Tessellations Penrose tiling represents one type of tessellation. (credit: "Penrose Tiling" by Inductiveload/Wikimedia Commons, Public Domain) Learning Objectives After completing this section, you should be able to: Apply translations, rotations, and reflections. Determine if a shape tessellates. The illustration shown above ( ) is an unusual pattern called a Penrose tiling. Notice that there are two types of shapes used throughout the pattern: smaller green parallelograms and larger blue parallelograms. What's interesting about this design is that although it uses only two shapes over and over, there is no repeating pattern. In this section, we will focus on patterns that do repeat. Repeated patterns are found in architecture, fabric, floor tiles, wall patterns, rug patterns, and many unexpected places as well. It may be a simple hexagon-shaped floor tile, or a complex pattern composed of several different motifs. These two-dimensional designs are called regular (or periodic) tessellations . There are countless designs that may be classified as regular tessellations, and they all have one thing in common—their patterns repeat and cover the plane. We will explore how tessellations are created and experiment with making some of our own as well. The topic of tessellations belongs to a field in mathematics called transformational geometry, which is a study of the ways objects can be moved while retaining the same shape and size. These movements are termed rigid motions and symmetries . M. C. Escher A good place to start the study of tessellations is with the work of M. C. Escher. The Dutch graphic artist was famous for the dimensional illusions he created in his woodcuts and lithographs, and that theme is carried out in many of his tessellations as well. Escher became obsessed with the idea of the “regular division of the plane.” He sought ways to divide the plane with shapes that would fit snugly next to each other with no gaps or overlaps, represent beautiful patterns, and could be repeated infinitely to fill the plane. He experimented with practically every geometric shape imaginable and found the ones that would produce a regular division of the plane. The idea is similar to dividing a number by one of its factors. When a number divides another number evenly, there are no remainders, like there are no gaps when a shape divides or fills the plane. Escher went far beyond geometric shapes, beyond triangles and polygons, beyond irregular polygons, and used other shapes like figures, faces, animals, fish, and practically any type of object to achieve his goal; and he did achieve it, beautifully, and left it for the ages to appreciate. M.C. Escher: How to Create a Tessellation The Mathematical Art of M.C. Escher Tessellation Properties and Transformations A regular tessellation means that the pattern is made up of congruent regular polygons, same size and shape, including some type of movement ; that is, some type of transformation or symmetry. Here we consider the rigid motions of translations, rotations, reflections, or glide reflections. A plane of tessellations has the following properties: Patterns are repeated and fill the plane. There are no gaps or overlaps. Shapes must fit together perfectly. (It was Escher who determined that a proper tessellation could have no gaps and no overlaps.) Shapes are combined using a transformation. All the shapes are joined at a vertex. In other words, if you were to draw a circle around a vertex, it would include a corner of each shape touching at that vertex. For a tessellation of regular congruent polygons, the sum of the measures of the interior angles that meet at a vertex equals 360 ∘ . In , the tessellation is made up of squares. There are four squares meeting at a vertex. An interior angle of a square is 90 ∘ and the sum of four interior angles is 360 ∘ . In , the tessellation is made up of regular hexagons. There are three hexagons meeting at each vertex. The interior angle of a hexagon is 120 ∘ , and the sum of three interior angles is 360 ∘ . Both tessellations will fill the plane, there are no gaps, the sum of the interior angle meeting at the vertex is 360 ∘ , and both are achieved by translation transformations. These tessellations work because all the properties of a tessellation are present. Tessellation – Squares Tessellation – Hexagons The movements or rigid motions of the shapes that define tessellations are classified as translations, rotations, reflections, or glide reflections. Let’s first define these movements and then look at some examples showing how these transformations are revealed. Translation A translation is a movement that shifts the shape vertically, horizontally, or on the diagonal. Consider the trapezoid A B C D in . We have translated it 3 units to the right and 3 units up. That means every corner is moved by the number of units and in the direction specified. Mathematicians will indicate this movement with a vector, an arrow that is drawn to illustrate the criteria and the magnitude of the translation. The location of the translated trapezoid is marked with the vertices, A ′ B ′ C ′ D ′ , but it is still the exact same shape and size as the original trapezoid A B C D . Translation Creating a Translation Suppose you have a hexagon on a grid as in . Translate the hexagon 5 units to the right and 3 units up. The best way to do this is by translating the individual points A , B , C , D , E , F . Once translated, the points become A ′ , B ′ , C ′ , D ′ , E ′ , F ′ . Rotation The rotation transformation occurs when you rotate a shape about a point and at a predetermined angle. In , the triangle is rotated around the rotation point by 90 ∘ , and then translated 7 units up and 4 units over to the right. That means that each corner is translated to the new location by the same number of units and in the same direction. Rotation We can see that Δ A is mapped to Δ A ′ by a rotation of 90 ∘ up and to the right. If rotated again by 90 ∘ , the triangle would be upside down. Applying a Rotation illustrates a tessellation begun with an equilateral triangle. Explain how this pattern is produced. A rotation to the right or to the left around the vertex by 60 ∘ , six times, produces the hexagonal shape. The sixth rotation brings the triangle back to its original position. Then, a reflection up and another one on the diagonal will reproduce the pattern. When a shape returns to its original position by a rotation, we say that it has rotational symmetry. Reflection A reflection is the third transformation. A shape is reflected about a line and the new shape becomes a mirror image. You can reflect the shape vertically, horizontally, or on the diagonal. There are two shapes in . The quadrilateral is reflected horizontally; the arrow shape is reflected vertically. Reflection Glide Reflection The glide reflection is the fourth transformation. It is a combination of a reflection and a translation. This can occur by first reflecting the shape and then gliding or translating it to its new location, or by translating first and then reflecting. The example in shows a trapezoid, which is reflected over the dashed line, so it appears upside down. Then, we shifted the shape horizontally by 6 units to the right. Whether we use the glide first or the reflection first, the end result is the same in most cases. However, the tessellation shown in the next example can only be achieved by a reflection first and then a translation. Glide Reflection Applying the Glide Reflection An obtuse triangle is reflected about the dashed line, and the two shapes are joined together. How does the tessellation shown in materialize? The new shape is reflected horizontally and joined with the original shape. It is then translated vertically and horizontally to make up the tessellation. Notice the blank spaces next to the vertical pattern. These areas are made up of the exact original shape rotated 180 ∘ , but with no line up the center. These rotated shapes are translated horizontally and vertically, and thus, the plane is tessellated with no gaps. This is an example of a glide reflection where the order of the transformations matters. Applying More Than One Tessellation Show how this tessellation ( ) can be achieved. This is a tessellation that has one color on the front of the trapezoid and a different color on the back. There is a translation on the diagonal, and a reflection vertically. These are two separate transformations resulting in two new placements of the trapezoid. We can call this a combination of two transformations or a glide reflection. Interior Angles The sum of the interior angles of a tessellation is 360 ∘ . In , the tessellation is made of six triangles formed into the shape of a hexagon. Each angle inside a triangle equals 60 ∘ , and the six vertices meet the sum of those interior angles, 6 ( 60 ° ) = 360 ° . Interior Angles at the Vertex of Triangles In , the tessellation is made up of trapezoids, such that two of the interior angles of each trapezoid equals 75 ° and the other two angles equal 105 ° . Thus, the sum of the interior angles where the vertices of four trapezoids meet equals 105 ° + 75 ° + 75 ° + 105 ° = 360 ° . Interior Angles at the Vertex of Trapezoids These tessellations illustrate the property that the shapes meet at a vertex where the interior angles sum to 360 ° . Tessellating Shapes We might think that all regular polygons will tessellate the plane by themselves. We have seen that squares do and hexagons do. The pattern of squares in is a translation of the shape horizontally and vertically. The hexagonal pattern in , is translated horizontally, and then on the diagonal, either to the right or to the left. This particular pattern can also be formed by rotations. Both tessellations are made up of congruent shapes and each shape fits in perfectly as the pattern repeats. Translation Horizontally and Vertically Translation Horizontally and Slide Diagonally We have also seen that equilateral triangles will tessellate the plane without gaps or overlaps, as shown in . The pattern is made by a reflection and a translation. The darker side is the face of the triangle and the lighter side is the back of the triangle, shown by the reflection. Each triangle is reflected and then translated on the diagonal. Reflection and Glide Translation Escher experimented with all regular polygons and found that only the ones mentioned, the equilateral triangle, the square, and the hexagon, will tessellate the plane by themselves. Let’s try a few other regular polygons to observe what Escher found. Tessellating the Plane Do regular pentagons tessellate the plane by themselves ( )? We can see that regular pentagons do not tessellate the plane by themselves. There is a gap, a gap in the shape of a parallelogram. We conclude that regular pentagons will not tessellate the plane by themselves. Tessellating Octagons Do regular octagons tessellate the plane by themselves ( )? Again, we see that regular octagons do not tessellate the plane by themselves. The gaps, however, are squares. So, two regular polygons, an octagon and a square, do tessellate the plane. Just because regular pentagons do not tessellate the plane by themselves does not mean that there are no pentagons that tessellate the plane, as we see in . Tessellation of Pentagons Another example of an irregular polygon that tessellates the plane is by using the obtuse irregular triangle from a previous example. What transformations should be performed to produce the tessellation shown in ? Tessellating with Obtuse Irregular Triangles First, the triangle is reflected over the tip at point A , and then translated to the right and joined with the original triangle to form a parallelogram. The parallelogram is then translated on the diagonal and to the right and to the left. Naming A tessellation of squares is named by choosing a vertex and then counting the number of sides of each shape touching the vertex. Each square in the tessellation shown in has four sides, so starting with square A , the first number is 4, moving around counterclockwise to the next square meeting the vertex, square B , we have another 4, square C adds another 4, and finally square D adds a fourth 4. So, we would name this tessellation a 4.4.4.4. The hexagon tessellation, shown in has six sides to the shape and three hexagons meet at the vertex. Thus, we would name this a 6.6.6. The triangle tessellation, shown in has six triangles meeting the vertex. Each triangle has three sides. Thus, we name this a 3.3.3.3.3.3. 4.4.4.4 6.6.6 3.3.3.3.3.3 Creating Your Own Tessellation Create a tessellation using two colors and two shapes. We used a parallelogram and an isosceles triangle. The parallelogram is reflected vertically and horizontally so that only every other corner touches. The triangles are reflected vertically and horizontally and then translated over the parallelogram. The result is alternating vertical columns of parallelograms and then triangles ( ). Check Your Understanding Key Terms tessellation translation reflection rotation glide reflection Key Concepts A tessellation is a particular pattern composed of shapes, usually polygons, that repeat and cover the plane with no gaps or overlaps. Properties of tessellations include rigid motions of the shapes called transformations. Transformations refer to translations, rotations, reflections, and glide reflections. Shapes are transformed in such a way to create a pattern. Videos M.C. Escher: How to Create a Tessellation The Mathematical Art of M.C. Escher

Math


Area The area of a regulation baseball diamond must adhere to specific measurements to be legal. (credit: "Diagram of Regulation Diamond" by Erica Fischer from "Baseball" The World Book, 1920/Flickr, Public Domain, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the area of triangles. Calculate the area of quadrilaterals. Calculate the area of other polygons. Calculate the area of circles. Some areas carry more importance than other areas. Did you know that in a baseball game, when the player hits the ball and runs to first base that he must run within a 6-foot wide path? If he veers off slightly to the right, he is out. In other words, a few inches can be the difference in winning or losing a game. Another example is real estate. On Manhattan Island, one square foot of real estate is worth far more than real estate in practically any other area of the country. In other words, we place a value on area. As the context changes, so does the value. Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. As opposed to linear measurements like perimeter, which in in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area. Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as in 2 or cm 2 . Note that linear measurements have no exponent above the units or we can say that the exponent is 1. Area of Triangles The formula for the area of a triangle is given as follows. The area of a triangle is given as A = 1 2 b h , where b represents the base and h represents the height. For example, consider the triangle in . Triangle 1 The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area: A = 1 2 ( 4 ) ( 5 ) = 1 2 ( 20 ) = 10 cm 2 In , the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then A = 1 2 ( 7 ) ( 3.5 ) = 12.25 cm 2 . Triangle 2 Finding the Area of a Triangle Find the area of this triangle that has a base of 4 cm and the height is 6 cm ( ). Using the formula, we have A = 1 2 ( 4 ) ( 6 ) = 24 2 = 12 cm 2 . Area of Quadrilaterals A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle. Square In , a 12 in × 12 in grid is represented with twelve 1 in × 1 in squares across each row, and twelve 1 in × 1 in squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where s = length of a side , is A = s ⋅ s . The area of the square in is A = 12 in × 12 in = 144 in 2 . Area of a Square The formula for the area of a square is A = s ⋅ s or A = s 2 . Rectangle Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in has width equal to 5 in and length equal to 12 in. The area is A = 5 ( 12 ) = 60 in 2 . Area of a Rectangle The area of a rectangle is given as A = l w . Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area? Finding the Area of a Rectangle You have a garden with an area of 196 square feet. How wide can the garden be if the length is 28 feet? The area of a rectangular region is A = l w . Letting the width equal w : 196 = 28 w 196 28 = w = 7 ft Determining the Cost of Floor Tile Jennifer is planning to install vinyl floor tile in her rectangular-shaped basement, which measures 29 ft by 16 ft. How many boxes of floor tile should she buy and how much will it cost if one box costs $50 and covers 20 ft 2 ? The area of the basement floor is A = 29 ( 16 ) = 464 ft 2 . We will divide this area by 20 ft . Thus, 464 20 = 23.2 . Therefore, Jennifer will have to buy 24 boxes of tile at a cost $1,200. Parallelogram The area of a parallelogram can be found using the formula for the area of a triangle. Notice in , if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram: A = 2 ( 1 2 b h ) A = b h Area of a Parallelogram The area of a parallelogram is A = b h . For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be A = ( 10 ) ( 5 ) = 50 in 2 . Finding the Area of a Parallelogram In the parallelogram ( ), if F B = 10 , A D = 15 , find the exact area of the parallelogram. Using the formula of A = b h , we have A = 10 ( 15 ) = 150 cm 2 . Finding the Area of a Parallelogram Park The boundaries of a city park form a parallelogram ( ). The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for $0.50 per square foot, installation is $1.50 per square foot, and the cost of the equipment for the day is $100. Step 1: As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft. Step 2: The park has the shape of a parallelogram, and the formula for the area is A = b h : A = 165 ( 117 ) = 19,305 ft 2 . Step 3: The city needs to purchase 19 , 305 ft 2 of sod. The cost will be $0.50 per square foot for the sod and $1.50 per square foot for installation, plus $100 for equipment: 19,305 ( $ 0.50 ) + 19,305 ( $ 1.50 ) = $ 9,652.5 + $ 28,957.5 + $ 100.00 = $ 38,710.00 Trapezoid Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases a and b and height h is given here. The formula for the area of a trapezoid is given as A = 1 2 h ( a + b ) . For example, find the area of the trapezoid in that has base a equal to 8 cm, base b equal to 6 cm, and height equal to 6 cm. Area of a Trapezoid The area is A = 1 2 ( 6 ) ( 6 + 8 ) = 42 cm 2 . Finding the Area of a Trapezoid A B C D ( ) is a regular trapezoid with A B ¯ ‖ C D ¯ . Find the exact perimeter of A B C D , and then find the area. The perimeter is the measure of the boundary of the shape, so we just add up the lengths of the sides. We have P = 31 + 13.5 + 11 + 13.5 = 69 in . Then, the area of the trapezoid using the formula is A = 1 2 ( 10 ) ( 11 + 31 ) = 210 in 2 . Rhombus The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals. The area of a rhombus is found using one of these formulas: A = d 1 d 2 2 , where d 1 and d 2 are the diagonals. A = 1 2 b h , where b is the base and h is the height. For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is 220 cm 2 , and the measure of d 2 = 11 , find the measure of d 1 . To solve this problem, we input the known values into the formula and solve for the unknown. See . Area of a Rhombus We have that 220 = 11 d 1 2 220 ( 2 ) = 11 d 1 440 11 = d 1 = 40 Finding the Area of a Rhombus Find the measurement of the diagonal d 1 if the area of the rhombus is 240 cm 2 , and the measure of d 2 = 24 cm . Use the formula with the known values: 240 = d 1 ( 24 ) 2 240 ( 2 ) = d 1 ( 24 ) 480 24 = d 1 = 20 Finding the Area of a Rhombus You notice a child flying a rhombus-shaped kite on the beach. When it falls to the ground, it falls on a beach towel measuring 36 in by 72 in . You notice that one of the diagonals of the kite is the same length as the 36 in width of the towel. The second diagonal appears to be 2 in longer. What is the area of the kite ( )? Using the formula, we have: A = d 1 d 2 2 = 36 ( 38 ) 2 = 684 in 2 Area of Polygons To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem a of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius r of a regular polygon is also a line segment that starts at the center but extends to a vertex. See . Apothem and Radius of a Polygon The area of a regular polygon is found with the formula A = 1 2 a p , where a is the apothem and p is the perimeter. For example, consider the regular hexagon shown in with a side length of 4 cm, and the apothem measures a = 2 3 . Area of a Hexagon We have the perimeter, p = 6 ( 4 ) = 24 cm . We have the apothem as a = 2 3 . Then, the area is: A = 1 2 ( 2 3 ) ( 24 ) = 24 3 = 41.57 cm 2 Finding the Area of a Regular Octagon Find the area of a regular octagon with the apothem equal to 18 cm and a side length equal to 13 cm ( ). Using the formula, we have the perimeter p = 8 ( 13 ) = 104 cm . Then, the area is A = 1 2 ( 10 ) ( 104 ) = 936 cm 2 . Changing Units Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency. Changing Units Carpeting comes in units of square yards. Your living room measures 21 ft wide by 24 ft long. How much carpeting do you buy? We must convert feet to yards. As there are 3 ft in 1 yd, we have 21 ft = 7 yds and 24 ft = 8 yds . Then, 7 ( 8 ) = 56 yd 2 . Area of Circles Just as the circumference of a circle includes the number π , so does the formula for the area of a circle. Recall that π is a non-terminating, non-repeating decimal number: π = 3.14159 … . It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area. The area of a circle is given as A = π r 2 , where r is the radius. For example, to find the of the circle with radius equal to 3 cm, as shown in , is found using the formula A = π r 2 . Circle with Radius 3 We have A = π r 2 = π ( 3 ) 2 = 9 π = 28.27 cm 2 Finding the Area of a Circle Find the area of a circle with diameter of 16 cm. The formula for the area of a circle is given in terms of the radius, so we cut the diameter in half. Then, the area is A = π ( 8 ) 2 = 201.1 cm 2 . Determining the Better Value for Pizza You decide to order a pizza to share with your friend for dinner. The price for an 8-inch diameter pizza is $7.99. The price for 16-inch diameter pizza is $13.99. Which one do you think is the better value? The area of the 8-inch diameter pizza is A = π ( 4 ) 2 = 50.3 in 2 . The area of the 16-inch diameter pizza is A = π ( 8 ) 2 = 201.1 in 2 . Next, we divide the cost of each pizza by its area in square inches. Thus, 13.99 201.1 = $ 0.07 per square inch and 7.99 50.3 = $ 0.16 per square inch. So clearly, the 16-inch pizza is the better value. Applying Area to the Real World You want to purchase a tinted film, sold by the square foot, for the window in . (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase. First, the rectangular portion has A = 5 ( 10 ) = 50 ft 2 . For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is A = 1 2 π ( 2.5 ) 2 = 9.8 ft 2 . Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is A = 50 + 9.8 = 59.8 ft 2 . Area within Area Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems. Finding the Area within an Area The patio in your backyard measures 20 ft by 10 ft ( ). On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over? The hot tub has a radius of 3 ft. That area is then A = π ( 3 ) 2 = 9 π = 28.27 ft 2 . The total square feet taken up with the table and chairs and the hot tub is 36 + 28.27 = 64.27 ft 2 . The area left over is equal to the total area of the patio, 200 ft 2 minus the area for the table and chairs and the hot tub. Thus, the area left over is 200 − 64.27 = 135.7 ft 2 . Finding the Cost of Fertilizing an Area A sod farmer wants to fertilize a rectangular plot of land 150 ft by 240 ft. A bag of fertilizer covers 5,000 ft 2 and costs $200. How much will it cost to fertilize the entire plot of land? The plot of land is 36 , 000 ft 2 . It will take 7.2 bags of fertilizer to cover the land area. Therefore, the farmer will have to purchase 8 bags of fertilizer at $200 a bag, which comes to $1,600. Heron of Alexandria Heron of Alexandria (credit: "Heron of Alexandria" from 1688 German translation of Hero's Pneumatics/Wikimedia Commons, Public Domain) Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, Mechanics and Optics . His contribution to the field of mathematics was enormous. Metrica , a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle A B C ( ), Heron’s formula is A = s ( s − a ) ( s − b ) ( s − c ) , where s is the semi-perimeter calculated as s = a + b + c 2 . Check Your Understanding Key Terms triangle square rectangle rhombus apothem radius circle Key Concepts The area A of a triangle is found with the formula A = 1 2 b h , where b is the base and h is the height. The area of a parallelogram is found using the formula A = b h , where b is the base and h is the height. The area of a rectangle is found using the formula A = l w , where l is the length and w is the width. The area of a trapezoid is found using the formula A = 1 2 h ( b 1 + b 2 ) , where h is the height, b 1 is the length of one base, and b 2 is the length of the other base. The area of a rhombus is found using the formula A = d 1 d 2 2 , where d 1 is the length of one diagonal and d 2 is the length of the other diagonal. The area of a regular polygon is found using the formula A = 1 2 a p , where a is the apothem and p is the perimeter. The area of a circle is found using the formula A = π r 2 , where r is the radius. Formulas The area of a triangle is given as A = 1 2 b h , where b represents the base and h represents the height. The formula for the area of a square is A = s ⋅ s or A = s 2 . The area of a rectangle is given as A = l w . The area of a parallelogram is A = b h . The formula for the area of a trapezoid is given as A = 1 2 h ( a + b ) . The area of a rhombus is found using one of these formulas: A = d 1 d 2 2 , where d 1 and d 2 are the diagonals. A = 1 2 b h , where b is the base and h is the height. The area of a regular polygon is found with the formula A = 1 2 a p , where a is the apothem and p is the perimeter. The area of a circle is given as A = π r 2 , where r is the radius.

Math


Volume and Surface Area Volume is illustrated in this 3-dimensional view of an interior space. This gives a buyer a more realistic interpretation of space. (credit: "beam render 10 with sun and cat tree" by monkeywing/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Calculate the surface area of right prisms and cylinders. Calculate the volume of right prisms and cylinders. Solve application problems involving surface area and volume. Volume and surface area are two measurements that are part of our daily lives. We use volume every day, even though we do not focus on it. When you purchase groceries, volume is the key to pricing. Judging how much paint to buy or how many square feet of siding to purchase is based on surface area. The list goes on. An example is a three-dimensional rendering of a floor plan. These types of drawings make building layouts far easier to understand for the client. It allows the viewer a realistic idea of the product at completion; you can see the natural space, the volume of the rooms. This section gives you practical information you will use consistently. You may not remember every formula, but you will remember the concepts, and you will know where to go should you want to calculate volume or surface area in the future. We will concentrate on a few particular types of three-dimensional objects: right prisms and right cylinders. The adjective “right” refers to objects such that the sides form a right angle with the base. We will look at right rectangular prisms, right triangular prisms, right hexagonal prisms, right octagonal prisms, and right cylinders. Although, the principles learned here apply to all right prisms. Three-Dimensional Objects In geometry, three-dimensional objects are called geometric solids . Surface area refers to the flat surfaces that surround the solid and is measured in square units. Volume refers to the space inside the solid and is measured in cubic units. Imagine that you have a square flat surface with width and length. Adding the third dimension adds depth or height, depending on your viewpoint, and now you have a box. One way to view this concept is in the Cartesian coordinate three-dimensional space. The x -axis and the y -axis are, as you would expect, two dimensions and suitable for plotting two-dimensional graphs and shapes. Adding the z -axis, which shoots through the origin perpendicular to the x y -plane, and we have a third dimension. See . Three-Dimensional Space Here is another view taking the two-dimensional square to a third dimension. See . Going from Two Dimensions to Three Dimensions To study objects in three dimensions, we need to consider the formulas for surface area and volume. For example, suppose you have a box ( ) with a hinged lid that you want to use for keeping photos, and you want to cover the box with a decorative paper. You would need to find the surface area to calculate how much paper is needed. Suppose you need to know how much water will fill your swimming pool. In that case, you would need to calculate the volume of the pool. These are just a couple of examples of why these concepts should be understood, and familiarity with the formulas will allow you to make use of these ideas as related to right prisms and right cylinders. Right Prisms A right prism is a particular type of three-dimensional object. It has a polygon-shaped base and a congruent, regular polygon-shaped top, which are connected by the height of its lateral sides, as shown in . The lateral sides form a right angle with the base and the top. There are rectangular prisms, hexagonal prisms, octagonal prisms, triangular prisms, and so on. Pentagonal Prism Generally, to calculate surface area, we find the area of each side of the object and add the areas together. To calculate volume, we calculate the space inside the solid bounded by its sides. The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, S A = 2 B + p h , where B is equal to the area of the base and top, p is the perimeter of the base, and h is the height. The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, V = B ⋅ h , where B is the area of the base and h is the height. Calculating Surface Area and Volume of a Rectangular Prism Find the surface area and volume of the rectangular prism that has a width of 10 cm, a length of 5 cm, and a height of 3 cm ( ). The surface area is S A = 2 ( 10 ) ( 5 ) + 2 ( 5 ) ( 3 ) + 2 ( 10 ) ( 3 ) = 190 cm 2 . The volume is V = 10 ( 5 ) ( 3 ) = 150 cm 3 . In , we have three views of a right hexagonal prism. The regular hexagon is the base and top, and the lateral faces are the rectangular regions perpendicular to the base. We call it a right prism because the angle formed by the lateral sides to the base is 90 ∘ . See . Right Hexagonal Prism The first image is a view of the figure straight on with no rotation in any direction. The middle figure is the base or the top. The last figure shows you the solid in three dimensions. To calculate the surface area of the right prism shown in , we first determine the area of the hexagonal base and multiply that by 2, and then add the perimeter of the base times the height. Recall the area of a regular polygon is given as A = 1 2 a p , where a is the apothem and p is the perimeter. We have that A b a s e = 1 2 ( 5.2 ) ( 36 ) = 93.6 cm 2 Then, the surface area of the hexagonal prism is S A = 2 ( 93.6 ) + 36 ( 20 ) = 907.2 in 2 To find the volume of the right hexagonal prism, we multiply the area of the base by the height using the formula V = B h . The base is 93.6 cm 2 , and the height is 20 cm . Thus, V = 93.6 ( 20 ) = 1872 cm 3 . Calculating the Surface Area of a Right Triangular Prism Find the surface area of the triangular prism ( ). The area of the triangular base is A b a s e = 1 2 ( 12 ) ( 6 ) = 36 in 2 . The perimeter of the base is p = 12 + 8.49 + 8.49 = 28.98 in . Then, the surface area of the triangular prism is S A = 2 ( 36 ) + 28.98 ( 10 ) = 361.8 in 2 . Finding the Surface Area and Volume Find the surface area and the volume of the right triangular prism with an equilateral triangle as the base and height ( ). The area of the triangular base is A b a s e = 1 2 ( 6 ) ( 10.39 ) = 31.17 cm 2 . Then, the surface area is S A = 2 ( 31.17 ) + 36 ( 10 ) = 422.34 cm 2 . The volume formula is found by multiplying the area of the base by the height. We have that V = B ⋅ h = ( 31.17 ) ( 10 ) = 311.7 cm 3 . Determining Surface Area Application Katherine and Romano built a greenhouse made of glass with a metal roof ( ). In order to determine the heating and cooling requirements, the surface area must be known. Calculate the total surface area of the greenhouse. The area of the long side measures 95 ft × 6.5 ft = 1,235 ft 2 . Multiplying by 2 gives 2,470 ft 2 . The front (minus the triangular area) measures 22 ft × 6.5 ft = 286 ft 2 . Multiplying by 2 gives 572 ft 2 . The floor measures 95 ft × 22 ft = 2,090 ft 2 . Each triangular region measures A = 1 2 ( 22 ) ( 5 ) = 55 ft 2 . Multiplying by 2 gives 110 ft 2 . Finally, one side of the roof measures 12.1 ft × 95 ft = 1,149.5 ft 2 . Multiplying by 2 gives 2299 ft 2 . Add them up and we have S A = 2,470 + 572 + 2,090 + 110 + 2,299 = 7,541 ft 2 . Right Cylinders There are similarities between a prism and a cylinder. While a prism has parallel congruent polygons as the top and the base, a right cylinder is a three-dimensional object with congruent circles as the top and the base. The lateral sides of a right prism make a 90 ∘ angle with the polygonal base, and the side of a cylinder, which unwraps as a rectangle, makes a 90 ∘ angle with the circular base. Right cylinders are very common in everyday life. Think about soup cans, juice cans, soft drink cans, pipes, air hoses, and the list goes on. In , imagine that the cylinder is cut down the 12-inch side and rolled out. We can see that the cylinder side when flat forms a rectangle. The S A formula includes the area of the circular base, the circular top, and the area of the rectangular side. The length of the rectangular side is the circumference of the circular base. Thus, we have the formula for total surface area of a right cylinder. Right Cylinder The surface area of a right cylinder is given as S A = 2 π r 2 + 2 π r h . To find the volume of the cylinder, we multiply the area of the base with the height. The volume of a right cylinder is given as V = π r 2 h . Finding the Surface Area and Volume of a Cylinder Given the cylinder in , which has a radius of 5 inches and a height of 12 inches, find the surface area and the volume. Step 1: We begin with the areas of the base and the top. The area of the circular base is A b a s e = π ( 5 ) 2 = 25 π = 78.5 in 2 . Step 2: The base and the top are congruent parallel planes. Therefore, the area for the base and the top is A = 2 ( 78.5 ) = 157 in 2 . Step 3: The area of the rectangular side is equal to the circumference of the base times the height: A = 2 π ( 5 ) ( 12 ) = 377 in 2 Step 4: We add the area of the side to the areas of the base and the top to get the total surface area. We have S A = 157 + 377 = 534 in 2 . Step 5: The volume is equal to the area of the base times the height. Then, V = π ( 5 ) 2 ( 12 ) = 942.48 in 3 . Applications of Surface Area and Volume The following are just a small handful of the types of applications in which surface area and volume are critical factors. Give this a little thought and you will realize many more practical uses for these procedures. Applying a Calculation of Volume A can of apple pie filling has a radius of 4 cm and a height of 10 cm. How many cans are needed to fill a pie pan ( ) measuring 22 cm in diameter and 3 cm deep? The volume of the can of apple pie filling is V = π ( 4 ) 2 ( 10 ) = 502.7 cm 3 . The volume of the pan is V = π ( 11 ) 2 ( 3 ) = 1140.4 cm 3 . To find the number of cans of apple pie filling, we divide the volume of the pan by the volume of a can of apple pie filling. Thus, 1140 ÷ 502.7 = 2.3. We will need 2.3 cans of apple pie filling to fill the pan. Optimization Problems that involve optimization are ones that look for the best solution to a situation under some given conditions. Generally, one looks to calculus to solve these problems. However, many geometric applications can be solved with the tools learned in this section. Suppose you want to make some throw pillows for your sofa, but you have a limited amount of fabric. You want to make the largest pillows you can from the fabric you have, so you would need to figure out the dimensions of the pillows that will fit these criteria. Another situation might be that you want to fence off an area in your backyard for a garden. You have a limited amount of fencing available, but you would like the garden to be as large as possible. How would you determine the shape and size of the garden? Perhaps you are looking for maximum volume or minimum surface area. Minimum cost is also a popular application of optimization. Let’s explore a few examples. Maximizing Area Suppose you have 150 meters of fencing that you plan to use for the enclosure of a corral on a ranch. What shape would give the greatest possible area? So, how would we start? Let’s look at this on a smaller scale. Say that you have 30 inches of string and you experiment with different shapes. The rectangle in measures 12 in long by 3 in wide. We have a perimeter of P = 2 ( 12 ) + 2 ( 3 ) = 30 in and the area calculates as A = 3 ( 12 ) = 36 in 2 . The rectangle in , measures 8 in long and 7 in wide. The perimeter is P = 2 ( 8 ) + 2 ( 7 ) = 30 in and the area is A = 8 ( 7 ) = 56 in 2 . In , the square measures 7.5 in on each side. The perimeter is then P = 4 ( 7.5 ) = 30 in and the area is A = 7.5 ( 7.5 ) = 56.25 in 2 . If you want a circular corral as in , we would consider a circumference of 30 = 2 π r , which gives a radius of 30 ÷ 2 π = 4.77 in and an area of A = π ( 4.77 ) 2 = 71.5 in 2 . We see that the circular shape gives the maximum area relative to a circumference of 30 in. So, a circular corral with a circumference of 150 meters and a radius of 23.87 meters gives a maximum area of 1,790.5 m 2 . Designing for Cost Suppose you want to design a crate built out of wood in the shape of a rectangular prism ( ). It must have a volume of 3 cubic meters. The cost of wood is $15 per square meter. What dimensions give the most economical design while providing the necessary volume? To choose the optimal shape for this container, you can start by experimenting with different sizes of boxes that will hold 3 cubic meters. It turns out that, similar to the maximum rectangular area example where a square gives the maximum area, a cube gives the maximum volume and the minimum surface area. As all six sides are the same, we can use a simplified volume formula: V = s 3 , where s is the length of a side. Then, to find the length of a side, we take the cube root of the volume. We have 3 = s 3 3 3 = s = 1.4422 m The surface area is equal to the sum of the areas of the six sides. The area of one side is A = ( 1.4422 ) 2 = 2.08 m 2 . So, the surface area of the crate is S A = 6 ( 2.08 ) = 12.5 m 2 . At $15 a square meter, the cost comes to 12.5 ( $ 15 ) = $ 187.50. Checking the volume, we have V = ( 1.4422 ) 3 = 2.99 m 3 . Check Your Understanding Key Terms surface area volume right prism right cylinder Key Concepts A right prism is a three-dimensional object that has a regular polygonal face and congruent base such that that lateral sides form a 90 ∘ angle with the base and top. The surface area S A of a right prism is found using the formula S A = 2 B + p h , where B is the area of the base, p is the perimeter of the base, and h is the height. The volume V of a right prism is found using the formula V = B h , where B is the area of the base and h is the height. A right cylinder is a three-dimensional object with a circle as the top and a congruent circle is the base, and the side forms a 90 ∘ angle to the base and top. The surface area of a right cylinder is found using the formula S A = 2 π r 2 + 2 π r h , where r is the radius and h is the height. The volume is found using the formula V = π r 2 h , where r is the radius and h is the height. Formulas The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, S A = 2 B + p h , where B is equal to the area of the base and top, p is the perimeter of the base, and h is the height. The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, V = B ⋅ h , where B is the area of the base and h is the height. The surface area of a right cylinder is given as S A = 2 π r 2 + 2 π r h . The volume of a right cylinder is given as V = π r 2 h .

Math


Right Triangle Trigonometry In the lower left corner of the fresco The School of Athens by Raphael, the figure in white writing in the book represents Pythagoras. Alongside him, to the right, the figure with the long, light-brown hair is said to depict Archimedes. (credit: modification of work “School of Athens” by Raphael (1483–1520), Vatican Museums/Wikimedia, Public Domain) Learning Objectives After completing this section, you should be able to: Apply the Pythagorean Theorem to find the missing sides of a right triangle. Apply the 30 ∘ - 60 ∘ - 90 ∘ and 45 ∘ - 45 ∘ - 90 ∘ right triangle relationships to find the missing sides of a triangle. Apply trigonometric ratios to find missing parts of a right triangle. Solve application problems involving trigonometric ratios. This is another excerpt from Raphael’s The School of Athens. The man writing in the book represents Pythagoras, the namesake of one of the most widely used formulas in geometry, engineering, architecture, and many other fields, the Pythagorean Theorem. However, there is evidence that the theorem was known as early as 1900–1100 BC by the Babylonians. The Pythagorean Theorem is a formula used for finding the lengths of the sides of right triangles. Born in Greece, Pythagoras lived from 569–500 BC. He initiated a cult-like group called the Pythagoreans, which was a secret society composed of mathematicians, philosophers, and musicians. Pythagoras believed that everything in the world could be explained through numbers. Besides the Pythagorean Theorem, Pythagoras and his followers are credited with the discovery of irrational numbers, the musical scale, the relationship between music and mathematics, and many other concepts that left an immeasurable influence on future mathematicians and scientists. The focus of this section is on right triangles. We will look at how the Pythagorean Theorem is used to find the unknown sides of a right triangle, and we will also study the special triangles, those with set ratios between the lengths of sides. By ratios we mean the relationship of one side to another side. When you think about ratios, you should think about fractions. A fraction is a ratio, the ratio of the numerator to the denominator. Finally, we will preview trigonometry. We will learn about the basic trigonometric functions, sine, cosine and tangent, and how they are used to find not only unknown sides but unknown angles, as well, with little information. Pythagorean Theorem The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle). The Pythagorean Theorem states a 2 + b 2 = c 2 where a and b are two sides (legs) of a right triangle and c is the hypotenuse, as shown in . Pythagorean Right Triangle For example, given that side a = 6 , and side b = 8 , we can find the measure of side c using the Pythagorean Theorem. Thus, a 2 + b 2 = c 2 ( 6 ) 2 + ( 8 ) 2 = c 2 36 + 64 = c 2 100 = c 2 100 = c 2 10 = c Using the Pythagorean Theorem Find the length of the missing side of the triangle ( ). Using the Pythagorean Theorem , we have ( 6 ) 2 + b 2 = ( 14 ) 2 36 + b 2 = 196 b 2 = 196 − 36 b 2 = 160 b = ± 160 = 4 10 = 12.65 When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root. Distance The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this: In , the problem is to find the distance between the points ( − 3 , − 1 ) and ( 3 , 2 ) . We call the length from point ( − 3 , − 1 ) to point ( 3 , − 1 ) side a , and the length from point ( 3 , − 1 ) to point ( 3 , 2 ) side b . To find side c , we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 , such that ( x 2 − x 1 ) is a substitute for a in the Pythagorean Theorem and is equal to 3 − ( − 3 ) = 6 ; and ( y 2 − y 1 ) is a substitute for b in the Pythagorean Theorem and is equal to 2 − ( − 1 ) = 3. When we plug in these numbers to the distance formula, we have d = ( 3 − ( − 3 ) ) 2 + ( 2 − ( − 1 ) ) 2 = ( 6 ) 2 + ( 3 ) 2 = 36 + 9 = 45 = 3 5 = 6.7 Thus, d = c , the hypotenuse, in the Pythagorean Theorem. Distance Calculating Distance Using the Distance Formula You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive ( ). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft. You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then 9 ( 200 ) + 7 ( 200 ) = 1,800 ft + 1,400 ft = 3,200 ft . As the crow flies, use the distance formula. We have d = ( 1,800 − 0 ) 2 + ( 1,400 − 0 ) 2 = 3,240,000 + 1,960,000 = 5,200,000 = 2280.4 ft Calculating Distance with the Pythagorean Theorem The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches ( ). What should the vertical height be? The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a , is 130 in. The length of c , or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b . Then, by the Pythagorean Theorem , we have: a 2 + b 2 = c 2 ( 130 ) 2 + b 2 = ( 132.4 ) 2 16,900 + b 2 = 17,529.76 b 2 = 17,529.76 − 16,900 b 2 = 629.8 b = 629.8 = 25 If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change? The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be 12 ( 25 ) = 300 in . So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length: ( 300 ) 2 + ( 25 ) 2 = c 2 90,625 = c 2 90,625 = c = 301 in The new ramp will look like . 30 ∘ - 60 ∘ - 90 ∘ Triangles In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like ( 2 x + 5 y ) 2 , we do not have to expand it the long way: ( 2 x + 5 y ) 2 = ( 2 x + 5 y ) ( 2 x + 5 y ) = ( 2 x ) 2 + 10 x y + 10 x y + ( 5 y ) 2 = 4 x 2 + 20 x y + 25 y 2 If we know the perfect square formula, given as ( a + b ) 2 = a 2 + 2 a b + b 2 , we can skip the middle step and just start writing down the answer. This may seem trivial with problems like ( a + b ) 2 . However, what if you have a problem like ( 2 3 + 3 31.8 c 3 ) 2 ? That is a different story. Nevertheless, we use the same perfect square formula. The same idea applies in geometry. There are special formulas and procedures to apply in certain types of problems. What is needed is to remember the formula and remember the kind of problems that fit. Sometimes we believe that because a formula is labeled special, we will rarely have use for it. That assumption is incorrect. So, let us identify the 30 ∘ - 60 ∘ - 90 ∘ triangle and find out why it is special. See . The 3 0 ∘ - 6 0 ∘ - 9 0 ∘ We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the 30 ∘ - 60 ∘ - 90 ∘ triangle given as 1 : 3 : 2 , or x : x 3 : 2 x . Thus, you only need to know the length of one side to find the other two sides in a 30 ∘ - 60 ∘ - 90 ∘ triangle. Finding Missing Lengths in a 3 0 ∘ - 6 0 ∘ - 9 0 ∘ Triangle Find the measures of the missing lengths of the triangle ( ). We can see that this is a 30 ∘ - 60 ∘ - 90 ∘ triangle because we have a right angle and a 30 ∘ angle. The remaining angle, therefore, must equal 60 ∘ . Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side a is the shortest side, as it is opposite the smallest angle 30 ∘ , and we can substitute a = x . The ratios are x : x 3 : 2 x . We have the hypotenuse equaling 10, which corresponds to side c , and side c is equal to 2 x . Now, we must solve for x : 2 x = 10 x = 5 Side b is equal to x 3 or 5 3 . The lengths are 5 , 5 3 , 10. Applying 3 0 ∘ - 6 0 ∘ - 9 0 ∘ Triangle to the Real World A city worker leans a 40-foot ladder up against a building at a 30 ∘ angle to the ground ( ). How far up the building does the ladder reach? We have a 30 ∘ - 60 ∘ - 90 ∘ triangle, and the hypotenuse is 40 ft. This length is equal to 2 x , where x is the shortest side. If 2 x = 40 , then x = 20 . The ladder is leaning on the wall 20 ft up from the ground. 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangles The 45 ∘ - 45 ∘ - 90 ∘ triangle is another special triangle such that with the measure of one side we can find the measures of all the sides. The two angles adjacent to the 90 ∘ angle are equal, and each measures 45 ∘ . If two angles are equal, so are their opposite sides. The ratio among sides is 1 : 1 : 2 , or x : x : x 2 , as shown in . 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangles Finding Missing Lengths of a 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangle Find the measures of the unknown sides in the triangle ( ). Because we have a 45 ∘ - 45 ∘ - 90 ∘ triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and 2 . One leg measures 3, so the other leg, a , measures 3. Remember the ratio of x : x : x 2 . Then, the hypotenuse, c , equals 3 2 . Trigonometry Functions Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications. NOTE: You will need either a scientific calculator or a graphing calculator for this section. It must have the capability to calculate trigonometric functions and express angles in degrees. Trigonometry is based on three functions. We title these functions using the following abbreviations: sin = sine cos = cosine tan = tangent Letting r = x 2 + y 2 , which is the hypotenuse of a right triangle, we have . The functions are given in terms of x , y , and r , and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse. sin θ = y r = o p p h y p cos θ = x r = a d j h y p tan θ = y x = o p p a d j Trigonometric Ratios We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In , we have an illustration mainly to identify r and the sides labeled x and y . Angle θ An angle θ sweeps out in a counterclockwise direction from the positive x -axis and stops when the angle reaches the desired measurement. That ray extending from the origin that marks θ ∘ is called the terminal side because that is where the angle terminates. Regardless of the information given in the triangle, we can find all missing sides and angles using the trigonometric functions. For example, in , we will solve for the missing sides. Solving for Missing Sides Let’s use the trigonometric functions to find the sides x and y . As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for y . We have sin θ = y r , and θ = 60 ∘ . Then, sin 60 ∘ = y 2 2 sin 60 ∘ = y 1.732 = y 3 = y Next, let’s find x . This is the cosine function. We have cos θ = x r . Then, cos 60 ∘ = x 2 2 cos 60 ∘ = x = 1 Now we have all sides, 1 , 3 , 2. You can also check the sides using the 30 ∘ - 60 ∘ - 90 ∘ ratio of 1 : 3 : 2. is a list of common angles, which you should find helpful. sin 0 ∘ = 0 cos 0 ∘ = 1 sin 30 ∘ = 1 2 cos 30 ∘ = 3 2 sin 45 ∘ = 2 2 cos 45 ∘ = 2 2 sin 60 ∘ = 3 2 cos 60 ∘ = 1 2 sin 90 ∘ = 1 cos 90 ∘ = 0 Common Angles Using Trigonometric Functions Find the lengths of the missing sides for the triangle ( ). We have a 55 ∘ angle, and the length of the triangle on the x -axis is 6 units. Step 1: To find the length of r , we can use the cosine function, as cos θ = x r . We manipulate this equation a bit to solve for r : cos ( 55 ∘ ) = 6 r r cos ( 55 ∘ ) = 6 r = 6 cos ( 55 ∘ ) r = 6 0.5736 = 10.46 Step 2: We can use the Pythagorean Theorem to find the length of y . Prove that your answers are correct by using other trigonometric ratios: 6 2 + y 2 = 10.46 2 y 2 = 109.4 − 36 y = 8.57 Step 3: Now that we have y , we can use the sine function to prove that r is correct. We have sin θ = y r . sin ( 55 ∘ ) = 8.57 r r sin ( 55 ∘ ) = 8.57 r = 8.57 sin ( 55 ∘ ) = 8.57 0.819 = 10.46 To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as sin − 1 , cos − 1 , or tan − 1 . The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had x = 6 and r = 10.46. To find what angle has these values, enter the values for the inverse cosine function cos − 1 ( x r ) in your calculator: cos − 1 ( 6 10.46 ) = 55 ∘ . You can also use the inverse sine function and enter the values of sin − 1 ( y r ) in your calculator given y = 8.57 and r = 10.46. We have sin − 1 ( 8.57 10.46 ) = 55 ∘ . Finally, we can also use the inverse tangent function. Recall tan θ = y x . We have tan − 1 ( 8.57 6 ) = 55 ∘ . Solving for Lengths in a Right Triangle Solve for the lengths of a right triangle in which θ = 30 ∘ and r = 6 ( ). Step 1: To find side a , we use the sine function: sin 30 ∘ = a 6 6 sin 30 ∘ = a = 3 Step 2: To find b , we use the cosine function: cos 30 ∘ = b 6 6 cos 30 ∘ = b = 5.196 Step 3: Since this is a 30 ∘ - 60 ∘ - 90 ∘ triangle and side b should equal x 3 , if we input 3 for x , we have b = 3 3 . Put this in your calculator and you will get 3 3 = 5.196. Finding Altitude A small plane takes off from an airport at an angle of 31.3 ∘ to the ground. About two-thirds of a mile (3,520 ft) from the airport is an 1,100-ft peak in the flight path of the plane ( ). If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak. To solve this problem, we use the tangent function: tan 31.3 ∘ = x 3,520 3,520 tan 31.3 ∘ = 2,140 The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak. Finding Unknown Sides and Angles Suppose you have two known sides, but do not know the measure of any angles except for the right angle ( ). Find the measure of the unknown angles and the third side. Step 1: We can find the third side using the Pythagorean Theorem: 6 2 + 4 2 = c 2 52 = c 2 2 13 = c Now, we have all three sides. Step 2: To find θ , we will first find sin θ . sin θ = o p p h y p = 4 2 13 = 2 13 . The angle θ is the angle whose sine is 2 13 . Step 3: To find θ , we use the inverse sine function: θ = sin − 1 ( 2 13 ) = 33.7 ∘ Step 4: To find the last angle, we just subtract: 180 ∘ − 90 ∘ − 33.7 ∘ = 56.3 ∘ . Angle of Elevation and Angle of Depression Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression . These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level. Finding the Angle of Elevation A guy wire of length 110 meters runs from the top of an antenna to the ground ( ). If the angle of elevation of an observer to the top of the antenna is 43 ∘ , how high is the antenna? We are looking for the height of the tower. This corresponds to the y -value, so we will use the sine function: sin 43 ∘ = y 110 110 sin 43 ∘ = y 75 = y The tower is 75 m high. Finding Angle of Elevation You are sitting on the grass flying a kite on a 50-foot string ( ). The angle of elevation is 60 ∘ . How high above the ground is the kite? We can solve this using the sine function, sin θ = o p p h y p . sin 60 ∘ = x 50 50 sin 60 ∘ = x = 43.3 ft Pythagoras and the Pythagoreans The Pythagorean Theorem is so widely used that most people assume that Pythagoras (570–490 BC) discovered it. The philosopher and mathematician uncovered evidence of the right triangle concepts in the teachings of the Babylonians dating around 1900 BC. However, it was Pythagoras who found countless applications of the theorem leading to advances in geometry, architecture, astronomy, and engineering. Among his accolades, Pythagoras founded a school for the study of mathematics and music. Students were called the Pythagoreans, and the school’s teachings could be classified as a religious indoctrination just as much as an academic experience. Pythagoras believed that spirituality and science coexist, that the intellectual mind is superior to the senses, and that intuition should be honored over observation. Pythagoras was convinced that the universe could be defined by numbers, and that the natural world was based on mathematics. His primary belief was All is Number. He even attributed certain qualities to certain numbers, such as the number 8 represented justice and the number 7 represented wisdom. There was a quasi-mythology that surrounded Pythagoras. His followers thought that he was more of a spiritual being, a sort of mystic that was all-knowing and could travel through time and space. Some believed that Pythagoras had mystical powers, although these beliefs were never substantiated. Pythagoras and his followers contributed more ideas to the field of mathematics, music, and astronomy besides the Pythagorean Theorem. The Pythagoreans are credited with the discovery of irrational numbers and of proving that the morning star was the planet Venus and not a star at all. They are also credited with the discovery of the musical scale and that different strings made different sounds based on their length. Some other concepts attributed to the Pythagoreans include the properties relating to triangles other than the right triangle, one of which is that the sum of the interior angles of a triangle equals 180 ∘ . These geometric principles, proposed by the Pythagoreans, were proven 200 years later by Euclid. A Visualization of the Pythagorean Theorem In , which is one of the more popular visualizations of the Pythagorean Theorem, we see that square a is attached to side a ; square b is attached to side b ; and the largest square, square c , is attached to side c . Side a measures 3 cm in length, side b measures 4 cm in length, and side c measures 5 cm in length. By definition, the area of square a measures 9 square units, the area of square b measures 16 square units, and the area of square c measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have a 2 + b 2 = c 2 3 2 + 4 2 = 5 2 9 + 16 = 25 Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem. Check Your Understanding Key Terms right triangle sine cosine tangent Key Concepts The Pythagorean Theorem is applied to right triangles and is used to find the measure of the legs and the hypotenuse according the formula a 2 + b 2 = c 2 , where c is the hypotenuse. To find the measure of the sides of a special angle, such as a 30 ∘ - 60 ∘ - 90 ∘ triangle, use the ratio x : x 3 : 2 x , where each of the three sides is associated with the opposite angle and 2 x is associated with the hypotenuse, opposite the 90 ∘ angle. To find the measure of the sides of the second special triangle, the 45 ∘ - 45 ∘ - 90 ∘ triangle, use the ratio x : x : x 2 , where each of the three sides is associated with the opposite angle and x 2 is associated with the hypotenuse, opposite the 90 ∘ angle. The primary trigonometric functions are sin θ = o p p h y p , cos θ = a d j h y p , and tan θ = o p p a d j . Trigonometric functions can be used to find either the length of a side or the measure of an angle in a right triangle, and in applications such as the angle of elevation or the angle of depression formed using right triangles. Formula The Pythagorean Theorem states a 2 + b 2 = c 2 where a and b are two sides (legs) of a right triangle and c is the hypotenuse. Projects One of the reasons so many formulas in geometry were discovered was because of the importance in finding measurements of lengths, areas, perimeter, and angles. Find at least five examples of how geometry can be used in practical applications today. Who were the Pythagoreans? Why did this society exist? Explore what they did and discuss some of their beliefs. Chapter Review Points, Lines, and Planes Angles Triangles Polygons, Perimeter, and Circumference Tessellations Area Volume and Surface Area Right Triangle Trigonometry Chapter Test

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Introduction Voters cast their ballots in one of the world’s many democracies. (credit: “Governor Votes Early” by Maryland GovPics/Flickr, CC BY 2.0) Suppose a friend asked you, “When did you last vote?” What would your answer be? Maybe you would tell your friend that the last time you voted was during the last presidential election, or perhaps you would tell your friend that you prefer not to vote. When thinking about voting, presidential campaigns or advertisements for reelections may come to mind, but you can cast your vote in many ways. Have you liked a post, followed a creator, friended a stranger, or clicked a heart online today? In the digital age, it's possible to vote several times a day. Voting systems are not only the machines that drive every democracy on Earth, but they are also the engines driving social media and many other aspects of life. A deeper understanding of these voting systems will enhance your ability to successfully engage with the world in which we live. In this chapter, you will become one of the founders of the new democratic country of Imaginaria. You have a great responsibility to the people of this fledgling democracy. You have been tasked with writing the portion of the constitution that lays out voting procedures. In preparation for this important task, you will explore the various types of voting systems, from school board elections to Twitter wars. You will see how these types are alike, how they differ, and how they might be applied in Imaginaria. Most importantly, you will learn about the mathematically inherent advantages and disadvantages of various voting systems so that you can make informed choices to better the lives of the Imaginarians.

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