interval [a, b]. 172 MATHEMATICS Fig 6.19 Working Rule Step 1: Find all critical points of f in the interval, i.e., find points x where either f x ′( ) 0 = or f is not differentiable. Step 2: Take the end points of the interval. Step 3: At all these points (listed in Step 1 and 2), calculate the values of f . Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f . Reprint 2025-26 Example 27 Find the absolute maximum and minimum values of a function f given by f (x) = 2x 3 – 15x 2 + 36x +1 on the interval [1, 5]. Solution We have f(x) = 2x 3 – 15x 2 + 36x + 1 or f ′(x) = 6x 2 – 30x + 36 = 6 (x – 3) (x – 2) Note that f ′(x) = 0 gives x = 2 and x = 3. We shall now evaluate the value of f at these points and at the end points of the interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So f(1) = 2(13 ) – 15(12 ) + 36 (1) + 1 = 24 f(2) = 2(23 ) – 15(22 ) + 36 (2) + 1 = 29 f(3) = 2(33 ) – 15(32 ) + 36 (3) + 1 = 28 f(5) = 2(53 ) – 15(52 ) + 36 (5) + 1 = 56 Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1. Example 28 Find absolute maximum and minimum values of a function f given by 4 1 3 3 f x x x x ( ) 12 6 , [ 1, 1] = − ∈ − Solution We have or f ′(x) = 1 3 2 2 3 3 2 2 (8 1) 16 x x f (x) = 4 1 3 3 12 6 x x − x x − − = APPLICATION OF DERIVATIVES 173 critical points are x = 0 and 1 8 x = . Now evaluating the value of f at critical points x = 0, 1 8 and at end points of the interval x = –1 and x = 1, we have Thus, f ′(x) = 0 gives 1 8 x = . Further note that f ′(x) is not defined at x = 0. So the f (–1) = 4 1 3 3 12 ( 1) 6 ( 1) 18 − − − = f (0) = 12 (0) – 6 (0) = 0 Reprint 2025-26 and absolute minimum value of f is 9 4 − that occurs at 1 8 x = . Example 29 An Apache helicopter of enemy is flying along the curve given by y = x 2 + 7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance. Solution For each value of x, the helicopter’s position is at point (x, x 2 + 7). Therefore, the distance between the helicopter and the soldier placed at (3,7) is Let f (x) = (x – 3)2 + x 4 or f ′(x) = 2(x – 3) + 4x 3 = 2 (x – 1) (2x 2 + 2x + 3) Also, there are no end points of the interval to be added to the set for which f ′ is zero, i.e., there is only one point, namely, x = 1. The value of f at this point is given by f (1) = (1 – 3)2 + (1)4 = 5. Thus, the distance between the solider and the helicopter is 174 MATHEMATICS f (1) 5 = . Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1 Thus, f ′(x) = 0 gives x = 1 or 2x 2 + 2x + 3 = 0 for which there are no real roots. 2 2 2 ( 3) ( 7 7) x x − + + − , i.e., 2 4 ( 3) x x − + . 1 8 f = f (1) = 4 1 3 3 12(1) 6(1) 6 − = 4 1 1 1 9 3 3 12 6 8 8 4 − − = Note that 5 is either a maximum value or a minimum value. Since it follows that 5 is the minimum value of f x( ) . Hence, 5 is the minimum distance between the soldier and the helicopter. 1. Find the maximum and minimum values, if any, of the following functions given by (i) f (x) = (2x – 1)2 + 3 (ii) f(x) = 9x 2 + 12x + 2 (iii) f(x) = – (x – 1)2 + 10 (iv) g(x) = x 3 + 1 f (0) = 2 4 (0 3) (0) 3 5 − + = > , EXERCISE 6.3 Reprint 2025-26 2. Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = | x + 2 | – 1 (ii) g(x) = – | x + 1| + 3 (iii) h(x) = sin(2x) + 5 (iv) f(x) = |sin 4x + 3| (v) h(x) = x + 1, x ∈ (– 1, 1) 3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i) f(x) = x 2 (ii) g(x) = x 3 – 3x (vii) 2 1 ( ) 2 g x x = + (viii) f x x x x ( ) 1 , 0 1 = − < < 4. Prove that the following functions do not have maxima or minima: (i) f(x) = e x (ii) g(x) = log x (iii) h (x) = x 3 + x 2 + x +1 5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f(x) = x 3 , x ∈ [– 2, 2] (ii) f (x) = sin x + cos x , x ∈ [0, π] (iv) f(x) = sin x – cos x, 0 2 < < π x (iii) h(x) = sin x + cos x, 0 2 x π < < (v) f(x) = x 3 – 6x 2 + 9x + 15 (vi) 2 ( ) , 0 2 x g x x x = + > APPLICATION OF DERIVATIVES 175 7. Find both the maximum value and the minimum value of 3x 4 – 8x 3 + 12x 2 – 48x + 25 on the interval [0, 3]. 8. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value? 9. What is the maximum value of the function sin x + cos x? 10. Find the maximum value of 2x 3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1]. 6. Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 – 72x – 18x 2 (iii) f (x) = 1 2 9 4 , 2, 2 2 x x x − ∈ − (iv) 2 f x x x ( ) ( 1) 3, [ 3,1] = − + ∈ − Reprint 2025-26 176 MATHEMATICS 11. It is given that at x = 1, the function x 4 – 62x 2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a. 12. Find the maximum and minimum values of x + sin 2x on [0, 2π]. 13. Find two numbers whose sum is 24 and whose product is as large as possible. 14. Find two positive numbers x and y such that x + y = 60 and xy3 is maximum. 15. Find two positive numbers x and y such that their sum is 35 and the product x 2 y 5 is a maximum. 16. Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. 17. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible. 18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ? 19. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. 20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base. 21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area? 22. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? 23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 27 of the volume of the sphere. 24. Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 time the radius of the base. 25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is 1 tan 2 − . 26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin− 1 1 3 . Reprint 2025-26 Choose the correct answer in Questions 27 and 29. Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in metres, covered by it, in t seconds is given by Find the time taken by it to reach Q and also find distance between P and Q. 27. The point on the curve x 2 = 2y which is nearest to the point (0, 5) is 28. For all real values of x, the minimum value of 2 29. The maximum value of 1 3 [ ( 1) 1] x x − + , 0 1 ≤ ≤x is (A) (2 2,4) (B) (2 2,0) (C) (0, 0) (D) (2, 2) (A) 0 (B) 1 (C) 3 (D) 1 3 (A) 1 3 1 3 (B) 1 2 (C) 1 (D) 0 Miscellaneous Examples x t t = − 2 2 3 APPLICATION OF DERIVATIVES 177 2 1 1 x x x x − + + + is Solution Let v be the velocity of the car at t seconds. Now x = 2 2 3 t t − Therefore v = dx dt = 4t – t 2 = t(4 – t) Thus, v = 0 gives t = 0 and/or t = 4. reach the point Q after 4 seconds. Also the distance travelled in 4 seconds is given by Now v = 0 at P as well as at Q and at P, t = 0. So, at Q, t = 4. Thus, the car will x] t = 4 = 2 4 2 32 4 2 16 m 3 3 3 − = = Reprint 2025-26 Example 31 A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan–1 (0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m. Solution Let r, h and α be as in Fig 6.20. Then . tan r h α = So α = 1 tan r h − . But α = tan–1 (0.5) (given) or r h = 0.5 or r = 2 h Let V be the volume of the cone. Then Therefore dV dt = 178 MATHEMATICS V = 2 3 1 1 2 3 3 2 12 h h r h h π π = π = = 2 4 12 d h dh dh dt π ⋅ (by Chain Rule) dh h dt π 3 Fig 6.20 Now rate of change of volume, i.e., V 5 d dt = m3 /h and h = 4 m. Therefore 5 = 2 (4) 4 or dh dt = 5 35 22 m/h 4 88 7 = π = π Thus, the rate of change of water level is 35 m/h 88 . Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases. Reprint 2025-26 dh dt π ⋅ Solution In Fig 6.21, Let AB be the lamp-post, the lamp being at the position B and let MN be the man at a particular time t and let AM = l metres. Then, MS is the shadow of the man. Let MS = s metres. Note that ∆MSN ~ ∆ASB or MS AS = MN AB or AS = 3s (as MN = 2 and AB = 6 (given)) Thus AM = 3s – s = 2s. But AM = l So l = 2s Therefore dl dt = 2 ds dt Since 5 dl dt = km/h. Hence, the length of the shadow increases at the rate 5 2 km/h. Example 33 Find intervals in which the function given by is (a) increasing (b) decreasing. f (x) = 3 4 36 4 3 2 3 11 10 5 5 x x x x − − + + APPLICATION OF DERIVATIVES 179 Fig 6.21 Solution We have Therefore f ′(x) = 3 4 3 2 36 (4 ) (3 ) 3(2 ) 10 5 5 x x x − − + f (x) = 3 4 36 4 3 2 3 11 10 5 5 x x x x − − + + Reprint 2025-26 = 6 ( 1)( 2)( 3) 5 x x x − + − (on simplification) Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3. The points x = 1, – 2, and 3 divide the real line into four disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3) and (3, ∞) (Fig 6.22). Consider the interval (– ∞, – 2), i.e., when – ∞ < x < – 2. In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0. (– 6) < 0) Therefore, f ′(x) < 0 when – ∞ < x < – 2. Thus, the function f is decreasing in (– ∞, – 2). Consider the interval (– 2, 1), i.e., when – 2 < x < 1. In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0 (In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3) = 6 > 0) So f ′(x) > 0 when – 2 < x < 1. Thus, f is increasing in (– 2, 1). Now consider the interval (1, 3), i.e., when 1 < x < 3. In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 < 0. So, f ′(x) < 0 when 1 < x < 3. Thus, f is decreasing in (1, 3). 180 MATHEMATICS (In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1) Finally, consider the interval (3, ∞), i.e., when x > 3. In this case, we have x – 1 > 0, Fig 6.22 x + 2 > 0 and x – 3 > 0. So f ′(x) > 0 when x > 3. Thus, f is increasing in the interval (3, ∞). Example 34 Show that the function f given by is always an increasing function in 0 4 , π . Solution We have f (x) = tan–1(sin x + cos x), x > 0 Therefore f ′(x) = 2 1 (cos sin ) 1 (sin cos ) x x x x − + + f(x) = tan–1(sin x + cos x), x > 0 Reprint 2025-26 Note that 2 + sin 2x > 0 for all x in 0, 4 π . Therefore f ′(x) > 0 if cos x – sin x > 0 or f ′(x) > 0 if cos x > sin x or cot x > 1 Now cot x > 1 if tan x < 1, i.e., if 0 4 x π < < Thus f ′(x) > 0 in 0 4 , π Hence f is increasing function in 0, 4 π . Example 35 A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm. Solution Let r be the radius of the given disc and A be its area. Then or dA dt = 2 dr r dt π (by Chain Rule) A = πr 2 = cos sin 2 sin 2 x x x − + (on simplification) APPLICATION OF DERIVATIVES 181 Now approximate rate of increase of radius = dr = 0.05 dr t dt ∆ = cm/s. Therefore, the approximate rate of increase in area is given by Example 36 An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box. dA = A ( ) d t dt ∆ = 2 dr r t dt π ∆ Reprint 2025-26 = 2π (3.2) (0.05) = 0.320π cm2 /s (r = 3.2 cm) Solution Let x metre be the length of a side of the removed squares. Then, the height of the box is x, length is 8 – 2x and breadth is 3 – 2x (Fig 6.23). If V(x) is the volume of the box, then Therefore 2 V ( ) 12 44 24 4( 3)(3 2) Now V′(x) = 0 gives 2 3, 3 x = . But x ≠ 3 (Why?) Thus, we have 2 3 x = . Now 2 2 V 24 44 28 0 3 3 ′′ = − = − < . Therefore, 2 3 x = is the point of maxima, i.e., if we remove a square of side 2 3 metre from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by 182 MATHEMATICS x x ′ = − + = − − ′′ = − V ( ) 24 44 x x x x x V(x) = x (3 – 2x) (8 – 2x) = 4x 3 – 22x 2 + 24x Fig 6.23 Example 37 Manufacturer can sell x items at a price of rupees 5 100 − x each. The cost price of x items is Rs x 5 + 500 . Find the number of items he should sell to earn maximum profit. 2 V 3 = Reprint 2025-26 = 200 3 m 27 3 2 2 2 2 4 22 24 3 3 3 − + Solution Let S(x) be the selling price of x items and let C(x) be the cost price of x items. Then, we have and C(x) = 500 5 x + Thus, the profit function P(x) is given by i.e. P(x) = 2 24 500 5 100 x x − − or P′(x) = 24 5 50 x − Now P′(x) = 0 gives x = 240. Also 1 P ( ) 50 x − ′′ = . So 1 P (240) 0 50 − ′′ = < profit, if he sells 240 items. 1. Show that the function given by log ( ) x f x x = has maximum at x = e. Thus, x = 240 is a point of maxima. Hence, the manufacturer can earn maximum Miscellaneous Exercise on Chapter 6 S(x) = 2 5 5 100 100 x x x x − = − P(x) = 2 S( ) C ( ) 5 500 100 5 x x x x x − = − − − APPLICATION OF DERIVATIVES 183 2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ? 3. Find the intervals in which the function f given by 4. Find the intervals in which the function f given by 3 3 1 f x x x ( ) , 0 x = + ≠ is 4sin 2 cos ( ) 2 cos x x x x f x x − − = + is (i) increasing (ii) decreasing. (i) increasing (ii) decreasing. Reprint 2025-26 184 MATHEMATICS 10. Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has (i) local maxima (ii) local minima (iii) point of inflexion 11. Find the absolute maximum and minimum values of the function f given by f (x) = cos2 x + sin x, x ∈ [0, π] 12. Show that the altitude of the right circular cone of maximum volume that can be 5. Find the maximum area of an isosceles triangle inscribed in the ellipse 2 2 with its vertex at one end of the major axis. 6. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3 . If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank? 7. The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle. 8. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. 9. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is 2 2 3 3 3 2 ( ) a b + . 2 2 1 x y a b + = 13. Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b). 14. Show that the height of the cylinder of maximum volume that can be inscribed in 15. Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the inscribed in a sphere of radius r is 4 3 r . a sphere of radius R is 2R 3 . Also find the maximum volume. cone and the greatest volume of cylinder is 4 3 2 tan 27 π α h . Reprint 2025-26 16. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h Æ If a quantity y varies with another quantity x, satisfying some rule y f x = ( ), 0 x x = . Æ If two variables x and y are varying with respect to another variable t, i.e., if x f t = ( ) and y g t = ( ), then by Chain Rule Æ A function f is said to be (a) increasing on an interval (a, b) if x 1 < x 2 in (a, b) ⇒ f(x 1 ) < f(x 2 ) for all x 1 , x 2 ∈ (a, b). Alternatively, if f ′(x) ≥ 0 for each x in (a, b) then dy dx (or f x ′( ) ) represents the rate of change of y with respect to x and x x0 dy dx = (or 0 f x ′( )) represents the rate of change of y with respect to x at dy dy dx dx dt dt = , if 0 dx dt ≠ . Summary APPLICATION OF DERIVATIVES 185 (b) decreasing on (a,b) if x1 < x2 in (a, b) ⇒ f(x1 ) > f(x2 ) for all x1 , x2 ∈ (a, b). (c) constant in (a, b), if f (x) = c for all x ∈ (a, b), where c is a constant. Æ A point c in the domain of a function f at which either f ′(c) = 0 or f is not differentiable is called a critical point of f. Æ First Derivative Test Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then (i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima. Reprint 2025-26 186 MATHEMATICS (ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima. (iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflexion. Æ Second Derivative Test Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then (i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0 In this case, we go back to the first derivative test and find whether c is a point of maxima, minima or a point of inflexion. Æ Working rule for finding absolute maxima and/or absolute minima Step 1: Find all critical points of f in the interval, i.e., find points x where either f ′(x) = 0 or f is not differentiable. Step 2:Take the end points of the interval. Step 3: At all these points (listed in Step 1 and 2), calculate the values of f . The values f (c) is local maximum value of f . (ii) x = c is a point of local minima if f ′(c) = 0 and f ″(c) > 0 In this case, f (c) is local minimum value of f . (iii) The test fails if f ′(c) = 0 and f ″(c) = 0. Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f . Reprint 2025-26 —v—" class_12,7,Integrals,ncert_books/class_12/lemh2dd/lemh201.pdf,"7.1 Introduction Differential Calculus is centred on the concept of the derivative. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions. If a function f is differentiable in an interval I, i.e., its derivative f ′exists at each point of I, then a natural question arises that given f ′at each point of I, can we determine the function? The functions that could possibly have given function as a derivative are called anti derivatives (or primitive) of the function. Further, the formula that gives all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration. Such type of problems arise in many practical situations. For instance, if we know the instantaneous velocity of an object at any instant, then there arises a natural question, i.e., can we determine the position of the object at any instant? There are several such practical and theoretical situations where the process of integration is involved. The development of integral calculus arises out of the efforts of solving the problems of the following types: (a) the problem of finding a function whenever its derivative is given, (b) the problem of finding the area bounded by the graph of a function under certain conditions. These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus. v Just as a mountaineer climbs a mountain – because it is there, so a good mathematics student studies new material because it is there. — JAMES B. BRISTOL v INTEGRALS Chapter 7 INTEGRALS 225 Reprint 2025-26 G .W. Leibnitz (1646 -1716) 226 MATHEMATICS There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability. In this Chapter, we shall confine ourselves to the study of indefinite and definite integrals and their elementary properties including some techniques of integration. 7.2 Integration as an Inverse Process of Differentiation Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation. Let us consider the following examples: We know that (sin ) d x dx = cos x ... (1) and ( ) d x e dx = e x ... (3) that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3), 3 e x are the anti derivatives (or integrals) of x 2 and e x , respectively. Again, we note that for any real number C, treated as constant function, its derivative is zero and hence, we can write (1), (2) and (3) as follows : We observe that in (1), the function cos x is the derived function of sin x. We say 3 ( ) 3 d x dx = x 2 ... (2) 3 x and (sin + C) cos = d x x dx , 3 2 ( + C) 3 = d x x dx and ( + C) = d x x e e dx Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by choosing C arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. More generally, if there is a function F such that F ( ) = ( ) d x f x dx , ∀ x ∈ I (interval), then for any arbitrary real number C, (also called constant of integration) [ ] F ( ) + C d x dx = f(x), x ∈ I Reprint 2025-26 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Remark Functions with same derivatives differ by a constant. To show this, let g and h be two functions having the same derivatives on an interval I. Consider the function f = g – h defined by f(x) = g(x) – h(x), ∀ x ∈ I Then df dx = f′ = g′ – h′ giving f′ (x) = g′ (x) – h′(x) ∀ x ∈ I or f′ (x) = 0, ∀ x ∈ I by hypothesis, i.e., the rate of change of f with respect to x is zero on I and hence f is constant. In view of the above remark, it is justified to infer that the family {F + C, C ∈ R} provides all possible anti derivatives of f. class of anti derivatives read as the indefinite integral of f with respect to x. Symbolically, we write f x dx x ( ) = F ( ) + C ∫ . Notation Given that ( ) dy f x dx = , we write y = f x dx ( ) ∫ . For the sake of convenience, we mention below the following symbols/terms/phrases with their meanings as given in the Table (7.1). We introduce a new symbol, namely, f x dx ( ) ∫ which will represent the entire Symbols/Terms/Phrases Meaning Table 7.1 INTEGRALS 227 f(x) in f x dx ( ) ∫ Integrand x in f x dx ( ) ∫ Variable of integration Integrate Find the integral An integral of f A function F such that F′(x) = f (x) Integration The process of finding the integral Constant of Integration Any real number C, considered as constant function f x dx ( ) ∫ Integral of f with respect to x Reprint 2025-26 228 MATHEMATICS We already know the formulae for the derivatives of many important functions. From these formulae, we can write down immediately the corresponding formulae (referred to as standard formulae) for the integrals of these functions, as listed below which will be used to find integrals of other functions. (iv) ( ) 2 tan sec d x x dx = ; 2 sec tan C x dx x = + ∫ (vi) ( ) sec sec tan d x x x dx = ; sec tan sec C x x dx x = + ∫ (iii) ( ) – cos sin d x x dx = ; sin cos C x dx – x = + ∫ (v) ( ) 2 – cot cosec d x x dx = ; 2 cosec cot C x dx – x = + ∫ (ii) ( ) sin cos d x x dx = ; cos sin C x dx x = + ∫ (i) 1 Derivatives Integrals (Anti derivatives) Particularly, we note that 1 n d x n x dx n ( ) 1 d x dx = ; dx x = + C ∫ + = + ; + = + + ∫ , n ≠ –1 1 C 1 n n x x dx n (viii) ( ) – 1 2 1 sin 1 d x dx – x = ; – 1 2 sin C 1 dx x – x = + ∫ (vii) ( ) – cosec cosec cot d x x x dx = ; cosec cot – cosec C x x dx x = + ∫ (ix) ( ) – 1 2 1 – cos 1 d x dx – x = ; – 1 2 cos C 1 dx – x – x = + ∫ (xi) ( ) d x x e e dx = ; C x x e dx e = + ∫ (x) ( ) – 1 2 1 tan 1 d x dx x = + ; – 1 2 tan C 1 dx x x = + + ∫ Reprint 2025-26 7.2.1 Some properties of indefinite integral In this sub section, we shall derive some properties of indefinite integrals. (I) The process of differentiation and integration are inverses of each other in the sense of the following results : ANote In practice, we normally do not mention the interval over which the various functions are defined. However, in any specific problem one has to keep it in mind. (xiii) (xii) ( ) 1 log | | d x dx x = ; 1 dx x log | | C x = + ∫ and f x dx ′( ) ∫ = f(x) + C, where C is any arbitrary constant. Proof Let F be any anti derivative of f, i.e., Then f x dx ( ) ∫ = F(x) + C Therefore ( ) d f x dx dx ∫ = ( ) F ( ) + C d x dx x d a x a dx log a = ; C x x a a dx log a = + ∫ ( ) d f x dx dx ∫ = f (x) F( ) d x dx = f (x) INTEGRALS 229 where C is arbitrary constant called constant of integration. (II) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. = F ( ) = ( ) d x f x dx Similarly, we note that and hence f x dx ′( ) ∫ = f (x) + C f ′(x) = ( ) d f x dx Reprint 2025-26 230 MATHEMATICS (III) [ f x g x dx f x dx g x dx ( ) + ( ) ( ) + ( ) ] = ∫ ∫ ∫ A Note The equivalence of the families {∫ f x dx ( ) + C ,C1 1 ∈R} and without mentioning the parameter. {∫ g x dx ( ) + C ,C2 2 ∈R} is customarily expressed by writing f x dx g x dx ( ) = ( ) ∫ ∫ , Proof Let f and g be two functions such that or ( ) ( ) d f x dx – g x dx dx ∫ ∫ = 0 Hence f x dx – g x dx ( ) ( ) ∫ ∫ = C, where C is any real number (Why?) or f x dx ( ) ∫ = g x dx ( ) C+ ∫ So the families of curves {∫ f x dx ( ) C , C R + ∈ 1 1 } and {∫ g x dx ( ) C , C R + ∈ 2 2 } are identical. Hence, in this sense, f x dx g x dx ( ) and ( ) ∫ ∫ are equivalent. ( ) d f x dx dx ∫ = ( ) d g x dx dx ∫ (IV) For any real number k, k f x dx k f x dx ( ) ( ) = ∫ ∫ Thus, in view of Property (II), it follows by (1) and (2) that Proof By Property (I), we have On the otherhand, we find that ( ) + ( ) d f x dx g x dx dx ∫ ∫ = ( ) + ( ) d d f x dx g x dx dx dx ∫ ∫ [ ( ) + ( )] d f x g x dx dx ∫ = f(x) + g (x) ... (1) ( f x g x dx ( ) ( ) + ) ∫ = f x dx g x dx ( ) ( ) + ∫ ∫ . Reprint 2025-26 = f(x) + g (x) ... (2) To find an anti derivative of a given function, we search intuitively for a function whose derivative is the given function. The search for the requisite function for finding an anti derivative is known as integration by the method of inspection. We illustrate it through some examples. Example 1 Write an anti derivative for each of the following functions using the method of inspection: Solution (i) We look for a function whose derivative is cos 2x. Recall that (V) Properties (III) and (IV) can be generalised to a finite number of functions f 1 , f 2 , ..., f n and the real numbers, k1 , k2 , ..., kn giving (i) cos 2x (ii) 3x 2 + 4x 3 (iii) 1 x , x ≠ 0 Therefore, using the Property (II), we have k f x dx k f x dx ( ) ( ) = ∫ ∫ . Proof By the Property (I), ( ) ( ) d k f x dx k f x dx = ∫ . Also ( ) d k f x dx dx ∫ = ( ) = ( ) d k f x dx k f x dx ∫ [k f x k f x ... k f x dx 1 1 2 2 ( ) ( ) ( ) + + + n n ] ∫ = 1 1 2 2 ( ) ( ) ( ) n n k f x dx k f x dx ... k f x dx + + + ∫ ∫ ∫ . INTEGRALS 231 (ii) We look for a function whose derivative is 3x 2 + 4x 3 . Note that or cos 2x = 1 2 Therefore, an anti derivative of cos 2x is 1 sin 2 2 x . Therefore, an anti derivative of 3x 2 + 4x 3 is x3 + x 4 . ( ) d 3 4 x x dx + = 3x 2 + 4x 3 . d dx (sin 2x) = 1 sin 2 2 d x dx d dx sin 2x = 2 cos 2x Reprint 2025-26 232 MATHEMATICS Example 2 Find the following integrals: Solution (i) We have (iii) We know that (i) 3 Combining above, we get ( ) 1 log 0 d x , x dx x = ≠ Therefore, 1 dx x log x = ∫ is one of the anti derivatives of 1 x . 2 x – 1 dx x ∫ (ii) 2 3 ( 1) x dx + ∫ (iii) ∫ 3 2 2 x – 1 – dx x dx – x dx x = ∫ ∫ ∫ (by Property V) 1 1 1 (log ) 0 and [log ( )] ( 1) 0 d d x , x – x – , x dx x dx – x x = > = = < = = 2 1 C C 1 2 2 1 – x x – – – + = 2 1 2 1 + C C 2 x – x + 1 1 2 1 C1 C2 1 1 2 1 – x x – – + + + + + + ; C1 , C2 are constants of integration 3 2 1 ( 2 – ) + x x e dx x ANote From now onwards, we shall write only one constant of integration in the final answer. (ii) We have 2 2 3 3 ( 1) x dx x dx dx + = + ∫ ∫ ∫ = 2 1 + C 2 x x + , where C = C1 – C2 is another constant of integration. = x x + + + + = 5 3 3 C 5 x x + + 2 1 3 C 2 1 3 Reprint 2025-26 Example 3 Find the following integrals: Solution (i) We have = – cot cosec C x – x + (iii) We have (sin cos ) sin cos x x dx x dx x dx + = + ∫ ∫ ∫ = – cos sin C x x + + (ii) We have (iii) We have 3 3 2 2 1 1 ( 2 ) 2 x x x e – dx x dx e dx – dx x x + = + ∫ ∫ ∫ ∫ (iii) 2 1 sin cos – x dx x ∫ (i) (sin cos ) x x dx + ∫ (ii) cosec (cosec cot ) x x x dx + ∫ 2 (cosec (cosec + cot ) cosec cosec cot x x x dx x dx x x dx = + ∫ ∫ ∫ = = x x e x + + + 5 2 2 2 – log + C 5 x x e x + 3 1 2 2 – log + C 3 1 2 INTEGRALS 233 Example 4 Find the anti derivative F of f defined by f (x) = 4x 3 – 6, where F (0) = 3 Solution One anti derivative of f (x) is x 4 – 6x since Therefore, the anti derivative F is given by 2 2 2 1 sin 1 sin cos cos cos – x x dx dx – dx x x x = ∫ ∫ ∫ = 2 sec tan sec x dx – x x dx ∫ ∫ = tan sec C x – x + 4 ( 6 ) d x – x dx = 4x 3 – 6 F(x) = x 4 – 6x + C, where C is constant. Reprint 2025-26 234 MATHEMATICS Remarks (i) We see that if F is an anti derivative of f, then so is F + C, where C is any constant. Thus, if we know one anti derivative F of a function f, we can write down an infinite number of anti derivatives of f by adding any constant to F expressed by F(x) + C, C ∈ R. In applications, it is often necessary to satisfy an additional condition which then determines a specific value of C giving unique anti derivative of the given function. (ii) Sometimes, F is not expressible in terms of elementary functions viz., polynomial, logarithmic, exponential, trigonometric functions and their inverses etc. We are (iii) When the variable of integration is denoted by a variable other than x, the integral formulae are modified accordingly. For instance Given that F(0) = 3, which gives, 3 = 0 – 6 × 0 + C or C = 3 Hence, the required anti derivative is the unique function F defined by F(x) = x 4 – 6x + 3. therefore blocked for finding f x dx ( ) ∫ . For example, it is not possible to find 2 – x e dx ∫ by inspection since we can not find a function whose derivative is 2 – x e 4 1 4 5 1 C C 4 1 5 y y dy y + = + = + + ∫ EXERCISE 7.1 Find an anti derivative (or integral) of the following functions by the method of inspection. 1. sin 2x 2. cos 3x 3. e 2x Find the following integrals in Exercises 6 to 20: 12. 3 x x3 4 dx x + + ∫ 13. 3 2 1 1 x x x – dx x – − + ∫ 14. (1 ) – x x dx ∫ 4. (ax + b) 2 5. sin 2x – 4 e 3x 6. 3 (4 + 1) x e dx ∫ 7. 2 2 1 x dx (1 – ) x ∫ 8. 2 ( ) ax bx c dx + + ∫ 9. 2 (2 ) x x e dx + ∫ 10. 2 1 x – dx x ∫ 11. Reprint 2025-26 2 x x – 5 4 dx x + ∫ 3 2 7.3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions. It was based on inspection, i.e., on the search of a function F whose derivative is f which led us to the integral of f. However, this method, which depends on inspection, is not very suitable for many functions. Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Prominent among them are methods based on: 1. Integration by Substitution 2. Integration using Partial Fractions 3. Integration by Parts 15. 2 x x x dx ( 3 2 3) + + ∫ 16. (2 3cos ) x x – x e dx + ∫ 17. 2 (2 3sin 5 ) x – x x dx + ∫ 18. sec (sec tan ) x x x dx + ∫ 19. 2 21. The anti derivative of 1 x x + equals 22. If 3 4 3 ( ) 4 d f x x dx x = − such that f (2) = 0. Then f (x) is (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 1 3 2 C 3 2 x x + + (A) 4 3 1 129 8 x x + − (B) 3 4 1 129 8 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + (C) 4 3 1 129 8 x x + + (D) 3 4 1 129 8 x x + − Choose the correct answer in Exercises 21 and 22. 2 sec cosec x dx x ∫ 20. 2 2 – 3sin cos x ∫ dx. x INTEGRALS 235 7.3.1 Integration by substitution In this section, we consider the method of integration by substitution. The given integral f x dx ( ) ∫ can be transformed into another form by changing the independent variable x to t by substituting x = g (t). Reprint 2025-26 236 MATHEMATICS This change of variable formula is one of the important tools available to us in the name of integration by substitution. It is often important to guess what will be the useful substitution. Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples. Example 5 Integrate the following functions w.r.t. x: Solution (i) We know that derivative of mx is m. Thus, we make the substitution mx = t so that mdx = dt. (iii) 4 2 tan sec x x x (iv) (ii) Derivative of x 2 + 1 is 2x. Thus, we use the substitution x 2 + 1 = t so that 2x dx = dt. (i) sin mx (ii) 2x sin (x 2 + 1) Consider I = f x dx ( ) ∫ Put x = g(t) so that dx dt = g′(t). We write dx = g′(t) dt Thus I = f x dx f g t g t dt ( ) ( ( )) ( ) = ′ ∫ ∫ Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C 2 sin (tan ) 1 – x + x 1 (iii) Derivative of x is 1 2 1 1 2 2 – x x = . Thus, we use the substitution Therefore, 2 2 sin ( 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x 2 + 1) + C Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx x t = ∫ ∫ = 4 2 2 tan sec t t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du 1 so that giving 2 x t dx dt x = = dx = 2t dt. Reprint 2025-26 Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference. (iv) Derivative of 1 2 1 tan 1 – x x = + . Thus, we use the substitution (i) ∫ tan = log sec + C x dx x Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + Hence, 4 2 tan sec x x dx x ∫ = 2 5 tan C 5 x + Alternatively, make the substitution tan x t = Therefore , 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos(tan–1x) + C 2 sin (tan ) sin 1 tan–1 x = t so that 2 1 dx + x = dt. = 2 5 tan C 5 t + (since u = tan t) = 2 5 tan C (since ) 5 x t x + = INTEGRALS 237 (ii) ∫ cot = log sin + C x dx x We have Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ We have cos cot sin x x dx dx x = ∫ ∫ sin tan cos x x dx dx x = ∫ ∫ Reprint 2025-26 238 MATHEMATICS (iii) ∫ sec = log sec + tan + C x dx x x (iv) ∫ cosec = log cosec – cot + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x x + = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x x + = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – – log | | – log |cosec cot | C dt x dx t x x t = = = + + ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log C t + = log sin C x + = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: Solution (i) We have (i) 3 2 sin cos x x dx ∫ (ii) sin sin ( ) x dx x a + ∫ (iii) 1 1 tan dx + x ∫ Put t = cos x so that dt = – sin x dx 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1 – cos ) cos (sin ) x x x dx ∫ Reprint 2025-26 (ii) Put x + a = t. Then dx = dt. Therefore Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 − (1 – ) t t dt ∫ Hence, sin sin ( ) x dx x a + ∫ = x cos a – sin a log |sin (x + a)| + C, sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t + = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a + + + = 1 x a a a – a x a – a cos cos (sin ) log sin ( ) C sin + + = 3 5 2 4 ( – ) C 3 5 t t – t t dt – – = + ∫ = 1 1 3 5 cos cos C 3 5 – x x + + INTEGRALS 239 (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ where, C = – C1 sin a + a cos a, is another arbitrary constant. = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C 1 cos sin 1 2 2 2 cos sin x x – x dx x x + + + ∫ ... (1) Reprint 2025-26 240 MATHEMATICS Integrate the functions in Exercises 1 to 37: 1. 2 2 4. sin sin (cos ) x x 5. sin ( ) cos ( ) ax b ax b + + 6. ax b + 7. x x + 2 8. 2 x x 1 2 + Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get C1 C2 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ 1 x + x 2. ( )2 log x x 3. 1 x x x + log = C C 1 2 1 + log cos sin 2 2 2 2 x x x + + + = C C 1 2 1 + log cos sin C C 2 2 2 2 x x x , + + = + EXERCISE 7.2 12. 1 3 5 3 ( 1) x – x 13. 2 15. 2 9 4 x – x 16. 2 3 x e + 17. 2 x x 18. 9. 2 (4 2) 1 x x x + + + 10. 1 2 1 – tan x e + x 19. 2 1 3 3 (2 3 ) x + x 14. 1 (log )m x x , x > 0, m ≠1 x e – e + 20. 2 2 x – x 11. 4 x 2 1 1 x Reprint 2025-26 e x – x e – e e e + 2 2 x – x x + , x > 0 Choose the correct answer in Exercises 38 and 39. 21. tan2 (2x – 3) 22. sec2 (7 – 4x) 23. 1 24. 2cos 3sin 6cos 4sin x – x x x + 25. 2 2 1 cos (1 tan ) x – x 26. cos x 27. sin 2 cos 2 x x 28. cos 1 sin x + x 29. cot x log sin x 30. sin 1 cos x + x 31. ( )2 sin 33. 1 1 tan – x 34. tan sin cos x x x 35. ( ) 2 1 log x x + 36. ( )2 ( 1) log x x x 38. 9 (A) 10x – x 10 + C (B) 10x + x 10 + C (C) (10x – x10) –1 + C (D) log (10x + x10) + C x + + ∫ equals 10 10 10 log 10 x + + 37. ( ) 3 1 4 sin tan 10 x e x x dx – x x 1 cos x + x 32. 1 1 cot + x 1 x 8 + 2 sin 1 – x – x x INTEGRALS 241 7.3.2 Integration using trigonometric identities When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following example. Example 7 Find (i) 2 cos x dx ∫ (ii) sin 2 cos 3 x x dx ∫ (iii) 3 sin x dx ∫ 39. 2 2 equals sin cos dx x x ∫ (A) tan x + cot x + C (B) tan x – cot x + C (C) tan x cot x + C (D) tan x – cot 2x + C Reprint 2025-26 242 MATHEMATICS Solution (iii) From the identity sin 3x = 3 sin x – 4 sin3 x, we find that (ii) Recall the identity sin x cos y = 1 2 [sin (x + y) + sin (x – y)] (Why?) (i) Recall the identity cos 2x = 2 cos2 x – 1, which gives Therefore, = 1 (1 + cos 2 ) 2 x dx ∫ = 1 1 cos 2 2 2 dx x dx + ∫ ∫ Then = sin3 x = 3sin sin 3 4 x – x cos2 x = 1 cos 2 2 + x = 1 sin 2 C 2 4 x + +x = 1 1 cos 5 cos C 2 5 – x x + + = 1 1 cos 5 cos C 10 2 – x x + + Remark It can be shown using trigonometric identities that both answers are equivalent. Therefore, 3 sin x dx ∫ = 3 1 sin sin 3 4 4 x dx – x dx ∫ ∫ = 3 1 – cos cos 3 C 4 12 x x + + Alternatively, 3 2 sin sin sin x dx x x dx = ∫ ∫ = 2 (1 – cos ) sin x x dx ∫ Put cos x = t so that – sin x dx = dt Therefore, 3 sin x dx ∫ = ( ) 2 − 1 – t dt ∫ = 3 2 C 3 t – dt t dt – t + = + + ∫ ∫ = 1 3 cos cos C 3 – x x + + Reprint 2025-26 Find the integrals of the functions in Exercises 1 to 22: 1. sin2 (2x + 5) 2. sin 3x cos 4x 3. cos 2x cos 4x cos 6x 4. sin3 (2x + 1) 5. sin3 x cos3 x 6. sin x sin 2x sin 3x Choose the correct answer in Exercises 23 and 24. 10. sin4 x 11. cos4 2x 12. 2 sin 1 cos x + x 13. cos 2 cos 2 cos cos x – x – 16. tan4x 17. 3 3 19. 3 1 sin cos x x 20. ( ) 2 cos 2 22. 1 cos ( ) cos ( ) x – a x – b 23. 2 2 7. sin 4x sin 8x 8. 1 cos 1 cos – x + x 9. cos 1 cos x + x 2 2 sin cos is equal to sin cos x x dx x x − ∫ α α 14. cos sin 1 sin 2 x – x + x 15. tan3 2x sec 2x 2 2 sin cos sin cos x x x x + 18. 2 cos sin x EXERCISE 7.3 x x + 21. sin – 1 (cos x) 2 cos 2 2sin cos x x x + INTEGRALS 243 7.4 Integrals of Some Particular Functions In this section, we mention below some important formulae of integrals and apply them for integrating many other related standard integrals: 24. 2 (1 ) equals cos ( ) x (1) ∫ 2 2 1 – = log + C – 2 + dx x a x a a x a (A) tan x + cot x + C (B) tan x + cosec x + C (C) – tan x + cot x + C (D) tan x + sec x + C (A) – cot (exx ) + C (B) tan (xex ) + C (C) tan (e x ) + C (D) cot (e x ) + C x e x dx e x + ∫ Reprint 2025-26 244 MATHEMATICS (2) ∫ 2 2 1 + = log + C – 2 – dx a x a x a a x (3) ∫ – 1 2 2 1 tan C dx x = + x + a a a (4) ∫ 2 2 2 2 = log + – + C – dx x x a x a (5) ∫ – 1 2 2 = sin + C – dx x a a x (6) ∫ 2 2 2 2 = log + + + C + dx x x a x a (1) We have 2 2 1 1 We now prove the above results: x – a ( ) ( ) x – a x a = + = 1 ( ) – ( ) 1 1 1 2 ( ) ( ) 2 x a x – a – a x – a x a a x – a x a + = + + (2) In view of (1) above, we have Therefore, 2 2 1 2 dx dx dx – x – a a x – a x a = + ∫ ∫ ∫ 2 2 1 1 ( ) ( ) – 2 ( ) ( ) a x a x a x a a x a x + + − = + − = 1 1 1 2a a x a x + − + = [ ] 1 log ( )| log ( )| C 2 | x – a – | x a a + + = 1 log C 2 x – a a x a + + Reprint 2025-26 Therefore, 2 2 – dx a x ∫ = 1 2 (3) Put x = a tan θ. Then dx = a sec2 θ dθ. = 1 1 1 1 θ θ C tan C – x d a a a a = + = + ∫ (4) Let x = a secθ. Then dx = a secθ tan θ dθ. a x a a x + + − ANote The technique used in (1) will be explained in Section 7.5. Therefore, 2 2 dx x a + ∫ = Therefore, 2 2 dx x a − ∫ = 2 2 2 secθ tanθ θ = 1 [ log | | log | |] C 2 a x a x a − − + + + = 1 log C 2 dx dx a a x a x + − + ∫ ∫ = 1 secθ θd = log secθ + tanθ + C ∫ = 2 = 2 2 1 log x x – a a + − + log C 2 1 log 1 C x x – a a + + a a − ∫ sec θ a d INTEGRALS 245 (5) Let x = a sinθ. Then dx = a cosθ dθ. = 1 θ = θ + C = sin C – x d a + ∫ (6) Let x = a tan θ. Then dx = a sec2θ dθ. Therefore, 2 2 dx Therefore, 2 2 dx x a + ∫ = 2 a x − ∫ = 2 2 2 θ θ Reprint 2025-26 = 1 secθ θd = log (sec tan ) C θ θ + + ∫ = 2 2 log x x – a + + C, whereC = C1 – log |a| a – a ∫ a a + ∫ tan a d sin a d 2 2 2 θ θ θ sec θ cos 246 MATHEMATICS (7) To find the integral 2 dx ax bx c + + ∫ , we write Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals. Now, put 2 b x t a + = so that dx = dt and writing 2 2 2 4 c b – k a a = ± . We find the integral reduced to the form 2 2 1 dt a t k ± ∫ depending upon the sign of 2 ax2 + bx + c = 2 2 2 2 2 4 b c b c b a x x a x – a a a a a + + = + + = 2 = 2 1 log x x a | a | log C 2 + + − + = 2 log x x a C 2 + + + , where C = C1 – log |a| 2 1 log 1 C x x a a + + + 2 4 c b – a a (8) To find the integral of the type , proceeding as in (7), we (9) To find the integral of the type 2 px q dx ax bx c + + + ∫ , where p, q, a, b, c are constants, we are to find real numbers A, B such that and hence can be evaluated. obtain the integral using the standard formulae. To determine A and B, we equate from both sides the coefficients of x and the constant terms. A and B are thus obtained and hence the integral is reduced to one of the known forms. 2 + = A ( ) + B = A (2 ) + B d px q ax bx c ax b dx + + + Reprint 2025-26 (10) For the evaluation of the integral of the type 2 ( ) px q dx Example 8 Find the following integrals: Solution Example 9 Find the following integrals : (ii) (i) 2 16 dx x − ∫ (ii) 2 2 dx (i) We have 2 2 2 16 4 dx dx x x – = − ∫ ∫ = 4 log C 8 4 x – x 1 + + [by 7.4 (1)] (i) 2 6 13 dx x x − + ∫ (ii) 2 3 13 10 dx x x + − ∫ (iii) 2 5 2 dx as in (9) and transform the integral into known standard forms. Let us illustrate the above methods by some examples. Put x – 1 = t. Then dx = dt. Therefore, 2 2 dx x x − ∫ = 2 1 dt x x − ∫ = 1 sin ( – 1) C – x + – t ∫ = 1 sin ( ) C – t + [by 7.4 (5)] + + ∫ , we proceed ax bx c + INTEGRALS 247 Solution (i) We have x 2 – 6x + 13 = x 2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4 So, 6 13 dx x x 2 − + ∫ = ( )2 2 1 Let x – 3 = t. Then dx = dt Therefore, 6 13 dx x x 2 − + ∫ = 1 2 2 1 tan C 2 2 2 dt – t t = + + ∫ [by 7.4 (3)] Reprint 2025-26 = 1 3 1 tan C 2 2 – x – + 3 2 dx x – + ∫ x x − ∫ 248 MATHEMATICS (ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand, Thus 3 13 10 dx x x 2 + − ∫ = 2 2 1 3 13 17 6 6 Put 13 6 x t + = . Then dx = dt. Therefore, 3 13 10 dx x x 2 + − ∫ = 2 2 1 3 17 6 2 3 13 10 x x – + = 2 13 10 3 3 3 x x – + = 2 2 13 17 3 6 6 x – + (completing the square) ∫ x + − = 1 17 1 6 log C 17 17 3 2 6 6 + × × + [by 7.4 (i)] ∫ t − dx dt t – t Reprint 2025-26 = 1 = 1 1 6 4 log C 17 6 30 x x = 1 1 3 2 1 1 log C log 17 5 17 3 x x = 1 3 2 log C 17 5 x x 13 17 1 6 6 log C 17 13 17 6 6 x + + + + x – − + + + − + + , where C = 1 1 1 C log 17 3 + − + + Example 10 Find the following integrals: (iii) We have 5 2 2 2 5 5 (i) 2 2 6 5 x dx x x 2 + + + ∫ (ii) 2 3 Put 1 5 x – t = . Then dx = dt. Therefore, 5 2 dx ∫ ∫ x x x x – 2 = − dx dx x x 2 − ∫ = 2 2 = 2 2 1 5 1 1 5 5 − ∫ ∫ (completing the square) = = 1 1 2 2 log C 5 5 5 x x – x – + + 5 4 x dx x – x + x – – 1 2 1 2 1 log C 5 5 t t – + + [by 7.4 (4)] 5 1 5 ∫ dx t – dt INTEGRALS 249 Solution (i) Using the formula 7.4 (9), we express x + 2 = ( ) 2 A 2 6 5 B d x x dx + + + = A (4 6) B x + + Equating the coefficients of x and the constant terms from both sides, we get 4A = 1 and 6A + B = 2 or A = 1 4 and B = 1 2 . Therefore, 2 2 6 5 x x x 2 + + + ∫ = 1 4 6 1 4 2 2 6 5 2 6 5 x dx dx x x x x 2 2 + + + + + + ∫ ∫ Reprint 2025-26 = 1 2 1 1 I I 4 2 + (say) ... (1) 250 MATHEMATICS In I1 , put 2x 2 + 6x + 5 = t, so that (4x + 6) dx = dt Therefore, I 1 = 1 log C dt t t = + ∫ and I 2 = 2 2 1 2 6 5 2 5 3 2 Put 3 2 x t + = , so that dx = dt, we get Using (2) and (3) in (1), we get I 2 = 2 2 1 2 1 2 = 1 2 3 tan 2 + C 2 – x + = ( ) 1 2 tan 2 3 + C – x + ... (3) ∫ = 1 2 1 tan 2 C 1 2 2 t + = 2 1 log | 2 6 5 | C x x + + + ... (2) = 2 2 1 2 3 1 2 2 dt dx dx x x x x = + + + + ∫ ∫ ∫ x + + – t + × [by 7.4 (3)] dx (ii) This integral is of the form given in 7.4 (10). Let us express where, C = C C 1 2 4 2 + x + 3 = 2 A (5 4 ) + B d – x – x dx = A (– 4 – 2x) + B Equating the coefficients of x and the constant terms from both sides, we get – 2A = 1 and – 4 A + B = 3, i.e., A = 1 2 – and B = 1 ( ) 2 1 2 1 1 log 2 6 5 tan 2 3 C 2 6 5 4 2 x – dx x x x x x 2 + = + + + + + + + ∫ Reprint 2025-26 Therefore, 2 3 In I1 , put 5 – 4x – x 2 = t, so that (– 4 – 2x) dx = dt. Therefore, I 1 = ( ) Now consider I 2 = 2 2 5 4 9 ( 2) dx dx Put x + 2 = t, so that dx = dt. Therefore, I 2 = 1 2 2 2 sin + C 3 3 dt t – Substituting (2) and (3) in (1), we obtain 2 3 2 5 – 4 – + sin C 3 5 4 x x – – x x – x – x + + = + ∫ , where 1 2 C C C 2 = – − − ∫ = ( ) 5 4 x dx x x + 2 1 = 1 2 – I1 + I2 ... (1) = 2 1 2 5 4 C – x – x + ... (2) = 1 2 2 sin C 3 – x + + ... (3) 2 2 1 4 2 2 5 4 5 4 – – x dx dx – x x x x + − − − − ∫ ∫ x x – x = − − + ∫ ∫ t = − ∫ [by 7.4 (5)] x x t − = − − ∫ ∫ = 1 2 C t + 5 4 – x dx dt 2 4 2 INTEGRALS 251 1. 2 4. 2 1 7. 2 1 Integrate the functions in Exercises 1 to 23. 6 3 1 x x + 2. 2 1 9 25 – x 5. 4 3 1 2 x + x 6. 2 1 x – x – 8. 2 EXERCISE 7.4 1 4 + x 3. ( )2 1 x a + 9. 2 6 6 x Reprint 2025-26 6 1 x − x tan 4 x 2 1 – x + 2 sec x + 252 MATHEMATICS Choose the correct answer in Exercises 24 and 25. 10. 2 1 13. ( )( ) 1 16. 2 4 1 19. ( )( ) 6 7 22. 2 3 2 5 x x – x + − 23. 2 5 3 24. 2 equals 2 2 dx x x + + ∫ (A) x tan–1 (x + 1) + C (B) tan–1 (x + 1) + C (C) (x + 1) tan–1x + C (D) tan–1x + C 25. 2 equals 9 4 dx x x − ∫ 2 3 x x x + + 2 2 11. 2 1 9 6 5 x x + + 12. 2 1 x – x – 1 2 14. 2 1 x – x – + 20. 2 2 x x – + + 17. 2 2 5 4 x 1 x 8 3 + x – x 15. ( )( ) 1 4 x x – + 18. 2 5 2 1 2 3 x x x − + + x x + x – x + 21. 2 2 2 3 x + + . 4 10 x 7 6 – x – x x x + x – a x – b + + 7.5 Integration by Partial Fractions Recall that a rational function is defined as the ratio of two polynomials in the form is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational functions can be reduced to the proper rational P( ) Q( ) x x , where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(x) (A) 1 9 8 –1 sin C 9 8 x − + (B) 1 8 9 –1 sin C 2 9 x − + (C) 1 9 8 –1 sin C 3 8 x − + (D) 1 9 8 –1 sin C 2 9 x − + Reprint 2025-26 functions by long division process. Thus, if P( ) Q( ) x x is improper, then P( ) P ( ) 1 T( ) Q( ) Q( ) x x x x x = + , where T(x) is a polynomial in x and P ( ) 1 Q( ) x x is a proper rational function. As we know how to integrate polynomials, the integration of any rational function is reduced to the integration of a proper rational function. The rational functions which we shall consider here for integration purposes will be those whose denominators can be factorised into linear and quadratic factors. Assume that we want to evaluate P( ) Q( ) x dx x ∫ , where P( ) Q( ) x x is proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods. The following Table 7.2 indicates the types of simpler partial fractions that are to be associated with various kind of rational functions. S.No. Form of the rational function Form of the partial fraction 1. ( – ) ( – ) px q x a x b + , a ≠ b A B x – a x – b + 2. 2 ( – ) px q x a + ( )2 A B x – a x – a + Table 7.2 INTEGRALS 253 3. 2 4. 2 5. 2 In the above table, A, B and C are real numbers to be determined suitably. where x 2 + bx + c cannot be factorised further ( – ) ( ) ( ) px qx r x a x – b x – c + + A B C x – a x – b x – c + + 2 ( – ) ( ) px qx r x a x – b + + 2 A B C x – a ( ) x – a x – b + + 2 ( – ) ( ) px qx r x a x bx c + + + + 2 A B + C x x – a x bx c + + + , Reprint 2025-26 254 MATHEMATICS Example 11 Find ( 1) ( 2) dx x x + + ∫ Solution The integrand is a proper rational function. Therefore, by using the form of partial fraction [Table 7.2 (i)], we write where, real numbers A and B are to be determined suitably. This gives 1 = A (x + 2) + B (x + 1). Equating the coefficients of x and the constant term, we get A + B = 0 and 2A + B = 1 Solving these equations, we get A =1 and B = – 1. Thus, the integrand is given by Therefore, ( 1) ( 2) dx x x + + ∫ = 1 2 dx dx – x x + + ∫ ∫ 1 ( 1) ( 2) x x + + = A B x x 1 2 + + + ... (1) 1 ( 1) ( 2) x x + + = 1 – 1 x x 1 2 + + + = log 1 log 2 C x x + − + + Remark The equation (1) above is an identity, i.e. a statement true for all (permissible) values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to indicate that the statement is true only for certain values of x. Example 12 Find 2 Solution Here the integrand 2 x 2 + 1 by x 2 – 5x + 6 and find that 2 1 5 6 x dx x x + − + ∫ 5 6 x x – x + + is not proper rational function, so we divide 2 1 Reprint 2025-26 = 1 log C 2 x x + + + Let 5 5 ( 2) ( 3) x – x – x – = A B x – x – 2 3 + So that 5x – 5 = A (x – 3) + B (x – 2) Equating the coefficients of x and constant terms on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10 Thus, 2 Therefore, 2 Example 13 Find 2 3 2 ( 1) ( 3) x dx x x − + + ∫ Solution The integrand is of the type as given in Table 7.2 (4). We write So that 3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2 = A (x 2 + 4x + 3) + B (x + 3) + C (x 2 + 2x + 1 ) Comparing coefficient of x 2 , x and constant term on both sides, we get A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2. Solving these equations, we get 11 5 11 A B and C 4 2 4 – – = = = , . Thus the integrand is given by 2 1 5 6 x dx x – x + + ∫ = 1 5 10 2 3 dx dx dx x – x – − + ∫ ∫ ∫ 2 3 2 ( 1) ( 3) x – x x + + = 2 A B C x x 1 ( 1) x 3 + + + + + 5 6 x x – x + + = 2 5 5 5 5 1 1 5 6 ( 2) ( 3) x – x – x – x x – x – + = + + 2 1 5 6 x x – x + + = 5 10 1 x – x – 2 3 − + 2 1 2 = x – 5 log | x – 2| + 10 log | x – 3| + C. INTEGRALS 255 Therefore, 2 3 2 ( 1) ( 3) x x x − + + ∫ = 2 11 5 11 4 1 2 ( 1) 4 3 dx dx dx – x x x − + + + ∫ ∫ ∫ 2 3 2 ( 1) ( 3) x x x − + + = 2 11 5 11 4 ( 1) 2 ( 1) 4 ( 3) – – x x + x + + Reprint 2025-26 = 11 5 11 log +1 log 3 C 4 2 ( +1) 4 x x x + − + + = 11 +1 5 log + C 4 + 3 2 ( + 1) x x x + 256 MATHEMATICS Example 14 Find 2 Solution Consider 2 Then 2 Write ( 1) ( 4) y y y + + = A B y y 1 4 + + + So that y = A (y + 4) + B (y + 1) Comparing coefficients of y and constant terms on both sides, we get A + B = 1 and 4A + B = 0, which give Thus, 2 Therefore, 2 2 2 ( 1) ( 4) x dx x x + + ∫ 2 2 ( 1) ( 4) x dx x x + + ∫ = 2 2 1 4 3 3 1 4 dx dx – x x + + + ∫ ∫ 2 2 ( 1) ( 4) x x x + + and put x 2 = y. 2 2 ( 1) ( 4) x x x + + = ( 1) ( 4) y y y + + 2 2 ( 1) ( 4) x x x + + = 2 2 1 4 3 ( 1) 3 ( 4) – x x + + + A = 1 4 and B 3 3 − = In the above example, the substitution was made only for the partial fraction part and not for the integration part. Now, we consider an example, where the integration involves a combination of the substitution method and the partial fraction method. Example 15 Find ( ) 2 3 sin 2 cos Solution Let y = sinφ Then dy = cosφ dφ 5 cos 4 sin – d – – φ φ φ φ φ ∫ Reprint 2025-26 = 1 4 1 1 1 tan tan C 3 3 2 2 – – x – x + × + = 1 2 1 1 tan tan C 3 3 2 – – x – x + + Therefore, ( ) 2 3 sin 2 cos Now, we write ( )2 3 2 Therefore, 3y – 2 = A (y – 2) + B Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2, which gives A = 3 and B = 4. Therefore, the required integral is given by I = 2 3 4 [ + ] 2 ( 2) dy y – y – ∫ = 2 3 + 4 2 ( 2) dy dy y – y – ∫ ∫ 5 cos 4 sin – d – – φ φ φ φ φ ∫ = 2 (3 – 2) = 1 3 log 2 4 C 2 y – y − + + − = 4 3 log sin 2 C 2 sin – φ − + + φ y – = 2 A B y 2 ( 2) y + − − [by Table 7.2 (2)] 2 y – = 2 3 2 4 4 y – dy y – y + ∫ = ( )2 3 2 I (say) 2 y – 5 (1 ) 4 y dy – – y – y ∫ y – = ∫ INTEGRALS 257 Example 16 Find 2 Solution The integrand is a proper rational function. Decompose the rational function into partial fraction [Table 2.2(5)]. Write Therefore, x 2 + x + 1 = A (x 2 + 1) + (Bx + C) (x + 2) = 4 3 log (2 sin ) + C 2 sin − φ + − φ (since, 2 – sinφ is always positive) 2 1 ( 2) ( 1) x x dx x x + + + + ∫ 2 1 ( 1) ( 2) x x x x + + 2 + + = 2 A B + C 2 ( 1) x x x + + + Reprint 2025-26 258 MATHEMATICS Equating the coefficients of x 2 , x and of constant term of both sides, we get A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get Integrate the rational functions in Exercises 1 to 21. 3 2 1 A , B and C 5 5 5 = = = 1. ( 1) ( 2) x x x + + 2. 2 1 x – 9 3. 3 1 ( 1) ( 2) ( 3) x – x – x – x – 4. ( 1) ( 2) ( 3) x x – x – x – 5. 2 2 3 2 x x x + + 6. 2 1 (1 2 ) – x x – x Thus, the integrand is given by Therefore, 2 2 1 ( +1) ( 2) x x dx x x + + + ∫ = 2 2 3 1 2 1 1 5 2 5 1 5 1 dx x dx dx x x x + + + + + ∫ ∫ ∫ 2 1 ( 1) ( 2) x x x x + + + + = 2 2 EXERCISE 7.5 = 3 1 2 1 1 log 2 log 1 tan C 5 5 5 – x + + + + + x x 2 1 3 5 5 5 ( 2) 1 x x + + + + = 2 3 1 2 1 5 ( 2) 5 1 x x x + + + + x 10. 2 2 3 ( 1) (2 3) x x – x − 13. 2 2 (1 ) (1 ) − + x x 14. 2 3 1 ( 2) x – x + 15. 4 1 x −1 16. 1 ( 1) n x x + [Hint: multiply numerator and denominator by x n – 1 and put x n = t ] 17. cos (1 – sin ) (2 – sin ) x x x [Hint : Put sin x = t] 7. 2 ( 1) ( – 1) x x x + 8. 2 ( 1) ( 2) x x – x + 9. 3 2 3 5 1 x x – x x + − + + 11. 2 5 ( 1) ( 4) x x x + − 12. 3 Reprint 2025-26 2 1 1 x x x + + − Choose the correct answer in each of the Exercises 22 and 23. 7.6 Integration by Parts In this section, we describe one more method of integration, that is found quite useful in integrating products of functions. If u and v are any two differentiable functions of a single variable x (say). Then, by the product rule of differentiation, we have 18. 2 2 21. 1 ( 1) x e – [Hint : Put e x = t] 22. ( 1) ( 2) x dx x x − − ∫ equals 23. 2 ( 1) dx x x + ∫ equals (A) 2 ( 1) log C 2 x x − + − (B) 2 ( 2) log C 1 x x − + − (A) 1 2 log log ( +1) + C 2 x x − (B) 1 2 log log ( +1) + C 2 x x + (C) 2 1 log C 2 x x − + − (D) log ( 1) ( 2) C x x − − + (C) 1 2 log log ( +1) + C 2 − + x x (D) 1 2 log log ( +1) + C 2 x x + 2 2 ( 1) ( 2) ( 3) ( 4) x x x x + + + + 19. 2 2 2 ( 1) ( 3) x x x + + 20. 4 1 x x – ( 1) INTEGRALS 259 Integrating both sides, we get or dv u dx dx ∫ = du uv – v dx dx ∫ ... (1) Let u = f(x) and dv dx = g(x). Then du dx = f ′(x) and v = g x dx ( ) ∫ ( ) d uv dx = dv du u v dx dx + uv = dv du u dx v dx dx dx + ∫ ∫ Reprint 2025-26 260 MATHEMATICS Therefore, expression (1) can be rewritten as i.e., f x g x dx ( ) ( ) ∫ = f x g x dx – f x g x dx dx ( ) ( ) [ ( ) ( ) ] ′ ∫ ∫ ∫ If we take f as the first function and g as the second function, then this formula may be stated as follows: “The integral of the product of two functions = (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]” Example 17 Find x x dx cos ∫ Solution Put f (x) = x (first function) and g (x) = cos x (second function). Then, integration by parts gives Suppose, we take f(x) = cos x and g(x) = x. Then f x g x dx ( ) ( ) ∫ = f x g x dx – g x dx f x dx ( ) ( ) [ ( ) ] ( ) ′ ∫ ∫ ∫ x x dx cos ∫ = cos [ ( ) cos ] d x x dx – x x dx dx dx ∫ ∫ ∫ x x dx cos ∫ = cos [ (cos ) ] d x x dx – x x dx dx dx ∫ ∫ ∫ = x x – x dx sin sin ∫ = x sin x + cos x + C = ( ) 2 2 cos sin 2 2 x x x x dx + ∫ complicated integral having more power of x. Therefore, the proper choice of the first function and the second function is significant. Remarks (i) It is worth mentioning that integration by parts is not applicable to product of (ii) Observe that while finding the integral of the second function, we did not add any constant of integration. If we write the integral of the second function cos x Thus, it shows that the integral x x dx cos ∫ is reduced to the comparatively more functions in all cases. For instance, the method does not work for x x dx sin ∫ . The reason is that there does not exist any function whose derivative is x sin x. Reprint 2025-26 Example 18 Find log x dx ∫ Solution To start with, we are unable to guess a function whose derivative is log x. We take log x as the first function and the constant function 1 as the second function. Then, the integral of the second function is x. Hence, (log .1) x dx ∫ = log 1 [ (log ) 1 ] d x dx x dx dx dx − ∫ ∫ ∫ Example 19 Find x x e dx ∫ Solution Take first function as x and second function as e x . The integral of the second function is e x . = x x k x – kx (sin ) cos C + − + = x x x sin cos C + + This shows that adding a constant to the integral of the second function is superfluous so far as the final result is concerned while applying the method of integration by parts. (iii) Usually, if any function is a power of x or a polynomial in x, then we take it as the first function. However, in cases where other function is inverse trigonometric function or logarithmic function, then we take them as first function. as sin x + k, where k is any constant, then x x dx cos ∫ = x x k x k dx (sin ) (sin ) + − + ∫ = x x k x dx k dx (sin ) (sin + − − ∫ ∫ = 1 (log ) – x x x dx x x – x log C x ⋅ = + ∫ . INTEGRALS 261 Therefore, x x e dx ∫ = 1 x x x e e dx − ⋅ ∫ = xex – e x + C. Example 20 Find Solution Let first function be sin – 1x and second function be 2 1 x First we find the integral of the second function, i.e., 2 1 x dx Put t =1 – x 2 . Then dt = – 2x dx – x x dx − x ∫ 2 sin 1 1 Reprint 2025-26 − x ∫ . − x . 262 MATHEMATICS Therefore, 2 1 x dx Hence, 1 Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and then integrating by parts. Example 21 Find sin x e x dx ∫ Solution Take e x as the first function and sin x as second function. Then, integrating by parts, we have I sin ( cos ) cos x x x = e x dx e – x e x dx = + ∫ ∫ = – ex cos x + I1 (say) ... (1) Taking e x and cos x as the first and second functions, respectively, in I1 , we get I 1 = sin sin x x e x – e x dx ∫ Substituting the value of I1 in (1), we get I = – e x cos x + e x sin x – I or 2I = e x (sin x – cos x) Hence, I = sin (sin cos ) + C 2 x x e e x dx x – x = ∫ – x x dx − x ∫ = ( ) 1 2 2 2 1 (sin ) 1 ( 1 ) 1 – x – x – x dx x − − − − ∫ − x ∫ = 1 2 2 sin 1 = 2 1 – x x x 1 sin C − − + + = 2 1 x – x x 1 sin C − − + dt – t ∫ = 2 – 1 t x = − − Alternatively, above integral can also be determined by taking sin x as the first function and e x the second function. 7.6.1 Integral of the type [ ( ) + ( )] x e f x f x dx ′ ∫ We have I = [ ( ) + ( )] x e f x f x dx ′ ∫ = ( ) + ( ) x x e f x dx e f x dx ′ ∫ ∫ integrating it by parts, we have I1 = f (x) e x – ( ) C x f x e dx ′ + ∫ Substituting I1 in (1), we get = 1 1 I ( ) , where I = ( ) x x + e f x dx ′ e f x dx ∫ ∫ ... (1) Taking f(x) and e x as the first function and second function, respectively, in I1 and I = ( ) ( ) ( ) C x x x e f x f x e dx e f x dx − ′ + ′ + ∫ ∫ = e x f (x) + C Reprint 2025-26 Thus, ′ ∫ [ ( ) ( )] x e f x + f x dx = ( ) C x e f x + Example 22 Find (i) 1 2 1 (tan ) 1 x – e x x + + ∫ dx (ii) 2 Solution (i) We have I = 1 2 1 (tan ) 1 x – e x dx x + + ∫ (ii) We have 2 Consider f(x) = tan– 1x, then f ′(x) = 2 1 1+ x Thus, the given integrand is of the form e x [ f (x) + f ′(x)]. Therefore, 1 2 1 I (tan ) 1 x – e x dx x = + + ∫ = e x tan– 1x + C Consider 1 ( ) 1 x f x x − = + , then 2 2 ( ) ( 1) f x x ′ = + 2 ( + 1) I ( +1) x x e x = ∫ dx 2 2 2 1 2 [ ] ( + 1) ( +1) x x – e dx x x = + ∫ 2 1 2 [ + ] +1 ( +1) x x – e dx x x = ∫ 2 2 1 +1+1) [ ] ( +1) x x – e dx x = ∫ x x e x ∫ dx 2 ( +1) ( +1) INTEGRALS 263 Integrate the functions in Exercises 1 to 22. 1. x sin x 2. x sin 3x 3. x 2 e x 4. x log x 5. x log 2x 6. x 2 log x 7. x sin– 1x 8. x tan–1 x 13. tan–1x 14. x (log x) 2 15. (x 2 + 1) log x 9. x cos–1 x 10. (sin–1x) 2 11. 1 Thus, the given integrand is of the form e x [f (x) + f ′(x)]. Therefore, 2 2 1 1 C ( 1) 1 x x x x e dx e x x + − = + + + ∫ EXERCISE 7.6 Reprint 2025-26 1 x x 2 cos − 12. x sec2 x − x 264 MATHEMATICS Choose the correct answer in Exercises 23 and 24. 7.6.2 Integrals of some more types Here, we discuss some special types of standard integrals based on the technique of integration by parts : 16. e x (sinx + cosx) 17. 2 (1 ) x x e + x 18. 1 sin 1 cos x x e x + + 19. 2 1 1 – x e x x 20. 3 ( 3) ( 1) 22. 1 2 2 sin 1 – x x + 23. 3 2 x x e dx ∫ equals 24. sec (1 tan ) x e x x dx + ∫ equals (A) 1 3 C 3 x e + (B) 1 2 C 3 x e + (A) e x cos x + C (B) e x sec x + C (C) 1 3 C 2 x e + (D) 1 2 C 2 x e + (C) e x sin x + C (D) e x tan x + C x x e x − − 21. e 2x sin x Taking constant function 1 as the second function and integrating by parts, we have (i) 2 2 x a dx − ∫ (ii) 2 2 x a dx + ∫ (iii) 2 2 a x dx − ∫ (i) Let 2 2 I = − x a dx ∫ I = 2 2 2 2 1 2 2 = 2 2 2 2 2 x x x a dx x a − − − ∫ = 2 2 2 2 2 2 2 x a a x x a dx x a − + − − − ∫ x x x a x dx x a − − − ∫ Reprint 2025-26 or 2I = 2 2 2 2 2 dx x x a a x a − − − ∫ or I = ∫ 2 2 x – a dx = 2 2 2 2 2 – – log + – + C 2 2 x a x a x x a Similarly, integrating other two integrals by parts, taking constant function 1 as the second function, we get Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively. Example 23 Find 2 x x dx + + 2 5 ∫ Solution Note that (iii) (ii) ∫ 2 2 2 2 2 1 2 2 + = + + log + + + C 2 2 a x a dx x x a x x a = 2 2 2 2 2 2 2 dx x x a x a dx a x a − − − − − ∫ ∫ = 2 2 2 2 2 I dx x x a a x a − − − − ∫ INTEGRALS 265 Example 24 Find 2 3 2 − −x x dx ∫ Solution Note that 2 2 3 2 − − = − + x x dx x dx 4 ( 1) ∫ ∫ 2 x x dx + + 2 5 ∫ = 2 ( 1) 4 x dx + + ∫ Put x + 1 = y, so that dx = dy. Then 2 x x dx + + 2 5 ∫ = 2 2 y dy + 2 ∫ = 1 2 4 2 4 log 4 C 2 2 y y + + + + + y y [using 7.6.2 (ii)] = 1 2 2 ( 1) 2 5 2 log 1 2 5 C 2 x x x + + + + + + + + + x x x Reprint 2025-26 266 MATHEMATICS Put x + 1 = y so that dx = dy. Thus 2 3 2 − −x x dx ∫ = 2 4 − y dy ∫ Integrate the functions in Exercises 1 to 9. Choose the correct answer in Exercises 10 to 11. 10. 2 1+ x dx ∫ is equal to 1. 2 4 − x 2. 2 1 4 − x 3. 2 x x + + 4 6 4. 2 x x + + 4 1 5. 2 1 4 − −x x 6. 2 x x + − 4 5 7. 2 1 3 + −x x 8. 2 x x + 3 9. 2 1 9 x + (A) ( ) 2 1 2 1 log 1 C 2 2 x + + + + + x x x (B) 3 2 2 2 (1 ) C 3 + + x (C) 3 2 2 2 (1 ) C 3 x x + + = 1 2 4 1 4 sin C 2 2 2 – y y y − + + [using 7.6.2 (iii)] = 1 2 1 1 ( 1) 3 2 2 sin C 2 2 – x x x x + + − − + + EXERCISE 7.7 11. 2 x x dx − + 8 7 ∫ is equal to (D) 2 2 2 2 1 1 log 1 C 2 2 x + + + + + x x x x (A) 1 2 2 ( 4) 8 7 9log 4 8 7 C 2 x x x x x x − − + + − + − + + (D) 1 2 9 2 ( 4) 8 7 log 4 8 7 C 2 2 x x x − − + − − + − + + x x x (B) 1 2 2 ( 4) 8 7 9log 4 8 7 C 2 x x x x x x + − + + + + − + + (C) 1 2 2 ( 4) 8 7 3 2 log 4 8 7 C 2 x x x − − + − − + − + + x x x Reprint 2025-26 7.7 Definite Integral In the previous sections, we have studied about the indefinite integrals and discussed few methods of finding them including integrals of some special functions. In this section, we shall study what is called definite integral of a function. The definite integral has a unique value. A definite integral is denoted by ( ) b lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) – F(a). 7.8 Fundamental Theorem of Calculus 7.8.1 Area function We have defined ( ) b the region bounded by the curve y = f(x), the ordinates x = a and x = b and x-axis. Let x be a given point in [a, b]. Then ( ) x represents the area of the light shaded region in Fig 7.1 [Here it is assumed that f(x) > 0 for x ∈ [a, b], the assertion made below is equally true for other functions as well]. The area of this shaded region depends upon the value of x. a f x dx ∫ as the area of a f x dx ∫ a f x dx ∫ , where a is called the INTEGRALS 267 In other words, the area of this shaded region is a function of x. We denote this function of x by A(x). We call the function A(x) as Area function and is given by Based on this definition, the two basic fundamental theorems have been given. However, we only state them as their proofs are beyond the scope of this text book. 7.8.2 First fundamental theorem of integral calculus Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′(x) = f (x), for all x ∈ [a, b]. A (x) = ∫ ( ) x Reprint 2025-26 a f x dx ... (1) Fig 7.1 268 MATHEMATICS 7.8.3 Second fundamental theorem of integral calculus We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative. Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then ∫ ( ) b Remarks (iv) In ( ) b a f x dx ∫ , the function f needs to be well defined and continuous in [a, b]. (iii) The crucial operation in evaluating a definite integral is that of finding a function whose derivative is equal to the integrand. This strengthens the relationship between differentiation and integration. (ii) This theorem is very useful, because it gives us a method of calculating the definite integral more easily, without calculating the limit of a sum. (i) In words, the Theorem 2 tells us that ( ) b a f x dx ∫ = (value of the anti derivative F of f at the upper limit b – value of the same anti derivative at the lower limit a). For instance, the consideration of definite integral 1 3 2 2 2 x x dx ( –1) ∫ − is erroneous since the function f expressed by f(x) = 1 2 2 x x( –1) is not defined in a portion – 1 < x < 1 of the closed interval [– 2, 3]. a f x dx = [F( )] =b a x F (b) – F(a). (ii) Evaluate F(b) – F(a) = [F ( )]b a x , which is the value of ( ) b (i) Find the indefinite integral f x dx ( ) ∫ . Let this be F(x). There is no need to keep integration constant C because if we consider F(x) + C instead of F(x), we get Steps for calculating ( ) b a f x dx ∫ . Thus, the arbitrary constant disappears in evaluating the value of the definite integral. We now consider some examples ( ) [F ( ) C] [F( ) C] – [F( ) C] F( ) – F( ) b b a a f x dx x b a b a = + = + + = ∫ . Reprint 2025-26 a f x dx ∫ . Example 25 Evaluate the following integrals: Solution (iii) 2 (ii) Let 9 3 4 2 2 I (i) 3 2 2 x dx ∫ (ii) 9 3 4 2 2 (30 – ) (i) Let 3 2 2 I = x dx ∫ . Since 3 2 F ( ) 3 x x dx x = = ∫ , Therefore, by the second fundamental theorem, we get Put 3 2 3 30 – . Then – 2 x t x dx dt = = or 2 – 3 x dx dt = Thus, 3 2 2 2 1 ( 1) ( 2) x dx x x + + ∫ (iv) 3 4 0 sin 2 cos 2 t t dt π ∫ x = ∫ . We first find the anti derivative of the integrand. = ∫ ∫ = 2 1 3 t = 3 2 2 – 3 (30 – ) (30 – ) x dt dx t x x dx I = 27 8 19 F (3) – F (2) – 3 3 3 = = x ∫ x dx 2 1 F ( ) 3 (30 – ) x = x INTEGRALS 269 (iii) Let 2 1 I ( 1) ( 2) x dx x x = + + ∫ Therefore, by the second fundamental theorem of calculus, we have Reprint 2025-26 I = = 2 1 1 3 (30 – 27) 30 – 8 − = 2 1 1 19 3 3 22 99 − = 2 1 F(9) – F(4) 3 (30 – ) x = 3 2 4 9 270 MATHEMATICS (iv) Let 4 3 0 I sin 2 cos 2 t t dt π = ∫ . Consider 3 sin 2 cos 2 ∫ t t dt Using partial fraction, we get –1 2 ( 1) ( 2) 1 2 x x x x x = + + + + + So ( 1) ( 2) x dx x x + + ∫ = – log 1 2log 2 F( ) x + + + = x x Therefore, by the second fundamental theorem of calculus, we have I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3] Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1 2 du So 3 sin 2 cos 2 ∫ t t dt = 1 3 2 u du ∫ Therefore, by the second fundamental theorem of integral calculus I = 1 4 4 1 F ( ) – F (0) [sin – sin 0] 4 8 2 8 π π = = = – 3 log 3 + log 2 + 2 log 4 = 32 log 27 = 1 1 4 4 [ ] sin 2 F ( ) say 8 8 u t t = = Evaluate the definite integrals in Exercises 1 to 20. 1. 1 4. sin 2 0 4 x dx π ∫ 5. cos 2 0 2 x dx π ∫ 6. 5 4 x e dx ∫ 7. 4 0 tan x dx π ∫ 8. 4 1 ( 1) x dx − + ∫ 2. 3 6 cosec x dx π ∫ π 9. 1 2 1 dx x ∫ 3. 2 3 2 1 (4 – 5 6 9) x x x dx + + ∫ x ∫ 10. 1 2 0 1 dx + x ∫ 11. 3 2 2 1 dx x − ∫ 0 2 1 – dx EXERCISE 7.8 Reprint 2025-26 Choose the correct answer in Exercises 21 and 22. 7.9 Evaluation of Definite Integrals by Substitution In the previous sections, we have discussed several methods for finding the indefinite integral. One of the important methods for finding the indefinite integral is the method of substitution. 21. 3 2 1 1 dx + x ∫ equals 22. 2 3 2 0 4 9 dx + x ∫ equals 12. 2 2 0 cos x dx π ∫ 13. 3 2 2 1 x dx x + ∫ 14. 1 2 0 2 3 5 1 x dx x + 16. 2 2 2 1 19. 2 2 0 6 3 4 x dx x (A) 3 π (B) 2 3 π (C) 6 π (D) 12 π (A) 6 π (B) 12 π (C) 24 π (D) 4 π x x + + ∫ 17. 4 2 3 0 (2sec 2) x x dx π + + ∫ 18. 2 2 0 (sin – cos ) 2 2 x x dx π ∫ + + ∫ 20. 1 0 ( sin ) 4 x x x e dx π + ∫ 5 4 3 x + ∫ 15. 1 2 INTEGRALS 271 0 x x e dx ∫ 1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form. 2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration. 3. Resubstitute for the new variable and write the answer in terms of the original variable. 4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits. To evaluate ( ) b a f x dx ∫ , by substitution, the steps could be as follows: Reprint 2025-26 272 MATHEMATICS Let us illustrate this by examples. Example 26 Evaluate 1 4 5 1 5 1 x x dx − + ∫ . Solution Put t = x5 + 1, then dt = 5x 4 dx. Therefore, 4 5 5 1 x x dx + ∫ = t dt ∫ = 3 2 2 3 t = 3 5 2 2 ( 1) 3 x + Hence, 1 4 5 1 5 1 x x dx − + ∫ = ANote In order to quicken this method, we can proceed as follows: After performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step. = ( ) 3 3 5 2 5 2 2 (1 1) – (– 1) 1 3 + + = 3 3 2 2 2 2 0 3 − = 2 4 2 (2 2) 3 3 = 2 ( 1) 3 x + 1 3 5 2 – 1 Alternatively, first we transform the integral and then evaluate the transformed integral with new limits. Let t = x 5 + 1. Then dt = 5 x 4 dx. Note that, when x = – 1, t = 0 and when x = 1, t = 2 Thus, as x varies from – 1 to 1, t varies from 0 to 2 Therefore 1 4 5 1 5 1 x x dx − + ∫ = 2 0 t dt ∫ Example 27 Evaluate – 1 1 2 0 tan 1 x dx + x ∫ Reprint 2025-26 = 2 2 2 – 0 3 3 t = = 2 4 2 (2 2) 3 3 = 2 3 3 3 2 2 2 0 Solution Let t = tan – 1x, then 2 1 1 dt dx x = + . The new limits are, when x = 0, t = 0 and when x = 1, 4 t π = . Thus, as x varies from 0 to 1, t varies from 0 to 4 π . Therefore –1 1 2 0 tan 1 Evaluate the integrals in Exercises 1 to 8 using substitution. x x − + + ∫ 8. 2 2 2 1 1 1 – 2 x e dx x x ∫ Choose the correct answer in Exercises 9 and 10. 1. 1 2 0 1 x dx x + ∫ 2. 2 5 0 sin cos d π φ φ φ ∫ 3. 1 – 1 2 0 2 sin 1 x dx x + ∫ 4. 2 6. 2 2 0 4 – dx 9. The value of the integral (A) 6 (B) 0 (C) 3 (D) 4 0 x x + 2 ∫ (Put x + 2 = t 2 ) 5. 2 2 0 sin 1 cos x dx x π x x + ∫ 7. 1 2 1 2 5 dx x dx + x ∫ = 2 4 4 0 0 2 t t dt ( ) x x dx x − ∫ is 1 3 3 1 1 4 3 EXERCISE 7.9 π π ∫ = 2 2 1 – 0 2 16 32 π π = + ∫ INTEGRALS 273 7.10 Some Properties of Definite Integrals We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily. 10. If f(x) = 0 sin x t t dt ∫ , then f ′(x) is P0 : ( ) ( ) b b a a f x dx f t dt = ∫ ∫ P1 : ( ) – ( ) b a P2 : ( ) ( ) ( ) b c b (A) cosx + x sin x (B) x sinx (C) x cosx (D) sinx + x cosx a b f x dx f x dx = ∫ ∫ . In particular, ( ) 0 a a a c f x dx f x dx f x dx = + ∫ ∫ ∫ Reprint 2025-26 a f x dx = ∫ 274 MATHEMATICS a f x dx − = ∫ , if f is an odd function, i.e., if f (– x) = – f (x). We give the proofs of these properties one by one. Proof of P0 It follows directly by making the substitution x = t. Proof of P1 Let F be anti derivative of f. Then, by the second fundamental theorem of calculus, we have ( ) F ( ) – F ( ) – [F ( ) F ( )] ( ) b a Here, we observe that, if a = b, then ( ) 0 a Proof of P2 Let F be anti derivative of f. Then P3 : ( ) ( ) b b P4 : 0 0 ( ) ( ) a a f x dx f a x dx = − ∫ ∫ P5 : 2 P6 : 2 0 0 ( ) 2 ( ) , if (2 ) ( ) a a f x dx f x dx f a x f x = − = ∫ ∫ and P7 : (i) 0 ( ) 2 ( ) a a (Note that P4 is a particular case of P3 ) (ii) ( ) 0 a a a f x dx f a b x dx = + − ∫ ∫ 0 0 0 ( ) ( ) (2 ) a a a f x dx f x dx f a x dx = + − ∫ ∫ ∫ a f x dx f x dx − = ∫ ∫ , if f is an even function, i.e., if f (– x) = f (x). 0 if f (2a – x) = – f (x) a b f x dx b a a b f x dx = = − = − ∫ ∫ a f x dx = ∫ . and ( ) b Proof of P3 Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a. Therefore Adding (2) and (3), we get ( ) ( ) F( ) – F( ) ( ) c b b a c a f x dx f x dx b a f x dx + = = ∫ ∫ ∫ This proves the property P2 . a f x dx ∫ = ( – ) a ( ) b b − + f a b t dt ∫ a f x dx ∫ = F(b) – F(a) ... (1) a f x dx ∫ = F(c) – F(a) ... (2) c f x dx ∫ = F(b) – F(c) ... (3) ( ) b ( ) c Reprint 2025-26 Proof of P4 Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in P3 . Proof of P5 Using P2 , we have 2 2 0 0 ( ) ( ) ( ) a a a a f x dx f x dx f x dx = + ∫ ∫ ∫ . Let t = 2a – x in the second integral on the right hand side. Then dt = – dx. When x = a, t = a and when x = 2a, t = 0. Also x = 2a – t. Therefore, the second integral becomes 2 ( ) a Hence 2 0 ( ) a f x dx ∫ = 0 0 ( ) (2 ) a a f x dx f a x dx + − ∫ ∫ Proof of P6 Using P5 , we have 2 0 0 0 ( ) ( ) (2 ) a a a f x dx f x dx f a x dx = + − ∫ ∫ ∫ ... (1) Now, if f(2a – x) = f(x), then (1) becomes and if f(2a – x) = – f(x), then (1) becomes a f x dx ∫ = 0 – (2 – ) a f a t dt ∫ = 0 (2 – ) a f a t dt ∫ = 0 (2 – ) a f a x dx ∫ = ( – ) b a f a b t dt + ∫ (by P1 ) = ( – ) b a f a b x + ∫ dx by P0 0 ( ) a f x dx ∫ = 0 0 0 ( ) ( ) 2 ( ) , a a a f x dx f x dx f x dx + = ∫ ∫ ∫ 0 ( ) a f x dx ∫ = 0 0 ( ) ( ) 0 a a f x dx f x dx − = ∫ ∫ 2 2 INTEGRALS 275 Proof of P7 Using P2 , we have Let t = – x in the first integral on the right hand side. Therefore ( ) a a f x dx ∫ − = 0 a f x dx ∫ − = 0 ( ) a dt = – dx. When x = – a, t = a and when x = 0, t = 0. Also x = – t. Reprint 2025-26 = 0 0 (– ) ( ) a a f x dx f x dx + ∫ ∫ (by P0 ) ... (1) a f x dx f x dx − + ∫ ∫ . Then 0 – (– ) ( ) a a f t dt f x dx + ∫ ∫ 0 ( ) ( ) a 276 MATHEMATICS Example 28 Evaluate 2 3 1 x x dx – ∫ − Solution We note that x 3 – x ≥ 0 on [– 1, 0] and x 3 – x ≤ 0 on [0, 1] and that x 3 – x ≥ 0 on [1, 2]. So by P2 we write (i) Now, if f is an even function, then f(–x) = f(x) and so (1) becomes (ii) If f is an odd function, then f(–x) = – f(x) and so (1) becomes 1 x x dx – ∫ − = 0 1 2 3 3 3 1 0 1 ( – ) – ( – ) ( – ) x x dx x x dx x x dx − + + ∫ ∫ ∫ 2 3 0 0 0 ( ) ( ) ( ) 2 ( ) a a a a a f x dx f x dx f x dx f x dx − = + = ∫ ∫ ∫ ∫ a f x dx f x dx f x dx − = − + = ∫ ∫ ∫ 0 0 ( ) ( ) ( ) 0 a a a = 0 1 2 3 3 3 1 0 1 ( – ) ( – ) ( – ) x x dx x x dx x x dx − + + ∫ ∫ ∫ = = ( ) 1 1 1 1 1 1 – – – 4 – 2 – – 4 2 2 4 4 2 + + – 1 0 1 – – – 4 2 2 4 4 2 x x x x x x + + 0 1 2 4 2 2 4 4 2 Example 29 Evaluate 4 2 – 4 sin x dx π ∫ π Solution We observe that sin2 x is an even function. Therefore, by P7 (i), we get sin x dx π ∫ π = 4 2 0 2 sin x dx π ∫ 24 – 4 = 1 1 1 1 1 1 – 2 4 2 2 4 4 2 + + − + − + = 3 3 11 2 2 4 4 − + = = 4 0 (1 cos 2 ) 2 2 x dx π − ∫ = 4 0 (1 cos 2 ) x dx π − ∫ Reprint 2025-26 Example 30 Evaluate 2 0 sin 1 cos x x dx x π Solution Let I = 2 0 sin 1 cos x x dx x π or 2 I = π π sin cos x dx 1 x 2 0 + ∫ or I = 2 0 sin 2 1 cos x dx x π π + ∫ Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1. Therefore, (by P1 ) we get + ∫ . Then, by P4 , we have I = 2 0 ( ) sin ( ) 1 cos ( ) x x dx x π π − π − + π − ∫ I = 1 2 1 – 2 1 dt = 4 = 2 0 ( ) sin 1 cos x x dx x π π − + ∫ = 2 0 sin I 1 cos x dx x π π − + ∫ π = 1 1 – sin – 0 – 4 2 2 4 2 π π π = + ∫ 0 1 – sin 2 2 x x t π − + ∫ = 1 2 2 1 1 dt t − π + ∫ INTEGRALS 277 Example 31 Evaluate 1 5 4 1 sin cos x x dx ∫− Solution Let I = 1 5 4 1 sin cos x x dx − ∫ . Let f(x) = sin5 x cos4 x. Then f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i.e., f is an odd function. Therefore, by P7 (ii), I = 0 = 1 2 0 1 dt t π + ∫ (by P7 , 2 1 since 1+ t is even function) = 2 1 – 1 – 1 1 0 tan tan 1 – tan 0 – 0 4 4 t − π π π = π = π = Reprint 2025-26 278 MATHEMATICS Example 32 Evaluate 4 2 4 4 0 sin sin cos x dx x x Solution Let I = 4 2 4 4 0 sin sin cos x dx x x Then, by P4 Adding (1) and (2), we get Hence I = 4 π Example 33 Evaluate 3 Solution Let I = 3 3 I = π π − + − ∫ = 4 2 4 4 0 cos cos sin x dx x x 2 0 4 4 π π − sin ( ) cos ( ) 2 2 + ∫ ... (1) 2I = 4 4 2 2 2 4 4 0 0 0 sin cos [ ] sin cos 2 x x dx dx x x x sin ( ) 2 π π = + + ∫ ∫ ... (1) π 6 6 x dx x x π π 4 1 tan cos sin dx x dx x x x π π π + π = = = + ∫ ∫ + ∫ π + ∫ 6 1 tan dx x π π cos + ∫ ... (2) π Then, by P3 I = 3 6 Adding (1) and (2), we get 2I = [ ] 3 3 = 3 ∫ π + ∫ ... (2) π π π π π = = − = ∫ . Hence I 12 π = π 6 6 6 3 6 6 dx x π π π π cos sin 3 6 3 6 sin cos x dx x x π π π π + − + + − sin Reprint 2025-26 cos 3 6 π π + − x x x dx Example 34 Evaluate 2 0 log sin x dx π ∫ Solution Let I = 2 0 log sin x dx π ∫ Then, by P4 Adding the two values of I, we get t = π. Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when 2 x π = , 2I = ( ) 2 0 log sin logcos x x dx π + ∫ = ( ) 2 0 log sin cos log 2 log 2 x x dx π + − ∫ (by adding and subtracting log2) = 2 2 0 0 log sin 2 log 2 x dx dx π π − ∫ ∫ (Why?) I = 2 2 0 0 log sin log cos 2 x dx x dx π π π − = ∫ ∫ INTEGRALS 279 Therefore 2I = 0 1 log sin log 2 2 2 t dt π π − ∫ Hence 2 0 log sin x dx π ∫ = – log 2 2 π . = 2 0 2 log sin log 2 2 2 t dt π π − ∫ [by P6 as sin (π – t) = sin t) = 2 0 log sin log 2 2 x dx π π − ∫ (by changing variable t to x) = I log 2 2 π − Reprint 2025-26 280 MATHEMATICS By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 10. 2 0 (2log sin log sin 2 ) x x dx π − ∫ 11. 22 – 2 12. 0 1 sin x dx x π 15. 2 0 18. 4 19. Show that 0 0 ( ) ( ) 2 ( ) a a f x g x dx f x dx = ∫ ∫ , if f and g are defined as f(x) = f(a – x) 1. 2 2 0 cos x dx π ∫ 2. 2 0 4. 5 2 5 5 0 cos 7. 1 + ∫ 5. 5 0 (1 )n x x dx − ∫ 8. 4 0 log (1 tan ) x dx π + ∫ 9. 2 + ∫ 13. 72 – 2 sin cos 1 sin cos x x dx x x π − + ∫ 16. 0 log (1 cos ) x dx π + ∫ 17. 0 a x dx x a x + − ∫ 0 x dx −1 ∫ π sin cos x dx x x EXERCISE 7.10 + ∫ 3. 5 | 2 | x dx − + ∫ 6. 8 sin x dx π ∫ π 14. 2 5 0 cos x dx π ∫ π sin sin cos x dx x x + ∫ 2 x dx − 5 ∫ 0 x x dx 2 − ∫ sin x dx π ∫ π 3 2 2 3 3 0 2 2 π sin cos sin x x x dx Choose the correct answer in Exercises 20 and 21. 20. The value of 2 3 5 21. The value of 2 0 (A) 0 (B) 2 (C) π (D) 1 (A) 2 (B) 3 4 (C) 0 (D) –2 and g(x) + g(a – x) = 4 −π + + + ∫ is 4 3 sin log 4 3 cos x dx x π + + ∫ is 2 ( cos tan 1) x x x x dx π Reprint 2025-26 Example 35 Find cos 6 1 sin 6 x x dx + ∫ Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx Therefore 1 2 1 cos 6 1 sin 6 6 x x dx t dt + = ∫ ∫ Example 36 Find Solution We have Put – 3 3 4 1 3 1 1 – , so t x t dx dt hat x x − = = = Therefore 1 4 1 4 4 5 ( ) 1 3 x x dx t dt x − = ∫ ∫ = 5 ( ) x x dx x − ∫ − − = ∫ ∫ 5 4 1 (1 ) ( ) x x x dx dx x x 1 4 4 1 1 4 4 4 3 Miscellaneous Examples = 3 3 2 2 1 2 1 ( ) C = (1 sin 6 ) C 6 3 9 × + + + t x 5 5 4 4 3 1 4 4 1 C = 1 C 3 5 15 t x × + − + INTEGRALS 281 Example 37 Find 4 Solution We have Now express 2 1 ( 1)( 1) x x − + = 2 A B C ( 1) ( 1) x x x 2 ( 1) ( 1) x dx x x − + ∫ 2 ( 1)( 1) x x x − + = 3 2 1 ( 1) 1 x x x x + + − + − 4 Reprint 2025-26 = 2 1 ( 1) ( 1) ( 1) x x x + + − + ... (1) + + − + ... (2) 282 MATHEMATICS So 1 = A (x 2 + 1) + (Bx + C) (x – 1) = (A + B) x 2 + (C – B) x + A – C Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1, which give 1 1 A , B C – 2 2 = = = . Substituting values of A, B and C in (2), we get Again, substituting (3) in (1), we have Therefore Example 38 Find 2 1 log (log ) (log ) x dx x + ∫ Solution Let 2 1 I log (log ) (log ) x dx x = + ∫ 4 2 2 – 1 2 1 1 1 log 1 – log ( 1) – tan C ( 1) ( 1) 2 2 4 2 x x dx x x x x x x x = + + − + + − + + ∫ = 2 1 log (log ) (log ) x dx dx x + ∫ ∫ 2 ( 1) ( 1) x x x x − + + = 2 2 1 1 1 ( 1) 2( 1) 2 ( 1) 2( 1) x x x x x + + − − − + + 2 1 ( 1) ( 1) x x − + = 2 2 1 1 1 2( 1) 2 ( 1) 2( 1) x x x x − − − + + ... (3) 4 In the first integral, let us take 1 as the second function. Then integrating it by parts, we get we have 2 1 1 – – log log (log ) dx x x dx x x x x = ∫ ∫ ... (2) Again, consider log dx x ∫ , take 1 as the second function and integrate it by parts, I = 2 1 log (log ) log (log ) dx x x x dx x x x − + ∫ ∫ = 2 log (log ) log (log ) dx dx x x x x − + ∫ ∫ ... (1) Reprint 2025-26 Putting (2) in (1), we get Example 39 Find cot tan x x dx + ∫ Solution We have I = cot tan x x dx + ∫ = + tan (1 cot ) x x dx ∫ Put tan x = t 2 , so that sec2 x dx = 2t dt or dx = 4 2 1 t dt + t Then I = 2 4 1 2 1 (1 ) t t dt t t + + ∫ Put 1 t t − = y, so that 2 1 1 t + dt = dy. Then 2 2 I log (log ) log (log ) (log ) x dx dx x x x x x = − − + ∫ ∫ = log (log ) C log x x x x − + = 2 2 2 1 1 1 1 ( 1) 2 = 2 = 2 1 1 1 2 ∫ ∫ ∫ dt dt t t t dt t t t t t 4 2 2 2 + + + + + − + INTEGRALS 283 Example 40 Find 4 sin 2 cos 2 Solution Let 4 sin 2 cos 2 I 9 – cos 2 x x dx x = ∫ I = ( ) x ∫ = 2 – 1 – 1 1 tan 1 2 tan C = 2 tan C 2 2 tan t x t x − − + + 2 2 tan C = 2 tan C 2 2 2 t dy y t 9 – cos (2 ) x x dx − = + + + ∫ y – 1 – 1 2 2 Reprint 2025-26 1 284 MATHEMATICS Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt Therefore –1 1 2 2 1 1 1 1 I – – sin C sin cos 2 C 4 4 3 4 3 9 – dt t x t Example 41 Evaluate 3 2 1 x x dx sin ( ) − π ∫ Solution Here f (x) = | x sin πx | = Therefore Integrating both integrals on righthand side, we get 3 2 1 | sin | x x dx − π ∫ = − = = + = − + ∫ 3 2 1 | sin | x x dx − π ∫ = 3 1 2 1 1 x x dx x x dx sin sin − π + − π ∫ ∫ = 2 2 2 1 1 3 1 − − − = + π π π π π π − ≤ ≤ − π ≤ ≤ x x x = x x x sin for 1 1 3 sin for1 2 3 1 2 1 1 x x dx x x dx sin sin − π − π ∫ ∫ Example 42 Evaluate 2 2 2 2 0 cos sin x dx a x b x π SolutionLet I = 2 2 2 2 2 2 2 2 0 0 ( ) cos sin cos ( ) sin ( ) x dx x dx a x b x a x b x π π π − = + π − + π − ∫ ∫ (using P4 ) Thus 2I = 2 2 2 2 0 cos sin dx = 2 2 2 2 2 2 2 2 0 0 cos sin cos sin dx x dx a x b x a x b x π π π − + + ∫ ∫ = 2 2 2 2 0 I cos sin dx a x b x π π − + ∫ a x b x π π + ∫ + ∫ Reprint 2025-26 or I = 2 2 2 2 2 2 2 2 2 0 0 2 2 cos sin 2 cos sin dx dx a x b x a x b x Integrate the functions in Exercises 1 to 23. 1. 3 1 x x − 2. 1 x a x b + + + 3. 2 1 1 = = ( ) 1 0 2 2 2 2 2 2 0 1 tan t cot π − = = + + ∫ ∫ dt du put x and x u a b t a u b = = 4 2 2 2 2 2 2 2 2 2 0 4 cos sin cos sin π π + + + ∫ ∫ x dx x dx a b x a x b 0 1 tan – tan π π bt au ab a ab b = –1 –1 tan tan π + b a ab a b = π π + + + ∫ ∫ dx dx a x b x a x b x π π π π = ⋅ + + ∫ ∫ (using P6 ) 2 2 4 2 2 2 2 2 2 2 0 4 π π π π Miscellaneous Exercise on Chapter 7 1 0 –1 –1 sec cosec tan cot 1 1 1 x ax x − [Hint:Put x= a t ] INTEGRALS 285 2 π ab 2 12. 3 15. cos3 x e log sinx 16. e 3 logx (x 4 + 1)– 1 17. f ′ (ax + b) [f (ax + b)]n 4. 3 2 4 4 6. 2 5 ( 1) ( 9) x x x + + 7. sin sin ( ) x x a − 8. 5 log 4 log 9. 2 cos x x( 1) + 8 1 x 4 sin x − x 13. (1 ) (2 ) x − x 10. 8 8 5. 1 1 2 3 x x + 2 2 sin cos 1 2sin cos x x x − − 11. 1 cos ( ) cos ( ) x a x b + + x x e + + e e 14. 2 2 1 Reprint 2025-26 [Hint: 11 1 1 2 3 3 6 x x x x 1 = + + , put x = t 6 ] x x e e e e − − ( 1) ( 4) x x + + 3 log 2 log x x 286 MATHEMATICS Evaluate the definite integrals in Exercises 24 to 31. Prove the following (Exercises 32 to 37) 18. 3 1 21. 2 23. 2 2 24. 27. 3 30. 2 1 0 sin 2 tan (sin ) x x dx π − ∫ 31. 4 32. 3 2 1 2 2 log ( 1) 3 3 dx x x = + + ∫ 33. 1 1 sin 1 cos π π − − ∫ x x e dx x 25. 4 4 4 0 sin cos cos sin x x dx x x π π + ∫ 28. 1 0 1 1 [| 1| | 2 | | 3|] x x x dx − + − + − ∫ ( 1) ( 2) x x x x + + + + 22. – 1 1 tan 1 x x − + 2 6 sin sin ( ) x x + α 19. 1 4 x x 1 log ( 1) 2 log x π sin cos sin 2 x x dx x x + + − 2 1 + ∫ 26. 2 2 2 2 0 cos cos 4 sin x dx x x dx + − x x ∫ 29. 4 0 1 x − + 20. 2 sin 2 1 cos 2 x x e x + + x + ∫ π + + ∫ 0 1 x x e dx = ∫ π sin cos 9 16 sin 2 x x dx x Choose the correct answers in Exercises 38 to 40 34. 1 17 4 1 x x dx cos 0 − = ∫ 35. 3 2 0 36. 4 3 0 2 tan 1 log2 x dx π = − ∫ 37. 1 1 0 sin 1 2 x dx − π = − ∫ 38. x x dx 39. 2 cos 2 (sin cos ) x dx x x + ∫ is equal to (A) tan–1 (e x ) + C (B) tan–1 (e –x) + C (C) log (e x – e –x) + C (D) log (e x + e –x) + C e e − + ∫ is equal to Reprint 2025-26 2 sin 3 x dx π = ∫ 40. If f (a + b – x) = f (x), then ( ) b Summary ® Integration is the inverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential of this function, but in the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation. Let F( ) ( ) d x f x dx = . Then we write f x dx x ( ) F ( ) C = + ∫ . These integrals are called indefinite integrals or general integrals, C is called constant of integration. All these integrals differ by a constant. ® Some properties of indefinite integrals are as follows: (A) –1 C sin cos x x + + (B) log |sin cos | C x x + + (A) ( ) 2 (C) log |sin cos | C x x − + (D) 2 1 (sin cos ) x x + (C) ( ) 2 a b a f x dx − ∫ (D) ( ) 2 a a b f b x dx + − ∫ (B) ( ) 2 b b a x f x dx ∫ is equal to a a b f b x dx + + ∫ a a b f x dx + ∫ b b INTEGRALS 287 = 1 1 2 2 ( ) ( ) ... ( ) n n k f x dx k f x dx k f x dx + + + ∫ ∫ ∫ ® Some standard integrals 1. [ ( ) ( )] ( ) ( ) f x g x dx f x dx g x dx + = + ∫ ∫ ∫ 2. For any real number k, k f x dx k f x dx ( ) ( ) = ∫ ∫ More generally, if f 1 , f 2 , f 3 , ... , f n are functions and k 1 , k 2 , ... ,k n are real numbers. Then 1 1 2 2 [ ( ) ( ) ... ( )] n n k f x k f x k f x dx + + + ∫ (i) 1 C 1 n n x x dx n + = + + ∫ , n ≠ – 1. Particularly, dx x = + C ∫ Reprint 2025-26 288 MATHEMATICS ® Integration by partial fractions (xiii) C log x x a a dx a = + ∫ (xiv) 1 dx x log | | C x = + ∫ Recall that a rational function is ratio of two polynomials of the form P( ) Q( ) x x , where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0. If degree of the polynomial P (x) is greater than the degree of the polynomial Q (x), then we may divide P (x) by Q (x) so that P( ) P ( ) 1 T ( ) Q( ) Q( ) x x x x x = + , where T(x) is a (vii) cosec cot – cosec C x x dx x = + ∫ (viii) 1 2 sin C 1 dx x x (iv) 2 sec tan C x dx x = + ∫ (v) 2 cosec – cot C x dx x = + ∫ (vi) sec tan sec C x x dx x = + ∫ (ix) 1 2 cos C 1 dx x x (xi) 1 2 cot C 1 dx x x − = − + + ∫ (xii) C x x e dx e = + ∫ (ii) cos sin C x dx x = + ∫ (iii) sin – cos C x dx x = + ∫ − = − + − ∫ (x) 1 2 tan C 1 dx x x − = + + ∫ − = + − ∫ polynomial in x and degree of P1 (x) is less than the degree of Q(x). T(x) being polynomial can be easily integrated. P ( ) 1 Q( ) x x can be integrated by expressing P ( ) 1 Q( ) x x as the sum of partial fractions of the following type: 1. ( ) ( ) px q x a x b + − − = A B x a x b + − − , a ≠ b 2. 2 ( ) px q x a + − = 2 A B x a ( ) x a + − − Reprint 2025-26 where x 2 + bx + c can not be factorised further. ® Integration by substitution A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. When the integrand involves some trigonometric functions, we use some well known identities to find the integrals. Using substitution technique, we obtain the following standard integrals. (iv) cosec log cosec cot C x dx = x x − + ∫ ® Integrals of some special functions 3. 2 4. 2 5. 2 (iii) sec log sec tan C x dx x x = + + ∫ (i) tan log sec C x dx x = + ∫ (ii) cot log sin C x dx x = + ∫ ( ) ( ) ( ) px qx r x a x b x c + + − − − = A B C x a x b x c + + − − − 2 ( ) ( ) px qx r 2 ( ) ( ) px qx r x a x bx c + + − + + = 2 A B + C x x a x bx c + − + + x a x b + + − − = 2 A B C x a ( ) x a x b + + − − − INTEGRALS 289 ® Integration by parts For given functions f 1 and f 2 , we have (iv) 2 2 2 2 log C dx x x a x a = + − + − ∫ (v) 1 2 2 sin C dx x a a x − = + − ∫ (vi) 2 2 2 2 log | | C dx x x a x a = + + + + ∫ (ii) 2 2 1 log C 2 dx a x a x a a x + = + − − ∫ (iii) 1 2 2 1 tan C dx x x a a a − = + + ∫ (i) 2 2 1 log C 2 dx x a x a a x a − = + − + ∫ Reprint 2025-26 290 MATHEMATICS ® [ ( ) ( )] ( ) C x x e f x f x dx e f x dx + = + ′ ∫ ∫ ® Some special types of integrals integral of the product of two functions = first function × integral of the second function – integral of {differential coefficient of the first function × integral of the second function}. Care must be taken in choosing the first function and the second function. Obviously, we must take that function as the second function whose integral is well known to us. (iv) Integrals of the types 2 2 or dx dx ax bx c + + ax bx c + + ∫ ∫ can be (iii) 2 2 2 2 2 1 sin C 2 2 x a x a x dx a x a − − = − + + ∫ (ii) 2 2 2 2 2 2 2 log C 2 2 x a x a dx x a x x a + = + + + + + ∫ (i) 2 2 2 2 2 2 2 log C 2 2 x a x a dx x a x x a − = − − + − + ∫ transformed into standard form by expressing 2 2 2 2 2 4 b c b c b a x x a x a a a a a + + = + + − , i.e., the ® We have defined ( ) b y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b. Let x be a (v) Integrals of the types 2 2 or px q dx px q dx ax bx c ax bx c + + + + + + ∫ ∫ can be ax2 + bx + c = transformed into standard form by expressing determined by comparing coefficients on both sides. 2 A ( ) B A (2 ) B d px q ax bx c ax b dx + = + + + = + + , where A and B are a f x dx ∫ as the area of the region bounded by the curve Reprint 2025-26 This concept of area function leads to the Fundamental Theorems of Integral Calculus. ® First fundamental theorem of integral calculus the function f is assumed to be continuous on [a, b]. Then A′ (x) = f (x) for all x ∈ [a, b]. ® Second fundamental theorem of integral calculus Let f be a continuous function of x defined on the closed interval [a, b] and given point in [a, b]. Then ( ) x a f x dx ∫ represents the Area function A (x). Let the area function be defined by A(x) = ( ) x let F be another function such that F( ) ( ) d x f x dx = for all x in the domain of f, then ( ) F( ) C F ( ) F ( ) [ ] b b a a f x dx x b a = + = − ∫ . This is called the definite integral of f over the range [a, b], where a and b are called the limits of integration, a being the lower limit and b the upper limit. —v— a f x dx ∫ for all x ≥ a, where INTEGRALS 291 Reprint 2025-26" class_12,8,Application of Integrals,ncert_books/class_12/lemh2dd/lemh202.pdf,"292 MATHEMATICS 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. v One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. – BIRKHOFF v APPLICATION OF INTEGRALS Chapter 8 A.L. Cauchy (1789-1857) 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f(x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f(x). Reprint 2025-26 Fig 8.1 This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express The area A of the region bounded by the curve x = g (y), y-axis and the lines y = c, y = d is given by Here, we consider horizontal strips as shown in the Fig 8.2 Remark If the position of the curve under consideration is below the x-axis, then since f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e., ( ) b a f x dx ∫ . A = A ( ) b b b a a a d ydx f x dx = = ∫ ∫ ∫ A = ( ) d d c c xdy g y dy = ∫ ∫ APPLICATION OF INTEGRALS 293 Fig 8.2 Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig 8.4. Here, A1 < 0 and A2 > 0. Therefore, the area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by A = |A1 | + A2 . Reprint 2025-26 Fig 8.3 294 MATHEMATICS Example 1 Find the area enclosed by the circle x 2 + y 2 = a 2 . Solution From Fig 8.5, the whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both x-axis and y-axis] = 0 4 a ydx ∫ (taking vertical strips) = 2 2 0 4 a a x dx − ∫ Fig 8.4 Since x 2 + y 2 = a 2 gives y = 2 2 ± − a x As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle = 2 2 2 –1 = = 2 2 4 2 2 a a π = π Fig 8.5 0 4 sin 2 2 a x a x a x a − + Reprint 2025-26 Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle Example 2 Find the area enclosed by the ellipse 2 2 Solution From Fig 8.7, the area of the region ABA′B′A bounded by the ellipse = 0 4 a xdy ∫ = 2 2 0 4 a a y dy − ∫ (Why?) = 2 2 2 1 = = 2 2 4 2 2 a a π = π = in 4 , 0, area of theregion AOBA the first quadrant bounded = 0 4 (taking verticalstrips) a ydx ∫ Now 2 2 0 4 sin 2 2 a y a y a y a − − + by thecurve x axis and theordinates x x a − = = (as the ellipse is symmetrical about both x-axis and y-axis) 2 2 x y a b + = 1 gives b 2 2 y a x a = ± − , but as the region AOBA lies in the first 2 2 1 x y a b + = APPLICATION OF INTEGRALS 295 Fig 8.6 quadrant, y is taken as positive. So, the required area is = 2 4 2 2 b a ab a = 2 2 0 4 a b a x dx a − ∫ = 2 2 2 –1 = 2 4 1 0 sin 1 0 2 2 b a a a − × + − 4 sin 2 2 a b x a x a x a a − + (Why?) π = π 0 Reprint 2025-26 Fig 8.7 296 MATHEMATICS Alternatively, considering horizontal strips as shown in the Fig 8.8, the area of the ellipse is Choose the correct answer in the following Exercises 3 and 4. 1. Find the area of the region bounded by the ellipse 2 2 1 16 9 x y + = . 2. Find the area of the region bounded by the ellipse 2 2 1 4 9 x y + = . = 4 0 xdy b ∫ = 4 2 2 = = = 2 4 2 2 a b ab b 2 4 –1 0 sin 1 0 2 2 a b b b × + − π = π 0 a b b y dy b − ∫ (Why?) EXERCISE 8.1 Fig 8.8 Example 3 Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = –1 and x = 1. 3. Area lying in the first quadrant and bounded by the circle x 2 + y 2 = 4 and the lines 4. Area of the region bounded by the curve y 2 = 4x, y-axis and the line y = 3 is (A) π (B) 2 π (C) 3 π (D) 4 π (A) 2 (B) 9 4 (C) 9 3 (D) 9 2 x = 0 and x = 2 is Miscellaneous Examples Reprint 2025-26 Solution As shown in the Fig 8.9, the line y = 3x + 2 meets x-axis at x = 2 3 − and its graph lies below x-axis for and above x-axis for . The required area = Area of the region ACBA + Area of the region ADEA Example 4 Find the area bounded by the curve y = cos x between x = 0 and x = 2π. Solution From the Fig 8.10, the required = = − − + + + ∫ ∫ + + + = 1 25 13 6 6 3 + = 3 3 2 2 2 2 x x x x − 2 1 3 1 2 3 (3 2) (3 2) x dx x dx − 2 1 2 2 3 − − 2 1 3 APPLICATION OF INTEGRALS 297 Fig 8.9 area = area of the region OABO + area of the region BCDB + area of the region DEFD. Thus, we have the required area = = [ ] [ ] [ ] 3 2 2 2 0 3 2 2 sin sin sin x x x π π π π π + + = 1 + 2 + 1 = 4 π 3π 2π 2 2 0 π 3π 2 2 cos cos cos x dx x dx xdx + + ∫ ∫ ∫ Reprint 2025-26 Fig 8.10 298 MATHEMATICS Choose the correct answer in the following Exercises from 4 to 5. 1. Find the area under the given curves and given lines: 2. Sketch the graph of y = x + 3 and evaluate . 3. Find the area bounded by the curve y = sin x between x = 0 and x = 2π. 4. Area bounded by the curve y = x 3 , the x-axis and the ordinates x = – 2 and x = 1 is 5. The area bounded by the curve y = x | x | , x-axis and the ordinates x = – 1 and x = 1 is given by Summary ® The area of the region bounded by the curve y = f (x), x-axis and the lines (A) – 9 (B) 15 4 − (C) 15 4 (D) 17 4 (A) 0 (B) 1 3 (C) 2 3 (D) 4 3 [Hint : y = x 2 if x > 0 and y = – x 2 if x < 0]. (i) y = x 2 , x = 1, x = 2 and x-axis (ii) y = x 4 , x = 1, x = 5 and x-axis Miscellaneous Exercise on Chapter 8 a a = = ydx f x dx ∫ ∫ . ® The area of the region bounded by the curve x = φ (y), y-axis and the lines The origin of the Integral Calculus goes back to the early period of development of Mathematics and it is related to the method of exhaustion developed by the mathematicians of ancient Greece. This method arose in the solution of problems on calculating areas of plane figures, surface areas and volumes of solid bodies etc. In this sense, the method of exhaustion can be regarded as an early method x = a and x = b (b > a) is given by the formula: Area ( ) b b y = c, y = d is given by the formula: Area ( ) d d Historical Note Reprint 2025-26 c c = = φ xdy y dy ∫ ∫ . of integration. The greatest development of method of exhaustion in the early period was obtained in the works of Eudoxus (440 B.C.) and Archimedes (300 B.C.) Systematic approach to the theory of Calculus began in the 17th century. In 1665, Newton began his work on the Calculus described by him as the theory of fluxions and used his theory in finding the tangent and radius of curvature at any point on a curve. Newton introduced the basic notion of inverse function called the anti derivative (indefinite integral) or the inverse method of tangents. During 1684-86, Leibnitz published an article in the Acta Eruditorum which he called Calculas summatorius, since it was connected with the summation of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’. In 1696, he followed a suggestion made by J. Bernoulli and changed this article to Calculus integrali. This corresponded to Newton’s inverse method of tangents. Both Newton and Leibnitz adopted quite independent lines of approach which was radically different. However, respective theories accomplished results that were practically identical. Leibnitz used the notion of definite integral and what is quite certain is that he first clearly appreciated tie up between the antiderivative and the definite integral. Conclusively, the fundamental concepts and theory of Integral Calculus and primarily its relationships with Differential Calculus were developed in the work of P.de Fermat, I. Newton and G. Leibnitz at the end of 17th century. However, this justification by the concept of limit was only developed in the works of A.L. Cauchy in the early 19th century. Lastly, it is worth mentioning the following quotation by Lie Sophie’s: “It may be said that the conceptions of differential quotient and integral which in their origin certainly go back to Archimedes were introduced in Science by the investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis .... The discovery that differentiation and integration are inverse operations belongs to Newton and Leibnitz”. APPLICATION OF INTEGRALS 299 Reprint 2025-26 —v—" class_12,9,Differential Equations,ncert_books/class_12/lemh2dd/lemh203.pdf,"9.1 Introduction In Class XI and in Chapter 5 of the present book, we discussed how to differentiate a given function f with respect to an independent variable, i.e., how to find f ′(x) for a given function f at each x in its domain of definition. Further, in the chapter on Integral Calculus, we discussed how to find a function f whose derivative is the function g, which may also be formulated as follows: An equation of the form (1) is known as a differential equation. A formal definition will be given later. 300 MATHEMATICS vHe who seeks for methods without having a definite problem in mind seeks for the most part in vain. – D. HILBERT v For a given function g, find a function f such that DIFFERENTIAL EQUATIONS dy dx = g(x), where y = f(x) ... (1) Chapter 9 Henri Poincare (1854-1912 ) These equations arise in a variety of applications, may it be in Physics, Chemistry, Biology, Anthropology, Geology, Economics etc. Hence, an indepth study of differential equations has assumed prime importance in all modern scientific investigations. In this chapter, we will study some basic concepts related to differential equation, general and particular solutions of a differential equation, formation of differential equations, some methods to solve a first order - first degree differential equation and some applications of differential equations in different areas. 9.2 Basic Concepts We are already familiar with the equations of the type: x 2 – 3x + 3 = 0 ... (1) sin x + cos x = 0 ... (2) x + y = 7 ... (3) Reprint 2025-26 Let us consider the equation: We see that equations (1), (2) and (3) involve independent and/or dependent variable (variables) only but equation (4) involves variables as well as derivative of the dependent variable y with respect to the independent variable x. Such an equation is called a differential equation. In general, an equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation, e.g., Of course, there are differential equations involving derivatives with respect to more than one independent variables, called partial differential equations but at this stage we shall confine ourselves to the study of ordinary differential equations only. Now onward, we will use the term ‘differential equation’ for ‘ordinary differential equation’. ANote 1. We shall prefer to use the following notations for derivatives: 2 3 , , dy d y d y y y y dx dx dx = = = ′ ′′ ′′′ 2 2 d y dy dx dx + = 0 is an ordinary differential equation .... (5) 2 3 dy x y dx + = 0 ... (4) 2 3 DIFFERENTIAL EQUATIONS 301 9.2.1. Order of a differential equation Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations: 2. For derivatives of higher order, it will be inconvenient to use so many dashes as supersuffix therefore, we use the notation y n for nth order derivative n dy dx = e x ... (6) Reprint 2025-26 n d y dx . The equations (6), (7) and (8) involve the highest derivative of first, second and third order respectively. Therefore, the order of these equations are 1, 2 and 3 respectively. 9.2.2 Degree of a differential equation To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the following differential equations: 302 MATHEMATICS 3 2 2 d y d y dy y dx dx dx + − + = 0 ... (9) 2 3 2 2 2 sin dy dy y dx dx + − = 0 ... (10) 3 3 2 2 3 2 d y d y x dx dx + = 0 ... (8) sin dy dy dx dx + = 0 ... (11) 2 d y y dx + = 0 ... (7) 2 We observe that equation (9) is a polynomial equation in y″′, y″ and y′, equation (10) is a polynomial equation in y′ (not a polynomial in y though). Degree of such differential equations can be defined. But equation (11) is not a polynomial equation in y′ and degree of such a differential equation can not be defined. By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation. In view of the above definition, one may observe that differential equations (6), (7), (8) and (9) each are of degree one, equation (10) is of degree two while the degree of differential equation (11) is not defined. ANote Order and degree (if defined) of a differential equation are always positive integers. Reprint 2025-26 Example 1 Find the order and degree, if defined, of each of the following differential equations: Solution (iii) 2 0 y y y e ′ ′′′ + + = (iii) The highest order derivative present in the differential equation is y′′′ , so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined. (ii) The highest order derivative present in the given differential equation is 2 (i) cos 0 dy x dx − = (ii) 2 2 (i) The highest order derivative present in the differential equation is dy dx , so its order is one. It is a polynomial equation in y′ and the highest power raised to dy dx is one, so its degree is one. its order is two. It is a polynomial equation in 2 power raised to 2 2 d y dx is one, so its degree is one. 2 0 d y dy dy xy x y dx dx dx + − = DIFFERENTIAL EQUATIONS 303 2 d y dx and dy dx and the highest 2 d y dx , so Determine order and degree (if defined) of differential equations given in Exercises 1 to 10. 1. 4 4. 6. 2 ( ) y′′′ + (y″) 3 + (y′) 4 + y 5 = 0 7. y′′′ + 2y″ + y′ = 0 2 cos 0 d y dy dx dx + = 5. 4 sin( ) 0 d y y dx + = ′′′ 2. y′ + 5y = 0 3. 4 2 2 2 EXERCISE 9.1 Reprint 2025-26 2 3 0 ds d s s dt dt + = 2 cos3 sin3 d y x x dx = + 2 9.3. General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the type: x 2 + 1 = 0 ... (1) sin2 x – cos x = 0 ... (2) Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown x in the given equation, L.H.S. becomes equal to the R.H.S.. In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. The curve y = φ (x) is called the solution curve (integral curve) of the given differential equation. Consider the function given by y = φ (x) = a sin (x + b), ... (4) where a, b ∈ R. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3). 304 MATHEMATICS (A) 3 (B) 2 (C) 1 (D) not defined 12. The order of the differential equation 8. y′ + y = e x 9. y″ + (y′) 2 + 2y = 0 10. y″ + 2y′ + sin y = 0 11. The degree of the differential equation Now consider the differential equation 2 (A) 2 (B) 1 (C) 0 (D) not defined 2 sin 1 0 d y dy dy dx dx dx + + + = is 2 2 2 2 3 0 d y dy x y dx dx − + = is 3 2 2 2 0 d y y dx + = ... (3) function y = φ1 (x) = 2sin 4 x π + ... (5) When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefore φ1 is also a solution of equation (3). Let a and b be given some particular values say a = 2 and 4 b π = , then we get a Reprint 2025-26 Function φ consists of two arbitrary constants (parameters) a, b and it is called general solution of the given differential equation. Whereas function φ1 contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. Example 2 Verify that the function y = e – 3x is a solution of the differential equation Solution Given function is y = e – 3x . Differentiating both sides of equation with respect to x , we get Now, differentiating (1) with respect to x, we have Substituting the values of 2 2 , d y dy dx dx and y in the given differential equation, we get L.H.S. = 9 e– 3x + (–3e – 3x ) – 6.e – 3x = 9 e– 3x – 9 e– 3x = 0 = R.H.S.. Therefore, the given function is a solution of the given differential equation. 2 6 0 d y dy y dx dx + − = 2 2 d y dx = 9 e – 3x 2 3 3 dy x e dx − = − ... (1) DIFFERENTIAL EQUATIONS 305 Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation 2 Solution The given function is y = a cos x + b sin x ... (1) Differentiating both sides of equation (1) with respect to x, successively, we get 2 0 d y y dx + = 2 d y dx = – a cos x – b sinx 2 dy dx = – a sinx + b cos x Reprint 2025-26 In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: 1. y = e x + 1 : y″ – y′ = 0 2. y = x 2 + 2x + C : y′ – 2x – 2 = 0 3. y = cos x + C : y′ + sin x = 0 306 MATHEMATICS 10. y = 2 2 a x − x ∈ (–a, a) : x + y dy dx = 0 (y ≠ 0) 11. The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4 12. The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0 4. y = 2 1+ x : y′ = 2 1 xy + x 5. y = Ax : xy′ = y (x ≠ 0) 6. y = x sin x : xy′ = y + x 2 2 x y − (x ≠ 0 and x > y or x < – y) 7. xy = log y + C : y′ = 2 8. y – cos y = x : (y sin y + cos y + x) y′ = y 9. x + y = tan–1y : y 2 y′ + y 2 + 1 = 0 Substituting the values of 2 L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S. Therefore, the given function is a solution of the given differential equation. 2 d y dx and y in the given differential equation, we get EXERCISE 9.2 1 y − xy (xy ≠ 1) 9.4. Methods of Solving First Order, First Degree Differential Equations In this section we shall discuss three methods of solving first order first degree differential equations. 9.4.1 Differential equations with variables separable A first order-first degree differential equation is of the form dy dx = F(x, y) ... (1) Reprint 2025-26 If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x and h(y) is a function of y, then the differential equation (1) is said to be of variable separable type. The differential equation (1) then has the form dy dx = h (y) . g(x) ... (2) If h(y) ≠ 0, separating the variables, (2) can be rewritten as Integrating both sides of (3), we get Thus, (4) provides the solutions of given differential equation in the form C is the arbitrary constant. Example 4 Find the general solution of the differential equation 1 2 dy x dx y + = − , (y ≠ 2) Solution We have Here, H (y) and G (x) are the anti derivatives of 1 h y( ) and g (x) respectively and 1 ( ) dy h y ∫ = g x dx ( ) ∫ ... (4) 1 h y( ) dy = g (x) dx ... (3) H(y) = G(x) + C dy dx = 1 2 x y + − ... (1) DIFFERENTIAL EQUATIONS 307 Separating the variables in equation (1), we get (2 – y) dy = (x + 1) dx ... (2) Integrating both sides of equation (2), we get or 2 2 2 y y − = 2 C1 2 x + +x or x 2 + y 2 + 2x – 4y + 2 C1 = 0 or x 2 + y 2 + 2x – 4y + C = 0, where C = 2C1 which is the general solution of equation (1). (2 ) − y dy ∫ = ( 1) x dx + ∫ Reprint 2025-26 Example 5 Find the general solution of the differential equation 2 Solution Since 1 + y 2 ≠ 0, therefore separating the variables, the given differential equation can be written as Integrating both sides of equation (1), we get or tan–1 y = tan–1x + C which is the general solution of equation (1). Example 6 Find the particular solution of the differential equation 2 4 dy xy dx = − given that y = 1, when x = 0. Solution If y ≠ 0, the given differential equation can be written as Integrating both sides of equation (1), we get 308 MATHEMATICS + y ∫ = 2 1 dx + x ∫ 2 1 dy + y = 2 1 dx + x ... (1) 2 1 dy y ∫ = − 4 x dx ∫ 2 dy y = – 4x dx ... (1) 2 dy 2 1 1 dy y dx x + = + . or 1 y − = – 2x 2 + C or y = 2 1 2 C x − ... (2) Substituting y = 1 and x = 0 in equation (2), we get, C = – 1. given differential equation as 2 1 2 1 y x = + . Example 7 Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x 2 + 1) dx (x ≠ 0). Now substituting the value of C in equation (2), we get the particular solution of the Reprint 2025-26 Solution The given differential equation can be expressed as or dy = 1 2x dx x + ... (1) Integrating both sides of equation (1), we get or y = x 2 + log |x | + C ... (2) Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0. Now substituting the value of C in equation (2) we get the equation of the required curve as y = x 2 + log | x |. Example 8 Find the equation of a curve passing through the point (–2, 3), given that the slope of the tangent to the curve at any point (x, y) is 2 2x y . Solution We know that the slope of the tangent to a curve is given by dy dx . dy ∫ = 1 2x dx x + ∫ dy* = DIFFERENTIAL EQUATIONS 309 so, dy dx = 2 2x y ... (1) Separating the variables, equation (1) can be written as y 2 dy = 2x dx ... (2) Integrating both sides of equation (2), we get 2 y dy ∫ = 2x dx ∫ or 3 dx due to Leibnitz is extremely flexible and useful in many calculation and formal transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers. By treating dx and dy like separate entities, we can give neater expressions to many calculations. Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant, Fritz John Spinger – Verlog New York. * The notation dy 3 y = x 2 + C ... (3) Reprint 2025-26 Substituting x = –2, y = 3 in equation (3), we get C = 5. Substituting the value of C in equation (3), we get the equation of the required curve as Example 9 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself? Solution Let P be the principal at any time t. According to the given problem, or dp dt = P 20 ... (1) separating the variables in equation (1), we get Integrating both sides of equation (2), we get or P = 20 C1 t e e⋅ or P = C 20 t e (where e C1 = C ) ... (3) 310 MATHEMATICS 3 2 5 3 y = + x or 1 2 3 y x = + (3 15) log P = C1 20 t + dp dt = 5 P 100 × P dp = 20 dt ... (2) Now P = 1000, when t = 0 Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3), gives For each of the differential equations in Exercises 1 to 10, find the general solution: 1. 1 cos 1 cos dy x dx x − = + 2. 2 4 ( 2 2) dy y y dx = − − < < Let t years be the time required to double the principal. Then 2000 = 1000 20 t e ⇒ t = 20 loge 2 EXERCISE 9.3 P = 1000 20 t e Reprint 2025-26 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition: 12. 2 ( 1) 1 dy x x dx − = ; y = 0 when x = 2 13. cos dy a dx = (a ∈ R); y = 1 when x = 0 14. tan dy y x dx = ; y = 1 when x = 0 15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e x sin x. 11. 3 2 ( 1) dy x x x dx + + + = 2x 2 + x; y = 1 when x = 0 3. 1 ( 1) dy y y dx + = ≠ 4. sec2 x tan y dx + sec2 y tan x dy = 0 5. (e x + e –x ) dy – (e x – e –x ) dx = 0 6. 2 2 (1 ) (1 ) dy x y dx = + + 7. y log y dx – x dy = 0 8. 5 dy 5 x y dx = − 9. 1 sin dy x dx − = 10. e x tan y dx + (1 – e x ) sec2 y dy = 0 DIFFERENTIAL EQUATIONS 311 16. For the differential equation ( 2) ( 2) dy xy x y dx = + + , find the solution curve passing through the point (1, –1). 17. Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point. 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1). 19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds. Reprint 2025-26 9.4.2 Homogeneous differential equations Consider the following functions in x and y If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get 312 MATHEMATICS 20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge 2 = 0.6931). 21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0.5 = 1.648). 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present? 23. The general solution of the differential equation dy x y e dx + = is F1 (x, y) = y 2 + 2xy, F2 (x, y) = 2x – 3y, F3 (x, y) = cos y x , F4 (x, y) = sin x + cos y F1 (λx, λy) = λ 2 (y 2 + 2xy) = λ 2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) (A) e x + e –y = C (B) e x + e y = C (C) e –x + e y = C (D) e –x + e –y = C F4 (λx, λy) = sin λx + cos λy ≠ λ n F4 (x, y), for any n ∈ N Here, we observe that the functions F1 , F2 , F3 can be written in the form F(λx, λy) = λ n F(x, y) but F4 can not be written in this form. This leads to the following definition: We note that in the above examples, F1 , F2 , F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function. F3 (λx, λy) = cos cos y y x x λ = λ = λ 0 F3 (x, y) A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λ n F(x, y) for any nonzero constant λ. Reprint 2025-26 We also observe that or F1 (x, y) = 2 2 2 2 1 x x y y h y y + = or F2 (x, y) = 1 1 4 2 3 x x y y h y y − = or F4 (x, y) ≠ 7 n x y h y , for any n ∈ N Therefore, a function F (x, y) is a homogeneous function of degree n if A differential equation of the form dy dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero. To solve a homogeneous differential equation of the type F4 (x, y) ≠ 6 n y x h x , for any n ∈ N F1 (x, y) = 2 2 2 2 1 y y y 2 x x h x x x + = F2 (x, y) = 1 1 3 3 2 y y x x h x x − = F3 (x, y) = 0 0 5 cos y y x x h x x = F(x, y) = or n n y x x g y h x y DIFFERENTIAL EQUATIONS 313 We make the substitution y = v . x ... (2) Differentiating equation (2) with respect to x, we get Substituting the value of dy dx from equation (3) in equation (1), we get F , ( ) dy x y dx = = y g x ... (1) dy dx = dv v x dx + ... (3) Reprint 2025-26 or dv x dx = g (v) – v ... (4) Separating the variables in equation (4), we get Integrating both sides of equation (5), we get we replace v by y x . 314 MATHEMATICS ANote If the homogeneous differential equation is in the form F( , ) dx x y dy = where, F (x, y) is homogenous function of degree zero, then we make substitution above by writing F( , ) . dx x x y h dy y = = x v y = i.e., x = vy and we proceed further to find the general solution as discussed Equation (6) gives general solution (primitive) of the differential equation (1) when ( ) dv g v v − ∫ = 1 dx C x + ∫ ... (6) ( ) dv g v v − = dx x ... (5) dv v x dx + = g (v) Example 10 Show that the differential equation (x – y) dy dx = x + 2y is homogeneous and solve it. Solution The given differential equation can be expressed as Let F(x, y) = x y 2 x y + − Now F(λx, λy) = 0 ( 2 ) ( , ) ( ) x y f x y x y λ + = λ ⋅ λ − dy dx = x y 2 x y + − ... (1) Reprint 2025-26 Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. Alternatively, function of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution Differentiating equation (3) with respect to, x we get Substituting the value of y and dy dx in equation (1) we get or dv x dx = 1 2 1 R.H.S. of differential equation (2) is of the form g y x and so it is a homogeneous y dy x dx y x + = − = y g x ... (2) dv v x dx + = 1 2 1 2 1 1 dy dx = dv v x dx + ... (4) y = vx ... (3) v v + − v v v + − − DIFFERENTIAL EQUATIONS 315 or dv x dx = or 2 1 1 v dv v v − + + = dx x − Integrating both sides of equation (5), we get or = – log | x| + C1 Reprint 2025-26 = 2 1 1 v v v + + − or or or or (Why?) Replacing v by y x , we get or or or 2 2 1 1 2 log ( ) 2 3 tan 2C 3 y x y xy x x − + + + = + 316 MATHEMATICS 2 2 1 2 1 1 2 log 1 3 tan C 2 3 y y y x x x x x − + + + = + or 2 2 1 2 log ( ) 2 3 tan C 3 − + + + = + x y x xy y x which is the general solution of the differential equation (1) Example 11 Show that the differential equation cos cos y dy y x y x x dx x = + is homogeneous and solve it. Solution The given differential equation can be written as dy dx = Reprint 2025-26 cos y y x x y x x cos + ... (1) It is a differential equation of the form F( , ) dy x y dx = . Here F(x, y) = Replacing x by λx and y by λy, we get Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution y = vx ... (2) Differentiating equation (2) with respect to x, we get Substituting the value of y and dy dx in equation (1), we get F(λx, λy) = 0 [ cos ] [F( , )] cos y y x x x y y x x dy dx = dv v x dx + ... (3) cos y y x x y x x λ + = λ λ cos + DIFFERENTIAL EQUATIONS 317 or dv x dx = cos 1 cos v v v v + − or dv x dx = 1 cos v or cos v dv = dx x Therefore cos v dv ∫ = 1 dx x ∫ dv v x dx + = cos 1 cos v v v + Reprint 2025-26 or sin v = log | x | + log |C| or sin v = log |Cx| Replacing v by y x , we get sin y x = log |Cx| which is the general solution of the differential equation (1). Example 12 Show that the differential equation 2 2 0 x x y y y e dx y x e dy + − = is homogeneous and find its particular solution, given that, x = 0 when y = 1. Solution The given differential equation can be written as Let F(x, y) = 2 318 MATHEMATICS dx dy = 2 λ − =λ λ x e y xe y 2 x y 2 y e − ... (1) ye − x y 2 x y x y x y Then F(λx, λy) = 0 2 Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution x = vy ... (2) Differentiating equation (2) with respect to y, we get dx dy = + dv v y dy Reprint 2025-26 xe y ye x y [F( , )] x y Substituting the value of and dx x dy in equation (1), we get or dv y dy = 2 1 2 v or dv y dy = 1 2 v e − or 2e v dv = dy y − or 2 v e dv ⋅ ∫ = dy y −∫ or 2 e v = – log |y| + C and replacing v by x y , we get Substituting x = 0 and y = 1 in equation (3), we get 2 e 0 + log |1| = C ⇒ C = 2 Substituting the value of C in equation (3), we get 2 x y e + log | y | = C ... (3) dv v y dy + = 2 1 2 v v v e e − v v e v e − − DIFFERENTIAL EQUATIONS 319 x y e + log | y | = 2 which is the particular solution of the given differential equation. Example 13 Show that the family of curves for which the slope of the tangent at any point (x, y) on it is 2 2 Solution We know that the slope of the tangent at any point on a curve is dy dx . Therefore, dy dx = 2 2 2 x y xy + , is given by x 2 – y 2 = cx. 2 Reprint 2025-26 2 x y xy + or dy dx = Clearly, (1) is a homogenous differential equation. To solve it we make substitution Differentiating y = vx with respect to x, we get or dv v x dx + = 2 1 2 v v + or dv x dx = or 2 2 1 v dv v − = dx x − 320 MATHEMATICS 1 v dv − v = dx x 2 2 dy dx = dv v x dx + y = vx 2 1 2 1 2 v v − + ... (1) 2 y x y x 2 Therefore 2 2 1 v dv v − ∫ = 1 dx x −∫ or log | v 2 – 1 | = – log | x | + log |C1 | or log |(v 2 – 1) (x)| = log |C1 | or (v 2 – 1) x = ± C1 Replacing v by y x , we get or (y 2 – x 2 ) = ± C1 x or x 2 – y 2 = Cx 2 1 y x x − = ± C1 2 Reprint 2025-26 In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them. For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 10. 12. x 2 dy + (xy + y 2 ) dx = 0; y = 1 when x = 1 11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 1. (x 2 + xy) dy = (x 2 + y 2 ) dx 2. x y y x + ′ = 3. (x – y) dy – (x + y) dx = 0 4. (x 2 – y 2 ) dx + 2xy dy = 0 5. 2 2 2 2 dy x x y xy dx = − + 6. x dy – y dx = 2 2 x y dx + 7. cos sin sin cos y y y y x y y dx y x x dy x x x x + = − 8. sin 0 dy y x y x dx x − + = 9. log 2 0 y y dx x dy x dy x + − = EXERCISE 9.4 DIFFERENTIAL EQUATIONS 321 13. when x = 1 14. cosec 0 dy y y dx x x − + = ; y = 0 when x = 1 15. 2 2 2 2 0 dy xy y x dx + − = ; y = 2 when x = 1 16. A homogeneous differential equation of the from dx x h dy y = can be solved by making the substitution. (A) y = vx (B) v = yx (C) x = vy (D) x = v Reprint 2025-26 9.4.3 Linear differential equations A differential equation of the from where, P and Q are constants or functions of x only, is known as a first order linear differential equation. Some examples of the first order linear differential equation are Another form of first order linear differential equation is 322 MATHEMATICS 17. Which of the following is a homogeneous differential equation? (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x 3 + y 3 ) dy = 0 (C) (x 3 + 2y 2 ) dx + 2xy dy = 0 (D) y 2 dx + (x 2 – xy – y 2 ) dy = 0 log dy y dx x x + = 1 x dy 1 y dx x + = e x P1 dx x dy + = Q1 P dy y dx + = Q dy y dx + = sin x where, P1 and Q1 are constants or functions of y only. Some examples of this type of differential equation are dx x dy + = cos y To solve the first order linear differential equation of the type Multiply both sides of the equation by a function of x say g (x) to get g(x) dy dx + P.(g(x)) y = Q . g (x) ... (2) dx x2 dy y − + = y 2e – y P dy y dx + = Q ... (1) Reprint 2025-26 Choose g (x) in such a way that R.H.S. becomes a derivative of y . g (x). i.e. g (x) dy dx + P. g(x) y = d dx [y . g (x)] or g (x) dy dx + P. g(x) y = g (x) dy dx + y g′ (x) ⇒ P. g (x) = g′ (x) or P = ( ) ( ) g x g x ′ Integrating both sides with respect to x, we get or P⋅dx ∫ = log (g (x)) or g (x) = P dx e ∫ of some function of x and y. This function g(x) = P dx e ∫ is called Integrating Factor (I.F.) of the given differential equation. Substituting the value of g (x) in equation (2), we get On multiplying the equation (1) by g(x) = P dx e ∫ , the L.H.S. becomes the derivative Pdx ∫ = ( ) ( ) g x dx g x ′ ∫ DIFFERENTIAL EQUATIONS 323 or d dx ye Pdx ∫ = Integrating both sides with respect to x, we get or y = e e dx − dx dx ∫ ∫ + ∫ P P . . Q C which is the general solution of the differential equation. Reprint 2025-26 = = Q P .e dx dx ∫ ∫ Steps involved to solve first order linear differential equation: where, P1 and Q1 are constants or functions of y only. Then I.F = P1 dy e and the solution of the differential equation is given by Example 14 Find the general solution of the differential equation cos dy y x dx − = . Solution Given differential equation is of the form Therefore I.F = Multiplying both sides of equation by I.F, we get 324 MATHEMATICS (iii) Write the solution of the given differential equation as (ii) Find the Integrating Factor (I.F) = . (i) Write the given differential equation in the form P Q dy y dx + = where P, Q are In case, the first order linear differential equation is in the form P Q 1 1 dx x dy + = , constants or functions of x only. P Q dy y dx + = , where P = –1 and Q = cos x x . (I.F) = ( ) Q × I.F C 1 dy + ∫ y (I.F) = or ( ) dy x ye dx − = e –x cos x On integrating both sides with respect to x, we get Let I = cos x e x dx − ∫ ye– x = cos C x e x dx − + ∫ ... (1) Reprint 2025-26 = e –x cos x = cos ( sin ) ( ) 1 x e x x x e dx − − − − − − ∫ = cos sin cos x x x x e x e x e dx − − − − + − ∫ or I = – e –x cos x + sin x e –x – I or 2I = (sin x – cos x) e –x or I = (sin cos ) 2 Substituting the value of I in equation (1), we get or y = sin cos C 2 x x x e − + which is the general solution of the given differential equation. Example 15 Find the general solution of the differential equation 2 2 ( 0) dy x y x x dx + = ≠ . Solution The given differential equation is Dividing both sides of equation (1) by x, we get 2 dy x y dx + = x 2 ... (1) ye– x = sin cos C 2 x x x e − − + = cos sin x x x e x e dx − − − −∫ = cos sin (– ) cos ( ) x x x x e x e x e dx − − − − − − − ∫ x x x e − − DIFFERENTIAL EQUATIONS 325 which is a linear differential equation of the type P Q dy y dx + = , where 2 P x = and Q = x. So I.F = 2 dx x e ∫ = e 2 log x = 2 log 2 x e x = log ( ) [ ( )] f x as e f x = Therefore, solution of the given equation is given by or y = 2 2 C 4 x x − + which is the general solution of the given differential equation. y . x 2 = 2 ( ) ( ) C x x dx + ∫ = 3 x dx + C ∫ dy 2 y dx x + = x Reprint 2025-26 Example 16 Find the general solution of the differential equation y dx – (x + 2y 2 ) dy = 0. Solution The given differential equation can be written as Q1 = 2y. Therefore 1 1 log log( ) 1 I.F dy y y y e e e y − − − ∫ = = = = Hence, the solution of the given differential equation is or x y = (2 ) C dy + ∫ or x y = 2y + C or x = 2y 2 + Cy which is a general solution of the given differential equation. Example 17 Find the particular solution of the differential equation 326 MATHEMATICS This is a linear differential equation of the type P Q 1 1 dx x dy + = , where 1 1 P y = − and dx x dy y − = 2y 1 x y = 1 (2 ) C y dy y + ∫ given that y = 0 when 2 x π = . Solution The given equation is a linear differential equation of the type P Q dy y dx + = , where P = cot x and Q = 2x + x 2 cot x. Therefore Hence, the solution of the differential equation is given by y . sin x = ∫(2x + x 2 cot x) sin x dx + C + cot dy y x dx = 2x + x 2 cot x (x ≠ 0) I.F = e e x x dx x cot log sin sin ∫ = = Reprint 2025-26 or y sin x = ∫ 2x sin x dx + ∫ x 2 cos x dx + C or y sin x = 2 2 2 2 2 sin cos cos C 2 2 x x x x dx x x dx − + + ∫ ∫ or y sin x = 2 2 2 x x x x dx x x dx sin cos cos C − + + ∫ ∫ or y sin x = x 2 sin x + C ... (1) Substituting y = 0 and 2 x π = in equation (1), we get or C = 2 Substituting the value of C in equation (1), we get or y = 2 2 (sin 0) 4 sin x x x π − ≠ which is the particular solution of the given differential equation. Example 18 Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. y sin x = 2 2 sin 4 x x π − 0 = 2 sin C 2 2 π π + 4 − π DIFFERENTIAL EQUATIONS 327 Solution We know that the slope of the tangent to the curve is dy dx . Therefore, dy dx = x + xy or dy xy dx − = x ... (1) This is a linear differential equation of the type P Q dy y dx + = , where P = – x and Q = x. Therefore, I.F = Reprint 2025-26 2 x x dx e e − −∫ = 2 Hence, the solution of equation is given by Let I = 2 ( ) 2 x x dx e − ∫ Let 2 Therefore, I = 2 – 2 x t t e dt e e − − = − = ∫ Substituting the value of I in equation (2), we get or y = 2 1 C x − + e ... (3) Now (3) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through (0, 1). Substituting x = 0 and y = 1 in equation (3) we get 1 = – 1 + C . e 0 or C = 2 Substituting the value of C in equation (3), we get 2 1 2 x − + e which is the equation of the required curve. 328 MATHEMATICS 2 x t − = , then – x dx = dt or x dx = – dt. 2 x y e − ⋅ = ( ) 2 ( ) C 2 x x dx e − + ∫ ... (2) 2 x y e − = 2 2 y = 2 + C − − x e 2 2 2 For each of the differential equations given in Exercises 1 to 12, find the general solution: 1. 2 sin dy y x dx + = 2. 2 3 dy x y e dx − + = 3. dy y 2 x dx x + = 4. (sec ) tan 0 2 dy x y x x dx 6. 2 2 log dy x y x x dx + = 7. 2 log log dy x x y x dx x + = 8. (1 + x 2 ) dy + 2xy dx = cot x dx (x ≠ 0) π + = ≤ < 5. 2 cos tan dy x y x dx + = 0 2 x π ≤ < EXERCISE 9.5 Reprint 2025-26 For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition: 13. 2 tan sin ; 0 when 3 dy y x x y x dx 14. 2 2 1 (1 ) 2 ; 0 when 1 1 dy x xy y x dx x + + = = = + 15. 3 cot sin 2 ; 2 when 2 dy y x x y x dx 16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. 17. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5. 18. The Integrating Factor of the differential equation 2 2 dy x y x dx − = is 11. y dx + (x – y 2 ) dy = 0 12. 2 ( 3 ) ( 0) dy x y y y dx + = > . 9. cot 0 ( 0) dy x y x xy x x dx + − + = ≠ 10. ( ) 1 dy x y dx + = (A) e –x (B) e –y (C) 1 x (D) x π + = = = π − = = = DIFFERENTIAL EQUATIONS 329 Example 19 Verify that the function y = c 1 e ax cos bx + c 2 e ax sin bx, where c 1 , c 2 are arbitrary constants is a solution of the differential equation 19. The Integrating Factor of the differential equation (A) 2 1 y −1 (B) 2 1 2 (1 ) dx y yx dy − + = ay y ( 1 1) − < < is ( ) 2 2 2 2 2 0 d y dy a a b y dx dx − + + = Miscellaneous Examples y −1 (C) 2 1 1− y (D) 2 1 Reprint 2025-26 1− y Solution The given function is y = e ax [c1 cosbx + c2 sinbx] ... (1) Differentiating both sides of equation (1) with respect to x, we get or dy dx = 2 1 2 1 [( )cos ( )sin ] ax e bc ac bx a c bc bx + + − ... (2) Differentiating both sides of equation (2) with respect to x, we get Substituting the values of 2 330 MATHEMATICS L.H.S. = 2 2 2 2 2 1 2 1 2 1 [ 2 )sin ( 2 )cos ] ax e a c abc b c bx a c abc b c bx − − + + − 2 d y dx = 2 1 2 1 [( ) ( sin ) ( ) ( cos )] ax e bc a c b bx ac bc b bx + − + − dy dx = [– sin cos cos sin 1 2 ] [ 1 2 ] ax ax e bc bx bc bx c bx c bx e a + + + ⋅ 2 = 2 2 2 2 2 1 2 1 2 1 [( 2 ) sin ( 2 ) cos ] ax e a c abc b c bx a c abc b c bx − − + + − + 2 1 2 1 [( ) cos ( ) sin ] . ax bc ac bx ac bc bx e a + + − 2 1 2 1 2 [( )cos ( )sin ] ax − + + − ae bc ac bx ac bc bx 2 2 1 2 ( ) [ cos sin ] ax + + a b e c bx c bx + 2 , d y dy dx dx and y in the given differential equation, we get Hence, the given function is a solution of the given differential equation. Example 20 Find the particular solution of the differential equation log 3 4 dy x y dx = + given that y = 0 when x = 0. Solution The given differential equation can be written as = ( ) 2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 2 2 1 2 1 2 1 1 1 = [0 sin 0cos ] ax e bx bx × + = e ax × 0 = 0 = R.H.S. ( 2 2 2 )cos ax a c abc b c a c abc a c b c bx e a c abc b c abc a c a c b c bx − − − + + + + + − − − + + dy dx = e (3x + 4y) 2 2 2 sin Reprint 2025-26 or dy dx = e 3x . e 4y ... (1) Separating the variables, we get Therefore 4 y e dy − ∫ = 3x e dx ∫ or 4 or 4 e 3x + 3 e – 4y + 12 C = 0 ... (2) Substituting x = 0 and y = 0 in (2), we get Substituting the value of C in equation (2), we get 4 e 3x + 3 e – 4y – 7 = 0, which is a particular solution of the given differential equation. Example 21 Solve the differential equation Solution The given differential equation can be written as (x dy – y dx) y sin y x = (y dx + x dy) x cos y x . 4 + 3 + 12 C = 0 or C = 7 12 − 4 y dy e = e 3x dx 4 y e − − = 3 C 3 x e + DIFFERENTIAL EQUATIONS 331 or dy dx = Dividing numerator and denominator on RHS by x 2 , we get 2 2 sin cos cos sin y y y y x y x dy xy y dx x x x x − = + dy dx = y y xy y x x y y xy x x x y y y y x x x x y y y x x x 2 cos sin cos sin sin cos sin cos + − + − 2 2 Reprint 2025-26 2 ... (1) Clearly, equation (1) is a homogeneous differential equation of the form dy y g dx x = . To solve it, we make the substitution y = vx ... (2) or dy dx = dv v x dx + or dv v x dx + = 2 cos sin sin cos v v v v v v v + − (using (1) and (2)) or dv x dx = 2 cos sin cos v v v v v − or sin cos cos v v v dv v v − = 2 dx x Therefore sin cos cos v v v dv v v − ∫ = 1 2 dx x ∫ or 1 tan v dv dv v − ∫ ∫ = 1 2 dx x ∫ or log sec log | | v v − = 2log | | log | C | 1 x + or 2 sec log v v x = log |C1 | 332 MATHEMATICS or 2 secv v x = ± C1 ... (3) Replacing v by y x in equation (3), we get or sec y x = C xy which is the general solution of the given differential equation. sec y x y x x ( ) 2 Reprint 2025-26 = C where, C = ± C1 Example 22 Solve the differential equation (tan–1y – x) dy = (1 + y 2 ) dx. Solution The given differential equation can be written as Now (1) is a linear differential equation of the form P1 dx dy + x = Q1 , Therefore, I .F = 2 1 1 1 tan dy y y e e − + ∫ = Thus, the solution of the given differential equation is Let I = 1 1 tan 2 tan 1 Substituting tan–1 y = t so that 2 1 1 dy dt y = + , we get where, P1 = 2 1 1+ y and 1 1 tan y xe − = 1 1 tan 2 tan C 1 − − + + ∫ ... (2) − − + ∫ 2 1 dx x dy y + + = y y e dy y y y e dy y 1 2 tan Q 1 y y − = + . 2 tan 1 + ... (1) y y − 1 DIFFERENTIAL EQUATIONS 333 or I = 1 tan y e − (tan–1y –1) Substituting the value of I in equation (2), we get or x = 1 1 tan (tan 1) C y y e − − − − + which is the general solution of the given differential equation. 1. For each of the differential equations given below, indicate its order and degree (if defined). I = t t e dt ∫ = t e t – ∫1 . e t dt = t et – e t = e t (t – 1) 1 1 tan tan 1 . (tan 1) C y y x e e y − − − = − + Miscellaneous Exercise on Chapter 9 Reprint 2025-26 334 MATHEMATICS 2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. 3. Prove that x 2 – y 2 = c (x 2 + y 2 ) 2 is the general solution of differential equation 4. Find the general solution of the differential equation 2 (x 3 – 3x y2 ) dx = (y 3 – 3x 2y) dy, where c is a parameter. (iv) x 2 = 2y 2 log y : 2 2 ( ) 0 dy x y xy dx + − = (iii) 4 3 (iii) y = x sin 3x : 2 (ii) y = e x (a cos x + b sin x) : 2 (i) 2 2 (i) xy = a ex + b e–x + x 2 : 2 2 2 2 2 0 d y dy x xy x dx dx + − + − = 2 5 6 log d y dy x y x dx dx + − = (ii) 4 3 sin 0 d y d y dx dx − = 2 2 2 0 d y dy y dx dx − + = 2 9 6cos3 0 d y y x dx + − = 3 2 4 7 sin dy dy y x dx dx − + = 2 1 0 1 dy y dx x − + = − . 5. Show that the general solution of the differential equation 2 6. Find the equation of the curve passing through the point 0, 4 π whose differential equation is sin x cos y dx + cos x sin y dy = 0. 7. Find the particular solution of the differential equation (1 + e 2x ) dy + (1 + y 2 ) e x dx = 0, given that y = 1 when x = 0. 8. Solve the differential equation 2 ( 0) x x y y y e dx x e y dy y = + ≠ . given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter. Reprint 2025-26 2 1 0 1 dy y y dx x x + + + = + + is 10. Solve the differential equation 2 1( 0) x e y dx x x x dy − − = ≠ . 12. Find a particular solution of the differential equation (x + 1) dy dx = 2 e –y – 1, given 13. The general solution of the differential equation 0 y dx x dy y − = is 14. The general solution of a differential equation of the type P Q 1 1 dx x dy + = is 11. Find a particular solution of the differential equation cot dy y x dx + = 4x cosec x 9. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t) (A) xy = C (B) x = Cy 2 (C) y = Cx (D) y = Cx 2 (A) ( ) P1 P1 Q1 C dy dy y e e dy ∫ ∫ = + ∫ (x ≠ 0), given that y = 0 when 2 x π = . that y = 0 when x = 0. DIFFERENTIAL EQUATIONS 335 15. The general solution of the differential equation e x dy + (y ex + 2x) dx = 0 is (D) ( ) P1 P1 Q1 C dx dx x e e dx ∫ ∫ = + ∫ (A) x ey + x 2 = C (B) x ey + y 2 = C (C) y ex + x 2 = C (D) y ey + x 2 = C (B) ( ) P1 P1 1 . Q C dx dx y e e dx ∫ ∫ = + ∫ (C) ( ) P1 P1 Q C 1 dy dy x e e dy ∫ ∫ = + ∫ Reprint 2025-26 336 MATHEMATICS Summary ® An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation. ® Order of a differential equation is the order of the highest order derivative occurring in the differential equation. ® Degree of a differential equation is defined if it is a polynomial equation in its derivatives. ® Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it. ® A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution. ® Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing x should remain with dx. ® A differential equation which can be expressed in the form ® A differential equation of the form +P Q dy y dx = , where P and Q are constants functions of degree zero is called a homogeneous differential equation. ( , ) or ( , ) dy dx f x y g x y dx dy = = where, f (x, y) and g(x, y) are homogenous One of the principal languages of Science is that of differential equations. Interestingly, the date of birth of differential equations is taken to be November, 11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black and white the identity 1 2 2 y dy y = ∫ , thereby introducing both the symbols ∫ and dy. Leibnitz was actually interested in the problem of finding a curve whose tangents were prescribed. This led him to discover the ‘method of separation of variables’ 1691. A year later he formulated the ‘method of solving the homogeneous or functions of x only is called a first order linear differential equation. Historical Note Reprint 2025-26 differential equations of the first order’. He went further in a very short time to the discovery of the ‘method of solving a linear differential equation of the first-order’. How surprising is it that all these methods came from a single man and that too within 25 years of the birth of differential equations! In the old days, what we now call the ‘solution’ of a differential equation, was used to be referred to as ‘integral’ of the differential equation, the word being coined by James Bernoulli (1654 - 1705) in 1690. The word ‘solution was first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost hundred years since the birth of differential equations. It was Jules Henri Poincare (1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the word ‘solution’ has found its deserved place in modern terminology. The name of the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748), a younger brother of James Bernoulli. Application to geometric problems were also considered. It was again John Bernoulli who first brought into light the intricate nature of differential equations. In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the differential equation x 2 y″ = 2y, which led to three types of curves, viz., parabolas, hyperbolas and a class of cubic curves. This shows how varied the solutions of such innocent looking differential equation can be. From the second half of the twentieth century attention has been drawn to the investigation of this complicated nature of the solutions of differential equations, under the heading ‘qualitative analysis of differential equations’. Now-a-days, this has acquired prime importance being absolutely necessary in almost all investigations. DIFFERENTIAL EQUATIONS 337 Reprint 2025-26 —v—" class_12,10,Vector Algebra,ncert_books/class_12/lemh2dd/lemh204.pdf,"10.1 Introduction In our day to day life, we come across many queries such as – What is your height? How should a football player hit the ball to give a pass to another player of his team? Observe that a possible answer to the first query may be 1.6 meters, a quantity that involves only one value (magnitude) which is a real number. Such quantities are called scalars. However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned). Such quantities are called vectors. In mathematics, physics and engineering, we frequently come across with both types of quantities, namely, scalar quantities such as length, mass, time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc. and vector quantities like displacement, velocity, acceleration, force, weight, momentum, electric field intensity etc. 338 MATHEMATICS vIn most sciences one generation tears down what another has built and what one has established another undoes. In Mathematics alone each generation builds a new story to the old structure. – HERMAN HANKEL v VECTOR ALGEBRA Chapter 10 In this chapter, we will study some of the basic concepts about vectors, various operations on vectors, and their algebraic and geometric properties. These two type of properties, when considered together give a full realisation to the concept of vectors, and lead to their vital applicability in various areas as mentioned above. 10.2 Some Basic Concepts Let ‘l’ be any straight line in plane or three dimensional space. This line can be given two directions by means of arrowheads. A line with one of these directions prescribed is called a directed line (Fig 10.1 (i), (ii)). Reprint 2025-26 W.R. Hamilton (1805-1865) Now observe that if we restrict the line l to the line segment AB, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed line segment (Fig 10.1(iii)). Thus, a directed line segment has magnitude as well as direction. Definition 1 A quantity that has magnitude as well as direction is called a vector. simply as , and read as ‘vector ’ or ‘vector ’. The point A from where the vector starts is called its initial point, and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as | |, or | |, or a. The arrow indicates the direction of the vector. ANote Since the length is never negative, the notation | | < 0 has no meaning. Notice that a directed line segment is a vector (Fig 10.1(iii)), denoted as or Fig 10.1 VECTOR ALGEBRA 339 Position Vector From Class XI, recall the three dimensional right handed rectangular coordinate system (Fig 10.2(i)). Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using distance formula (from Class XI), the magnitude of (or ) is given by are denoted by , , , etc., respectively (Fig 10.2 (ii)). | |= 2 2 2 x y z + + In practice, the position vectors of points A, B, C, etc., with respect to the origin O Reprint 2025-26 Direction Cosines Consider the position vector of a point P(x, y, z) as in Fig 10.3. The angles α, β, γ made by the vector with the positive directions of x, y and z-axes respectively, are called its direction angles. The cosine values of these angles, i.e., cosα, cosβ and cos γ are called direction cosines of the vector , and usually denoted by l, m and n, respectively. 340 MATHEMATICS O B C z Z r Fig 10.2 y P( ) x,y,z P( ) x,y,z Y Fig 10.3 From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we have . Similarly, from the right angled triangles OBP and OCP, we may write cos and cos y z r r β = γ = . Thus, the coordinates of the point P may also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector , and denoted as a, b and c, respectively. X A x Reprint 2025-26 O 90° A X P 10.3 Types of Vectors Zero VectorA vector whose initial and terminal points coincide, is called a zero vector (or null vector), and denoted as . Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors represent the zero vector, Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector in the direction of a given vector is denoted by aˆ . Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors. Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions. Equal Vectors Two vectors are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as . Negative of a Vector A vector whose magnitude is the same as that of a given vector (say, ), but direction is opposite to that of it, is called negative of the given vector. For example, vector is negative of the vector , and written as = – . ANote One may note that l 2 + m2 + n 2 = 1 but a 2 + b 2 + c 2 ≠ 1, in general. VECTOR ALGEBRA 341 Remark The vectors defined above are such that any of them may be subject to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors. Throughout this chapter, we will be dealing with free vectors only. Example 1 Represent graphically a displacement of 40 km, 30° west of south. Solution The vector represents the required displacement (Fig 10.4). Example 2 Classify the following measures as scalars and vectors. (i) 5 seconds (ii) 1000 cm3 Fig 10.4 Reprint 2025-26 Solution (i) Time-scalar (ii) Volume-scalar (iii) Force-vector (iv) Speed-scalar (v) Density-scalar (vi) Velocity-vector Example 3 In Fig 10.5, which of the vectors are: Solution 342 MATHEMATICS 1. Represent graphically a displacement of 40 km, 30° east of north. 2. Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 meters north-west (iii) 40° (iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2 3. Classify the following as scalar and vector quantities. (iii) 10 Newton (iv) 30 km/hr (v) 10 g/cm3 (vi) 20 m/s towards north (iii) Coinitial vectors : (ii) Equal vectors : (i) Collinear (ii) Equal (iii) Coinitial (i) Collinear vectors : . EXERCISE 10.1 Fig 10.5 (i) time period (ii) distance (iii) force (iv) velocity (v) work done 4. In Fig 10.6 (a square), identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal 5. Answer the following as true or false. (iv) Two collinear vectors having the same magnitude are equal. (i) and – are collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. Reprint 2025-26 Fig 10.6 10.4 Addition of Vectors A vector simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig 10.7). The net displacement made by the girl from point A to the point C, is given by the vector and expressed as In general, if we have two vectors and (Fig 10.8 (i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig 10.8(ii)). = This is known as the triangle law of vector addition. b (i) (iii) a A a b + (ii) a C b B B A VECTOR ALGEBRA 343 a b – Fig 10.7 a C b –b C’ Fig 10.8 For example, in Fig 10.8 (ii), we have shifted vector without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of . Then, the vector + , represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors and i.e., in triangle ABC (Fig 10.8 (ii)), we have This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig 10.8(iii)). = Now again, since , from the above equation, we have Reprint 2025-26 Now, construct a vector so that its magnitude is same as the vector , but the direction opposite to that of it (Fig 10.8 (iii)), i.e., The vector is said to represent the difference of . Now, consider a boat in a river going from one bank of the river to the other in a direction perpendicular to the flow of the river. Then, it is acted upon by two velocity vectors–one is the velocity imparted to the boat by its engine and other one is the velocity of the flow of river water. Under the simultaneous influence of these two velocities, the boat in actual starts travelling with a different velocity. To have a precise idea about the effective speed and direction (i.e., the resultant velocity) of the boat, we have the following law of vector addition. Ifwe havetwo vectors represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig 10.9), then their sum is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition. 344 MATHEMATICS = Then, on applying triangle law from the Fig 10.8 (iii), we have = Fig 10.9 Properties of vector addition Property 1 For any two vectors , which is parallelogram law. Thus, we may say that the two laws of vector addition are equivalent to each other. ANote From Fig 10.9, using the triangle law, one may note that = or = (since ) Reprint 2025-26 = (Commutative property) the triangle law, from triangle ABC, we have parallelogram are equal and parallel, from Fig 10.10, we have, and triangle ADC, we have Hence = Property 2 For any three vectors a b c , and Proof Let the vectors be represented by , respectively, as shown in Fig 10.11(i) and (ii). (Fig 10.10). Let then using Proof Consider the parallelogram ABCD Now, since the opposite sides of a . Again using triangle law, from = (Associative property) Fig 10.10 VECTOR ALGEBRA 345 Then = and = So = Reprint 2025-26 Fig 10.11 and = Hence = Remark The associative property of vector addition enables us to write the sum of three vectors without using brackets. Note that for any vector a , we have = Here, the zero vector is called the additive identity for the vector addition. 10.5 Multiplication of a Vector by a Scalar Let be a given vector and λ a scalar. Then the product of the vector by the scalar λ, denoted as λ , is called the multiplication of vector by the scalar λ. Note that, λ is also a vector, collinear to the vector . The vector λ has the direction same (or opposite) to that of vector according as the value of λ is positive (or negative). Also, the magnitude of vector λ is |λ| times the magnitude of the vector , i.e., |λ | = |λ | | | A geometric visualisation of multiplication of a vector by a scalar is given in Fig 10.12. 346 MATHEMATICS a a 1 2 1 2 –2 a a2 a When λ = –1, then λ = – , which is a vector having magnitude equal to the magnitude of and direction opposite to that of the direction of . The vector – is called the negative (or additive inverse) of vector and we always have Also, if 1 = | | a λ , provided ≠ 0 i.e. is not a null vector, then + (– ) = (– ) + = |λ | =|λ|| | = Reprint 2025-26 Fig 10.12 So, λ represents the unit vector in the direction of . We write it as 10.5.1 Components of a vector Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. Then, clearly The vectors , each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by ˆ ˆ ˆ i j k , and , respectively (Fig 10.13). Now, consider the position vector of a point P(x, y, z) as in Fig 10.14. Let P1 be the foot of the perpendicular from P on the plane XOY. ANote For any scalar k, 0 = 0. r r k | |= 1, | |= 1 and | |= 1 aˆ = 1 | | r r a a VECTOR ALGEBRA 347 Fig 10.13 We, thus, see that P1 P is parallel to z-axis. As ˆ ˆ ˆ i j k , and are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have . Similarly, and . Reprint 2025-26 Fig 10.14 Therefore, it follows that = and = Hence, the position vector of P with reference to O is given by as the scalar components of , and ˆ ˆ ˆ xi yj zk , and are called the vector components of along the respective axes. Sometimes x, y and z are also termed as rectangular components. The length of any vector = ˆ ˆ ˆ r xi yj zk = + + , is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP1 (Fig 10.14) and in the right angle triangle OP1 P, we have 348 MATHEMATICS 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively, then = This form of any vector is called its component form. Here, x, y and z are called Hence, the length of any vector = ˆ ˆ ˆ r xi yj zk = + + is given by If are any two vectors given in the component form 1 2 3 ˆ ˆ ˆ a i a j a k + + and | | = = , = a1 = b1 , a2 = b2 and a3 = b3 (iv) the multiplication of vector by any scalar λ is given by (iii) the vectors are equal if and only if (ii) the difference of the vector is given by (i) the sum (or resultant) of the vectors is given by λ = 1 2 3 ˆ ˆ ˆ ( ) ( ) ( ) λ + λ + λ a i a j a k = 1 1 2 2 3 3 ˆ ˆ ˆ ( ) ( ) ( ) a b i a b j a b k + + + + + = 1 1 2 2 3 3 ˆ ˆ ˆ ( ) ( ) ( ) a b i a b j a b k − + − + − Reprint 2025-26 The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws: Let be any two vectors, and k and m be any scalars. Then (i) (ii) (iii) Remarks (iii) In case if it is given that l, m, n are direction cosines of a vector, then ˆ ˆ ˆ li mj nk + + (ii) If = 1 2 3 ˆ ˆ ˆ a a i a j a k = + + , then a1 , a2 , a3 are also called direction ratios of . (i) One may observe that whatever be the value of λ, the vector λ is always collinear to the vector . In fact, two vectors are collinear if and only if there exists a nonzero scalar λ such that . If the vectors are given in the component form, i.e. = 1 2 3 ˆ ˆ ˆ a a i a j a k = + + and , then the two vectors are collinear if and only if ⇔ 1 2 3 ˆ ˆ ˆ b i b j b k + + = 1 2 3 ˆ ˆ ˆ ( ) ( ) ( ) λ + λ + λ a i a j a k ⇔ b a 1 1 = λ , 2 2 3 3 b a b a = λ = λ , ⇔ 1 1 b a = 2 3 2 3 b b a a = = λ 1 2 3 ˆ ˆ ˆ b i b j b k + + = 1 2 3 ˆ ˆ ˆ λ + + (a i a j a k) VECTOR ALGEBRA 349 Example 4 Find the values of x, y and z so that the vectors and Solution Note that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors will be equal if and only if x = 2, y = 2, z = 1 = ˆ ˆ ˆ (cos ) (cos ) (cos ) α + β + γ i j k is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with x, y and z axes respectively. are equal. Reprint 2025-26 Example 5 Let and . Is ? Are the vectors equal? Solution We have and So, . But, the two vectors are not equal since their corresponding components are distinct. Example 6 Find unit vector in the direction of vector Solution The unit vector in the direction of a vector is given by . Now = 2 2 2 2 3 1 14 + + = Therefore 1 ˆ ˆ ˆ ˆ (2 3 ) 14 a i j k = + + = 2 3 1 ˆ ˆ ˆ 14 14 14 i j k + + Example 7 Find a vector in the direction of vector that has magnitude 7 units. Solution The unit vector in the direction of the given vector is Therefore, the vector having magnitude equal to 7 and in the direction of is 350 MATHEMATICS = 1 1 2 ˆ ˆ ˆ ˆ ( 2 ) 5 5 5 i j i j − = − Example 8 Find the unit vector in the direction of the sum of the vectors, and . Solution The sum of the given vectors is and = 2 2 2 4 3 ( 2) 29 + + − = 7a ∧ = 1 2 7 5 5 i j ∧ ∧ − = 7 14 ˆ ˆ 5 5 i j − Reprint 2025-26 Thus, the required unit vector is Example 9 Write the direction ratio’s of the vector and hence calculate its direction cosines. Solution Note that the direction ratio’s a, b, c of a vector are just the respective components x, y and z of the vector. So, for the given vector, we have a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given vector, then Thus, the direction cosines are 1 1 2 , , – 6 6 6 . 10.5.2 Vector joining two points If P1 (x1 , y1 , z 1 ) and P2 (x2 , y2 , z 2 ) are any two points, then the vector joining P1 and P2 is the vector (Fig 10.15). Joining the points P1 and P2 with the origin O, and applying triangle law, from the triangle OP1 P2 , we have 1 4 3 2 ˆ ˆ ˆ ˆ ˆ ˆ (4 3 2 ) 29 29 29 29 = + − = + − i j k i j k VECTOR ALGEBRA 351 = Using the properties of vector addition, the above equation becomes i.e. = 2 2 2 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ (x i y j z k x i y j z k + + − + + ) ( ) The magnitude of vector is given by | | = 2 2 2 2 1 2 1 2 1 ( ) ( ) ( ) x x y y z z − + − + − = = 2 1 2 1 2 1 ˆ ˆ ˆ ( ) ( ) ( ) x x i y y j z z k − + − + − Reprint 2025-26 Fig 10.15 Example 10 Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed from P to Q. Solution Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point. So, the required vector joining P and Q is the vector , given by i.e. = ˆ ˆ ˆ − − − 3 5 4 . i j k 10.5.3 Section formula Let P and Q be two points represented by the position vectors , respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways – internally (Fig 10.16) and externally (Fig 10.17). Here, we intend to find the position vector for the point R with respect to the origin O. We take the two cases one by one. Case I When R divides PQ internally (Fig 10.16). If R divides such that = , where m and n are positive scalars, we say that the point R divides internally in the ratio of m : n. Now from triangles ORQ and OPR, we have 352 MATHEMATICS = ˆ ˆ ˆ ( 1 2) ( 2 3) ( 4 0) − − + − − + − − i j k = Fig 10.16 and = , Therefore, we have = (Why?) or = (on simplification) ratio of m : n is given by Hence, the position vector of the point R which divides P and Q internally in the Reprint 2025-26 = Case II When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment PQ externally in the ratio m : n PR i.e. QR = m n is given by Remark If R is the midpoint of PQ , then m = n. And therefore, from Case I, the midpoint R of , will have its position vector as Example 11 Consider two points P and Q with position vectors and . Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally. Solution (i) The position vector of the point R dividing the join of P and Q internally in the ratio 2:1 is and = ˆ ˆ ˆ (2 3) ( 1 4) (1 4) − + − + + + i j k ˆ ˆ ˆ = − + + i j k 3 5 Fig 10.17 (ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2:1 is = = = VECTOR ALGEBRA 353 Example 12 Show that the points A(2 ), B( 3 5 ), C(3 4 4 ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k i j k − + − − − − are the vertices of a right angled triangle. Solution We have = ˆ ˆ ˆ (3 1) ( 4 3) ( 4 5) − + − + + − + i j k ˆ ˆ ˆ = − + 2i j k = = ˆ ˆ ˆ (1 2) ( 3 1) ( 5 1) − + − + + − − i j k ˆ ˆ ˆ = − − − i j k 2 6 Reprint 2025-26 Further, note that = Hence, the triangle is a right angled triangle. 354 MATHEMATICS 1. Compute the magnitude of the following vectors: 2. Write two different vectors having same magnitude. 3. Write two different vectors having same direction. 4. Find the values of x and y so that the vectors 2 3 and ˆ ˆ ˆ ˆ i j xi yj + + are equal. 5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7). 6. Find the sum of the vectors = ˆ ˆ ˆ i j k − + 2 , = ˆ ˆ ˆ − + + 2 4 5 i j k and = ˆ ˆ ˆ c i j k = − 6 – 7 . 7. Find the unit vector in the direction of the vector = ˆ ˆ ˆ a i j k = + + 2 . 8. Find the unit vector in the direction of vector , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively. 9. For given vectors, = ˆ ˆ ˆ 2 2 i j k − + and = ˆ ˆ ˆ − + − i j k , find the unit vector in the direction of the vector . = ˆ ˆ i j k + + ; = ˆ ˆ ˆ 2 7 3 ; i j k − − = 1 1 1 ˆ ˆ ˆ 3 3 3 i j k + − EXERCISE 10.2 10. Find a vector in the direction of vector ˆ ˆ ˆ 5 2 i j k − + which has magnitude 8 units. 12. Find the direction cosines of the vector ˆ ˆ ˆ i j k + + 2 3 . 13. Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1), directed from A to B. 14. Show that the vector ˆ ˆ ˆ i j k + + is equally inclined to the axes OX, OY and OZ. 15. Find the position vector of a point R which divides the line joining two points P 11. Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ 2 3 4 and 4 6 8 i j k i j k − + − + − are collinear. and Q whose position vectors are ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k + − + + 2 and – respectively, in the ratio 2 : 1 (i) internally (ii) externally Reprint 2025-26 10.6 Product of Two Vectors So far we have studied about addition and subtraction of vectors. An other algebraic operation which we intend to discuss regarding vectors is their product. We may recall that product of two numbers is a number, product of two matrices is again a matrix. But in case of functions, we may multiply them in two ways, namely, multiplication of two functions pointwise and composition of two functions. Similarly, multiplication of two vectors is also defined in two ways, namely, scalar (or dot) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, they have found various applications in geometry, mechanics and engineering. In this section, we will discuss these two types of products. 16. Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2). 17. Show that the points A, B and C with position vectors, = ˆ ˆ ˆ 3 4 4 , i j k − − = ˆ ˆ ˆ 2i j k − + and = ˆ ˆ ˆ i j k − − 3 5 , respectively form the vertices of a right angled triangle. 18. In triangle ABC (Fig 10.18), which of the following is not true: 19. If are two collinear vectors, then which of the following are incorrect: (A) (D) (A) (B) (C) the respective components of are not proportional (D) both the vectors have same direction, but different magnitudes. (B) (C) VECTOR ALGEBRA 355 Fig 10.18 10.6.1 Scalar (or dot) product of two vectors Definition 2 The scalar product of two nonzero vectors , denoted by , is Reprint 2025-26 where, θ is the angle between (Fig 10.19). If either then θ is not defined, and in this case, we define Observations 356 MATHEMATICS 1. is a real number. 2. Let be two nonzero vectors, then if and only if are 3. If θ = 0, then 4. If θ = π, then ⋅ = −| | | | r r r r a b a b 5. In view of the Observations 2 and 3, for mutually perpendicular unit vectors defined as = perpendicular to each other. i.e. In particular, as θ in this case is 0. In particular, , as θ in this case is π. ˆ ˆ ˆ i j k , and , we have ˆ ˆ ˆ ˆ i j j k ⋅ = ⋅ = k i ˆ ⋅ =ˆ 0 ˆ ˆ ˆ ˆ i i j j ⋅ = ⋅ = ˆ ˆ k k⋅ =1, Fig 10.19 (Why?) Two important properties of scalar product Property 1 (Distributivity of scalar product over addition) Let be any three vectors, then 6. The angle between two nonzero vectors is given by 7. The scalar product is commutative. i.e. Reprint 2025-26 or Thus = 1 1 2 2 3 3 a b a b a b + + 10.6.2 Projection of a vector on a line Suppose a vector makes an angle θ with a given directed line l (say), in the anticlockwise direction (Fig 10.20). Then the projection of on l is a vector (say) with magnitude , and the direction of being the same (or opposite) to that of the line l, depending upon whether cosθ is positive or negative. The vector 1 2 3 ˆ ˆ ˆ b i b j b k + + , then their scalar product is given as If two vectors are given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and Property 2 Let be any two vectors, and l be any scalar. Then = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ (a i a j a k b i b j b k + + ⋅ + + ) ( ) = 1 1 2 3 2 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a i b i b j b k a j b i b j b k ⋅ + + + ⋅ + + ( ) ( ) + 3 1 2 3 ˆ ˆ ˆ ˆ a k b i b j b k ⋅ + + ( ) = 1 1 1 2 1 3 2 1 2 2 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b i i a b i j a b i k a b j i a b j j a b j k ( ) ( ) ( ) ( ) ( ) ( ) ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ a b k i a b k j a b k k ( ) ( ) ( ) ⋅ + ⋅ + ⋅ (Using the above Properties 1 and 2) = a1 b1 + a2 b2 + a3 b3 (Using Observation 5) θ θ p p a a (i) B VECTOR ALGEBRA 357 B A C l (270 < < 360 ) 0 0 (180 < < 270 ) θ 0 0 θ (90 < < 180 ) 0 0 (0 < < 90 ) θ 0 0 θ (iii) C l B p p a A θ θ Reprint 2025-26 Fig 10.20 A a l C A (iv) C l (ii) B is called the projection vector, and its magnitude | | is simply called as the projection of the vector on the directed line l. For example, in each of the following figures (Fig 10.20(i) to (iv)), projection vector of along the line l is vector . Observations Remark If α, β and γ are the direction angles of vector 1 2 3 ˆ ˆ ˆ = + + a i a j a k , then its direction cosines may be given as 358 MATHEMATICS 1. If pˆ is the unit vector along a line l, then the projection of a vector on the line 2. Projection of a vector on other vector r b , is given by 3. If θ = 0, then the projection vector of will be itself and if θ = π, then the 4. If = 2 π θ or 3 = 2 π θ , then the projection vector of will be zero vector. l is given by a p⋅ ˆ . projection vector of will be . ˆ a b⋅ , or Also, note that are respectively the projections of along OX, OY and OZ. i.e., the scalar components a1 , a2 and a3 of the vector , are precisely the projections of along x-axis, y-axis and z-axis, respectively. Further, if is a unit vector, then it may be expressed in terms of its direction cosines as ˆ ˆ ˆ = α + β + γ cos cos cos i j k Example 13 Find the angle between two vectors with magnitudes 1 and 2 respectively and when =1. Solution Given . We have Reprint 2025-26 Example 14 Find angle ‘θ’ between the vectors . Solution The angle θ between two vectors is given by Now = ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 1 1 1 1 i j k i j k + − ⋅ − + = − − = − . Therefore, we have cosθ = 1 3 − hence the required angle is θ = Example 15 If , then show that the vectors Solution We know that two nonzero vectors are perpendicular if their scalar product is zero. Here = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 6 2 8 i j k i j k i j k − − + + − = + − and = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 4 4 2 i j k i j k i j k − − − + − = − + So ( ) . ( ) ˆ ˆ ˆ ˆ ˆ ˆ = + − ⋅ − + = − − = (6 2 8 ) (4 4 2 ) 24 8 16 0. i j k i j k Hence are perpendicular vectors. Example 16 Find the projection of the vector ˆ ˆ ˆ = + + 2 3 2 i j k on the vector are perpendicular. cosθ = VECTOR ALGEBRA 359 Solution The projection of vector on the vector r b is given by Example 17 Find , if two vectors are such that and . Solution We have ˆ ˆ ˆ = + + i j k 2 . = 2 2 2 (2 1 3 2 2 1) 10 5 6 6 3 (1) (2) (1) × + × + × = = + + Reprint 2025-26 = Therefore = 5 Example 18 If is a unit vector and , then find . Solution Since is a unit vector, . Also, or = 8 or Therefore = 3 (as magnitude of a vector is non negative). Example 19 For any two vectors , we always have (CauchySchwartz inequality). Solution The inequality holds trivially when either = = 0 or 0 r r r r a b . Actually, in such a situation we have . So, let us assume that . Then, we have Therefore 360 MATHEMATICS = | cos | 1 θ ≤ = = 2 2 (2) 2(4) (3) − + C Example 20 For any two vectors , we always have (triangle inequality). Solution The inequality holds trivially in case either (How?). So, let . Then, = = (scalar product is commutative) ≤ (since x x x ≤ ∀ ∈ | | R ) ≤ (from Example 19) = Reprint 2025-26 A a Fig 10.21 a + b B b Hence Remark If the equality holds in triangle inequality (in the above Example 20), i.e. then = showing that the points A, B and C are collinear. Example 21 Show that the points A( 2 3 5 ), B( 2 3 ) ˆ ˆ ˆ ˆ ˆ ˆ − + + + + i j k i j k and C(7 ) ˆ ˆ i k − are collinear. Solution We have Therefore Hence the points A, B and C are collinear. ANote In Example 21, one may note that although but the points A, B and C do not form the vertices of a triangle. = ˆ ˆ ˆ ˆ ˆ ˆ (1 2) (2 3) (3 5) 3 2 + + − + − = − − i j k i j k , = ˆ ˆ ˆ ˆ ˆ ˆ (7 1) (0 2) ( 1 3) 6 2 4 − + − + − − = − − i j k i j k , = ˆ ˆ ˆ ˆ ˆ ˆ (7 2) (0 3) ( 1 5) 9 3 6 + + − + − − = − − i j k i j k = = , VECTOR ALGEBRA 361 1. Find the angle between two vectors with magnitudes 3 and 2 , 2. Find the angle between the vectors ˆ ˆ i j k i j k ˆ ˆ ˆ ˆ − + − + 2 3 and 3 2 and ˆ ˆ i j k i j k ˆ ˆ ˆ ˆ − + − + 2 3 and 3 2 3. Find the projection of the vector ˆ ˆ i j − on the vector ˆ ˆ i j + . 4. Find the projection of the vector ˆ ˆ ˆ i j k + + 3 7 on the vector ˆ ˆ ˆ 7 8 i j k − + . 5. Show that each of the given three vectors is a unit vector: respectively having . Also, show that they are mutually perpendicular to each other. 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 3 6 ), (3 6 2 ), (6 2 3 ) 7 7 7 i j k i j k i j k + + − + + − EXERCISE 10.3 Reprint 2025-26 362 MATHEMATICS 10. If are such that is perpendicular to , then find the value of λ. 12. If , then what can be concluded about the vector ? 13. If are unit vectors such that , find the value of 14. If either vector . But the converse need not be 15. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors and ]. 11. Show that is perpendicular to , for any two nonzero 6. Find , if . 7. Evaluate the product . 8. Find the magnitude of two vectors , having the same magnitude and such 9. Find , if for a unit vector , . that the angle between them is 60o and their scalar product is 1 2 . vectors . true. Justify your answer with an example. . 10.6.3 Vector (or cross) product of two vectors In Section 10.2, we have discussed on the three dimensional right handed rectangular coordinate system. In this system, when the positive x-axis is rotated counterclockwise 16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear. 17. Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 , 3 5 and 3 4 4 i j k i j k i j k − + − − − − form the vertices of a right angled triangle. 18. If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit vector if (A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ| Reprint 2025-26 into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig 10.22(i)). In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig 10.22(ii)). Definition 3 The vector product of two nonzero vectors , is denoted by and defined as where, θ is the angle between , 0 ≤ θ ≤ π and nˆ is a unit vector perpendicular to both , such that Fig 10.22 (i), (ii) = , VECTOR ALGEBRA 363 form a right handed system (Fig 10.23). i.e., the right handed system rotated from moves in the direction of nˆ . If either , then θ is not defined and in this case, we define . Observations 1. is a vector. 2. Let be two nonzero vectors. Then if and only if are parallel (or collinear) to each other, i.e., Reprint 2025-26 = Fig 10.23 364 MATHEMATICS 3. If 2 π θ = then . 4. In view of the Observations 2 and 3, for mutually perpendicular 5. In terms of vector product, the angle between two vectors may be given as 6. It is always true that the vector product is not commutative, as = . In particular, and , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0. unit vectors ˆ ˆ ˆ i j k , and (Fig 10.24), we have Indeed, , where form a right handed system, i.e., θ is traversed from , Fig 10.25 (i). While, , where Fig 10.25(ii). form a right handed system i.e. θ is traversed from , ˆ ˆ i i × = ˆ ˆ i j × = ˆ ˆ ˆ ˆ ˆ ˆ ˆ k j k i k i j , , × = × = sin θ = Fig 10.24 Thus, if we assume to lie in the plane of the paper, then 1 n n ˆ ˆ and both will be perpendicular to the plane of the paper. But, nˆ being directed above the paper while 1 nˆ directed below the paper. i.e. 1 n n ˆ ˆ = − . Fig 10.25 (i), (ii) Reprint 2025-26 = 7. In view of the Observations 4 and 6, we have 8. If represent the adjacent sides of a triangle then its area is given as 9. If represent the adjacent sides of a parallelogram, then its area is given Hence = By definition of the area of a triangle, we have from Fig 10.26, Area of triangle ABC = 1 AB CD. 2 ⋅ But (as given), and CD = sin θ. Thus, Area of triangle ABC = by . From Fig 10.27, we have Area of parallelogram ABCD = AB. DE. . ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ j i k k j i i k j × = − × = − × = − , and . VECTOR ALGEBRA 365 Fig 10.26 (i) (ii) But (as given), and Thus, Area of parallelogram ABCD = We now state two important properties of vector product. Property 3 (Distributivity of vector product over addition): If are any three vectors and λ be a scalar, then . Reprint 2025-26 Fig 10.27 Explanation We have 366 MATHEMATICS 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively. Then their cross product may be given by Let be two vectors given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as i i j j k k i k k i j i i j k j j k × = × = × = × = − × × = − × × = − × 0 and , and ) = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ (a i a j a k b i b j b k + + × + + ) ( ) = 1 1 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b i i a b i j a b i k a b j i ( ) ( ) ( ) ( ) × + × + × + × = 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ a b i j a b k i a b i j ( ) ( ) ( ) × − × − × = 1 2 1 3 2 1 2 3 3 1 3 2 a b k a b j a b k a b i a b j a b i ˆ − − + + − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as , and ) i j k j k i k i j × = × = × = = 2 3 3 2 1 3 3 1 1 2 2 1 ˆ ˆ ˆ ( ) ( ) ( ) a b a b i a b a b j a b a b k − − − + − + 2 2 2 3 ˆ ˆ ˆ ˆ a b j j a b j k ( ) ( ) × + × + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ a b k i a b k j a b k k ( ) ( ) ( ) × + × + × (by Property 1) + 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ a b j k a b k i a b j k ( ) ( ) ( ) × + × − × ˆ ˆ ˆ i j k a a a = 1 2 3 ˆ ˆ ˆ i j k a a a b b b 1 2 3 Example 22 Find Solution We have Hence = 2 2 2 ( 17) (13) (7) 507 − + + = = = 1 2 3 = ˆ ˆ ˆ i( 2 15) ( 4 9) (10 – 3) − − − − − +j k ˆ ˆ ˆ = − + + 17 13 7 i j k b b b 2 1 3 3 5 2 i j k ˆ ˆ ˆ 1 2 3 − Reprint 2025-26 Example 23 Find a unit vector perpendicular to each of the vectors and Solution We have A vector which is perpendicular to both + − and r r r r a b a b is given by Now = 4 16 4 24 2 6 + + = = Therefore, the required unit vector is Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices. ANote There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to will be 1 2 1 ˆ ˆ ˆ . 6 6 6 i j k − + But that will be a consequence of . where . | | r r c c = 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + − = VECTOR ALGEBRA 367 Solution We have . The area of the given triangle is . Now, = Therefore = 16 4 1 21 + + = Thus, the required area is 1 21 2 Reprint 2025-26 ˆ ˆ ˆ ˆ ˆ ˆ 0 1 2 4 2 1 2 0 i j k = − + − i j k Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors Solution The area of a parallelogram with as its adjacent sides is given by . Now = Therefore = 25 1 16 42 + + = and hence, the required area is 42 . 368 MATHEMATICS 1. Find . 2. Find a unit vector perpendicular to each of the vector , where 3. If a unit vector makes angles with , with ˆ ˆ 3 4 i j π π and an acute angle θ with ˆk , then find θ and hence, the components of . 4. Show that EXERCISE 10.4 ˆ ˆ ˆ ˆ ˆ ˆ 3 1 4 5 4 1 1 1 i j k = + − i j k . − 5. Find λ and µ if . 6. Given that and . What can you conclude about the vectors 7. Let the vectors be given as 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ a i a j a k b i b j b k + + + + , , 8. If either then . Is the converse true? Justify your answer with an example. 9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). 1 2 3 ˆ ˆ ˆ c i c j c k + + . Then show that . ? Reprint 2025-26 Example 26 Write all the unit vectors in XY-plane. Solution Let be a unit vector in XY-plane (Fig 10.28). Then, from the figure, we have x = cos θ and y = sin θ (since | | = 1). So, we may write the vector as Clearly, | | = 2 2 cos sin 1 θ + θ = 10. Find the area of the parallelogram whose adjacent sides are determined by the vectors and . vector, if the angle between is (A) π/6 (B) π/4 (C) π/3 (D) π/2 12. Area of a rectangle having vertices A, B, C and D with position vectors 1 1 ˆ ˆ ˆ ˆ ˆ ˆ – 4 , 4 2 2 i j k i j k + + + + , 1 ˆ ˆ ˆ 4 2 i j k − + and 1 ˆ ˆ ˆ – 4 2 i j k − + , respectively is 11. Let the vectors be such that , then is a unit (A) 1 2 (B) 1 (C) 2 (D) 4 Miscellaneous Examples = cos sin ˆ ˆ θ + θ i j ... (1) VECTOR ALGEBRA 369 Also, as θ varies from 0 to 2π, the point P (Fig 10.28) traces the circle x 2 + y 2 = 1 counterclockwise, and this covers all possible directions. So, (1) gives every unit vector in the XY-plane. Reprint 2025-26 Fig 10.28 Example 27 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i j k i j i j k i j k + + + + − − − , 2 5 , 3 2 3 and 6 are the position vectors of points A, B, C and D respectively, then find the angle between and . Deduce that and are collinear. Solution Note that if θ is the angle between AB and CD, then θ is also the angle between and . Now = Position vector of B – Position vector of A Therefore | | = 2 2 2 (1) (4) ( 1) 3 2 + + − = Similarly = ˆ ˆ ˆ − − + 2 8 2 and |CD | 6 2 = uuur i j k Thus cos θ = Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that and are collinear. Alternatively, which implies that and are collinear vectors. Example 28 Let be three vectors such that and each one of them being perpendicular to the sum of the other two, find . 370 MATHEMATICS = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 5 ) ( ) 4 i j i j k i j k + − + + = + − = 1( 2) 4( 8) ( 1)(2) 36 1 (3 2)(6 2) 36 − + − + − − = = − Solution Given Now Therefore = 50 5 2 = Reprint 2025-26 = = = 9 + 16 + 25 = 50 + Example 29 Three vectors satisfy the condition . Evaluate the quantity . Solution Since , we have or = 0 Therefore ... (1) Again, = 0 or ... (2) Similarly = – 4. ... (3) Adding (1), (2) and (3), we have or 2µ = – 29, i.e., µ = 29 2 − Example 30 If with reference to the right handed system of mutually perpendicular = 0 = – 29 VECTOR ALGEBRA 371 unit vectors , then express in the form Solution Let is a scalar, i.e., . Now = ˆ ˆ ˆ (2 3 ) (1 ) 3 − λ + + λ − i j k. Now, since β2 r is to be perpendicular to α r , we should have . i.e., or λ = 1 2 Therefore 1 = 3 1 ˆ ˆ 2 2 i j − and 3(2 3 ) (1 ) − λ − + λ = 0 is parallel to is perpendicular to Reprint 2025-26 372 MATHEMATICS 10. The two adjacent sides of a parallelogram are ˆ ˆ ˆ ˆ ˆ ˆ 2 4 5 and 2 3 i j k i j k − + − − . Find the unit vector parallel to its diagonal. Also, find its area. 11. Show that the direction cosines of a vector equally inclined to the axes OX, OY 1. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis. 2. Find the scalar components and magnitude of the vector joining the points P(x1 , y1 , z 1 ) and Q(x2 , y2 , z 2 ). 3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure. 4. If , then is it true that ? Justify your answer. 5. Find the value of x for which ˆ ˆ ˆ x i j k ( ) + + is a unit vector. 6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors 7. If , find a unit vector parallel to the vector . 8. Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC. 9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ. Miscellaneous Exercise on Chapter 10 . 12. Let . Find a vector which 13. The scalar product of the vector ˆ ˆ ˆ i j k + + with a unit vector along the sum of 14. If are mutually perpendicular vectors of equal magnitudes, show that the and OZ are ± 1 is perpendicular to both and , and . vectors ˆ ˆ ˆ 2 4 5 i j k + − and ˆ ˆ ˆ λ + + i j k 2 3 is equal to one. Find the value of λ. vector is equally inclined to . 3 3 , , . 1 3 1 Reprint 2025-26 . Choose the correct answer in Exercises 16 to 19. 15. Prove that , if and only if are perpendicular, given 16. If θ is the angle between two vectors , then only when 17. Let be two unit vectors and θ is the angle between them. Then is a unit vector if 18. The value of ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i j k j i k k i j .( ) ( ) ( ) × + ⋅ × + ⋅ × is (A) 0 (B) –1 (C) 1 (D) 3 19. If θ is the angle between any two vectors , then when θ (A) 0 2 π < θ < (B) 0 2 π ≤ θ ≤ (A) 4 π θ = (B) 3 π θ = (C) 2 π θ = (D) 2 3 π θ = (A) 0 (B) 4 π (C) 2 π (D) π (C) 0 < θ < π (D) 0 ≤ θ ≤ π is equal to VECTOR ALGEBRA 373 ® Position vector of a point P(x, y, z) is given as , and its magnitude by 2 2 2 x y z + + . ® The scalar components of a vector are its direction ratios, and represent its projections along the respective axes. ® The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of any vector are related as: , , a b c l m n r r r = = = Reprint 2025-26 Summary 374 MATHEMATICS ® The vector sum of the three sides of a triangle taken in order is . ® The vector sum of two coinitial vectors is given by the diagonal of the parallelogram whose adjacent sides are the given vectors. ® The multiplication of a given vector by a scalar λ, changes the magnitude of the vector by the multiple |λ|, and keeps the direction same (or makes it opposite) according as the value of λ is positive (or negative). ® For a given vector , the vector gives the unit vector in the direction of . ® The position vector of a point R dividing a line segment joining the points ® The scalar product of two given vectors having angle θ between them is defined as P and Q whose position vectors are respectively, in the ratio m : n (i) internally, is given by . (ii) externally, is given by . Also, when is given, the angle ‘θ’ between the vectors may be determined by . ® If θ is the angle between two vectors , then their cross product is given as ® If we have two vectors , given in component form as where nˆ is a unit vector perpendicular to the plane containing . Such that form right handed system of coordinate axes. cosθ = Reprint 2025-26 and λ any scalar, The word vector has been derived from a Latin word vectus, which means “to carry”. The germinal ideas of modern vector theory date from around 1800 when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described that how a complex number a + ib could be given a geometric interpretation with the help of a directed line segment in a coordinate plane. William Rowen Hamilton (1805-1865) an Irish mathematician was the first to use the term vector for a directed line segment in his book Lectures on Quaternions (1853). Hamilton’s method of quaternions (an ordered set of four real numbers given as: problem of multiplying vectors in three dimensional space. Though, we must mention here that in practice, the idea of vector concept and their addition was known much earlier ever since the time of Aristotle (384-322 B.C.), a Greek philosopher, and pupil of Plato (427-348 B.C.). That time it was supposed to be known that the combined action of two or more forces could be seen by adding them according to parallelogram law. The correct law for the composition of forces, that forces add vectorially, had been discovered in the case of perpendicular forces by Stevin-Simon (1548-1620). In 1586 A.D., he analysed the principle of geometric addition of forces in his treatise DeBeghinselen der Weeghconst (“Principles of the Art of Weighing”), which caused a major breakthrough in the development of mechanics. But it took another 200 years for the general concept of vectors to form. In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created what we now know as vector analysis, essentially by separating the real (scalar) ˆ ˆ ˆ ˆ ˆ ˆ a bi cj dk i j k + + + , , , following certain algebraic rules) was a solution to the then = 1 1 2 2 3 3 ˆ ˆ ˆ ( ) ( ) ( ) a b i a b j a b k + + + + + ; and = 1 1 1 λ = 1 2 3 ˆ ˆ ˆ ( ) ( ) ( ) λ + λ + λ a i a j a k ; = 1 1 2 2 3 3 a b a b a b + + ; . i j k a b c a b c ˆ ˆ ˆ 2 2 2 Historical Note VECTOR ALGEBRA 375 Reprint 2025-26 376 MATHEMATICS part of quaternion from its imaginary (vector) part. In 1881 and 1884, Gibbs printed a treatise entitled Element of Vector Analysis. This book gave a systematic and concise account of vectors. However, much of the credit for demonstrating the applications of vectors is due to the D. Heaviside and P.G. Tait (1831-1901) who contributed significantly to this subject. —v— Reprint 2025-26" class_12,11,Three Dimensional Geometry,ncert_books/class_12/lemh2dd/lemh205.pdf,"11.1 Introduction In Class XI, while studying Analytical Geometry in two dimensions, and the introduction to three dimensional geometry, we confined to the Cartesian methods only. In the previous chapter of this book, we have studied some basic concepts of vectors. We will now use vector algebra to three dimensional geometry. The purpose of this approach to 3-dimensional geometry is that it makes the study simple and elegant*. In this chapter, we shall study the direction cosines and direction ratios of a line joining two points and also discuss about the equations of lines and planes in space under different conditions, angle between two lines, two planes, a line and a plane, shortest distance between two skew lines and distance of a point from a plane. Most of the above results are obtained in vector form. Nevertheless, we shall also translate these results in the Cartesian form which, at times, presents a more clear geometric and analytic picture of the situation. THREE DIMENSIONAL GEOMETRY v The moving power of mathematical invention is not reasoning but imagination. – A.DEMORGAN v THREE DIMENSIONAL GEOMETRY 377 Chapter 11 Leonhard Euler (1707-1783) 11.2 Direction Cosines and Direction Ratios of a Line From Chapter 10, recall that if a directed line L passing through the origin makes angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine of these angles, namely, cos α, cos β and cos γ are called direction cosines of the directed line L. If we reverse the direction of L, then the direction angles are replaced by their supplements, i.e., , and . Thus, the signs of the direction cosines are reversed. * For various activities in three dimensional geometry, one may refer to the Book “A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005 Reprint 2025-26 Note that a given line in space can be extended in two opposite directions and so it has two sets of direction cosines. In order to have a unique set of direction cosines for a given line in space, we must take the given line as a directed line. These unique direction cosines are denoted by l, m and n. Remark If the given line in space does not pass through the origin, then, in order to find its direction cosines, we draw a line through the origin and parallel to the given line. Now take one of the directed lines from the origin and find its direction cosines as two parallel line have same set of direction cosines. Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R. ANote Some authors also call direction ratios as direction numbers. 378 MATHEMATICS Fig 11.1 Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines (d.c’s) of the line. Then Therefore l = ak, m = bk, n = ck ... (1) But l 2 + m2 + n 2 = 1 Therefore k 2 (a 2 + b 2 + c 2 ) = 1 or k = 2 2 2 1 l a = m b = n k c = (say), k being a constant. Reprint 2025-26 a b c ± + + Hence, from (1), the d.c.’s of the line are where, depending on the desired sign of k, either a positive or a negative sign is to be taken for l, m and n. For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a set of direction ratios. So, any two sets of direction ratios of a line are also proportional. Also, for any line there are infinitely many sets of direction ratios. 11.2.1 Direction cosines of a line passing through two points Since one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points P(x1 , y1 , z 1 ) and Q(x2 , y2 , z 2 ) as follows (Fig 11.2 (a)). 2 2 2 2 2 2 2 2 2 , , a b c l m n a b c a b c a b c = ± = ± = ± + + + + + + THREE DIMENSIONAL GEOMETRY 379 Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ with the x, y and z-axis, respectively. Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠PQN= γ (Fig 11.2 (b). Therefore, cos γ = NQ 2 1 PQ PQ z z − = Similarly cosα = 2 1 2 1 and cos PQ PQ x x − y y − β= Hence, the direction cosines of the line segment joining the points P(x 1 , y1 , z1 ) and Q(x 2 , y2 , z2 ) are Reprint 2025-26 Fig 11.2 where PQ = ( ) 2 2 2 2 1 2 1 2 1 ( ) ( ) x x y y z z − + − + − Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and z-axis respectively, find its direction cosines. Solution Let the d . c .'s of the lines be l , m, n. Then l = cos 900 = 0, m = cos 600 = 1 2 , n = cos 300 = 2 3 . Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines. Solution Direction cosines are or 2 1 2 , , 3 3 3 − − Example 3 Find the direction cosines of the line passing through the two points (– 2, 4, – 5) and (1, 2, 3). Solution We know the direction cosines of the line passing through two points P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) are given by 380 MATHEMATICS ANote The direction ratios of the line segment joining P(x1 , y1 , z 1 ) and Q(x2 , y2 , z 2 ) may be taken as x2 – x1 , y2 – y1 , z 2 – z1 or x1 – x2 , y1 – y2 , z 1 – z2 2 2 2 )2()1(2 2 −+−+ , 2 2 2 )2()1(2 1 2 1 PQ x x − , 2 1 PQ y y − , 2 1 PQ z z − −+−+ − , ( ) 2 2 2 )2(12 2 −+−+ − where PQ = ( )2 12 2 12 2 12 )()( −+−+− zzyyxx Here P is (– 2, 4, – 5) and Q is (1, 2, 3). So PQ = 2 2 2 (1 ( 2)) (2 4) (3 ( 5)) − − + − + − − = 77 Thus, the direction cosines of the line joining two points is 2 1 2 1 2 1 , , PQ PQ PQ x x y y z z − − − 3 2 8 , , 77 77 77 − Reprint 2025-26 Example 4 Find the direction cosines of x, y and z-axis. Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis. Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0. Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively. Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear. Solution Direction ratios of line joining A and B are 1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7. The direction ratios of line joining B and C are 3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14. It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points. 1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines. 2. Find the direction cosines of a line which makes equal angles with the coordinate axes. 3. If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ? 4. Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear. 5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2). EXERCISE 11.1 THREE DIMENSIONAL GEOMETRY 381 11.3 Equation of a Line in Space We have studied equation of lines in two dimensions in Class XI, we shall now study the vector and cartesian equations of a line in space. 11.3.1 Equation of a line through a given point and parallel to given vector Let be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector . Let be the position vector of an arbitrary point P on the line (Fig 11.3). A line is uniquely determined if (i) it passes through a given point and has given direction, or (ii) it passes through two given points. Reprint 2025-26 AP uuur = λ , where λ is some real number. But AP uuur = OP – OA uuur uuur i.e. λ = − r r r a Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by Remark If ˆ ˆ ˆ = + + r b ai bj ck , then a, b, c are direction ratios of the line and conversely, if a, b, c are direction ratios of a line, then ˆ ˆ ˆ = + + r b ai bj ck will be the parallel to the line. Here, b should not be confused with | |. Derivation of cartesian form from vector form Let the coordinates of the given point A be (x1 , y1 , z 1 ) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then and ˆ ˆ ˆ = + + r b a i b j c k Substituting these values in (1) and equating the coefficients of ˆ ˆ i j , and k ˆ , we get x = x1 + λ a; y = y1 + λ b; z = z 1 + λ c ... (2) These are parametric equations of the line. Eliminating the parameter λ from (2), we get 382 MATHEMATICS Then AP uuur is parallel to the vector , i.e., ˆ ˆ ˆ r xi yj zk = + + r ; 1 1 1 ˆ ˆ ˆ a x i y j z k = + + r = » r r a + b ... (1) Fig 11.3 Example 6 Find the vector and the Cartesian equations of the line through the point (5, 2, – 4) and which is parallel to the vector ˆ ˆ ˆ 3 2 8 i j k + − . Solution We have 1 x – x a = 1 1 y – y z – z = b c ... (3) This is the Cartesian equation of the line. ANote If l, m, n are the direction cosines of the line, the equation of the line is 1 x – x l = 1 1 y – y z – z = m n = ˆ ˆ ˆ ˆ ˆ ˆ 5 2 4 and 3 2 8 + − = + − r i j k b i j k Reprint 2025-26 Therefore, the vector equation of the line is = ˆ ˆ ˆ ˆ ˆ ˆ 5 2 4 ( 3 2 8 ) i j k i j k + − + λ + − Now, is the position vector of any point P(x, y, z) on the line. Therefore, ˆ ˆ ˆ xi y j z k + + = ˆ ˆ ˆ ˆ ˆ ˆ 5 2 4 ( 3 2 8 ) i j k i j k + − + λ + − = ˆ ˆ ˆ (5 3 ) (2 2 ) ( 4 8 ) + λ + + λ + − − λ i j k Eliminating λ , we get 5 3 x− = 2 4 2 8 y z − + = − which is the equation of the line in Cartesian form. 11.4 Angle between Two Lines Let L1 and L2 be two lines passing through the origin and with direction ratios a1 , b1 , c1 and a2 , b2 , c2 , respectively. Let P be a point on L1 and Q be a point on L2 . Consider the directed lines OP and OQ as given in Fig 11.6. Let θ be the acute angle between OP and OQ. Now recall that the directed line segments OP and OQ are vectors with components a1 , b1 , c1 and a2 , b2 , c2 , respectively. Therefore, the angle θ between them is given by cos θ = 1 2 1 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 a b c a b c + + + + + + ... (1) a a b b c c THREE DIMENSIONAL GEOMETRY 383 Fig 11.4 The angle between the lines in terms of sin θ is given by sin θ = 2 1 cos − θ = ( )( ) 2 1 2 1 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 ( ) 1 a a b b c c = ( )( ) ( ) = 2 2 2 1 2 2 1 1 2 2 1 1 2 2 1 2 2 2 2 2 2 1 1 1 2 2 2 a b c a b c a a b b c c ( − + − + − ) ( ) ( ) a b c a b c + + − + + + + a b a b b c b c c a c a 2 2 2 2 2 2 2 1 1 1 2 2 2 1 2 1 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 + + + + − + + ( ) ( ) a b c a b c a b c a b c ... (2) + + + + + + + + Reprint 2025-26 If instead of direction ratios for the lines L1 and L2 , direction cosines, namely, l 1 , m1 , n1 for L1 and l 2 , m2 , n2 for L2 are given, then (1) and (2) takes the following form: and sin θ = ( )2 2 2 1 2 2 1 1 2 2 1 1 2 2 1 l m l m m n m n n l n l − − − + − ( ) ( ) ... (4) Two lines with direction ratios a1 , b1 , c1 and a2 , b2 , c2 are (i) perpendicular i.e. if θ = 90° by (1) a1 a2 + b1 b2 + c1 c2 = 0 Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines then cosθ = 1 2 1 2 b b In Cartesian form, if θ is the angle between the lines 384 MATHEMATICS ANote In case the lines L1 and L2 do not pass through the origin, we may take lines L and L 1 2 ′ ′ which are parallel to L1 and L2 respectively and pass through the origin. (ii) parallel i.e. if θ = 0 by (2) cos θ = |l 1 l 2 + m1m2 + n1 n2 | (as 2 2 2 1 1 1 l m n + + =1 2 2 2 2 2 2 = + + l m n ) ... (3) 2 a a = 1 1 1 = 1 1 a b + λ and = 2 2 + µ r r a b 2 2 b c b c = ⋅ r r r r b b and 2 2 x x a − = 2 2 2 2 y y z z b c − − = ... (2) where, a1 , b1, c1 and a2, b2 , c2 are the direction ratios of the lines (1) and (2), respectively, then Example 7 Find the angle between the pair of lines given by = ˆ ˆ ˆ ˆ ˆ ˆ 3 2 4 ( 2 2 ) i j k i j k + − + λ + + 1 x x a − = 1 1 1 1 y y z z b c − − = ... (1) cos θ = 1 2 1 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 1 Reprint 2025-26 a b c a b c + + + + + + a a b b c c and = ˆ ˆ ˆ ˆ ˆ 5 2 (3 2 6 ) i j i j k − + µ + + Solution Here 1 r b = ˆ ˆ ˆ i j k + + 2 2 and 2 r b = ˆ ˆ ˆ 3 2 6 i j k + + The angle θ between the two lines is given by Hence θ = cos–1 19 21 Example 8 Find the angle between the pair of lines 3 3 x + = 1 3 5 4 y z − + = and 1 1 x + = 4 5 1 2 y z − − = Solution The direction ratios of the first line are 3, 5, 4 and the direction ratios of the second line are 1, 1, 2. If θ is the angle between them, then Hence, the required angle is cos –1 8 3 15 . cos θ = 1 2 1 2 cos θ = 2 2 2 2 2 2 3.1 5.1 4.2 16 16 8 3 50 6 5 2 6 15 3 5 4 1 1 2 + + = = = + + + + = 3 4 12 19 3 7 21 + + = × r r r r b b i j k i j k b b ˆ ˆ ˆ ˆ ˆ ˆ ( 2 2 ) (3 2 6 ) 1 4 4 9 4 36 ⋅ + + ⋅ + + = + + + + THREE DIMENSIONAL GEOMETRY 385 11.5 Shortest Distance between Two Lines If two lines in space intersect at a point, then the shortest distance between them is zero. Also, if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line. Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines. For example, let us consider a room of size 1, 3, 2 units along x, y and z-axes respectively Fig 11.5. Reprint 2025-26 Fig 11.5 The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet. By the shortest distance between two lines we mean the join of a point in one line with one point on the other line so that the length of the segment so obtained is the smallest. For skew lines, the line of the shortest distance will be perpendicular to both the lines. 11.5.1 Distance between two skew lines We now determine the shortest distance between two skew lines in the following way: Let l 1 and l 2 be two skew lines with equations (Fig. 11.6) and = 2 2 + µ r r a b ... (2) Take any point S on l 1 with position vector 1 a r and T on l 2 , with position vector 2 a r . Then the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance (See 10.6.2). If PQ uuur is the shortest distance vector between l 1 and l 2 , then it being perpendicular to both 1 r b and 2 r b , the unit vector nˆ along PQ uuur would therefore be 386 MATHEMATICS nˆ = 1 2 1 2 | | × × = 1 1 + λ r r a b ... (1) r r r r b b b b ... (3) S T Q Fig 11.6 P l 1 l 2 Then PQ uuur = d nˆ where, d is the magnitude of the shortest distance vector. Let θ be the angle between ST uur and PQ uuur . Then PQ = ST | cos θ| But cos θ = PQ ST | PQ | | ST | ⋅ uuur uur uuuur uur Reprint 2025-26 = 2 1 ˆ ( ) ST ⋅ − r r d n a a d (since ST ) = − 2 1 uur r r a a = 1 2 2 1 1 2 ( ) ( ) r r r r r r b b a a ST × ⋅ − × b b [From (3)] Hence, the required shortest distance is d = PQ = ST |cos θ| or d = ( × ) . ( ) | × | 1 2 2 1 Cartesian form The shortest distance between the lines and l 2 : x x a − 2 2 = y y b is 11.5.2 Distance between parallel lines If two lines l 1 and l 2 are parallel, then they are coplanar. Let the lines be given by ... (1) and … (2) where, 1 a r is the position vector of a point S on l 1 and 2 a r ( − + − + ) ( ) ( − a b2 1 2 ) b c b c c a c a a b 1 2 2 1 2 1 2 2 1 2 1 2 l 1 : x x a − 1 1 = y y b x x y y z z 2 1 2 1 2 1 a b c a b c − − − 1 1 1 2 2 2 r r r r r r b b a a b b z z c − = 1 − z z c − = 2 − 1 2 THREE DIMENSIONAL GEOMETRY 387 1 2 1 1 2 2 × is the position vector of a point T on l 2 Fig 11.7. As l 1 , l 2 are coplanar, if the foot of the perpendicular from T on the line l 1 is P, then the distance between the lines l 1 and l 2 = |TP|. Let θ be the angle between the vectors ST uur and . Then where nˆ is the unit vector perpendicular to the plane of the lines l 1 and l 2. But ST uur = 2 1 − r r a a × = ST r uur b ˆ ... (3) Reprint 2025-26 Fig 11.7 Therefore, from (3), we get i.e., 2 1 | ( )| × − r r r b a a = | | PT 1⋅ r b (as | nˆ | = 1) Hence, the distance between the given parallel lines is Example 9 Find the shortest distance between the lines l 1 and l 2 whose vector equations are and = ˆ ˆ ˆ ˆ ˆ ˆ 2 i j k i j k + − + µ − + (3 5 2 ) ... (2) Solution Comparing (1) and (2) with = 1 1 + λ r r a b and 2 2 r a b = + µ r r r respectively, we get 1 a r = 1 ˆ ˆ ˆ ˆ ˆ + = − , 2 + r i j b i j k Therefore 2 1 − r r a a = ˆ ˆ i k − and 1 2 × r r b b = ˆ ˆ ˆ ˆ ˆ ˆ ( 2 ) ( 3 5 2 ) i j k i j k − + × − + 388 MATHEMATICS 2 1 × − ( ) r r r b a a = | PT ˆ r b n (since PT = ST sin θ) 2 a r = 2 ˆi + ˆj – ˆk and 2 r b = 3 ˆi – 5 ˆj + 2 ˆk d = = ˆ ˆ ˆ ˆ ˆ i j i j k + + λ − + (2 ) ... (1) ˆ ˆ ˆ So 1 2 × || r r b b = 9 1 49 59 + + = Hence, the shortest distance between the given lines is given by Example 10 Find the distance between the lines l 1 and l 2 given by and = ˆ ˆ ˆ ˆ ˆ ˆ 3 3 5 ( 2 3 6 ) i j k i j k + − + µ + + d = 1 2 2 1 = ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k + − + λ + + 2 4 ( 2 3 6 ) 1 2 ( b b ) . ( a a ) | b b | × − × r r r r r r 59 10 59 Reprint 2025-26 = ˆ ˆ ˆ 2 1 1 3 7 3 5 2 i j k − = − − i j k − |703| = +− = Solution The two lines are parallel (Why? ) We have 1 a r = ˆ ˆ ˆ i j k + − 2 4 , 2 a r = ˆ ˆ ˆ 3 3 5 i j k + − and = ˆ ˆ ˆ 2 3 6 i j k + + Therefore, the distance between the lines is given by or = ˆ ˆ ˆ | 9 14 4 | 293 293 49 49 7 − + − i j k = = 1. Show that the three lines with direction cosines 2. Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). 3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5). 4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector ˆ ˆ ˆ 3 2 2 i j k + − . 12 3 4 4 12 3 3 4 12 , , ; , , ; , , 13 13 13 13 13 13 13 13 13 − − − are mutually perpendicular. d = 2 1 ( ) | | × − r r r r b a a b = EXERCISE 11.2 4 9 36 2 3 6 2 1 1 i j k ˆ ˆ ˆ + + THREE DIMENSIONAL GEOMETRY 389 − 5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector ˆ ˆ 2 4 i j k − + and is in the direction ˆ ˆ ˆ i j k + − 2 . 6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) 7. The cartesian equation of a line is 5 4 6 3 7 2 x y z − + − = = . Write its vector form. 8. Find the angle between the following pairs of lines: (i) ˆ ˆ ˆ ˆ ˆ ˆ = − + + λ + + 2 5 (3 2 6 ) r r i j k i j k and ˆ ˆ ˆ ˆ ˆ = − + µ + + 7 6 ( 2 2 ) r r i k i j k and parallel to the line given by 3 4 8 3 5 6 x y z + − + = = . Reprint 2025-26 390 MATHEMATICS 10. Find the values of p so that the lines 1 7 14 3 3 2 2 x y z p − − − = = each other. 12. Find the shortest distance between the lines ˆ ˆ ˆ ˆ ˆ ˆ = − − + µ + + 2 (2 2 ) r r i j k i j k 13. Find the shortest distance between the lines 1 1 1 7 6 1 x y z + + + = = − and 3 5 7 1 2 1 x y z − − − = = − 14. Find the shortest distance between the lines whose vector equations are 11. Show that the lines 5 2 7 5 1 x y z − + = = − and 1 2 3 x y z = = are perpendicular to (ii) ˆ ˆ ˆ ˆ ˆ ˆ = + − + λ − − 3 2 ( 2 ) r r i j k i j k and ˆ ˆ ˆ ˆ ˆ ˆ = − − + µ − − 2 56 (3 5 4 ) r r i j k i j k 9. Find the angle between the following pair of lines: and 7 7 5 6 3 1 5 x y z p − − − = = are at right angles. ˆ ˆ ˆ = + + ( 2 ) r r i j k + ˆ ˆ ˆ λ − + ( ) i j k and (ii) 5 2 3 and 2 2 1 4 1 8 x y z x y z − − − = = = = (i) 2 1 3 2 4 5 and 2 5 3 1 8 4 x y z x y z − − + + − − = = = = − − and ˆ ˆ ˆ ˆ ˆ ˆ = + + + µ + + 4 5 6 (2 3 ) r r i j k i j k 15. Find the shortest distance between the lines whose vector equations are 1. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b. 2. Find the equation of a line parallel to x-axis and passing through the origin. ˆ ˆ ˆ = + + ( 2 3 ) r r i j k + ˆ ˆ ˆ λ − + ( 3 2 ) i j k ˆ ˆ ˆ = − + − + − (1 ) ( 2) (3 2 ) r r t i t j t k and ˆ ˆ ˆ = + + − − + ( 1) (2 1) (2 1) r r s i s j s k Miscellaneous Exercise on Chapter 11 Reprint 2025-26 3. If the lines 1 2 3 1 1 6 and 3 2 2 3 1 5 x y z x y z k k − − − − − − = = = = − − are perpendicular, 4. Find the shortest distance between lines ˆ ˆ ˆ ˆ ˆ ˆ = + + + λ − + 6 2 2 ( 2 2 ) r r i j k i j k and ˆ ˆ ˆ ˆ ˆ = − − + µ − − 4 (3 2 2 ) r r i k i j k . 5. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: Summary ® Direction cosines of a line are the cosines of the angles made by the line with the positive directions of the coordinate axes. ® If l, m, n are the direction cosines of a line, then l 2 + m2 + n 2 = 1. ® Direction cosines of a line joining two points P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) are where PQ = ( )2 12 2 12 2 12 )()( −+−+− zzyyxx ® Direction ratios of a line are the numbers which are proportional to the direction cosines of a line. ® If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then find the value of k. 2 1 2 1 2 1 , , PQ PQ PQ x x y y z z − − − 7 10 16 19 3 8 − = − + = − zyx and 15 3 x − = 29 5 8 5 y z − − = − . THREE DIMENSIONAL GEOMETRY 391 ++ ® Skew lines are lines in space which are neither parallel nor intersecting. They lie in different planes. ® Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines. ® If l1 , m1 , n1 and l 2 , m2 , n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then cosθ = |l 1 l 2 + m1m2 + n1 n2 | l = 222 cba a ++ ; m = 222 cba b Reprint 2025-26 ++ ; n = 222 cba c 392 MATHEMATICS ® If a1 , b1 , c1 and a2 , b2 , c2 are the direction ratios of two lines and θ is the acute angle between the two lines; then ® Vector equation of a line that passes through the given point whose position vector is and parallel to a given vector is = + λ r r r r a b . ® Equation of a line through a point (x1 , y1 , z 1 ) and having direction cosines l, m, n is ® The vector equation of a line which passes through two points whose position ® If θ is the acute angle between = + λ 1 1 r r r r a b and = + λ 2 2 r r r r a b , then ® If 1 are the equations of two lines, then the acute angle between the two lines is given by cos θ = |l 1 l 2 + m1m2 + n1 n2 |. ® Shortest distance between two skew lines is the line segment perpendicular to both the lines. vectors are and is = + λ − ( ) r r r r r a b a . 1 2 cos | | | | ⋅ θ = 1 1 1 x x y y z z l m n − − − = = xx − = − = − and 2 zz m yy l 1 1 n b b r r r r b b 1 2 cosθ = 1 2 1 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 1 1 a b c a b c + + 1 + + + + a a b b c c xx − = − = − 2 n zz m yy l 2 2 2 2 ® Shortest distance between = + λ 1 1 r r r r a b and = + µ 2 2 r r r r a b is ® Shortest distance between the lines: 1 1 1 1 1 1 x x y y z z a b c − − − = = and 2 2 x x y y a b − − = = 2 2 z z c − is 2 2 1 2 ( ) ( – ) Reprint 2025-26 r r r r r r b b a a 1 2 2 1 | | × ⋅ × b b ® Distance between parallel lines = + λ 1 r r r r a b and = + µ 2 r r r r a b is 2 2 2 1 2 2 1 1 2 2 1 1 2 2 1 ( ) ( ) ( ) b c b c c a c a a b a b − + − + − x x y y z z 2 1 2 1 2 1 a b c a b c − − − 1 1 1 2 2 2 2 1 ( ) | | × − r r r r b a a b —v— THREE DIMENSIONAL GEOMETRY 393 Reprint 2025-26" class_12,12,Linear Programming,ncert_books/class_12/lemh2dd/lemh206.pdf,"394 MATHEMATICS 12.1 Introduction In earlier classes, we have discussed systems of linear equations and their applications in day to day problems. In Class XI, we have studied linear inequalities and systems of linear inequalities in two variables and their solutions by graphical method. Many applications in mathematics involve systems of inequalities/equations. In this chapter, we shall apply the systems of linear inequalities/equations to solve some real life problems of the type as given below: A furniture dealer deals in only two items–tables and chairs. He has Rs 50,000 to invest and has storage space of at most 60 pieces. A table costs Rs 2500 and a chair Rs 500. He estimates that from the sale of one table, he can make a profit of Rs 250 and that from the sale of one chair a profit of Rs 75. He wants to know how many tables and chairs he should buy from the available money so as to maximise his total profit, assuming that he can sell all the items which he buys. Such type of problems which seek to maximise (or, minimise) profit (or, cost) form a general class of problems called optimisation problems. Thus, an optimisation problem may involve finding maximum profit, minimum cost, or minimum use of resources etc. A special but a very important class of optimisation problems is linear programming problem. The above stated optimisation problem is an example of linear programming problem. Linear programming problems are of much interest because of their wide applicability in industry, commerce, management science etc. In this chapter, we shall study some linear programming problems and their solutions by graphical method only, though there are many other methods also to solve such problems. vThe mathematical experience of the student is incomplete if he never had the opportunity to solve a problem invented by himself. – G. POLYA v LINEAR PROGRAMMING Chapter 12 L. Kantorovich Reprint 2025-26 12.2 Linear Programming Problem and its Mathematical Formulation We begin our discussion with the above example of furniture dealer which will further lead to a mathematical formulation of the problem in two variables. In this example, we observe Suppose he decides to buy tables only and no chairs, so he can buy 50000 ÷ 2500, i.e., 20 tables. His profit in this case will be Rs (250 × 20), i.e., Rs 5000. Suppose he chooses to buy chairs only and no tables. With his capital of Rs 50,000, he can buy 50000 ÷ 500, i.e. 100 chairs. But he can store only 60 pieces. Therefore, he is forced to buy only 60 chairs which will give him a total profit of Rs (60 × 75), i.e., Rs 4500. There are many other possibilities, for instance, he may choose to buy 10 tables and 50 chairs, as he can store only 60 pieces. Total profit in this case would be Rs (10 × 250 + 50 × 75), i.e., Rs 6250 and so on. We, thus, find that the dealer can invest his money in different ways and he would earn different profits by following different investment strategies. Now the problem is : How should he invest his money in order to get maximum profit? To answer this question, let us try to formulate the problem mathematically. (i) The dealer can invest his money in buying tables or chairs or combination thereof. Further he would earn different profits by following different investment strategies. (ii) There are certain overriding conditions or constraints viz., his investment is limited to a maximum of Rs 50,000 and so is his storage space which is for a maximum of 60 pieces. LINEAR PROGRAMMING 395 12.2.1 Mathematical formulation of the problem Let x be the number of tables and y be the number of chairs that the dealer buys. Obviously, x and y must be non-negative, i.e., The dealer is constrained by the maximum amount he can invest (Here it is Rs 50,000) and by the maximum number of items he can store (Here it is 60). Stated mathematically, 2500x + 500y ≤ 50000 (investment constraint) or 5x + y ≤ 100 ... (3) and x + y ≤ 60 (storage constraint) ... (4) 0 ... (1) (Non-negative constraints) 0 ... (2) x y ≥ ≥ Reprint 2025-26 396 MATHEMATICS The dealer wants to invest in such a way so as to maximise his profit, say, Z which stated as a function of x and y is given by Z = 250x + 75y (called objective function) ... (5) Mathematically, the given problems now reduces to: Maximise Z = 250x + 75y subject to the constraints: 5x + y ≤ 100 So, we have to maximise the linear function Z subject to certain conditions determined by a set of linear inequalities with variables as non-negative. There are also some other problems where we have to minimise a linear function subject to certain conditions determined by a set of linear inequalities with variables as non-negative. Such problems are called Linear Programming Problems. Thus, a Linear Programming Problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear function (called objective function) of several variables (say x and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). The term linear implies that all the mathematical relations used in the problem are linear relations while the term programming refers to the method of determining a particular programme or plan of action. Before we proceed further, we now formally define some terms (which have been used above) which we shall be using in the linear programming problems: Objective function Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function. In the above example, Z = 250x + 75y is a linear objective function. Variables x and y are called decision variables. Constraints The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions. In the above example, the set of inequalities (1) to (4) are constraints. Optimisation problem A problem which seeks to maximise or minimise a linear function (say of two variables x and y) subject to certain constraints as determined by a set of linear inequalities is called an optimisation problem. Linear programming problems are special type of optimisation problems. The above problem of investing a x ≥ 0, y ≥ 0 x + y ≤ 60 Reprint 2025-26 given sum by the dealer in purchasing chairs and tables is an example of an optimisation problem as well as of a linear programming problem. We will now discuss how to find solutions to a linear programming problem. In this chapter, we will be concerned only with the graphical method. 12.2.2 Graphical method of solving linear programming problems In Class XI, we have learnt how to graph a system of linear inequalities involving two variables x and y and to find its solutions graphically. Let us refer to the problem of investment in tables and chairs discussed in Section 12.2. We will now solve this problem graphically. Let us graph the constraints stated as linear inequalities: 5x + y ≤ 100 ... (1) The graph of this system (shaded region) consists of the points common to all half planes determined by the inequalities (1) to (4) (Fig 12.1). Each point in this region represents a feasible choice open to the dealer for investing in tables and chairs. The region, therefore, is called the feasible region for the problem. Every point of this region is called a feasible solution to the problem. Thus, we have, Feasible region The common region determined by all the constraints including non-negative constraints x, y ≥ 0 of a linear programming problem is called the feasible region (or solution region) for the problem. In Fig 12.1, the region OABC (shaded) is the feasible region for the problem. The region other than feasible region is called an infeasible region. Feasible solutions Points within and on the boundary of the feasible region represent feasible solutions of the constraints. In Fig 12.1, every point within and on the boundary of the feasible region OABC represents feasible solution to the problem. For example, the point (10, 50) is a feasible solution of the problem and so are the points (0, 60), (20, 0) etc. Any point outside the feasible region is called an infeasible solution. For example, the point (25, 40) is an infeasible solution of the problem. Fig 12.1 x + y ≤ 60 ... (2) x ≥ 0 ... (3) y ≥ 0 ... (4) LINEAR PROGRAMMING 397 Reprint 2025-26 398 MATHEMATICS Optimal (feasible) solution: Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. Now, we see that every point in the feasible region OABC satisfies all the constraints as given in (1) to (4), and since there are infinitely many points, it is not evident how we should go about finding a point that gives a maximum value of the objective function Z = 250x + 75y. To handle this situation, we use the following theorems which are fundamental in solving linear programming problems. The proofs of these theorems are beyond the scope of the book. Theorem 1 Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point* (vertex) of the feasible region. Theorem 2 Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded**, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. Remark If R is unbounded, then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R. (By Theorem 1). In the above example, the corner points (vertices) of the bounded (feasible) region are: O, A, B and C and it is easy to find their coordinates as (0, 0), (20, 0), (10, 50) and (0, 60) respectively. Let us now compute the values of Z at these points. We have * A corner point of a feasible region is a point in the region which is the intersection of two boundary lines. ** A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle. Otherwise, it is called unbounded. Unbounded means that the feasible region does extend indefinitely in any direction. Vertex of the Corresponding value Feasible Region of Z (in Rs) O (0,0) 0 C (0,60) 4500 B (10,50) 6250 A (20,0) 5000 Reprint 2025-26 ← Maximum We observe that the maximum profit to the dealer results from the investment strategy (10, 50), i.e. buying 10 tables and 50 chairs. This method of solving linear programming problem is referred as Corner Point Method. The method comprises of the following steps: We will now illustrate these steps of Corner Point Method by considering some examples: Example 1 Solve the following linear programming problem graphically: 1. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point. 2. Evaluate the objective function Z = ax + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points. 3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z. 4. (a) M is the maximum value of Z, if the open half plane determined by ax + by > M has no point in common with the feasible region. Otherwise, Z has no maximum value. (ii) In case, the feasible region is unbounded, we have: (b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value. LINEAR PROGRAMMING 399 Maximise Z = 4x + y ... (1) subject to the constraints: Solution The shaded region in Fig 12.2 is the feasible region determined by the system of constraints (2) to (4). We observe that the feasible region OABC is bounded. So, we now use Corner Point Method to determine the maximum value of Z. The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and (0, 50) respectively. Now we evaluate Z at each corner point. x ≥ 0, y ≥ 0 ... (4) 3x + y ≤ 90 ... (3) x + y ≤ 50 ... (2) Reprint 2025-26 400 MATHEMATICS Fig 12.2 Hence, maximum value of Z is 120 at the point (30, 0). Example 2 Solve the following linear programming problem graphically: Minimise Z = 200 x + 500 y ... (1) subject to the constraints: x + 2y ≥ 10 ... (2) 3x + 4y ≤ 24 ... (3) x ≥ 0, y ≥ 0 ... (4) Solution The shaded region in Fig 12.3 is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points Corner Point Corresponding value of Z (0, 0) 0 (30, 0) 120 ← Maximum (20, 30) 110 (0, 50) 50 Reprint 2025-26 Fig 12.3 Corner Point Corresponding value of Z (0, 5) 2500 (4, 3) 2300 (0, 6) 3000 ← Minimum A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y at these points. Hence, minimum value of Z is 2300 attained at the point (4, 3) Example 3 Solve the following problem graphically: Minimise and Maximise Z = 3x + 9y ... (1) subject to the constraints: x + 3y ≤ 60 ... (2) x + y ≥ 10 ... (3) x ≤ y ... (4) Solution First of all, let us graph the feasible region of the system of linear inequalities (2) to (5). The feasible region ABCD is shown in the Fig 12.4. Note that the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15) and (0, 20) respectively. x ≥ 0, y ≥ 0 ... (5) Corner Corresponding value of Point Z = 3x + 9y A (0, 10) 90 B (5, 5) 60 C (15, 15) 180 D (0, 20) 180 LINEAR PROGRAMMING 401 ← }← Minimum Maximum (Multiple optimal solutions) We now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B (5, 5) of the feasible region. The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case. Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180. In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case if the two points produce same minimum value. Reprint 2025-26 Fig 12.4 402 MATHEMATICS Example 4 Determine graphically the minimum value of the objective function Z = – 50x + 20y ... (1) subject to the constraints: 2x – y ≥ – 5 ... (2) 3x + y ≥ 3 ... (3) 2x – 3y ≤ 12 ... (4) x ≥ 0, y ≥ 0 ... (5) Solution First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the Fig 12.5. Observe that the feasible region is unbounded. We now evaluate Z at the corner points. Corner Point Z = – 50x + 20y (0, 5) 100 (0, 3) 60 (1, 0) –50 (6, 0) – 300 ← smallest From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality i.e., – 5x + 2y < – 30 and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z. Otherwise, –300 will be the minimum value of Z. – 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.) Reprint 2025-26 Fig 12.5 As shown in the Fig 12.5, it has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints. In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0,5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region. (Why?) Example 5 Minimise Z = 3x + 2y subject to the constraints: Solution Let us graph the inequalities (1) to (3) (Fig 12.6). Is there any feasible region? Why is so? From Fig 12.6, you can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution. Remarks From the examples which we have discussed so far, we notice some general features of linear programming problems: 3x + 5y ≤ 15 ... (2) x ≥ 0, y ≥ 0 ... (3) x + y ≥ 8 ... (1) LINEAR PROGRAMMING 403 Solve the following Linear Programming Problems graphically: (ii) The maximum (or minimum) solution of the objective function occurs at the vertex (corner) of the feasible region. If two corner points produce the same maximum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maximum (or minimum) value. (i) The feasible region is always a convex region. 1. Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0. EXERCISE 12.1 Reprint 2025-26 Fig 12.6 404 MATHEMATICS Show that the minimum of Z occurs at more than two points. 10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0. 2. Minimise Z = – 3x + 4 y 3. Maximise Z = 5x + 3y 4. Minimise Z = 3x + 5y 5. Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. 6. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. 7. Minimise and Maximise Z = 5x + 10 y 8. Minimise and Maximise Z = x + 2y 9. Maximise Z = – x + 2y, subject to the constraints: subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0. subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0. subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0. subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0. x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0. Summary ® A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables (called objective function) subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative. In the World War II, when the war operations had to be planned to economise expenditure, maximise damage to the enemy, linear programming problems came to the forefront. The first problem in linear programming was formulated in 1941 by the Russian mathematician, L. Kantorovich and the American economist, F. L. Hitchcock, Historical Note Reprint 2025-26 both of whom worked at it independently of each other. This was the well known transportation problem. In 1945, an English economist, G.Stigler, described yet another linear programming problem – that of determining an optimal diet. In 1947, the American economist, G. B. Dantzig suggested an efficient method known as the simplex method which is an iterative procedure to solve any linear programming problem in a finite number of steps. L. Katorovich and American mathematical economist, T. C. Koopmans were awarded the nobel prize in the year 1975 in economics for their pioneering work in linear programming. With the advent of computers and the necessary softwares, it has become possible to apply linear programming model to increasingly complex problems in many areas. —v— LINEAR PROGRAMMING 405 Reprint 2025-26" class_12,13,Probability,ncert_books/class_12/lemh2dd/lemh207.pdf,"406 MATHEMATICS 13.1 Introduction In earlier Classes, we have studied the probability as a measure of uncertainty of events in a random experiment. We discussed the axiomatic approach formulated by Russian Mathematician, A.N. Kolmogorov (1903-1987) and treated probability as a function of outcomes of the experiment. We have also established equivalence between the axiomatic theory and the classical theory of probability in case of equally likely outcomes. On the basis of this relationship, we obtained probabilities of events associated with discrete sample spaces. We have also studied the addition rule of probability. In this chapter, we shall discuss the important concept of conditional probability of an event given that another event has occurred, which will be helpful in understanding the Bayes' theorem, multiplication rule of probability and independence of events. We shall also learn an important concept of random variable and its probability distribution and also the mean and variance of a probability distribution. In the last section of the chapter, we shall study an important discrete probability distribution called Binomial distribution. Throughout this chapter, we shall take up the experiments having equally likely outcomes, unless stated otherwise. 13.2 Conditional Probability Uptill now in probability, we have discussed the methods of finding the probability of events. If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random experiment in which the outcomes are equally likely to occur. Consider the experiment of tossing three fair coins. The sample space of the experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} vThe theory of probabilities is simply the Science of logic quantitatively treated. – C.S. PEIRCE v PROBABILITY Chapter 13 Pierre de Fermat (1601-1665) Reprint 2025-26 E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’. Then E = {HHH, HHT, HTH, THH} and F = {THH, THT, TTH, TTT} Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH}) and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT}) Also E ∩ F = {THH} with P(E ∩ F) = P({THH}) = 1 8 Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is the probability of occurrence of E? With the information of occurrence of F, we are sure that the cases in which first coin does not result into a tail should not be considered while finding the probability of E. This information reduces our sample space from the set S to its subset F for the event E. In other words, the additional information really amounts to telling us that the situation may be considered as being that of a new random experiment for which the sample space consists of all those outcomes only which are favourable to the occurrence of the event F. Now, the sample point of F which is favourable to event E is THH. Since the coins are fair, we can assign the probability 1 8 to each sample point. Let = 1 1 1 1 1 8 8 8 8 2 + + + = (Why ?) = 1 1 1 1 1 8 8 8 8 2 + + + = PROBABILITY 407 Thus, Probability of E considering F as the sample space = 1 4 , or Probability of E given that the event F has occurred = 1 4 This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F). Thus P(E|F) = 1 4 Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩ F. Reprint 2025-26 408 MATHEMATICS Thus, we can also write the conditional probability of E given that F has occurred as Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that P(E|F) can also be written as Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ (Why?) Thus, we can define the conditional probability as follows : Definition 1 If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given by 13.2.1 Properties of conditional probability P(E|F) = Numberof elementaryeventsfavourableto E F Numberof elementaryevents whicharefavourable to F ∩ = (E F) (F) n n ∩ P(E|F) = P(E|F) = P(E F) P(F) ∩ provided P(F) ≠ 0 n n n n ∩ ∩ = ... (1) (E F) (S) P(E F) (F) P(F) (S) Let E and F be events of a sample space S of an experiment, then we have Property 1 P(S|F) = P(F|F) = 1 We know that Also P(F|F) = P(F F) P(F) 1 P(F) P(F) ∩ = = Thus P(S|F) = P(F|F) = 1 Property 2 If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F) P(S|F) = P(S F) P(F) 1 P(F) P(F) ∩ = = Reprint 2025-26 In particular, if A and B are disjoint events, then P((A∪B)|F) = P(A|F) + P(B|F) We have When A and B are disjoint events, then P((A ∩ B)|F) = 0 ⇒ P((A ∪ B)|F) = P(A|F) + P(B|F) Property 3 P(E′|F) = 1 − P(E|F) From Property 1, we know that P(S|F) = 1 ⇒ P(E ∪ E′|F) = 1 since S = E ∪ E′ ⇒ P(E|F) + P (E′|F) = 1 since E and E′ are disjoint events Thus, P(E′|F) = 1 − P(E|F) Let us now take up some examples. P((A∪B)|F) = P[(A B) F] P(F) ∪ ∩ (by distributive law of union of sets over intersection) = P[(A F) (B F)] P(F) ∩ ∪ ∩ = P(A F)+ P(B F) – P(A B F) P(F) ∩ ∩ ∩ ∩ = P(A F) P(B F) P[(A B) F] P(F) P(F) P(F) ∩ ∩ ∩ ∩ + − = P(A|F) + P(B|F) – P((A ∩B)|F) PROBABILITY 409 Example 1 If P(A) = 7 13 , P(B) = 9 13 and P(A ∩ B) = 4 13 , evaluate P(A|B). Solution We have Example 2 A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ? 4 P(A B) 4 13 P(A|B)= P(B) 9 9 13 ∩ = = Reprint 2025-26 410 MATHEMATICS Solution Let b stand for boy and g for girl. The sample space of the experiment is S = {(b, b), (g, b), (b, g), (g, g)} Let E and F denote the following events : E : ‘both the children are boys’ F : ‘at least one of the child is a boy’ Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)} Now E ∩ F = {(b,b)} Thus P(F) = 3 4 and P (E ∩ F )= 1 4 Therefore P(E|F) = Example 3 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number? Solution Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card drawn is greater than 3’. We have to find P(A|B). Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} and A ∩ B = {4, 6, 8, 10} 1 P(E F) 1 4 P( F) 3 3 4 ∩ = = Also P(A) = 5 7 4 , P(B) = and P(A B) 10 10 10 ∩ = Then P(A|B) = Example 4 In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl? Solution Let E denote the event that a student chosen randomly studies in Class XII and F be the event that the randomly chosen student is a girl. We have to find P (E|F). Reprint 2025-26 4 P(A B) 4 10 P(B) 7 7 10 ∩ = = Now P(F) = 430 0.43 1000 = and 43 P(E F)= 0.043 1000 ∩ = (Why?) Then P(E|F) = P(E F) 0.043 0.1 P(F) 0.43 ∩ = = Example 5 A die is thrown three times. Events A and B are defined as below: A : 4 on the third throw B : 6 on the first and 5 on the second throw Find the probability of A given that B has already occurred. Solution The sample space has 216 outcomes. Now A = and A ∩ B = {(6,5,4)}. Now P(B) = 6 216 and P (A ∩ B) = 1 216 Then P(A|B) = B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} 1 P(A B) 1 216 P(B) 6 6 216 (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4) (3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4) (5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,6,4) ∩ = = PROBABILITY 411 Example 6 A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once? Solution Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’. Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)} and F = {(1,5), (2,4), (3,3), (4,2), (5,1)} We have P(E) = 11 36 and P(F) = 5 36 Also E∩F = {(2,4), (4,2)} Reprint 2025-26 412 MATHEMATICS Therefore P(E∩F) = 2 36 Hence, the required probability For the conditional probability discussed above, we have considered the elementary events of the experiment to be equally likely and the corresponding definition of the probability of an event was used. However, the same definition can also be used in the general case where the elementary events of the sample space are not equally likely, the probabilities P(E∩F) and P(F) being calculated accordingly. Let us take up the following example. Example 7 Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’. Solution The outcomes of the experiment can be represented in following diagrammatic manner called the ‘tree diagram’. P(E|F) = 2 P(E F) 2 36 P(F) 5 5 36 ∩ = = The sample space of the experiment may be described as where (H, H) denotes that both the tosses result into head and (T, i) denote the first toss result into a tail and the number i appeared on the die for i = 1,2,3,4,5,6. Thus, the probabilities assigned to the 8 elementary events are 1 1 1 1 1 1 1 1 , , , , , , , 4 4 12 12 12 12 12 12 respectively which is clear from the Fig 13.2. (H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6) S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)} Reprint 2025-26 Fig 13.1 Fig 13.2 Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows a number greater than 4’. Then F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)} E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)} Now P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)}) + P ({(T,4)}) + P({(T,5)}) + P({(T,6)}) and P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1 1 1 12 12 6 + = Hence P(E|F) = 1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E) 2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32 3. If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B) = 1 1 1 1 1 1 1 3 4 12 12 12 12 12 12 4 + + + + + + = 1 P(E F) 2 6 P(F) 9 3 4 ∩ = = EXERCISE 13.1 PROBABILITY 413 4. Evaluate P(A ∪ B), if 2P(A) = P(B) = 5 13 and P(A|B) = 2 5 5. If P(A) = 6 11 , P(B) = 5 11 and P(A ∪ B) 7 11 = , find 6. A coin is tossed three times, where (i) E : head on third toss , F : heads on first two tosses (i) P(A∩B) (ii) P(A|B) (iii) P(B|A) Determine P(E|F) in Exercises 6 to 9. (ii) E : at least two heads , F : at most two heads (iii) E : at most two tails , F : at least one tail Reprint 2025-26 414 MATHEMATICS 7. Two coins are tossed once, where (i) E : tail appears on one coin, F : one coin shows head (ii) E : no tail appears, F : no head appears 8. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses 9. Mother, father and son line up at random for a family picture E : son on one end, F : father in middle 10. A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. 11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} Find (i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E) (iii) P((E ∪ F)|G) and P((E ∩ F)|G) 12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl? 13. An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question? 14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’. 15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’. In each of the Exercises 16 and 17 choose the correct answer: 16. If P(A) = 1 2 , P(B) = 0, then P(A|B) is (A) 0 (B) 1 2 (C) not defined (D) 1 Reprint 2025-26 13.3 Multiplication Theorem on Probability Let E and F be two events associated with a sample space S. Clearly, the set E ∩ F denotes the event that both E and F have occurred. In other words, E ∩ F denotes the simultaneous occurrence of the events E and F. The event E ∩ F is also written as EF. Very often we need to find the probability of the event EF. For example, in the experiment of drawing two cards one after the other, we may be interested in finding the probability of the event ‘a king and a queen’. The probability of event EF is obtained by using the conditional probability as obtained below : We know that the conditional probability of event E given that F has occurred is denoted by P(E|F) and is given by From this result, we can write P(E ∩ F) = P(F) . P(E|F) ... (1) Also, we know that or P(F|E) = P(E F) P(E) ∩ (since E ∩ F = F ∩ E) 17. If A and B are events such that P(A|B) = P(B|A), then (A) A ⊂ B but A ≠ B (B) A = B (C) A ∩ B = φ (D) P(A) = P(B) P(F|E) = P(F E),P(E) 0 P(E) ∩ ≠ P(E|F) = P(E F),P(F) 0 P(F) ∩ ≠ PROBABILITY 415 Thus, P(E ∩ F) = P(E). P(F|E) .... (2) Combining (1) and (2), we find that P(E ∩ F) = P(E) P(F|E) = P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0. The above result is known as the multiplication rule of probability. Let us now take up an example. Example 8 An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black? Solution Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P (EF). Reprint 2025-26 416 MATHEMATICS Now P(E) = P (black ball in first draw) = 10 15 Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred. i.e. P(F|E) = 9 14 By multiplication rule of probability, we have P (E ∩ F) = P(E) P(F|E) Multiplication rule of probability for more than two events If E, F and G are three events of sample space, we have P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF) Similarly, the multiplication rule of probability can be extended for four or more events. The following example illustrates the extension of multiplication rule of probability for three events. Example 9 Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace? = 10 9 3 15 14 7 × = Solution Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace. Clearly, we have to find P (KKA) Now P(K) = 4 52 Also, P (K|K) is the probability of second king with the condition that one king has already been drawn. Now there are three kings in (52 − 1) = 51 cards. Therefore P(K|K) = 3 51 Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards. Reprint 2025-26 Therefore P(A|KK) = 4 50 By multiplication law of probability, we have P(KKA) = P(K) P(K|K) P(A|KK) 13.4 Independent Events Consider the experiment of drawing a card from a deck of 52 playing cards, in which the elementary events are assumed to be equally likely. If E and F denote the events 'the card drawn is a spade' and 'the card drawn is an ace' respectively, then Also E and F is the event ' the card drawn is the ace of spades' so that Hence P(E|F) = Since P(E) = 1 4 = P (E|F), we can say that the occurrence of event F has not P(E ∩F) = 1 52 P(E) = 13 1 4 1 and P(F) 52 4 52 13 = = = = 4 3 4 2 52 51 50 5525 × × = 1 P(E F) 1 52 P(F) 4 1 13 ∩ = = PROBABILITY 417 affected the probability of occurrence of the event E. We also have Again, P(F) = 1 13 = P(F|E) shows that occurrence of event E has not affected the probability of occurrence of the event F. Thus, E and F are two events such that the probability of occurrence of one of them is not affected by occurrence of the other. Such events are called independent events. P(F|E) = Reprint 2025-26 1 P(E F) 1 52 P(F) P(E) 13 1 4 ∩ = = = 418 MATHEMATICS Definition 2 Two events E and F are said to be independent, if P(F|E) = P (F) provided P (E) ≠ 0 and P (E|F) = P (E) provided P (F) ≠ 0 Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0 Now, by the multiplication rule of probability, we have P(E ∩ F) = P(E) . P (F|E) ... (1) If E and F are independent, then (1) becomes P(E ∩ F) = P(E) . P(F) ... (2) Thus, using (2), the independence of two events is also defined as follows: Definition 3 Let E and F be two events associated with the same random experiment, then E and F are said to be independent if Remarks (i) Two events E and F are said to be dependent if they are not independent, i.e. if P(E ∩ F ) ≠ P(E) . P (F) (ii) Sometimes there is a confusion between independent events and mutually exclusive events. Term ‘independent’ is defined in terms of ‘probability of events’ whereas mutually exclusive is defined in term of events (subset of sample space). Moreover, mutually exclusive events never have an outcome common, but independent events, may have common outcome. Clearly, ‘independent’ and ‘mutually exclusive’ do not have the same meaning. In other words, two independent events having nonzero probabilities of occurrence can not be mutually exclusive, and conversely, i.e. two mutually exclusive events having nonzero probabilities of occurrence can not be independent. (iii) Two experiments are said to be independent if for every pair of events E and F, where E is associated with the first experiment and F with the second experiment, the probability of the simultaneous occurrence of the events E and F when the two experiments are performed is the product of P(E) and P(F) calculated separately on the basis of two experiments, i.e., P (E ∩ F) = P (E) . P(F) (iv) Three events A, B and C are said to be mutually independent, if P(A ∩ B) = P(A) P(B) P(A ∩ C) = P(A) P(C) P(E ∩ F) = P(E) . P (F) P (B ∩ C) = P(B) P(C) and P(A ∩ B ∩ C) = P(A) P (B) P(C) Reprint 2025-26 If at least one of the above is not true for three given events, we say that the events are not independent. Example 10 A die is thrown. If E is the event ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’ then find whether E and F are independent ? Solution We know that the sample space is S = {1, 2, 3, 4, 5, 6} Now E = { 3, 6}, F = { 2, 4, 6} and E ∩ F = {6} Then P(E) = 2 1 3 1 1 , P(F) and P(E F) 6 3 6 2 6 = = = ∩ = Clearly P(E ∩ F) = P(E). P (F) Hence E and F are independent events. Example 11 An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event ‘odd number on the second throw’. Check the independence of the events A and B. Solution If all the 36 elementary events of the experiment are considered to be equally likely, we have Also P(A ∩ B) = P (odd number on both throws) P(A) = 18 1 36 2 = and 18 1 P(B) 36 2 = = = 9 1 36 4 = PROBABILITY 419 Now P(A) P(B) = 1 1 1 2 2 4 × = Clearly P(A ∩ B) = P(A) × P(B) Thus, A and B are independent events Example 12 Three coins are tossed simultaneously. Consider the event E ‘three heads or three tails’, F ‘at least two heads’ and G ‘at most two heads’. Of the pairs (E,F), (E,G) and (F,G), which are independent? which are dependent? Solution The sample space of the experiment is given by Clearly E = {HHH, TTT}, F= {HHH, HHT, HTH, THH} S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Reprint 2025-26 420 MATHEMATICS and G = {HHT, HTH, THH, HTT, THT, TTH, TTT} Also E ∩ F = {HHH}, E ∩ G = {TTT}, F ∩ G = {HHT, HTH, THH} Therefore P(E) = 2 1 4 1 7 , P(F) , P(G) 8 4 8 2 8 = = = = and P(E∩F) = 1 1 3 , P(E G) , P(F G) 8 8 8 ∩ = ∩ = Also P(E) . P(F) = 1 1 1 1 7 7 , P(E) P(G) 4 2 8 4 8 32 × = ⋅ = × = and P(F) . P(G) = 1 7 7 2 8 16 × = Thus P(E ∩ F) = P(E) . P(F) P(E ∩ G) ≠ P(E) . P(G) and P(F ∩ G) ≠ P (F) . P(G) Hence, the events (E and F) are independent, and the events (E and G) and (F and G) are dependent. Example 13 Prove that if E and F are independent events, then so are the events E and F′. Solution Since E and F are independent, we have P(E ∩ F) = P(E) . P(F) ....(1) From the venn diagram in Fig 13.3, it is clear that E ∩ F and E ∩ F′ are mutually exclusive events and also E =(E ∩ F) ∪ (E ∩ F′). Therefore P(E) = P(E ∩ F) + P(E ∩ F′) or P(E ∩ F′) = P(E) − P(E ∩ F) Hence, E and F′ are independent = P(E) − P(E) . P(F) (by (1)) = P(E) (1−P(F)) = P(E). P(F′) Reprint 2025-26 (E F ) ∩ ’ (E F) ’∩ E F S Fig 13.3 (E F) ∩ (E F ) ’ ’ ∩ Example 14 If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′) Solution We have P(at least one of A and B) = P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A) P(B) = P(A) + P(B) [1−P(A)] = P(A) + P(B). P(A′) ANote In a similar manner, it can be shown that if the events E and F are independent, then 1. If P(A) 3 5 = and P (B) 1 5 = , find P (A ∩ B) if A and B are independent events. 2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black. 3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale. 4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not. 5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent? (a) E′ and F are independent, (b) E′ and F′ are independent EXERCISE 13.2 = 1− P(A′) + P(B) P(A′) = 1− P(A′) [1− P(B)] = 1− P(A′) P (B′) PROBABILITY 421 6. Let E and F be events with P(E) 3 5 = , P(F) 3 10 = and P (E ∩ F) = 1 5 . Are E and F independent? Reprint 2025-26 422 MATHEMATICS 10. Events A and B are such that P (A) = 1 2 , P(B) = 7 12 and P(not A or not B) = 1 4 . State whether A and B are independent ? 11. Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B) 12. A die is tossed thrice. Find the probability of getting an odd number at least once. 13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red. 7. Given that the events A and B are such that P(A) = 1 2 , P(A ∪ B) = 3 5 and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent. 8. Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P (A|B) (iv) P (B|A) 9. If A and B are two events such that P(A) = 1 4 , P (B) = 1 2 and P(A ∩ B) = 1 8 , find P (not A and not B). 14. Probability of solving specific problem independently by A and B are 1 2 and 1 3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem. 15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ? (i) E : ‘the card drawn is a spade’ F : ‘the card drawn is an ace’ (ii) E : ‘the card drawn is black’ F : ‘the card drawn is a king’ (iii) E : ‘the card drawn is a king or queen’ F : ‘the card drawn is a queen or jack’. Reprint 2025-26 16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper. Choose the correct answer in Exercises 17 and 18. 17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is 13.5 Bayes' Theorem Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags. We can find the probability of selecting any of the bags (i.e. 1 2 ) or probability of (A) 0 (B) 1 3 (C) 1 12 (D) 1 36 18. Two events A and B will be independent, if (A) A and B are mutually exclusive (B) P(A′B′) = [1 – P(A)] [1 – P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1 PROBABILITY 423 drawing a ball of a particular colour (say white) from a particular bag (say Bag I). In other words, we can find the probability that the ball drawn is of a particular colour, if we are given the bag from which the ball is drawn. But, can we find the probability that the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is given? Here, we have to find the reverse probability of Bag II to be selected when an event occurred after it is known. Famous mathematician, John Bayes' solved the problem of finding reverse probability by using conditional probability. The formula developed by him is known as ‘Bayes theorem’ which was published posthumously in 1763. Before stating and proving the Bayes' theorem, let us first take up a definition and some preliminary results. 13.5.1 Partition of a sample space A set of events E1 , E2 , ..., En is said to represent a partition of the sample space S if (a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, ..., n Reprint 2025-26 424 MATHEMATICS (c) P(Ei )> 0 for all i = 1, 2, ..., n. In other words, the events E1 , E2 , ..., En represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities. As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S. From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′} is a partition of the sample space S. It may be mentioned that the partition of a sample space is not unique. There can be several partitions of the same sample space. 13.5.2 Theorem of total probability Let {E1 , E2 ,...,En } be a partition of the sample space S, and suppose that each of the events E1 , E2 ,..., En has nonzero probability of occurrence. Let A be any event associated with S, then Proof Given that E1 , E2 ,..., En is a partition of the sample space S (Fig 13.4). Therefore, S = E1 ∪ E2 ∪ ... ∪ En ... (1) and Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, ..., n Now, we know that for any event A, A = A ∩ S We shall now prove a theorem known as Theorem of total probability. (b) E1 ∪ Ε2 ∪ ... ∪ En = S and P(A) = P(E1 ) P(A|E1 ) + P(E2 ) P(A|E2 ) + ... + P(En ) P(A|En ) = 1 P(E ) P(A|E ) n j j j= ∑ Ei and Ej are disjoint, for i ≠ j, therefore, A ∩ Ei and A ∩ Ej are also disjoint for all i ≠ j, i, j = 1, 2, ..., n. Thus, P(A) = P [(A ∩ E1 ) ∪ (A ∩ E2 )∪ .....∪ (A ∩ En )] = P (A ∩ E1 ) + P (A ∩ E2 ) + ... + P (A ∩ En ) Now, by multiplication rule of probability, we have P(A ∩ Ei ) = P(Ei ) P(A|Ei ) as P (Ei ) ≠ 0∀i = 1,2,..., n = A ∩ (E1 ∪ E2 ∪ ... ∪ En ) = (A ∩ E1 ) ∪ (A ∩ E2 ) ∪ ...∪ (A ∩ En ) Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej . We know that Reprint 2025-26 Fig 13.4 Therefore, P (A) = P (E1 ) P (A|E1 ) + P (E2 ) P (A|E2 ) + ... + P (En )P(A|En ) or P(A) = 1 P(E ) P(A|E ) n j j j= ∑ Example 15 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time. Solution Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike. We have to find P(A). We have P(B) = 0.65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0.65 = 0.35 P(A|B) = 0.32, P(A|B′) = 0.80 Since events B and B′ form a partition of the sample space S, therefore, by theorem on total probability, we have P(A) = P(B) P(A|B) + P(B′) P(A|B′) = 0.65 × 0.32 + 0.35 × 0.8 = 0.208 + 0.28 = 0.488 Thus, the probability that the construction job will be completed in time is 0.488. We shall now state and prove the Bayes' theorem. Bayes’ Theorem If E1 , E2 ,..., En are n non empty events which constitute a partition of sample space S, i.e. E1 , E2 ,..., En are pairwise disjoint and E1∪ E2∪ ... ∪ En = S and A is any event of nonzero probability, then PROBABILITY 425 Proof By formula of conditional probability, we know that P(Ei |A) = P(Ei |A) = P(A E ) P(A) ∩ i = P(E )P(A|E ) P(A) i i (by multiplication rule of probability) = P(E ) P(A|E ) i i n j j j= ∑ for any i = 1, 2, 3, ..., n P(E )P(A|E ) i i n j j j= ∑ (by the result of theorem of total probability) P(E ) P(A|E ) 1 P(E ) P(A|E ) 1 Reprint 2025-26 426 MATHEMATICS Remark The following terminology is generally used when Bayes' theorem is applied. The events E1 , E2 , ..., En are called hypotheses. The probability P(Ei ) is called the priori probability of the hypothesis Ei The conditional probability P(Ei |A) is called a posteriori probability of the hypothesis Ei . Bayes' theorem is also called the formula for the probability of ""causes"". Since the Ei 's are a partition of the sample space S, one and only one of the events Ei occurs (i.e. one of the events Ei must occur and only one can occur). Hence, the above formula gives us the probability of a particular Ei (i.e. a ""Cause""), given that the event A has occurred. The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples. Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II. Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing a red ball. Then P(E1 ) = P(E2 ) = 1 2 Also P(A|E1 ) = P(drawing a red ball from Bag I) = 3 7 and P(A|E2 ) = P(drawing a red ball from Bag II) = 5 11 Now, the probability of drawing a ball from Bag II, being given that it is red, is P(E2 |A) By using Bayes' theorem, we have Example 17 Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? P(E2 |A) = 2 2 1 1 2 2 P(E )P(A|E ) P(E )P(A|E )+ P(E )P(A|E ) = Reprint 2025-26 1 5 2 11 35 1 3 1 5 68 2 7 2 11 × = × + × Solution Let E1 , E2 and E3 be the events that boxes I, II and III are chosen, respectively. Then P(E1 ) = P(E2 ) = P(E3 ) = 1 3 Also, let A be the event that ‘the coin drawn is of gold’ Then P(A|E1 ) = P(a gold coin from bag I) = 2 2 = 1 P(A|E3 ) = P(a gold coin from bag III) = 1 2 Now, the probability that the other coin in the box is of gold = the probability that gold coin is drawn from the box I. = P(E1 |A) By Bayes' theorem, we know that Example 18 Suppose that the reliability of a HIV test is specified as follows: Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as showing HIV+ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV+ive. What is the probability that the person actually has HIV? P(A|E2 ) = P(a gold coin from bag II) = 0 P(E1 |A) = 1 1 1 1 2 2 3 3 P(E )P(A|E ) P(E )P(A|E )+ P(E )P(A|E )+ P(E )P(A|E ) = 1 1 3 2 1 1 1 1 3 1 0 3 3 3 2 × = × + × + × PROBABILITY 427 Solution Let E denote the event that the person selected is actually having HIV and A the event that the person's HIV test is diagnosed as +ive. We need to find P(E|A). Also E′ denotes the event that the person selected is actually not having HIV. Clearly, {E, E′} is a partition of the sample space of all people in the population. We are given that P(E) = 0.1% 0.1 0.001 100 = = Reprint 2025-26 428 MATHEMATICS and P(A|E′) = P(Person tested as HIV +ive given that he/she is actually not having HIV) = 1% = 1 100 = 0.01 Now, by Bayes' theorem Thus, the probability that a person selected at random is actually having HIV given that he/she is tested HIV+ive is 0.083. Example 19 In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B? Solution Let events B1 , B2 , B3 be the following : P(E′) = 1 – P(E) = 0.999 P(A|E) = P(Person tested as HIV+ive given that he/she is actually having HIV) P(E|A) = P(E)P(A|E) P(E)P(A|E)+ P(E )P(A|E ) ′ ′ = 90% 90 0.9 100 = = = 0.001 0.9 90 0.001 0.9 0.999 0.01 1089 × = × + × = 0.083 approx. B1 : the bolt is manufactured by machine A B2 : the bolt is manufactured by machine B B3 : the bolt is manufactured by machine C Clearly, B1 , B2 , B3 are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space. Again P(E|B1 ) = Probability that the bolt drawn is defective given that it is manufactured by machine A = 5% = 0.05 Similarly, P(E|B2 ) = 0.04, P(E|B3 ) = 0.02. Let the event E be ‘the bolt is defective’. The event E occurs with B1 or with B2 or with B3 . Given that, P(B1 ) = 25% = 0.25, P (B2 ) = 0.35 and P(B3 ) = 0.40 Reprint 2025-26 Hence, by Bayes' Theorem, we have Example 20 A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3 1 1 2 , , and 10 5 10 5 . The probabilities that he will be late are 1 1 1 , , and 4 3 12 , if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train? Solution Let E be the event that the doctor visits the patient late and let T1 , T2 , T3 , T4 be the events that the doctor comes by train, bus, scooter, and other means of transport respectively. Then P(T1 ) = 2 3 4 3 1 1 2 , P(T ) ,P(T ) and P(T ) 10 5 10 5 = = = (given) Similarly, P(E|T2 ) = 1 3 , P(E|T3 ) = 1 12 and P(E|T4 ) = 0, since he is not late if he P(B2 |E) = 2 2 1 1 2 2 3 3 P(B )P(E|B ) P(B )P(E|B )+ P(B )P(E|B )+P(B )P(E|B ) P(E|T1 ) = Probability that the doctor arriving late comes by train = 1 4 = 0.35 0.04 0.25 0.05 0.35 0.04 0.40 0.02 × × + × + × = 0.0140 28 0.0345 69 = PROBABILITY 429 comes by other means of transport. Therefore, by Bayes' Theorem, we have P(T1 |E) = Probability that the doctor arriving late comes by train Hence, the required probability is 1 2 . = 1 1 1 1 2 2 3 3 4 4 P(T )P(E|T ) P(T )P(E|T )+ P(T )P(E|T )+ P(T )P(E|T )+ P(T )P(E|T ) = 3 1 10 4 3 1 1 1 1 1 2 0 10 4 5 3 10 12 5 × + × + × + × Reprint 2025-26 × = 3 120 1 40 18 2 × = 430 MATHEMATICS Example 21 A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. Solution Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur. Then P(S1 ) = Probability that six occurs = 1 6 P(S2 ) = Probability that six does not occur = 5 6 P(E|S1 ) = Probability that the man reports that six occurs when six has actually occurred on the die = Probability that the man speaks the truth = 3 4 P(E|S2 ) = Probability that the man reports that six occurs when six has not actually occurred on the die Thus, by Bayes' theorem, we get P(S1 |E) = Probability that the report of the man that six has occurred is actually a six = Probability that the man does not speak the truth 3 1 1 4 4 = − = = 1 1 1 1 2 2 P(S )P(E |S ) P(S )P(E|S )+P(S )P(E|S ) Hence, the required probability is 3 . 8 Remark A random variable is a real valued function whose domain is the sample space of a random experiment. For example, let us consider the experiment of tossing a coin two times in succession. The sample space of the experiment is S = {HH, HT, TH, TT}. = 1 3 6 4 1 24 3 1 3 5 1 8 8 8 6 4 6 4 × = × = × + × Reprint 2025-26 If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below : X(HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0. More than one random variables can be defined on the same sample space. For example, let Y denote the number of heads minus the number of tails for each outcome of the above sample space S. Then Y(HH) = 2, Y (HT) = 0, Y (TH) = 0, Y (TT) = – 2. Thus, X and Y are two different random variables defined on the same sample space S. 1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red? 2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. 3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier? 4. In answering a question on a multiple choice test, a student either knows the EXERCISE 13.3 PROBABILITY 431 dent knows the answer given that he answered it correctly? 5. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population answer or guesses. Let 3 4 be the probability that he knows the answer and 1 4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1 4 . What is the probability that the stuReprint 2025-26 432 MATHEMATICS actually has the disease, what is the probability that a person has the disease given that his test result is positive ? 6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ? 7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? 8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B? 9. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group. 10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die? 11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A? 12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond. 13. Probability that A speaks truth is 4 5 . A coin is tossed. A reports that a head appears. The probability that actually there was head is Reprint 2025-26 Example 22 Coloured balls are distributed in four boxes as shown in the following table: A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III? (A) 4 5 (B) 1 2 (C) 1 5 (D) 2 5 14. If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct? (A) P(B) P(A | B) P(A) = (B) P(A|B) < P(A) (C) P(A|B) ≥ P(A) (D) None of these Box Colour III 1 2 3 1 IV 4 3 1 5 II 2 2 2 2 I 3 4 5 6 Black White Red Blue Miscellaneous Examples PROBABILITY 433 Solution Let A, E1 , E2 , E3 and E4 be the events as defined below : Since the boxes are chosen at random, Therefore P(E1 ) = P(E2 ) = P(E3 ) = P(E4 ) = 1 4 Also P(A|E1 ) = 3 18 , P(A|E2 ) = 2 8 , P(A|E3 ) = 1 7 and P(A|E4 ) = 4 13 P(box III is selected, given that the drawn ball is black) = P(E3 |A). By Bayes' theorem, A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Reprint 2025-26 434 MATHEMATICS Example 23 A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first. Solution Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’). Thus, P(S) = 1 5 , P(F) 6 6 = P(A wins in the first throw) = P(S) = 1 6 A gets the third throw, when the first throw by A and second throw by B result into failures. Therefore, P(A wins in the 3rd throw) = P(FFS) = 5 5 1 P(F)P(F)P(S)= 6 6 6 × × P(E3 |A) = 3 3 1 1 2 2 3 3 4 4 P(E ) P(A|E ) P(E )P(A|E ) P(E )P(A|E )+P(E )P(A|E ) P(E )P(A|E ) ⋅ + + P(A wins in the 5th throw) = P (FFFFS) = 5 6 = 1 1 4 7 0.165 1 3 1 1 1 1 1 4 4 18 4 4 4 7 4 13 × = × + × + × + × = 2 5 1 6 6 × 1 6 4 and so on. Hence, P(A wins) = 1 6 Remark If a + ar + ar2 + ... + arn–1 + ..., where |r| < 1, then sum of this infinite G.P. is given by . 1 a − r (Refer A.1.3 of Class XI Text book). P(B wins) = 1 – P (A wins) = 6 5 1 11 11 − = Reprint 2025-26 = 1 6 25 1 36 − = 6 11 1 6 2 4 + + + ... 5 6 1 6 5 6 Example 24 If a machine is correctly set up, it produces 90% acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Past experience shows that 80% of the set ups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup. Solution Let A be the event that the machine produces 2 acceptable items. Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup. Now P(B1 ) = 0.8, P(B2 ) = 0.2 Therefore P(B1 |A) = 1 1 1 1 2 2 P(B ) P(A|B ) P(B ) P(A|B ) + P(B ) P(A|B ) 1. A and B are two events such that P (A) ≠ 0. Find P(B|A), if (i) A is a subset of B (ii) A ∩ B = φ 2. A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. P(A|B1 ) = 0.9 × 0.9 and P(A|B2 ) = 0.4 × 0.4 Miscellaneous Exercise on Chapter 13 = 0.8× 0.9 × 0.9 648 0.95 0.8× 0.9 × 0.9 + 0.2 × 0.4 × 0.4 680 = = PROBABILITY 435 3. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females. 4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed? 5. If a leap year is selected at random, what is the chance that it will contain 53 tuesdays? 6. Suppose we have four boxes A,B,C and D containing coloured marbles as given below: (ii) Find the probability that both children are females, if it is known that the elder child is a female. Reprint 2025-26 436 MATHEMATICS D 0 6 4 One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C? 7. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga? 8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being 9. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15 Evaluate the following probabilities (i) P(A fails|B has failed) (ii) P(A fails alone) assumed with probability 1 2 ). Box Marble colour A 1 6 3 B 6 2 2 C 8 1 1 Red White Black 10. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. Choose the correct answer in each of the following: 11. If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then (A) A ⊂ B (B) B ⊂ A (C) B = φ (D) A = φ Reprint 2025-26 12. If P(A|B) > P(A), then which of the following is correct : 13. If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then The salient features of the chapter are – ® The conditional probability of an event E, given the occurrence of the event F ® 0 ≤ P (E|F) ≤ 1, P (E′|F) = 1 – P (E|F) P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G) ® P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0 P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0 ® If E and F are independent, then P (E ∩ F) = P (E) P (F) P (E|F) = P (E), P (F) ≠ 0 P (F|E) = P (F), P(E) ≠ 0 ® Theorem of total probability (A) P(B|A) < P(B) (B) P(A ∩ B) < P(A) . P(B) (A) P(B|A) = 1 (B) P(A|B) = 1 (C) P(B|A) > P(B) (D) P(B|A) = P(B) (C) P(B|A) = 0 (D) P(A|B) = 0 is given by P(E F) P(E | F) P(F) ∩ = , P(F) ≠ 0 Summary PROBABILITY 437 then P(A) = P(E1 ) P (A|E1 ) + P (E2 ) P (A|E2 ) + ... + P (En ) P(A|En ) ® Bayes' theorem If E1 , E2 , ..., En are events which constitute a partition of sample space S, i.e. E1 , E2 , ..., En are pairwise disjoint and E1 4 E2 4 ... 4 En = S and A be any event with nonzero probability, then Let {E1 , E2 , ...,En ) be a partition of a sample space and suppose that each of E1 , E2 , ..., En has nonzero probability. Let A be any event associated with S, P(E A P(E ) P(A|E ) P(E )P(A|E ) i i i j j j n | )= Reprint 2025-26 = ∑ 1 438 MATHEMATICS The earliest indication on measurement of chances in game of dice appeared in 1477 in a commentary on Dante's Divine Comedy. A treatise on gambling named liber de Ludo Alcae, by Geronimo Carden (1501-1576) was published posthumously in 1663. In this treatise, he gives the number of favourable cases for each event when two dice are thrown. Galileo (1564-1642) gave casual remarks concerning the correct evaluation of chance in a game of three dice. Galileo analysed that when three dice are thrown, the sum of the number that appear is more likely to be 10 than the sum 9, because the number of cases favourable to 10 are more than the number of cases for the appearance of number 9. Apart from these early contributions, it is generally acknowledged that the true origin of the science of probability lies in the correspondence between two great men of the seventeenth century, Pascal (1623-1662) and Pierre de Fermat (1601-1665). A French gambler, Chevalier de Metre asked Pascal to explain some seeming contradiction between his theoretical reasoning and the observation gathered from gambling. In a series of letters written around 1654, Pascal and Fermat laid the first foundation of science of probability. Pascal solved the problem in algebraic manner while Fermat used the method of combinations. Great Dutch Scientist, Huygens (1629-1695), became acquainted with the content of the correspondence between Pascal and Fermat and published a first book on probability, ""De Ratiociniis in Ludo Aleae"" containing solution of many interesting rather than difficult problems on probability in games of chances. The next great work on probability theory is by Jacob Bernoulli (1654-1705), in the form of a great book, ""Ars Conjectendi"" published posthumously in 1713 by his nephew, Nicholes Bernoulli. To him is due the discovery of one of the most important probability distribution known as Binomial distribution. The next remarkable work on probability lies in 1993. A. N. Kolmogorov (1903-1987) is credited with the axiomatic theory of probability. His book, ‘Foundations of probability’ published in 1933, introduces probability as a set function and is considered a ‘classic!’. Historical Note Reprint 2025-26 —v—"