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are called exterior angles. • Angles 4 and 5 are interior angles on opposite sides of the transversal and do not have the same vertex. They are called alternate interior angles. Angles 3 and 6 are another pair of alternate interior angles. Angles and Parallel Lines 259 • Angles 1 and 8 are exterior angles on opposite sides of the transversal and do not have the same vertex. They are called alternate exterior angles. Angles 2 and 7 are another pair of alternate exterior angles. • Angles 4 and 6 are interior angles on the same side of the transversal. Angles 3 and 5 are another pair of interior angles on the same side of the transversal. • Angles 1 and 5 are on the same side of the transversal, one interior and one exterior, and at different vertices. They are called corresponding angles. Other pairs of corresponding angles are 2 and 6, 3 and 7, 4 and 8. Alternate Interior Angles and Parallel Lines In the figure, a transversal intersects two parallel lines, forming a pair of alternate interior angles, 3 and 6. If we measure 3 and 6 with a protractor, we will find that each angle measures 60°. Here, alternate interior angles 3 and 6 have equal measures, and 3 6. If we draw other pairs of parallel lines intersected by transversals, we will find again that pairs of alternate interior angles have equal measures. Yet, we would be hard-pressed to prove that this is always true. Therefore, we accept, without proof, the following statement: If two parallel lines are cut by a transversal, then the alternate interior angles that are formed have equal measures, that is, they are congruent. Note that 4 and 5 are another pair of alternate interior angles formed by a transversal that intersects the parallel lines. Therefore, 4 5. Corresponding Angles and Parallel Lines If two lines are cut by a transversal, four pairs of corresponding angles are formed. One such pair of corresponding angles is 2 and 6, as shown in the figure on the left. If the original two lines are parallel, do these corresponding angles have equal measures? We are ready to prove in an informal manner that they do. (1) Let m2 x. (2) If m2 x, then m3 x (because 2 and 3 are vertical angles, and vertical angles are congruent). (3) If m3 x, then m6 x (because 3 |
and 6 are alternate interior angles of parallel lines, and alternate interior angles of parallel lines are congruent). (4) Therefore m2 m6 (because the measure of each angle is x). 3 4 5 6 2 3 6 260 Geometric Figures, Areas, and Volumes The four steps on page 259 serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then the corresponding angles formed have equal measures, that is, they are congruent. Note that this theorem is true for each pair of corresponding angles: 1 5; 2 6; 3 7; 4 8 Alternate Exterior Angles and Parallel Lines If two parallel lines are cut by a transversal, we can prove informally that the alternate exterior angles formed have equal measures. One such pair of alternate exterior angles consists of 2 and 7, as shown in the figure on the left. (1) Let m2 x. (2) If m2 x, then m6 x (because 2 and 6 are corresponding angles of parallel lines, proved to have the same measure). (3) If m6 x, then m7 x (because 6 and 7 are vertical angles, previ- ously proved to have the same measure). (4) Therefore m2 m7 (because the measure of each angle is x). These four steps serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then the alternate exterior angles formed have equal measures, that is, they are congruent. Note that this theorem is true for each pair of alternate exterior angles: 1 8; 2 7. Interior Angles on the Same Side of the Transversal When two parallel lines are cut by a transversal, we can prove informally that the sum of the measures of the interior angles on the same side of the transversal is 180°. One such pair of interior angles on the same side of the transversal consists of 3 and 5, as shown in the figure on the left. (1) m5 m7 180 (5 and 7 are supplementary angles). (2) m7 m3 (7 and 3 are corresponding angles). (3) m5 m3 180 (by substituting m3 for its equal, m7). Angles and Parallel Lines 261 These three steps serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then |
the sum of the measures of the interior angles on the same side of the transversal is 180°. EXAMPLE 1 In the figure, the parallel lines are cut by a transversal. If ml (5x 10) and m2 (3x 60), find the measures of 1 and 2. Solution Since the lines are parallel, the alternate interior angles, 1 and 2, have equal measures. Write and solve an equation using the algebraic expressions for the measures of these angles: 5x 10 3x 60 5x 3x 70 2x 70 x 35 5x 10 5(35) 10 175 10 165 3x 60 3(35) 60 105 60 165 1 2 Answer m1 165 and m2 165. EXERCISES Writing About Mathematics 1. Give an example of a pair of lines that are neither parallel nor intersecting. 2. If a transversal is perpendicular to one of two parallel lines, is it perpendicular to the other? Prove your answer using definitions and theorems given in this chapter. Developing Skills In 3–10, the figure below shows two parallel lines cut by a transversal. For each given measure, find the measures of the other seven angles. 4. m6 150 6. m1 75 8. m4 10 10. m8 179 3. m3 80 5. m5 60 7. m2 55 9. m7 2 3 4 2 1 6 5 7 8 262 Geometric Figures, Areas, and Volumes In 11–15, the figure below shows two parallel lines cut by a transversal. In each exercise, find the measures of all eight angles under the given conditions. 11. m3 2x 40 and m7 3x 27 12. m4 4x 10 and m6 x 80 13. m4 3x 40 and m5 2x 14. m4 2x 10 and m2 x 60 15. 8 1 16. If g AB g i CD, m5 40, and m4 30, find the measures of the other angles in the figure 10 5 11 In 17–22, tell whether each statement is always, sometimes, or never true. 17. If two distinct lines intersect, then they are parallel. 18. If two distinct lines do not intersect, then they are parallel. 19. If two angles are alternate interior angles, then they are on opposite sides of the transversal. 20. If two parallel lines are cut by a transversal, then the alternate |
interior angles are congruent. 21. If two parallel lines are cut by a transversal, then the alternate interior angles are comple- mentary. 22. If two parallel lines are cut by a transversal, then the corresponding angles are supplemen- tary. 23. In the figure on the right, two parallel lines are cut by a transversal. Write an informal proof that 1 and 2 have equal measures. 1 2 7-4 TRIANGLES A polygon is a plane figure that consists of line segments joining three or more points. Each line segment is a side of the polygon and the endpoints of each side are called vertices. Each vertex is the endpoint of exactly two sides and no two sides have any other points in common. When three points that are not all on the same line are joined in pairs by line segments, the figure formed is a triangle. Each of the given points is the vertex of an angle of the triangle, and each line segment is a side of the triangle. There are many practical uses of the triangle, especially in construction work such as the building of bridges, radio towers, and airplane wings, because the tri- angle is a rigid figure. The shape of a triangle cannot be changed without changing the length of at least one of its sides. Triangles 263 We begin our discussion of the triangle by considering triangle ABC, shown below, which is symbolized as ABC. In ABC, points A, B, and C are the verBC tices, and are the sides. Angles A, B, and C of the triangle are symbolized as A, B, and C., and AB CA, We make the following observations: 1. Side AB 2. Side BC 3. Side CA is included between A and B. is included between B and C. is included between C and A. 4. Angle A is included between sides AB and CA. 5. Angle B is included between sides AB and BC. 6. Angle C is included between sides CA and BC. C A B Classifying Triangles According to Angles B F G J A C Acute triangle 90° E D Equiangular triangle H I Right triangle K L Obtuse triangle • An acute triangle has three acute angles. • An equiangular triangle has three angles equal in measure. • A right triangle has one right angle. • An obtuse triangle has one obtuse angle. In right triangle GHI above, the two sides |
of the triangle that form the right, are called the legs of the right triangle. The side opposite the GH angle, right angle, and GI HI, is called the hypotenuse. Sum of the Measures of the Angles of a Triangle When we change the shape of a triangle, changes take place also in the measures of its angles. Is there any relationship among the measures of the angles of a triangle that does not change? Let us see. 264 Geometric Figures, Areas, and Volumes Draw several triangles of different shapes. Measure the three angles of each triangle and find the sum of these measures. For example, in ABC on the right, mA + mB mC = 65 45 70 180. C 70° If you measured accurately, you should have found that in each triangle that you drew and measured, the sum of the measures of the three angles is 180°, regardless of its size or shape. We can write an informal algebraic proof of the following statement: A B 65° 45° The sum of the measures of the angles of a triangle is 180°. (1) In ABC, let mA x, mACB y, and mB z. g DCE (2) Let (3) Since DCE is a straight angle, be a line parallel to g AB. mDCE 180. C y x z D (4) mDCE mDCA mACB mBCE 180. x A (5) mDCA mA x. (6) mACB y. (7) mBCE mB z. (8) Substitute from statements (5), (6), and (7) in statement (4): x y z 180. E z B EXAMPLE 1 Find the measure of the third angle of a triangle if the measures of two of the angles are 72.6° and 84.2°. Solution Subtract the sum of the known measures from 180: 180 (72.6 84.2) 23.2 Answer 23.2° EXAMPLE 2 In ABC, the measure of B is twice the measure of A, and the measure of C is 3 times the measure of A. Find the number of degrees in each angle of the triangle. Triangles 265 Solution Let x the number of degrees in A. Then 2x the number of degrees in B. Then 3x the number of degrees in C. The sum of the measures of the angles of a triangle is 180°. B 2 |
x 3x C x A x 2x 3x 180 6x 180 x 30 2x 60 3x 90 Check 60 2(30) 90 3(30) 30 60 90 180 ✔ Answer mA 30, mB 60, mC 90 Classifying Triangles According to Sides T Z N L Scalene triangle M R S Isosceles triangle X Equilateral triangle Y • A scalene triangle has no sides equal in length. • An isosceles triangle has two sides equal in length. • An equilateral triangle has three sides equal in length. C A B Isosceles Triangles In isosceles triangle ABC, shown on the left, the two sides that are equal in measure,, are called the legs. The third side,, is the base. and AB AC BC Two line segments that are equal in measure are said to be congruent. The angle formed by the two congruent sides, C, is called the vertex angle. The two angles at the endpoints of the base, A and B, are the base angles. In isosceles triangle ABC, if we measure the base angles, A and B, we find that each angle contains 65°. Therefore, mA mB. Similarly, if we measure the base angles in any other isosceles triangle, we find that they are equal in measure. Thus, we will accept the truth of the following statement: 266 Geometric Figures, Areas, and Volumes The base angles of an isosceles triangle are equal in measure; that is, they are congruent. This statement may be rephrased in a variety of ways. For example: 1. If a triangle is isosceles, then its base angles are congruent. 2. If two sides of a triangle are equal in measure, then the angles opposite these sides are equal in measure. The following statement is also true: If two angles of a triangle are equal in measure, then the sides opposite these angles are equal in measure and the triangle is an isosceles triangle. This statement may be rephrased as follows: 1. If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 2. If two angles of a triangle have equal measures, then the sides opposite these angles have equal measures. Equilateral Triangles Triangle ABC is an equilateral triangle. Since AB BC, mC mA; also, since AC |
BC, mB mA. Therefore, mA mB mC. In an equilateral triangle, the measures of all of the angles are equal. In DEF, all of the angles are equal in measure. Since mD mE, EF DF; also, since mD mF, EF DE. Therefore, DE EF DF, and DEF is equilateral. If a triangle is equilateral, then it is equiangular. Properties of Special Triangles 1. If two sides of a triangle are equal in measure, the angles opposite these sides are also equal in measure. (The base angles of an isosceles triangle are equal in measure.) 2. If two angles of a triangle are equal in measure, the sides opposite these angles are also equal in measure. 3. All of the angles of an equilateral triangle are equal in measure. (An equi- lateral triangle is equiangular.) 4. If three angles of a triangle are equal in measure, the triangle is equilat- eral. (An equiangular triangle is equilateral.) C F B E A D Triangles 267 EXAMPLE 3 In isosceles triangle ABC, the measure of vertex angle C is 30° more than the measure of each base angle. Find the number of degrees in each angle of the triangle. Solution Let x number of degrees in one base angle, A. Then x number of degrees in the other base angle, B, and x 30 number of degrees in the vertex angle, C. C x + 30 The sum of the measures of the angles of a triangle is 180°. x A x B x x x 30 180 3x 30 180 3x 150 x 50 x 30 80 Check 50 50 80 180 ✔ Answer mA 50, mB 50, mC 80 EXERCISES Writing About Mathematics 1. Ayyam said that if the sum of the measures of two angles of a triangle is equal to the measure of the third angle, the triangle is a right triangle. Prove or disprove Ayyam’s statement. 2. Janice said that if two angles of a triangle each measure 60°, then the triangle is equilateral. Prove or disprove Janice’s statement. Developing Skills In 3–5, state, in each case, whether the angles with the given measures can be the three angles of the same triangle. 3. 30°, 70°, 80° 4. 70°, 80°, 90° |
5. 30°, 110°, 40° In 6–9, find, in each case, the measure of the third angle of the triangle if the measures of two angles are: 6. 60°, 40° 7. 100°, 20° 8. 54.5°, 82.3° 9. 241 48, 811 28 268 Geometric Figures, Areas, and Volumes 10. What is the measure of each angle of an equiangular triangle? 11. Can a triangle have: a. two right angles? obtuse angle? Explain why or why not. b. two obtuse angles? c. one right and one 12. What is the sum of the measures of the two acute angles of a right triangle? 13. In ABC, AC 4, CB 6, and AB 6. a. What type of triangle is ABC? b. Name two angles in ABC whose measures C are equal. B c. Why are these angles equal in measure? d. Name the legs, base, base angles and vertex angle of this triangle. 14. In RST, mR 70 and mT 40. a. Find the measure of S. b. Name two sides in RST that are congruent. c. Why are the two sides congruent? d. What type of triangle is RST? e. Name the legs, base, base angles, and vertex angle of this triangle. A T R S 15. Find the measure of the vertex angle of an isosceles triangle if the measure of each base angle is: a. 80° b. 55° c. 42° d. 221 28 e. 51.5° 16. Find the measure of each base angle of an isosceles triangle if the measure of the vertex angle is: a. 20° b. 50° c. 76° d. 100° e. 65° 17. What is the number of degrees in each acute angle of an isosceles right triangle? 18. If a triangle is equilateral, what is the measure of each angle? Applying Skills 19. The measure of each base angle of an isosceles triangle is 7 times the measure of the vertex angle. Find the measure of each angle of the triangle. 20. The measure of each of the congruent angles of an isosceles triangle is one-half of the mea- sure of the vertex angle. Find the measure of each angle of the triangle. 21 |
. The measure of the vertex angle of an isosceles triangle is 3 times the measure of each base angle. Find the number of degrees in each angle of the triangle. 22. The measure of the vertex angle of an isosceles triangle is 15° more than the measure of each base angle. Find the number of degrees in each angle of the triangle. Triangles 269 23. The measure of each of the congruent angles of an isosceles triangle is 6° less than the mea- sure of the vertex angle. Find the measure of each angle of the triangle. 24. The measure of each of the congruent angles of an isosceles triangle is 9° less than 4 times the vertex angle. Find the measure of each angle of the triangle. 25. In ABC, mA x, mB x 45, and mC 3x 15. a. Find the measures of the three angles. b. What kind of triangle is ABC? 26. The measures of the three angles of DEF can be represented by (x 30)°, 2x°, and (4x 60)°. a. What is the measure of each angle? b. What kind of triangle is DEF? 27. In a triangle, the measure of the second angle is 3 times the measure of the first angle, and the measure of the third angle is 5 times the measure of the first angle. Find the number of degrees in each angle of the triangle. 28. In a triangle, the measure of the second angle is 4 times the measure of the first angle. The measure of the third angle is equal to the sum of the measures of the first two angles. Find the number of degrees in each angle of the triangle. What kind of triangle is it? 29. In a triangle, the measure of the second angle is 30° more than the measure of the first angle, and the measure of the third angle is 45° more than the measure of the first angle. Find the number of degrees in each angle of the triangle. 30. In a triangle, the measure of the second angle is 5° more than twice the measure of the first angle. The measure of the third angle is 35° less than 3 times the measure of the first angle. Find the number of degrees in each angle of the triangle. 31. is a straight line, mAEG 130, and d–——S AEFB mBFG 140. a. Find mx, my, and |
mz. b. What kind of a triangle is EFG? 32. In RST, mR x, mS x 30, mT x 30. a. Find the measures of the three angles of the triangle. b. What kind of a triangle is RST? 33. In KLM, mK 2x, mL x 30, mM 3x 30. a. Find the measures of the three angles of the triangle. b. What kind of a triangle is KLM? G z x 130° E A y 140° F B 270 Geometric Figures, Areas, and Volumes Hands-On Activity 1: Constructing a Line Segment Congruent to a Given Segment To construct a geometric figure means that a specific design is accurately made by using only two instruments: a compass used to draw a complete circle or part of a circle and a straightedge used to draw a straight line. In this activity, you will learn how to construct a line segment congruent to a given segment, that is, construct a copy of a line segment. STEP 1. Use the straightedge to draw a line segment. Label the endpoints A and B. STEP 2. Use the straightedge to draw a ray and label the endpoint C. STEP 3. Place the compass so that the point of the compass is at A and the point of the pencil is at B. STEP 4. Keeping the opening of the compass unchanged, place the point at C and draw an arc that intersects the ray. Label this intersection D. Result: AB CD a. Now that you know how to construct congruent line segments, explain how you can construct a line segment that is three times the length of a given segment. b. Explain how to construct a line segment with length equal to the difference of two given seg- ments. c. Explain how to construct a line segment whose length is the sum of the lengths of two given line segments. Hands-On Activity 2: Constructing an Angle Congruent to a Given Angle In this activity, you will learn how to construct an angle congruent to a given angle, that is, construct a copy of an angle. STEP 1. Use the straightedge to draw an acute angle. Label the vertex S. STEP 2. Use the straightedge to draw a ray and label the endpoint M. STEP 3. With the point of the compass at S, draw an arc that intersects each ray of S. Label the point of intersection on one |
ray R and the point of intersection on the other ray T. STEP 4. Using the same opening of the compass as was used in step 3, place the point of the compass at M and draw an arc that intersects the ray and extends beyond the ray. (Draw at least half of a circle.) Label the point of intersection L. STEP 5. Place the point of the compass at R and the point of the pencil at T. STEP 6. Without changing the opening of the compass, place the point at L and draw an arc that intersects the arc drawn in step 4. Label this point of intersection N. h MN. STEP 7. Draw Result: RST LMN a. Now that you know how to construct congruent angles, explain how you can construct an angle that is three times the measure of a given angle. b. Explain how to construct an angle with a measure equal to the difference of two given angles. c. Explain how to construct a triangle congruent to a given triangle using two sides and the included angle. Triangles 271 Hands-On Activity 3: Constructing a Perpendicular Bisector In this activity, you will learn how to construct a perpendicular bisector of a line segment. A perpendicular bisector of a line segment is the line that divides a segment into two equal parts and is perpendicular to the segment. STEP 1. Use the straightedge to draw a line segment. Label one endpoint A and the other C. STEP 2. Open the compass so that the distance between the point and the pencil point is more than half of the length of AC. STEP 3. With the point of the compass at A, draw an arc above AC and an arc below AC. STEP 4. With the same opening of the compass and the point of the compass at C, draw an arc AC and an arc below above these intersections E and the other F. AC that intersect the arcs drawn in step 3. Call one of STEP 5. Use the straightedge to draw g EF, intersecting AC at B. g EF is perpendicular to Result: a. What is true about EAB and FAB? What is true about EAB and ECB? b. Explain how to construct an isosceles triangle with a vertex angle that is twice the measure of a, and B is the midpoint of AC AC. given angle. c. Explain how to construct an isosceles right triangle. Hands-On Activity 4: Constructing an Angle Bisector In this activity, you |
will learn how to construct an angle bisector. An angle bisector is the line that divides an angle into two congruent angles. STEP 1. Use the straightedge to draw an acute angle. Label the vertex S. STEP 2. With any convenient opening of the compass, place the point at S and draw an arc that intersects both rays of S. Call one of the intersections R and the other T. STEP 3. Place the point of the compass at R and draw an arc in the interior of S. STEP 4. With the same opening of the compass, place the point of the compass at T and draw an arc that intersects the arc drawn in step 3. Label the intersection of the arcs P. STEP 5. Draw h SP. Result: RSP PST; a. Now that you know how to construct an angle bisector, explain how you can construct an angle bisects angle S. h SP that is one and a half times the measure of a given angle. b. Is it possible to use an angle bisector to construct a 90° angle? Explain. c. Explain how to construct an isosceles triangle with a vertex angle that is congruent to a given angle. 272 Geometric Figures, Areas, and Volumes 7-5 QUADRILATERALS In your study of mathematics you have learned many facts about polygons that have more than three sides. In this text you have frequently solved problems using the formulas for the perimeter and area of a rectangle. In this section we will review what you already know and use that knowledge to demonstrate the truth of many of the properties of polygons. C B D A A quadrilateral is a polygon that has four sides. A point at which any two sides of the quadrilateral meet is a vertex of the quadrilateral. At each vertex, the two sides that meet form an angle of the quadrilateral. Thus, ABCD on the AB left is a quadrilateral whose sides are. Its vertices are A,, B, C, and D. Its angles are ABC, BCD, CDA, and DAB., and BC, DA CD In a quadrilateral, two angles whose vertices are the endpoints of a side are called consecutive angles. For example, in quadrilateral ABCD, A and B are consecutive angles because their vertices are the endpoints of a side,. Other pairs of consecutive angles are B and C, C and D, and |
D and A. Two angles that are not consecutive angles are called opposite angles; A and C are opposite angles, and B and D are opposite angles. AB Special Quadrilaterals When we vary the shape of the quadrilateral by making some of its sides parallel, by making some of its sides equal in length, or by making its angles right angles, we get different members of the family of quadrilaterals, as shown and named below Trapezoid F E Parallelogram J K Q R W X Rectangle Rhombus Square A trapezoid is a quadrilateral in which two and only two opposite sides are are called the. Parallel sides AB i CD and CD AB parallel. In trapezoid ABCD, bases of the trapezoid. A parallelogram is a quadrilateral in which both pairs of opposite sides are. The symbol for a EH i FG EF i GH and parallel. In parallelogram EFGH, ~. parallelogram is A rectangle is a parallelogram in which all four angles are right angles. Rectangle JKLM is a parallelogram in which J, K, L, and M are right angles. Quadrilaterals 273 A rhombus is a parallelogram in which all sides are of equal length. Rhombus QRST is a parallelogram in which QR = RS = ST TQ. A square is a rectangle in which all sides are of equal length. It is also correct to say that a square is a rhombus in which all angles are right angles. Therefore, square WXYZ is also a parallelogram in which W, X, Y, and Z are right angles, and WX XY YZ ZW. The Angles of a Quadrilateral C D Draw a large quadrilateral like the one shown at the left and measure each of its four angles. Is the sum 360°? It should be. If we do the same with several other quadrilaterals of different shapes and sizes, is the sum of the four measures 360° in each case? It should be. Relying on what you have just verified by experimentation, it seems reasonable to make the following statement: A B The sum of the measures of the angles of a quadrilateral is 360°. D AC To prove informally that this statement is true, we draw diagonal endpoints are the vertices of the two opposite angles, A and C. Then:, whose C (1) |
Diagonal ADC. AC divides quadrilateral ABCD into two triangles, ABC and (2) The sum of the measures of the angles of ABC is 180°, and the sum of the measures of the angles of ADC is 180°. (3) The sum of the measures of all the angles of ABC and ADC together A B is 360°. (4) Therefore, mA mB mC mD 360. The Family of Parallelograms Listed below are some relationships that are true for the family of parallelograms that includes rectangles, rhombuses, and squares. Parallelogram 1. All rectangles, rhombuses, and squares are parallelograms. Therefore any property of a parallelogram must also be a property of a rectangle, a rhombus, or a square. 2. A square is also a rectangle. Therefore, any property of a rectangle must also be a property of a square. 3. A square is also a rhombus. Therefore, any property of a rhombus must also be a property of a square. Rectangle Rhombus Square 274 Geometric Figures, Areas, and Volumes D C A B In parallelogram ABCD at the left, opposite sides are parallel.Thus, AD i BC. The following statements are true for any parallelograms. and AB i DC Opposite sides of a parallelogram are congruent. Here, AB > DC and AD > BC. Therefore AB DC and AD BC. Opposite angles of a parallelogram are congruent. Here, A C and B D. Therefore mA mC and mB mD. Consecutive angles of a parallelogram are supplementary. Here, mA mB 180, mB mC 180, and so forth. Since rhombuses, rectangles, and squares are also parallelograms, these statements will be true for any rhombus, any rectangle, and any square. Informal Proofs for Statements About Angles in a Parallelogram The statements about angles formed by parallel lines cut by a transversal that were shown to be true in Section 3 of this chapter can now be used to establish the relationships among the measures of the angles of a parallelogram. (1) In parallelogram ABCD, AB and DC ments of parallel lines cut by transversal are segAD. (2) Let mA x. Then mD 180 x because |
A and D are interior angles on the same side of a transversal, and these angles have been shown to be supplementary. (3) In parallelogram ABCD, AD and BC ments of parallel lines cut by transversal are segAB. D 180 – x C x x A 180 – x B (4) Since mA x, mB 180 x because A and B are interior angles on the same side of a transversal, and these angles have been shown to be supplementary. (5) Similarly, AD and DC. Since mD 180 x, mC 180 (180 x) x because D and C are interior angles on the same side of the transversal. are segments of parallel lines cut by transversal BC Therefore, the consecutive angles of a parallelogram are supplementary: mA mD x (180 x) 180 mA mB x (180 x) 180 mC mB x (180 x) 180 mC mD x (180 x) 180 Quadrilaterals 275 Also, the opposite angles of a parallelogram have equal measures: mA x and mC x m B 180 x and mD 180 x or mA mC or mB mD Polygons and Angles D C We can use the sum of the measures of the interior angles of a triangle to find the sum of the measures of the interior angles of any polygon. Polygon ABCDE is a pentagon, a polygon with five sides. From vertex A, we draw diagonals to vertices C and D, the vertices that are not adjacent to A. These diagonals divide the pentagon into three triangles. The sum of the measures of the interior angles of ABCDE is the sum of the measures of ABC, ACD, and ADE. The sum of the measures of the interior angles of ABCDE 3(180°) 540°. We can use this same method to find the sum of the interior angles of any polygon of more than three sides. E A B Hexagon 4(180°) = 720° Octagon 6(180°) = 1,080° Decagon 8(180°) = 1,440° In each case, the number of triangles into which the polygon can be divided is 2 fewer than the number of sides of the polygon. In general: The sum of the measures of the interior angles of an n-sided polygon is 180(n 2). Tra |
pezoids A trapezoid has one and only one pair of parallel lines. Each of the two parallel sides is called a base of the trapezoid. Therefore, we can use what we know about angles formed when parallel lines are cut by a transversal to demonstrate some facts about the angles of a trapezoid. Quadrilateral ABCD is a trapezoid with. Since DAB and AB i CD CDA are interior angles on the same side of transversal DA, they are supplementary. Also, CBA and DCB are interior angles on the same side of transversal, and they are supplementary. CB D A C B 276 Geometric Figures, Areas, and Volumes In a trapezoid, the parallel sides can never be congruent. But the nonparallel sides can be congruent. A trapezoid in which the nonparallel sides are congruent is called an isosceles trapezoid. In an isosceles trapezoid, the base angles, the two angles whose vertices are the endpoints of the same base, are congruent. To write a formula for the area of AC. Since AFCE is trapezoid ABCD, draw diagonal and altitudes and a rectangle AE CF. AE CF Let AB b1, CD b2, and AE CF h. b2 D E Area of triangle ABC Area of triangle ACD Area of trapezoid ABCD 1 2b1h 1 2b2h 1 2b1h 1 1 2b2h 1 2h(b1 1 b2) A B b1 C h F EXAMPLE 1 ABCD is a parallelogram where mA 2x 50 and mC 3x 40. D C 3x + 40 a. Find the value of x. b. Find the measure of each angle. 2x + 50 A B Solution a. In ~ABCD, mC mA because the opposite angles of a parallelogram have equal measure. Thus: b. By substitution: 3x 40 2x 50 x 10 mA 2x 50 2(10) 50 70 mB 180 mA 180 70 110 mC 3x 40 3(10) 40 70 mD 180 mC 180 70 110 Answer a. x 10 b. mA 70 mB 110 mC 70 mD 110 Quadrilaterals 277 EXERCISES Writing About Mathematics 1. Adam said that if a quadrilateral has four equal |
angles then the parallelogram is a rectangle. Do you agree with Adam? Explain why or why not. 2. Emmanuel said that if a parallelogram has one right angle then the parallelogram is a rec- tangle. Do you agree with Emmanuel? Explain why or why not. Developing Skills In 3–8, in each case, is the statement true or false? 3. If a polygon is a trapezoid, it is a quadrilateral. 4. If a polygon is a rectangle, it is a parallelogram. 5. If a polygon is a rhombus, it is a parallelogram. 6. If a polygon is a rhombus, it is a square. 7. If a polygon is a square, then it is a rhombus. 8. If two angles are opposite angles of a parallelogram, they are congruent. In 9–12, the angle measures are represented in each quadrilateral. In each case: a. Find the value of x. b. Find the measure of each angle of the quadrilateral. 9. D C 80° 120° 110° A x B 11. N M 2x 3x – 50 x K x – 10 L 10. G H x + 20 x – 20 x E x F 12. S 2x + 20 T 2x 3x – 20 R x + 10 Q 278 Geometric Figures, Areas, and Volumes In 13 and 14, polygon ABCD is a parallelogram. A D 13. AB 3x 8 and DC x 12. Find AB and DC. 14. mA 5x 40 and mC 3x 20. Find mA, mB, mC, and mD. B C In 15 and 16, polygon ABCD is a rectangle. 15. BC 4x 5, AD 2x 3. Find BC and AD. 16. mA 5x 10. Find the value of x. 17. ABCD is a square. If AB 8x 6 and BC 5x 12, find the length of each side of the square. D A 18. In rhombus KLMN, KL 3x and LM 2(x 3). Find the length of each side of the rhombus. Applying Skills 19. One side of a barn is in the shape of a seven-sided polygon called a heptagon. The heptagon, ABCDEFG, can be |
divided into rectangle ABFG, isosceles trapezoid BCEF, and isosceles triangle CDE, as shown in the diagram. a. Find the sum of the measures of the interior angles of ABCDEFG by using the sum of the measures of the angles of the two quadrilaterals and the triangle. b. Sketch the heptagon on your answer paper and show how it can be divided into triangles by drawing diagonals from A. Use these triangles to find the sum of the measures of the interior angles of ABCDEFG. If the measure of each lower base angle of trapezoid BCEF is 45° and the measure of each base angle of isosceles triangle CDE is 30°, find the measure of each angle of heptagon ABCDEFG. d. Find the sum of the measures of the interior angles of ABCDEFG by adding the angle measures found in c. 20. Cassandra had a piece of cardboard that was in the shape of an equilateral triangle. She cut an isosceles triangle from each vertex of the cardboard. The length of each leg of the triangles that she cut was one third of the length of a side of the original triangle. a. Show that each of the triangles that Cassandra cut off is an equilateral triangle. b. What is the measure of each angle of the remaining piece of cardboard? c. What is the shape of the remaining piece of cardboard? Areas of Irregular Polygons 279 21. A piece of land is bounded by two parallel roads and two roads that are not parallel forming a trapezoid. Along the parallel roads the land measures 1.3 miles and 1.7 miles. The distance between the parallel roads, measured perpendicular to the roads, is 2.82 miles. a. Find the area of the land. Express the area to the nearest tenth of a mile. b. An acre is a unit of area often used to measure land. There are 640 acres in a square mile. Express, to the nearest hundred acres, the area of the land. 22. If possible, draw the following quadrilaterals. If it is not possible, state why. a. A quadrilateral with four acute angles. b. A quadrilateral with four obtuse angles. c. A quadrilateral with one acute angle and three obtuse angles. d. A quadrilateral with three acute angles and one obtuse angle. e. A quadrilateral with exactly |
three right angles. 7-6 AREAS OF IRREGULAR POLYGONS Many polygons are irregular figures for which there is no formula for the area. However, the area of such figures can often be found by separating the figure into regions with known area formulas and adding or subtracting these areas to find the required area. EXAMPLE 1 and F is a point. If AB 12, FC 3, and DE 8, find the area ABCD is a square. E is a point on BC on of ABFE. DC Solution Since ABCD is a square and AB 12, then BC 12, CD 12, and DA 12. Therefore, CE CD DE 12 8 4. D 8 E C 3 F A 12 B Area of ABFE Area of ABCD Area of FCE Area of EDA 2(3)(4) 2 1 2(8)(12) 12(12) 2 1 144 6 48 90 Answer 90 square units 280 Geometric Figures, Areas, and Volumes EXAMPLE 2 ABCD is a quadrilateral. AB 5.30 centimeters, BC 5.5 centimeters, CD 3.2 centimeters, and DA 3.52 centimeters. Angle A and angle C are right angles. Find the area of ABCD to the correct number of significant digits. D 3.52 cm C c m 3. 2 5.5 cm Solution Draw BD, separating the quadrilateral into two right triangles. In a right triangle, either leg is the base and the other leg is the altitude. A 5.30 cm B In ABD, b 5.30, and h 3.52: Area of ABD 1 2(5.30)(3.52) 9.328 In BCD, b 5.5, and h 3.2: Area of ABD 1 2(5.5)(3.2) 8.80 The first area should be given to three significant digits, the least number of significant digits involved in the calculation. Similarly, the second area should be given to two significant digits. However, the area of ABCD should be no more precise than the least precise measurement of the areas that are added. Since the least precise measurement is 8.8, given to the nearest tenth, the area of ABCD should also be written to the nearest tenth: Area of ABCD Area of ABD Area of BCD 9.328 8.80 18.128 18.1 Answer 18.1 square centimeters Note: When working with significant digits involving multiple |
operations, make sure to keep at least one extra digit for intermediate calculations to avoid roundoff error. Alternatively, when working with a calculator, you can round at the end of the entire calculation. EXERCISES Writing About Mathematics 1. ABCD is a trapezoid with AB i CD. The area of ABD is 35 square units. What is the area of ABC? Explain your answer. Areas of Irregular Polygons 281 2. ABCD is a quadrilateral. The area of ABD is 57 square inches, of BDC is 62 square inches, and of ABC is 43 square inches. What is the area of ADC? Explain your answer. Developing Skills In 3–10, find each measure to the correct number of significant digits. D C C 10.00 ft.00 ft E O 10.00 ft B A 10.0 cm B Ex. 3 Ex Ex. 5 3. ABCD is a square and arc BEC is a semicircle. If AB 10.0 cm, find the area of the figure. 4. ABCD is a trapezoid. The vertices of the trapezoid are on a circle whose center is at O. DE'AB, OB DC = 10.00 ft, DE 8.66 ft, and AE 5.00 ft. Find the area of EBCD. 5. ABCDEF is a regular hexagon. The vertices of the hexagon are on a circle whose center is at O. OG'BC, AB 15.0 m and OG 13.0 m. Find the area of ABCDEF. E 18.5 in. D 2.4 ft C 24.0 in. 9.60 in. C A 8.25 in. F B Ex. 6 D 2. 4 ft ft A Ex. 7 D 24.0 cm C 7.00 cm E 5.00 cm A 12.0 cm B Ex. 8 6. ABCDE is a pentagon. DC'ED, AC 24.0 in., and BF 8.25 in., find the area of ABCDE. ED i AC BF'AC, and. If ED 18.5 in., DC 9.60 in., 7. ABCD is a quadrilateral. The diagonals of the quadrilateral are perpendicular to each other at E. If AE 8.7 ft, BE 5.6 ft, CE 2.4 ft, and DE 2.4 ft, find the area of ABCD. 8 |
. ABCD is a trapezoid with BC'AB BE 5.00 cm, and EC 7.00 cm. Find the area of triangle AED. and E a point on BC. AB 12.0 cm, DC = 24.0 cm, 282 Geometric Figures, Areas, and Volumes D C 15 yd 13 yd 18 yd B E 38 yd Ex. 9 A D 6.0 mm E 4.0 mm C 8.0 mm A Ex. 10 F 5.0 mm B 9. ABCD is a quadrilateral with DE'AB BC 13 yd, and DE 15 yd. Find the area of ABCD. BC'AB and. If AB 38 yd, EB 18 yd, 10. ABCD is a rectangle. AD 8 mm, DE 6 mm, EC 4 mm, and BF 5 mm. Find the area of ABFE. Applying Skills 11. Mason has a square of fabric that measures one yard on each side. He makes a straight cut from the center of one edge to the center of an adjacent edge. He now has one piece that is an isosceles triangle and another piece that is a pentagon. Find the area of each of these pieces in square inches. 12. A park is in the shape of isosceles trapezoid ABCD. The bases of the trapezoid, and measure 20.0 meters and 5.00 meters respectively. The measure of each base angle at is 60°, and the height of the trapezoid is 13.0 meters. A straight path from A to E, a CD AB point on trapezoid into two regions; a quadrilateral planted with grass and a triangle planted with shrubs and trees. Find the area of the region planted with grass to the nearest meter. and BE 10.0 meters. The path separates the, makes an angle of 30° with AB BC AB 7-7 SURFACE AREAS OF SOLIDS A right prism is a solid with congruent bases and with a height that is perpendicular to these bases. Some examples of right prisms, as seen in the diagrams, include solids whose bases are triangles, trapezoids, and rectangles. The prism is named for the shape of its base. The two bases may be any polygons, as long as they have the same size and shape. The remaining sides, or faces, are rectangles. B B h h Triangular right prism Trapezoidal |
right prism The surface area of a solid is the sum of the areas of the surfaces that make up the solid. The surface area of a right prism is the sum of the areas of the bases and the faces of the solid. The number of faces of a prism is equal to the number of sides of a base. Surface Areas of Solids 283 • If the base of the prism is a triangle, it has 2 bases 3 faces or 5 surfaces. • If the base of the prism is a quadrilateral, it has 2 bases 4 faces or 6 sur- faces. • If the base of the prism has n sides, it has 2 bases n faces or 2 n sur- faces. Two or more of the faces are congruent if and only if two or more of the sides of a base are equal in length. For example, if the bases are isosceles triangles, two of the faces will be congruent rectangles and if the bases are equilateral triangles, all three of the faces will be congruent rectangles. If the bases are squares, then the four faces will be congruent rectangles. The most common right prism is a rectangular solid that has rectangles as bases and as faces. Any two surfaces that have no edge in common can be the bases and the other four surfaces are the faces. If the dimensions of the rectangular solid are 3 by 5 by 4, then there are two rectangles that are 3 by 5, two that are 5 by 4 and two that are 3 by 4. The surface area of the rectangular solid is: 2(3)(5) 2(5)(4) 2(3)(4) 30 40 24 94 square units B h Rectangular solid (a prism) In general, when the dimensions of a rectangular solid are represented by l, w, and h, then the formula for the surface area is: Surface Area of a Rectangular Solid 2lw 2lh 2wh If the rectangular solid has six squares as bases and sides, the solid is a cube. If s represents the length of each side, then l s, w s, and h s. If we substitute in the formula for the area of a rectangular solid, then: S 2lw 2lh 2wh 2s(s) 2s(s) 2s(s) 2s2 2s2 2s2 6s2 Surface Area of a Cube 6s2 A right circular cylinder is a solid with two bases that are circles of |
the same size, and with a height that is perpendicular to these bases. Fruits and vegetables are often purchased in “cans” that are in the shape of a right circular cylinder. The label on the can corresponds to the surface area of the curved portion of the cylinder. That label is a rectangle whose length is the circumference of the can, 2pr, and whose width is the height of the can, h. Therefore, the area of the curved portion of the cylinder is 2prh and the area of each base is pr2. Surface Area of a Cylinder 2 the area of a base r r h B Right circular cylinder Surface Area of a Cylinder 2pr 2 2prh the area of the curved portion 284 Geometric Figures, Areas, and Volumes EXAMPLE 1 A rectangular solid has a square base. The height of the solid is 2 less than twice the length of one side of the square. The height of the solid is 22.2 centimeters. Find the surface area of the figure to the correct number of significant digits. Solution Let x the length of a side of the base. 2x 2 the height of the figure. 2x 2 22.2 2x 24.2 x 12.1 2x – 2 x x The surface area consists of two square bases with sides that measure 12.1 centimeters and four rectangular faces with width 12.1 and length 22.2. Surface area of the solid surface area of the bases surface area of the faces 2(12.1)2 4(12.1)(22.2) 292.82 1,074.48 1,367.3 1,370 The areas of the bases should be given to three significant digits, the least number of significant digits involved in the calculation. Similarly, the areas of the faces should be given to three significant digits. However, the surface area of the solid should be no more precise than the least precise measurement of the areas that are added. Since the least precise measurement is 1,074.48, given to the nearest ten, the surface area of the solid should also be written to the nearest ten: 1,370 square centimeters. Answer Surface area 1,370 cm2 Note: Recall that when working with significant digits involving multiple operations, you should keep at least one extra digit for intermediate calculations or wait until the end of the entire calculation to round properly. Surface Areas of Solids 285 EXERCISES Writing About Mathematics 1. A right prism has bases that are |
regular hexagons. The measure of each of the six sides of the hexagon is represented by a and the height of the solid by 2a. a. How many surfaces make up the solid? b. Describe the shape of each face c. Express the dimensions and the area of each face in terms of a. 2. A regular hexagon can be divided into six equilateral triangles. If the length of a side of an equilateral triangle is a, the height is 3 2 a ". For the rectangular solid described in exercise 1: a. Express the area of each base in terms of a. b. Express the surface area in terms of a. Developing Skills In 3–9, find the surface area of each rectangular prism or cylinder to the nearest tenth of a square unit. 3. Bases are circles with a radius of 18 inches. The height of the cylinder is 48 inches. 4. Bases are squares with sides that measure 27 inches. The height is 12 inches. 5. Bases are rectangles with dimensions of 8 feet by 12 feet. The height is 3 feet. 6. Bases are isosceles trapezoids with parallel sides that measure 15 centimeters and 25 cen- timeters. The distance between the parallel sides is 12 centimeters and the length of each of the equal sides is 13 centimeters. The height of the prism is 20 centimeters. 7. Bases are isosceles right triangles with legs that measure 5 centimeters. The height is 7 cen- timeters. 8. Bases are circles with a diameter of 42 millimeters. The height is 3.4 centimeters. 9. Bases are circles each with an area of 314.16 square feet. The height is 15 feet. Applying Skills 10. Agatha is using scraps of wallpaper to cover a box that is a rectangular solid whose base measures 8 inches by 5 inches and whose height is 3 inches. The box is open at the top. How many square inches of wallpaper does she need to cover the outside of the box? 11. Agatha wants to make a cardboard lid for the box described in exercise 10. Her lid will be a rectangular solid that is open at the top, with a base that is slightly larger than that of the box. She makes the base of the lid 8.1 inches by 5.1 inches with a height of 2.0 inches. To the nearest tenth of a square inch, how much wallpaper does she need to cover the outside of the lid? 286 |
Geometric Figures, Areas, and Volumes 12. Sandi wants to make a pillow in the shape of a right circular cylinder. The diameter of the circular ends is 10.0 inches and the length of the pillow (the height of the cylinder) is 16.0 inches. Find the number of square inches of fabric Sandi needs to make the pillow to the correct number of significant digits. 13. Mr. Breiner made a tree house for his son. The front and back walls of the house are trapezoids to allow for a slanted roof. The floor, roof, and remaining two sides are rectangles. The tree house is a rectangular solid. The front and back walls are the bases of this solid. The dimensions of the floor are 8 feet by 10 feet and the roof is 10 feet by 10 feet. The height of one side wall is 7 feet and the height of the other is 13 feet. The two side walls each contain a window measuring 3 feet by 3 feet. Mr. Breiner is going to buy paint for the floor and the exterior of the house, including the roof, walls, and the door. How many square feet must the paint cover, excluding the windows? 7-8 VOLUMES OF SOLIDS 10 ft 10 ft 13 ft 7 ft 3 ft 3 ft 10 ft 8 ft The volume of a solid is the number of unit cubes (or cubic units) that it contains. Both the volume V of a right prism and the volume V of a right circular cylinder can be found by multiplying the area B of the base by the height h. This formula is written as V Bh 2 4 3 V = Bh V = (lw)h = (4 · 2) · 3 = 8 · 3 = 24 cm3 To find the volume of a solid, all lengths must be expressed in the same unit of measure. The volume is then expressed in cubic units of this length. For example, in the rectangular solid or right prism shown at left, the base is a rectangle, 4 centimeters by 2 centimeters. The area, B, of this base is lw. Therefore, B 4(2) 8 cm2. Then, using a height h of 3 centimeters, the volume V of the rectangular solid Bh (8)(3) 24 cm3. For a rectangular solid, note that the volume formula can be written in two ways: V Bh or V lwh Volumes of Solids 287 To understand volume, count the cubes in the diagram on the previous page. |
There are 3 layers, each containing 8 cubes, for a total of 24 cubes. Note that 3 corresponds to the height, h, that 8 corresponds to the area of the base, B, and that 24 corresponds to the volume V in cubic units. A cube that measures 1 foot on each side represents 1 cubic foot. Each face of the cube is 1 square foot. Since each foot can be divided into 12 inches, the area of each face is 12 12 or 144 square inches and the volume of the cube is 12 12 12 or 1,728 cubic inches. 1 square foot 144 square inches 1 cubic foot 1,728 cubic inches A cube that measures 1 meter on each side represents 1 cubic meter. Each face of the cube is 1 square meter. Since each meter can be divided into 100 centimeters, the area of each face is 100 100 or 10,000 square centimeters and the volume of the cube is 100 100 100 or 1,000,000 cubic centimeters. 1 square meter 10,000 square centimeters 1 cubic meter 1,000,000 cubic centimeters EXAMPLE 1 A cylindrical can of soup has a radius of 1.5 inches and a height of 5 inches. Find the volume of this can: r = 1.5 in. a. in terms of p. b. to the nearest cubic inch. Solution This can is a right circular cylinder. Use the formula V Bh. Since the base is a circle whose area equals pr2, the area B of the base can be replaced by pr2. a. V Bh (pr2)h p(1.5)2(5) 11.25p O h = 5 in. b. When we use a calculator to evaluate 11.25p, the calculator gives 35.34291735 as a rational approximation. This answer rounded to the nearest integer is 35. Answers a. 11.25p cu in. b. 35 cu in. 288 Geometric Figures, Areas, and Volumes A pyramid is a solid figure with a base that is a polygon and triangular faces or sides that meet at a common point. The formula for the volume V of a pyramid is: V 5 1 3 Bh In the pyramid and prism shown here, the bases are the same in size and shape, and the heights are equal in measure. If the pyramid could be filled with water and that water poured into the prism, exactly three pyramids of water would be needed to fill the prism. In other words, the volume of a pyramid is one- |
third the volume of a right prism with the same base and same height as those of the pyramid. The volume of a cone is onethird of the volume of a right circular cylinder, when both the cone and the cylinder have circular bases and heights that are equal in measure. The formula for the volume of the cone is ) h ( t h g i e H Base (B) Base (B) Pyramid = prism Base (B) r Base (B) V 5 1 3Bh or V 5 1 3pr 2h As in the case of the pyramid and prism, if the cone could be filled with water and that water poured into the right circular cylinder, exactly three cones of water would be needed to fill the cylinder. C A B O Sphere A sphere is a solid figure whose points are all equally distant from one fixed point in space called its center. A sphere is not a circle, which is drawn on a flat surface or plane but similar terminology is used for a sphere. A line segment,, that joins the center O to any point on the sphere is called a radius such as of the sphere. A line segment, such as, that joins two points of the sphere and passes through its center is called a diameter of the sphere. AB OC The volume of a sphere with radius r is found by using the formula: V 5 4 3pr 3 Volumes of Solids 289 EXAMPLE 2 An ice cream cone that has a diameter of 6.4 centimeters at its top and a height of 12.2 centimeters is filled with ice cream and topped with a scoop of ice cream that is approximately in the shape of half of a sphere. Using the correct number of significant digits, how many cubic centimeters of ice cream are needed? Solution The amount of ice cream needed is approximately the volume of the cone plus one-half the volume of a sphere with a radius equal to the radius of the top of the cone. Volume of ice cream 4 3pr3 3pr2h 1 1 5 1 2 A 3p(3.2)2(12.2) 1 1 5 1 2 130.824... 68.629... 199.453... 4 3 A B B p(3.2)3 The least precise volume is the volume of the cone, given to the nearest ten. Therefore, the answer should be rounded to the nearest ten. Note that in the answer, the zero in the tens place is significant. The zero in the ones place is not significant |
. Answer The volume of ice cream is 200 cubic centimeters. Error in Geometric Calculations When a linear measure is used to find area or volume, any error in the linear value will be increased in the higher-dimension calculations. EXAMPLE 3 The length of a side of a cube that is actually 10.0 centimeters is measured to be 10.5 centimeters. Find the percent error in: a. the linear measure. b. the surface area. c. the volume. d. compare the results in a–c above. Solution a. The true length is 10.0 centimeters. The measured value is 10.5 centimeters. Percent error |measured value 2 true value| true value 10.0 3 100% 5 0.5 0.05 100% 5% Answer |10.5 – 10.0| 100% 10.0 3 100% 290 Geometric Figures, Areas, and Volumes b. For a cube, S 6s2. Using the measured value: S 6(10.5)2 661.5 Using the true value: S 6(10.0)2 600.0 Percent error |661.5 – 600.0| 600.0 3 100% 5 61.5 600.0 3 100% 0.1025 100% 10.25% Answer c. For a cube, V s3. Using the measured value: V (10.5)3 1,157.625 Using the true value: V (10.0)3 1,000 Percent error |1,157.625 – 1,000| 1,000 3 100% 5 157.625 1,000 3 100% 0.157625 100% ≈ 15.76% Answer d. The error in the linear measure is 5%. The error increases to 10.25% when the surface area is calculated. The error increased even more to 15.76% when the volume is calculated. Answer EXERCISES Writing About Mathematics 1. A chef purchases thin squares of dough that she uses for the top crust of pies. The squares measure 8 inches on each side. From each square of dough she cuts either one circle with a diameter of 8 inches or four circles, each with a diameter of 4 inches. a. Compare the amount of dough left over when she makes one 8-inch circle with that left over when she makes four 4-inch circles. b. The chef uses the dough to form the top crust of pot pies that are 1.5 inches deep |
. Each pie is approximately a right circular cylinder. Compare the volume of one 8-inch pie to that of one 4-inch pie. 2. Tennis balls can be purchased in a cylindrical can in which three balls are stacked one above the other. How does the radius of each ball compare with the height of the can in which they are packaged? Developing Skills In 3–6, use the formula V lwh and the given dimensions to find the volume of each rectangular solid using the correct number of significant digits. 3. l 5.0 ft, w 4.0 ft, h 7.0 ft 31 5. l m, w cm, h 85 cm 4 21 2 4. l 8.5 cm, w 4.2 cm, h 6.0 cm 6. l 7.25 in., w 6.40 in., h 0.25 ft Volumes of Solids 291 7. Find the volume of a cube if each edge measures 8. The measure of each edge of a cube is represented by (2y 3) centimeters. Find the volume centimeters. 83 5 of the cube when y 7.25. 9. The base of a right prism is a triangle. One side of the triangular base measures 8 centime- ters and the altitude to that side measures 6 centimeters. The height h of the prism measures 35 millimeters. a. Find the area of the triangular base of the prism. b. Find the volume of the prism. 10. The base of a right prism is a trapezoid. This trapezoid has bases that measure 6 feet and 10 feet and an altitude that measures 4 feet. The height h of the prism is 2 yards. a. Find the area of the trapezoidal base of the prism. b. Find the volume of the prism. 11. The base of a pyramid has an area B of 48 square millimeters and a height h of 13 millime- ters. What is the volume of the pyramid? 12. The height of a pyramid is 4 inches, and the base is a rectangle 6 inches long and wide. What is the volume of the pyramid? 31 2 inches 13. A right circular cylinder has a base with a radius of 24.1 centimeters and a height of 17.3 centimeters. a. Express the volume of the cylinder in terms of p. b. Find the volume of the cylinder to the nearest hundred cubic centimeters. 14. A right circular cylinder has a base with a diameter of 25 meters and a height |
of 15 meters. a. Express the volume of the cylinder in terms of p. b. Find the volume of the cylinder to the nearest ten cubic meters. 15. The base of a cone has a radius of 7 inches. The height of the cone is 5 inches. a. Find the volume of the cone in terms of p. b. Find the volume of the cone to the nearest cubic inch. 16. The base of a cone has a radius of 7 millimeters. The height of the cone is 2 centimeters. a. Find the volume of the cone in terms of p. b. Express the volume of the cone to the nearest cubic centimeter. 17. A sphere has a radius of 12.5 centimeters. a. Find the volume of the sphere in terms of p. b. Express the volume of the sphere to the nearest cubic centimeter. 18. A sphere has a diameter of 3 feet. a. Find the volume of the sphere in terms of p. b. Express the volume of the sphere to the nearest cubic foot. 292 Geometric Figures, Areas, and Volumes 19. The side of a cube that is actually 8 inches is measured to be 7.6 inches. Find, to the nearest tenth of a percent, the percent error in: a. the length of the side. b. the surface area. c. the volume. Applying Skills 20. An official handball has a diameter of 4.8 centimeters. Find its volume: a. to the nearest cubic centimeter b. to the nearest cubic inch. 21. A tank in the form of a right circular cylinder is used for storing water. It has a diameter of 12 feet and a height of 14 feet. How many gallons of water will it hold? (1 cubic foot contains 7.5 gallons.) 22. Four pieces of cardboard that are 8.0 inches by 12 inches and two pieces that are 12 inches by 12 inches are used to form a rectangular solid. a. What is the surface area of the rectangular solid formed by the six pieces of card- board? b. What is the volume of the rectangular solid in cubic inches? c. What is the volume of the rectangular solid in cubic feet? 23. A can of soda is almost in the shape of a cylinder with a diameter of 6.4 centimeters and a height of 12.3 centimeters. a. What is the volume of the can? b. If there are 1,000 cubic centimeters in a liter, find how many liters of |
soda the can holds. In 24–26, express each answer to the correct number of significant digits. 24. Cynthia used a shipping carton that is a rectangular solid measuring 12.0 inches by 15.0 inches by 3.20 inches. What is the volume of the carton? 25. The highway department stores sand in piles that are approximately the shape of a cone. What is the volume of a pile of sand if the diameter of the base is 7.0 yards and the height of the pile is 8.0 yards? 26. The largest pyramid in the world was built around 2500 B.C. by Khufu, or Cheops, a king of ancient Egypt. The pyramid had a square base 230˙ meters (756 feet) on each side, and a height of 147 meters (482 feet). (The length of a side of the base is given to the nearest meter. The zero in the ones place is significant.) Find the volume of Cheops’ pyramid using: a. cubic meters b. cubic feet CHAPTER SUMMARY Chapter Summary 293 Point, line, and plane are undefined terms that are used to define other terms. A line segment is a part of a line consisting of two points on the line, called endpoints, and all of the points on the line between the endpoints. A ray is a part of a line that consists of a point on the line and all of the points on one side of that point. An angle is the union of two rays with a common endpoint. Two angles are complementary if the sum of their measures is 90°. If the measure of an angle is x°, the measure of its complement is (90 x)°. Two angles are supplementary if the sum of their measures is 180°. If the measure of an angle is x°, the measure of its supplement is (180 x)°. A linear pair of angles are adjacent angles that are supplementary. Two angles are vertical angles if the sides of one are opposite rays of the sides of the other. Vertical angles are congruent. If two parallel lines are cut by a transversal, then: • The alternate interior angles are congruent. • The alternate exterior angles are congruent. • The corresponding angles are congruent. • Interior angles on the same side of the transversal are supplementary. The sum of the measures of the angles of a triangle is 180°. The base angles of an isosceles triangle are congruent. |
An equilateral triangle is equiangular. The sum of the measures of the angles of a quadrilateral is 360°. The sum of the measures of the angles of any polygon with n sides is 180(n 2). If the measures of the bases of a trapezoid are b1 and b2 and the measure of 1 2h(b11b2). the altitude is h, then the formula for the area of the trapezoid is A Formulas for Surface Area Formulas for Volume Rectangular solid: S 2lw 2lh 2wh Cube: S 6s2 Cylinder: S 2pr2 2prh Any right prism: V Bh Rectangular solid: V lwh Right circular cylinder: V pr2h 1 3Bh 1 3pr2h Pyramid: V Cone: V Cube: V s3 Sphere: V 4 3pr3 294 Geometric Figures, Areas, and Volumes VOCABULARY 7-1 Undefined term • Point • Line • Straight line • Plane • Axiom (postulate) • Line segment (segment) • Endpoints of a segment • Measure of a line segment (length of a line segment) • Half-line • Ray • Endpoint of a ray • Opposite rays • Angle • Vertex of an angle • Sides an angle • Degree • Right angle • Acute angle • Obtuse angle • Straight angle • Perpendicular 7-2 Adjacent angles • Complementary angles • Complement • Supplementary angles • Linear pair of angles • Vertical angles • Congruent angles • Theorem 7-3 Parallel lines • Transversal • Interior angles • Alternate interior angles • Exterior angles • Alternate exterior angles • Interior angles on the same side of the transversal • Corresponding angles 7-4 Polygon • Sides of a polygon • Vertices • Triangle • Acute triangle • Equiangular triangle • Right triangle • Obtuse triangle • Legs of a right triangle • Hypotenuse • Scalene triangle • Isosceles triangle • Equilateral triangle • Legs of an isosceles triangle • Base of an isosceles triangle • Congruent line segments • Vertex angle of an isosceles triangle • Base angles of an isosceles triangle • Construction • Compass • Straightedge • Perpendicular bisector • Angle bisector 7-5 Quadrilateral • Consecutive angles • Opposite angles • Trapezoid |
• Bases of a trapezoid • Parallelogram • Rectangle • Rhombus • Square • Isosceles trapezoid 7-7 Right prism • Face • Surface area • Rectangular solid • Right circular cylinder 7-8 Volume • Pyramid • Cone • Sphere • Center of a sphere • Radius of a sphere • Diameter of a sphere REVIEW EXERCISES 1. If g AB g CD find mAEC. and intersect at E, mAEC x 10, and mDEB 2x 30, 2. If two angles of a triangle are complementary, what is the measure of the third angle? 3. The measure of the complement of an angle is 20° less than the measure of the angle. Find the number of degrees in the angle. 4. If each base angle of an isosceles triangle measures 55°, find the measure of the vertex angle of the triangle. 5. The measures of the angles of a triangle are consecutive even integers. What are the measures of the angles? Review Exercises 295 g CD, and these lines are cut by E In 6–8, g AB transversal is parallel to g EF at points G and H, respectively. 6. If mAGH 73, find mGHD. 7. If mEGB 70 and mGHD 3x 2, find x. 8. If mHGB 2x 10 and mGHD x 20, find mGHD. A G C H F B D 9. In ART, mA y 10, mR 2y, and mT 2y 30. a. Find the measure of each of the three angles. b. Is ART acute, right, or obtuse? c. Is ART scalene, isosceles but not equilateral or equiangular? 10. The measure of each base angle of an isosceles triangle is 15° more than the measure of the vertex angle. Find the measure of each angle. 11. The measure of an angle is 20° less than 3 times the measure of its supple- ment. What is the measure of the angle and its supplement? 12. In ABC, the measure of B is the measure of A, and the measure of C is 5 2 the measure of A. What are the measures of the three angles? 3 2 13. The measure of one angle is 3 times that of another angle, and the sum of these measures is 120°. What are |
the measures of the angles? 14. The measure of the smaller of two supplementary angles is of the mea- 4 5 sure of the larger. What are the measures of the angles? 15. The vertices of a trapezoid are A(3, 1), B(7, 1), C(5, 5), and D(7, 5). a. Draw ABCD on graph paper. b. E is the point on CD such that CD'AE. What are the coordinates of E? c. Find AB, CD, and AE. d. Find the area of trapezoid ABCD. 16. In parallelogram ABCD, AB 3x 4, BC 2x 5, and CD x 18. Find the measure of each side of the parallelogram. 17. A flowerpot in the shape of a right circular cylinder has a height of 4.5 inches. The diameter of the base of the pot is 4.1 inches. Find the volume of the pot to the nearest tenth. 18. Natali makes a mat in the shape of an octagon (an eight-sided polygon) by cutting four isosceles right triangles of equal size from the corners of a 9 by 15 inch rectangle. If the measure of each of the equal sides of the triangles is 2 inches, what is the area of the octagonal mat? 296 Geometric Figures, Areas, and Volumes 19. Marvin measures a block of wood and records the dimensions as 5.0 centimeters by 3.4 centimeters by 4.25 centimeters. He places the block of wood in a beaker that contains 245 milliliters of water. With the block completely submerged, the water level rises to 317 milliliters. a. Use the dimensions of the block to find the volume of the block of wood. b. What is the volume of the block of wood based on the change in the water level? c. Marvin knows that 1 ml 1 cm3. Can the answers to parts a and b both be correct? Explain your answer. 20. A watering trough for cattle is in the shape of a prism whose ends are the bases of the prism and whose length is the height of the prism. The ends of the trough are trapezoids with parallel edges 31 centimeters and 48 centimeters long and a height of 25 centimeters. The length of the trough is 496 centimeters. a. Find the surface area of one end of the trough. b. If the trough is filled to capacity, how many cubic |
centimeters of water does it hold? c. How many liters of water does the trough hold? (1 liter 1,000 cm3) Exploration A sector of a circle is a fractional part of the interior of a circle, determined by an angle whose vertex is at the center of the circle (a central angle). The area of a sector depends on the measure of its central angle. For example, • If the central angle equals 90°, then the area of the sector is one-fourth the area of the circle, or 90 360 pr2. • If the central angle equals 180°, then the area of the sector is one-half the area of the circle, or 180 360 pr2. 0 r C The shaded region represents a sector with central angle u. • If the central angle equals 270°, then the area of the sector is three-fourths the area of the circle, or 270 360 pr2. In general, if the measure of the central angle is u (theta), then the area of the sector is Area of a sector u 360 pr 2 For this Exploration, use your knowledge of Geometry and the formula for the area of a sector to find the areas of the shaded regions. Express your answers in terms of p. Assume that all the arcs that are drawn are circular. Cumulative Review 297 30° 3 1 2 100° 1 2 2 a. c. e. 1 1 1 1 1 1 b. d. f. 1 1 1 45° 1 45° 45° 1 1 1 √3 2 60° 45° 1 1 1 √3 2 60° 1 1 CUMULATIVE REVIEW CHAPTERS 1–7 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following inequalities is true? (1) 4 (3) 5 (2) 4 5 (3) (3) (3) 4 5 (4) 5 4 (3) 298 Geometric Figures, Areas, and Volumes 2. Which of the following is an example of the use of the associative property? (1) 2(x 5) 2(5 x) (2) 2(x 5) 2x 2(5) (3) 2 (x 5) 2 (5 x) (4) 2 (5 x) (2 5) x 3. Which of the following is not a rational number? (1) |
0.09 " (2) 0.9 " 4. The product (a2b)(a3b) is equivalent to (3) 2–3 (4) 0.15 (1) a6b 5. The solution set of (1) {24} (2) a5b 3x 1 7 5 1 2 (2) {24} 6x 2 5 6. The sum of b2 7 and b2 3b is (3) a5b2 (4) a6b2 is (3) {6} (4) {6} (1) b4 4b (2) b4 3b 7 (3) 2b2 4b (4) 2b2 3b 7 7. Two angles are supplementary. If the measure of one angle is 85°, the mea- sure of the other is (1) 5° (2) 85° (3) 95° (4) 180° 8. In trapezoid ABCD, AB i CD. If mA is 75°, then mD is (1) 15° (2) 75° (3) 105° (4) 165° 9. The graph at the right is the solution set of (1) 2 x 3 (2) 2 x 3 (3) 2 x 3 (4) (2 x) or (x 3) –4 –3 –2 –1 0 1 2 3 4 10. When a 1.5, 3a2 a equals (1) 5.25 (2) 5.25 (3) 18.75 (4) 21.75 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The area of a rectangle is (x2 6x 8) square inches and its length is (x 4) inches. Express the width of the rectangle in terms of x. 12. Solve the following equation for x. Cumulative Review 299 Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Mr. Popowich mailed |
two packages. The larger package weighed 12 ounces more than the smaller. If the total weight of the packages was 17 pounds, how much did each package weigh? 14. a. Solve for x in terms of a and b: ax 3b 7. b. Find, to the nearest hundredth, the value of x when a b. 5 " Part IV and 3 " Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Calvin traveled 600 miles, averaging 40 miles per hour for part of the trip and 60 miles per hour for the remainder of the trip. The entire trip took 11 hours. How long did Calvin travel at each rate? 16. A box used for shipping is in the shape of a rectangular prism. The bases are right triangles. The lengths of the sides of the bases are 9.0, 12, and 15 feet. The height of the prism is 4.5 feet. a. Find the surface area of the prism. Express the answer using the correct number of significant digits. b. Find the volume of the prism. Express the answer using the correct number of significant digits. CHAPTER 8 CHAPTER TABLE OF CONTENTS 8-1 The Pythagorean Theorem 8-2 The Tangent Ratio 8-3 Applications of the Tangent Ratio 8-4 The Sine and Cosine Ratios 8-5 Applications of the Sine and Cosine Ratios 8-6 Solving Problems Using Trigonometric Ratios Chapter Summary Vocabulary Review Exercises Cumulative Review 300 TRIGONOMETRY OF THE RIGHT TRIANGLE The accurate measurement of land has been a critical challenge throughout the history of civilization. Today’s land measurement problems are not unlike those George Washington might have solved by using measurements made with a transit, but the modern surveyor has available a total workstation including EDM (electronic distance measuring) and a theodolite for angle measurement. Although modern instruments can perform many measurements and calculations, the surveyor needs to understand the principles of indirect measurement and trigonometry to correctly interpret and apply these results. In this chapter, we will begin the study of a branch of mathematics called trigonometry. The word trigonometry is Greek in origin and means “measurement of triangles.” Although the trig |
onometric functions have applications beyond the study of triangles, in this chapter we will limit the applications to the study of right triangles. 8-1 THE PYTHAGOREAN THEOREM The Pythagorean Theorem 301 The solutions of many problems require the measurement of line segments and angles. When we use a ruler or tape measure to determine the length of a segment, or a protractor to find the measure of an angle, we are taking a direct measurement of the segment or the angle. In many situations, however, it is inconvenient or impossible to make a measurement directly. For example, it is difficult to make the direct measurements needed to answer the following questions: What is the height of a 100-year-old oak tree? What is the width of a river? What is the distance to the sun? We can answer these questions by using methods that involve indirect measurement. Starting with some known lengths of segments or angle measures, we apply a formula or a mathematical relationship to indirectly find the measurement in question. Engineers, surveyors, physicists, and astronomers frequently use these trigonometric methods in their work. B The figure at the left represents a right triangle. Recall that such a triangle, which is contains one and only one right angle. In right triangle ABC, side opposite the right angle, is called the hypotenuse. AB The hypotenuse is the longest side of the triangle. The other two sides of the, form the right angle. They are called the legs of the right and AC BC triangle, triangle. C A More than 2,000 years ago, the Greek mathematician Pythagoras demonstrated the following property of the right triangle, which is called the Pythagorean Theorem: B a c C b A The Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If we represent the length of the hypotenuse of right triangle ABC by c and the lengths of the other two sides by a and b, the Theorem of Pythagoras may be written as the following formula: c2 a2 b2 To show that this relationship is true for any right triangle ABC with the length of the hypotenuse represented by c and the lengths of the legs represented by a and b, consider a square with sides (a b). The area of the square is (a b)2. However, since it is divided into four triangles and one smaller square, its |
area can also be expressed as 302 Trigonometry of the Right Triangle Area of the square area of the four triangles area of the smaller square 4 1 2 ab c2 A Although the area is written in two different ways, both expressions are equal. B Thus, (a b)2 4 1 2 ab c2. A If we simplify, we obtain the relationship of the Pythagorean Theorema b)2 4 ab c2 1 2 A B a2 2ab b2 2ab c2 Expand the binomial term (a b)2. a2 2ab 2ab b2 2ab 2ab c2 Subtract 2ab from both sides of the equality. a2 b2 c2 Statements of the Pythagorean Theorem Two statements can be made for any right triangle where c represents the length of the hypotenuse (the longest side) and a and b represent the lengths of the other two sides. 1. If a triangle is a right triangle, then the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If a triangle is a right triangle, then c2 a2 b2. 2. If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. If c2 a2 b2 in a triangle, then the triangle is a right triangle. If we know the lengths of any two sides of a right triangle, we can find the length of the third side. For example, if the measures of the legs of a right triangle are 7 and 9, we can write: c2 a2 b2 c2 72 92 c2 49 81 c2 130 The Pythagorean Theorem 303 To solve this equation for c, we must do the opposite of squaring, that is, we must find the square root of 130. There are two square roots of 130, and 2 130. 130 There are two things that we must consider here when finding the value of c. which we write as 130 " " " 1 1. Since c represents the length of a line segment, only the positive number is 1 an acceptable value. Therefore, c 130. 2. There is no rational number that has a square of 130. The value of c is an irrational number. However, we usually use a calculator to find a rational approximation for the irrational number. " ENTER: 2nd |
¯ 130 ENTER DISPLAY Therefore, to the nearest tenth, the length of the hypotenuse is 11.4. Note that the calculator gives only the positive rational approximation of the square root of 130. EXAMPLE 1 A ladder is placed 5 feet from the foot of a wall. The top of the ladder reaches a point 12 feet above the ground. Find the length of the ladder. Solution The ladder, the wall, and the ground form a right triangle. The length of the ladder is c, the length of the hypotenuse of the right triangle. The distance from the foot of the ladder to the wall is a 5, and the distance from the ground to the top of the ladder is b 12. Use the Theorem of Pythagoras. c2 a2 b2 c2 52 122 c2 25 144 c2 169 c 6 " 169 5 613 Reject the negative value. Note that in this case the exact value of c is a rational number because 169 is a perfect square. Answer The length of the ladder is 13 feet. c b = 12 a = 5 304 Trigonometry of the Right Triangle EXAMPLE 2 The hypotenuse of a right triangle is 36.0 centimeters long and one leg is 28.5 centimeters long. a. Find the length of the other leg to the nearest tenth of a centimeter. b. Find the area of the triangle using the correct number of significant digits. Solution a. The length of the hypotenuse is c = 36.0 and the length of one leg is a = 28.5. The length of the other leg is b. Substitute the known values in the Pythagorean Theorem. c2 a2 b2 36.02 28.52 b2 1,296 812.25 b2 483.75 b2 b 483.75 " 6 Reject the negative value. Use a calculator to find a rational approximation of the value of b. A calculator displays 21.99431745. Round the answer to the nearest tenth. Answer The length of the other leg is 22.0 centimeters. b. Area of ABC 1 2bh 1 2(28.5)(22.0) 313.5 Since the lengths are given to three significant digits, we will round the area to three significant digits. Answer The area of ABC is 314 square centimeters. EXAMPLE 3 Is a triangle whose sides measure 8 centimeters, 7 centimeters, and 4 centimeters a right triangle? Solution If the triangle is |
a right triangle, the longest side, whose measure is 8, must be the hypotenuse. Then: 82 c2 a2 b2 72 1 42 5? 5? 64 64 65 ✘ 49 1 16 Answer The triangle is not a right triangle. The Pythagorean Theorem 305 EXERCISES Writing About Mathematics 1. A Pythagorean triple is a set of three positive integers that make the equation c2 a2 b2 true. Luz said that 3, 4, and 5 is a Pythagorean triple, and, for any positive integer k, 3k, 4k, and 5k is also a Pythagorean triple. Do you agree with Luz? Explain why or why not. 2. Regina said that if n is a positive integer, 2n 1, 2n2 2n, and 2n2 2n 1 is a Pythagorean triple. Do you agree with Regina? Explain why or why not. Developing Skills In 3–11, c represents the length of the hypotenuse of a right triangle and a and b represent the lengths of the legs. For each right triangle, find the length of the side whose measure is not given. 3. a 3, b 4 6. c 13, a 12 9. a, b 2 " 2 " 4. a 8, b 15 7. c 17, b 15 10. a 1, b 3 " 5. c 10, a 6 8. c 25, b 20 11. a, c 3 8 " In 12–17, c represents the length of the hypotenuse of a right triangle and a and b represent the lengths of the legs. For each right triangle: a. Express the length of the third side in radical form. b. Express the length of the third side to the nearest hundredth. 12. a 2, b 3 15. a 7, b 2 13. a 3, b 3 16. b, c 3 " 14 " 14. a 4, c 8 17. a, c 6 7 " In 18–21, find x in each case and express irrational results in radical form. 18. 19. 20. 21. √ 2 0 0 4x 3x 90° x 90° 8 x 2x 9 x 90° x 6 90° x1 2 In 22–27, find, in each case, the length of the diagonal of a rectangle whose sides have the given measurements. 22. 7 inches by 24 inches 23. 9 centimeters |
by 40 centimeters 24. 28 feet by 45 feet 25. 17 meters by 144 meters 26. 15 yards by 20 yards 27. 18 millimeters by 24 millimeters 306 Trigonometry of the Right Triangle 28. The diagonal of a rectangle measures 65 centimeters. The length of the rectangle is 33 cen- timeters. Consider the measurements to be exact. a. Find the width of the rectangle. b. Find the area of the rectangle. 29. Approximate, to the nearest inch, the length of a rectangle whose diagonal measures 25.0 inches and whose width is 18.0 inches. 30. The altitude to the base of a triangle measures 17.6 meters. The altitude divides the base into two parts that are 12.3 meters and 15.6 meters long. What is the perimeter of the triangle to the nearest tenth of a meter? Applying Skills 31. A ladder 39 feet long leans against a building and reaches the ledge of a window. If the foot of the ladder is 15 feet from the foot of the building, how high is the window ledge above the ground to the nearest foot? 32. Mr. Rizzo placed a ladder so that it reached a window 15.0 feet above the ground when the foot of the ladder was 5.0 feet from the wall. Find the length of the ladder to the nearest tenth of a foot. 33. Mrs. Culkowski traveled 24.0 kilometers north and then 10.0 kilometers east. How far was she from her starting point? BC AB and 34. One day, Ronnie left his home at A and reached his, the sides school at C by walking along of a rectangular open field that was muddy. The dimensions of the field are 1,212 feet by 885 feet. When he was ready to return home, the field was dry and Ronnie decided to take a shortcut by walking diagonally across the field, along nearest whole foot, how much shorter was the trip home than the trip to school?. To the AC D A 885 ft C B 35. Corry and Torry have a two-way communication device that has a range of one-half mile (2,640 feet). Torry lives 3 blocks west and 2 blocks north of Corry. If the length of each block is 600 feet, can Corry and Torry communicate using this device when each is home? Explain your answer. 36. A baseball diamond has the shape of a square with the bases at the vertices of the square. If the |
distance from home plate to first base is 90.0 feet, approximate, to the nearest tenth of a foot, the distance from home plate to second base. 8-2 THE TANGENT RATIO Naming Sides The Tangent Ratio 307 In a right triangle, the hypotenuse, which is the longest side, is opposite the right angle. The other two sides in a right triangle are called the legs. However, in trigonometry of the right triangle, we call these legs the opposite side and the adjacent side to describe their relationship to one of the acute angles in the triangle. Notice that ABC is the same right triangle in both figures below, but the position names we apply to the legs change with respect to the angles. B B h y p o te n u se side opposite ∠A h y p o te n u se side adjacent to ∠B A side adjacent to ∠A C In ABC: BC AC is opposite A; is adjacent to A. A side opposite ∠B C In ABC: AC BC is opposite B; is adjacent to B. Similar Triangles Three right triangles are drawn to coincide at vertex A. Since each triangle contains a right angle as well as A, we know that the third angles of each triangle are congruent. When three angles of one triangle are congruent to the three angles of another, the triangles are similar. The corresponding sides of similar triangles are in proportion. Therefore: CB BA 5 ED DA 5 GF FA G E C A B D F 308 Trigonometry of the Right Triangle The similar triangles shown in the previous page, ABC, ADE, and AFG, are separated and shown below. G E C A B In ABC: CB BA is opposite A; is adjacent to A. A D A F In ADE: ED DA is opposite A; is adjacent to A. In AFG: GF FA is opposite A; is adjacent to A. CB BA 5 ED DA 5 GF FA a constant for ABC, Therefore, ADE, AFG and for any right triangle similar to these triangles, that is, for any right triangle with an acute angle congruent to A. This ratio is called the tangent of the angle. length of side opposite /A length of side adjacent to /A DEFINITION The tangent of an acute angle of a right triangle is the ratio of the length of the side opposite the acute angle to the length of the side adjacent to the acute angle. For right triangle ABC |
, with mC 90, the definition of the tangent of A is as follows: tangent A length of side opposite /A length of side adjacent to /A BC AC 5 a b By using “tan A” as an abbreviation for tangent A, “opp” as an abbreviation for the length of the leg opposite A, and “adj” as an abbreviation for the length of the leg adjacent to A, we can shorten the way we write the relationship given above as follows: B a c A b C tan A opp adj 5 BC AC 5 a b Finding Tangent Ratios on a Calculator The length of each side of equilateral triangle ABD is 2. The altitude B to hypotenuse rule to find BC. from forms two congruent right triangles with AC 1. We can use the AD BC (AC)2 (BC)2 (AB)2 12 (BC)2 22 1 (BC)2 4 (BC)2 3 BC 3 " The measure of each angle of an equilateral triangle is BC 60°. Therefore we can use the lengths of to find the exact value of the tangent of a 60° angle. and AC The Tangent Ratio 309 B 2 2 A 1 C 1 D tan 60° opp adj 5 BC 3 1 5 AC 5 " 3 " But how can we find the constant value of the tangent ratio when the right triangle has an angle of 40° or 76°? Since we want to work with the value of this ratio for any right triangle, no matter what the measures of the acute angles may be, mathematicians have compiled tables of the tangent values for angles with measures from 0° to 90°. Also, a calculator has the ability to display the value of this ratio for any angle. We will use a calculator to determine these values. The measure of an angle can be given in degrees or in radians. In this book, we will always express the measure of an angle in degrees. A graphing calculator can use either radians or degrees. To place the calculator in degree mode,, then use the down arrow and the right arrow keys to highlight press MODE Deg. Press and each time you turn it on. ENTER 2nd QUIT. Your calculator will be in degree mode CASE 1 Given an angle measure, find the tangent ratio. We saw that tan 60° is equal to 3. The calculator will display this value as " an approximate decimal. To find tan 60° |
, enter the sequence of keys shown below. ENTER: TAN 60 ) ENTER DISPLAY, The value given in the calculator display is the rational approximation of " the value of tan 60° that we found using the ratio of the lengths of the legs of a right triangle with a 60° angle. Therefore, to the nearest ten-thousandth, tan 60° 1.7321. 310 Trigonometry of the Right Triangle CASE 2 Given a tangent ratio, find the angle measure. The value of the tangent ratio is different for each different angle measure from 0° to 90°. Therefore, if we know the value of the tangent ratio, we can find the measure of the acute angle that has this tangent ratio.The calculator key used to do this is labeled. We can think of tan1 as “the angle whose tangent is.” Therefore, tan1 (0.9004) can be read as “the angle whose tangent is 0.9004.” and is accessed by first pressing TAN1 2nd To find the measure of A from the calculator, we use the following sequences of keys. ENTER: 2nd TAN1 0.9004 ) ENTER DISPLAY The measure of A to the nearest degree is 42°. EXAMPLE 1 In ABC, C is a right angle, BC = 3, AC = 4, and AB = 5. a. Find: (1) tan A (2) tan B (3) mA to the nearest degree (4) mB to the nearest degree b. Show that the acute angles of the triangle are complementary. Solution a. (1) tan A AC 5 3 4 BC 5 4 (2) tan B 3 Use a calculator to find the measures of A and B. opp adj 5 BC opp adj 5 AC Answer Answer (3) ENTER: 2nd TAN1 3 4 ) ENTER DISPLAY To the nearest degree, mA 37. Answer B 3 C 5 4 A The Tangent Ratio 311 (4) ENTER: 2nd TAN1 4 3 ) ENTER DISPLAY To the nearest degree, mB = 53. Answer b. mA mB 36.869889765 53.13010235 90.000000. Therefore, the acute angles of ABC are complementary. Answer Note: In a right triangle, the tangents of the two acute angles are reciprocals. EXERCISES Writing About Mathematics 1. Explain why the tangent of a 45° |
angle is 1. 2. Use one of the right triangles formed by drawing an altitude of an equilateral triangle to find tan 30°. Express the answer that you find to the nearest hundred-thousandth and compare this result to the valued obtained from a calculator. Developing Skills In 3–6, find: a. tan A b. tan B 3. √18 4. B 3 90° A 3 C A B 5 C 90° 13 12 5. B 6. 5 C 90° √61 6 A B A k 90° C p t 7. In ABC, mC 90, AC 6, and AB 10. Find tan A. 8. In RST, mT 90, RS 13, and ST 12. Find tan S. In 9–16, use a calculator to find each of the following to the nearest ten-thousandth: 9. tan 10° 13. tan 1° 10. tan 25° 14. tan 89° 11. tan 70° 15. tan 36° 12. tan 55° 16. tan 67° In 17–28, in each of the following, use a calculator to find the measure of A to the nearest degree. 17. tan A 0.0875 20. tan A 1.0000 18. tan A 0.3640 21. tan A 2.0503 19. tan A 0.5543 22. tan A 3.0777 312 Trigonometry of the Right Triangle 23. tan A 0.3754 26. tan A 0.3500 24. tan A 0.7654 27. tan A 0.1450 25. tan A 1.8000 28. tan A 2.9850 29. Does the tangent of an angle increase or decrease as the degree measure of the angle increases from 1° to 89°? 30. a. Use a calculator to find tan 20° and tan 40° to the nearest ten-thousandth. b. Is the tangent of the angle doubled when the measure of the angle is doubled? Applying Skills 31. In ABC, mC = 90, AC = 6, and BC = 6. a. Find tan A. b. Find the measure of A. 32. In ABC, mC = 90, BC = 4, and AC = 9. a. Find tan A. b. Find the measure of A to the nearest degree. c. Find tan B. d. Find the measure of B to the nearest degree. 33. |
In rectangle ABCD, AB 10 and BC 5. a. Find tan CAB. b. Find the measure of CAB to the nearest degree. c. Find tan CAD. d. Find the measure of CAD to the nearest degree. 34. In ABC, C is a right angle, mA = 45, AC = 4, BC = 4, and AB a. Using the given lengths, write the ratio for tan A. 4 " 2. b. Use a calculator to find tan 45°. 35. In RST, T is a right angle and r, s, and t are lengths of sides. Using these lengths: a. Write the ratio for tan R. b. Write the ratio for tan S. c. Use parts a and b to find the numerical value of the product R (tan R)(tan S). B 4 C S r T 4√2 45° 4 A t s Applications of the Tangent Ratio 313 8-3 APPLICATIONS OF THE TANGENT RATIO The tangent ratio is often used to make indirect measurements when the measures of a leg and an acute angle of a right triangle are known. Angle of Elevation and Angle of Depression is the line of sight and When a telescope or some similar instrument is used to sight the top of a telephone pole, the instrument is elevated (tilted upward) from a horizontal position. Here, g OT is the horizontal line. The angle of elevation is the angle determined by the rays that are parts of the horizontal line and the line of sight when looking upward. Here, TOA is the angle of elevation. g OA When an instrument is used to sight a boat from a cliff, the instrument is depressed (tilted downward) g OB g OH from a horizontal position. Here, is the line of sight is the horizontal line. The angle of depression and is the angle determined by the rays that are parts of the horizontal line and of the line of sight when looking downward. Here, HOB is the angle of depression. T line of sight angle of elevation horizontal line A horizontal line O angle of depression line of sight O H B A g BA g BO Note that, if is a horizontal line and is the line of sight from the boat to the top of the cliff, ABO is called the angle of elevation from the boat to the top of the cliff. Since is a transversal, alternate interior and angles are congruent, namely, HOB ABO. Thus, the angle of elevation |
measured from B to O is congruent to the angle of depression measured from O to B. g HO g BA g OB 7 Using the Tangent Ratio to Solve Problems Procedure To solve a problem by using the tangent ratio: 1. For the given problem, make a diagram that includes a right triangle. Label the known measures of the sides and angles. Identify the unknown quantity by a variable. 2. If for the right triangle either (1) the lengths of two legs or (2) the length of one leg and the measure of one acute angle are known, write a formula for the tangent of an acute angle. 3. Substitute known values in the formula and solve the resulting equation for the unknown value. 314 Trigonometry of the Right Triangle EXAMPLE 1 Find to the nearest degree the measure of the angle of elevation of the sun when a vertical pole 6.5 meters high casts a shadow 8.3 meters long. Solution The angle of elevation of the sun is the same as A, the angle of elevation to the top of the pole from A, the tip of the shadow. Since the vertical pole and the shadow are the legs of a right triangle opposite and adjacent to A, use the tangent ratio. B 6.5 m tan A opp adj 5 BC AC 5 6.5 8.3 A 8.3 m C ENTER: 2nd TAN1 6.5 8.3 ) ENTER DISPLAY Answer To the nearest degree, the measure of the angle of elevation of the sun is 38°. EXAMPLE 2 At a point on the ground 39 meters from the foot of a tree, the measure of the angle of elevation of the top of the tree is 42°. Find the height of the tree to the nearest meter. Solution Let T be the top of the tree, A be the foot of the tree, and B be the point on the ground 39 meters from A. Draw ABT, and label the diagram: mB 42, AB 39. Let x height of tree (AT). The height of the tree is the length of the perpendicular from the top of the tree to the ground. Since the problem involves the measure of an acute angle and the measures of the legs of a right triangle, use the tangent ratio: T x tan B tan B tan 42° opp adj AT BA x 39 42° B 39 m A Substitute the given values. x 39 tan 42° Use a calculator for the computation: x 35.11575773 Sol |
ve for x. Answer To the nearest meter, the height of the tree is 35 meters. Applications of the Tangent Ratio 315 EXAMPLE 3 From the top of a lighthouse 165 feet above sea level, the measure of the angle of depression of a boat at sea is 35.0°. Find to the nearest foot the distance from the boat to the foot of the lighthouse. Solution Let L be the top of the lighthouse, LA be the length of the perpendicular from L to sea level, and B be the position of the boat. Draw right triangle ABL, and h LH draw, the horizontal line through L. Since HLB is the angle of depression, mHLB = 35.0, mLBA 35.0, and mBLA = 90 35.0 55.0. Let x distance from the boat to H B L 35.0° 55.0° 165 ft x A the foot of the lighthouse (BA). METHOD 1 METHOD 2 is the opposite is the adjacent side. BA Using BLA, LA side and Form the tangent ratio: tan BLA tan 55.0° BA LA x 165 x 165 tan 55.0° Use a calculator to perform the computation. The display will read 235.644421. is the opposite is the adjacent side. LA Using LBA, BA side and Form the tangent ratio: tan LBA tan 35.0° x tan 35.0° 165 LA BA 165 x x 165 tan 35.0o Use a calculator to perform the computation. The display will read 235.644421. Answer To the nearest foot, the boat is 236 feet from the foot of the lighthouse. EXERCISES Writing About Mathematics 1. Zack is solving a problem in which the measure of the angle of depression from the top of a building to a point 85 feet from the foot of the building is 64°. To find the height of the building, Zack draws the diagram shown at the right. Explain why Zack’s diagram is incorrect. 64° h 85 ft 316 Trigonometry of the Right Triangle 2. Explain why the angle of elevation from point A to point B is always congruent to the angle of depression from point B to point A. Developing Skills In 3–11, in each given triangle, find the length of the side marked x to the nearest foot or the measure of the angle marked x to the nearest degree. 3. 6. 9. x 42° 25 ft x 4. 7. 65 |
° 13 ft 10 ft 50 ft 55° x 6.0 ft 9.0 ft x 10. 60° x 68° x 20 ft 5. 40° 18 ft x 8. 12 ft 24 ft x 11. 8.0 ft x 8.0 ft Applying Skills 12. At a point on the ground 52 meters from the foot of a tree, the measure of the angle of ele- vation of the top of the tree is 48°. Find the height of the tree to the nearest meter. 13. A ladder is leaning against a wall. The foot of the ladder is 6.25 feet from the wall. The ladder makes an angle of 74.5° with the level ground. How high on the wall does the ladder reach? Round the answer to the nearest tenth of a foot. The Sine and Cosine Ratios 317 14. From a point, A, on the ground that is 938 feet from the foot, C, of the Empire State Building, the angle of elevation of the top, B, of the building has a measure of 57.5°. Find the height of the building to the nearest ten feet. 15. Find to the nearest meter the height of a building if its shadow is 18 meters long when the angle of elevation of the sun has a measure of 38°. 16. From the top of a lighthouse 50.0 meters high, the angle of depression of a boat out at sea has a measure of 15.0°. Find, to the nearest meter, the distance from the boat to the foot of the lighthouse, which is at sea level. 17. From the top of a school 61 feet high, the measure of the angle of depression to the road in front of the school is 38°. Find to the nearest foot the distance from the road to the school. 18. Find to the nearest degree the measure of the angle of elevation of the sun when a student 170 centimeters tall casts a shadow 170 centimeters long. 19. Find to the nearest degree the measure of the angle of elevation of the sun when a woman 150 centimeters tall casts a shadow 43 centimeters long. 20. A ladder leans against a building. The top of the ladder reaches a point on the building that is 18 feet above the ground. The foot of the ladder is 7.0 feet from the building. Find to the nearest degree the measure of the angle that the ladder makes with the level ground. 21. In any rhombus, the diagonals are perpendicular to each |
other and bisect each other. In BD rhombus ABCD, diagonals sure of each angle to the nearest degree. a. mBCM b. mMBC and AC meet at M. If BD 14 and AC 20, find the mea- c. mABC d. mBCD 8-4 THE SINE AND COSINE RATIOS Since the tangent is the ratio of the lengths of the two legs of a right triangle, it is not directly useful in solving problems in which the hypotenuse is involved. In trigonometry of the right triangle, two ratios that involve the hypotenuse are called the sine of an angle and the cosine of an angle. D F As in our discussion of the tangent ratio, we recognize that the figure at the right shows three similar triangles. Therefore, the ratios of corresponding sides are equal. B A C E G The Sine Ratio From the figure, we see that BC AB 5 DE This ratio is called the sine of A. AD 5 FG AF a constant 318 Trigonometry of the Right Triangle B a c A b C DEFINITION The sine of an acute angle of a right triangle is the ratio of the length of the side opposite the acute angle to the length of the hypotenuse. In right triangle ABC, with mC 90, the definition of the sine of A is: sine A length of side opposite /A length of hypotenuse BC AB 5 a c By using “sin A” as an abbreviation for sine A, “opp” as an abbreviation for the length of the leg opposite A, and “hyp” as an abbreviation for the length of the hypotenuse, we can shorten the way we write the definition of sine A as follows: sin A opp hyp BC AB 5 a c The Cosine Ratio From the preceding figure on page 317, which shows similar triangles, ABC, ADE, and AFG, we see that AC AB 5 AE a constant. AD 5 AG AF This ratio is called the cosine of /A. DEFINITION The cosine of an acute angle of a right triangle is the ratio of the length of the side adjacent to the acute angle to the length of the hypotenuse. In right triangle ABC, with mC 90, the definition of the cosine of A is: cosine A length of side adjacent to /A length of hypotenuse AC AB 5 |
b c By using “cos A” as an abbreviation for cosine A, “adj” as an abbreviation for the length of the leg adjacent to A, and “hyp” as an abbreviation for the length of the hypotenuse, we can shorten the way we write the definition of cosine A as follows: B a c A b C cos A adj hyp AC AB 5 b c The Sine and Cosine Ratios 319 Finding Sine and Cosine Ratios on a Calculator CASE 1 Given an angle measure, find the sine or cosine ratio. On a calculator we use the keys labeled to display the values of the sine and cosine of an angle. The sequence of keys that a calculator requires for tangent will be the same as the sequence for sine or cosine. For example, to find sin 50° and cos 50°, we use the following: and COS SIN ENTER: SIN 50 ) ENTER ENTER: COS 50 ) ENTER DISPLAY: s i n ( 5 0 ) DISPLAY CASE 2 Given a sine or cosine ratio, find the angle measure. A calculator will also find the measure of A when sin A or cos A is given.. These are the second func- To do this we use the keys labeled COS1 and SIN1 SIN COS and tions of. We can think of the meaning of sin1 as “the angle whose sine is.” Therefore, if sin A 0.2588, then sin1(0.2588) can be read as “the angle whose sine is 0.2588.” To find the measure of A from the calculator, we use the following and are accessed by first pressing 2nd sequences of keys: ENTER: 2nd SIN1 0.2588 ) ENTER DISPLAY The measure of A to the nearest degree is 15°. 320 Trigonometry of the Right Triangle EXAMPLE 1 In ABC, C is a right angle, BC 7, AC 24, and AB 25. Find: a. sin A b. cos A c. sin B d. cos B e. mB, to the nearest degree Solution a. sin A b. cos A c. sin B d. cos B Answers opp hyp 5 BC adj hyp 5 AC opp hyp 5 AC adj hyp 5 BC AB 5 7 25 AB 5 24 25 AB 5 24 25 AB 5 7 25 25 24 A B 7 |
C e. Use a calculator. Start with the ratio in part c and use SIN1 or start with the ratio in part d and use COS1. METHOD 1 sin B = 24 25 METHOD 2 cos B = 7 25 ENTER: 2nd SIN1 24 25 ENTER: 2nd COS1 7 25 ) ENTER ) ENTER DISPLAY ) DISPLAY mB 74 to the nearest degree. Answer EXERCISES Writing About Mathematics 1. If A and B are the acute angles of right triangle ABC, show that sin A cos B. 2. If A is an acute angle of right triangle ABC, explain why it is always true that sin A 1 and cos A 1. The Sine and Cosine Ratios 321 Developing Skills In 3–6, find: a. sin A b. cos A c. sin B d. cos B 3. B 4. B 5. A 6. C 10 A 8 6 90° C 5 C 90° 13 12 29 A 20 A p B r 90° k 7. In ABC, mC 90, AC 4, and BC 3. Find sin A. 8. In RST, mS 90, RS 5, and ST 12. Find cos T. B 21 90° C In 9–20, for each of the following, use a calculator to find the trigonometric function value to the nearest ten-thousandth. 9. sin 18° 13. sin 1° 17. cos 40° 10. sin 42° 14. sin 89° 18. cos 59° 11. sin 58° 15. cos 21° 19. cos 74° 12. sin 76° 16. cos 35° 20. cos 88° In 21–38, for each of the following, use a calculator to find the measure of A to the nearest degree. 21. sin A 0.1908 24. cos A 0.9397 27. sin A 0.8910 30. sin A 0.1900 33. cos A 0.8545 36. cos A 0.2968 22. sin A 0.8387 25. cos A 0.0698 28. sin A 0.9986 31. cos A 0.9750 34. sin A 0.5800 37. sin A 0.1275 23. sin A 0.3420 26. cos A 0.8910 29. cos A 0.9986 32. sin A 0.8740 35. cos A 0.5934 38. |
cos A 0.8695 39. a. Use a calculator to find sin 25° and sin 50°. b. If the measure of an angle is doubled, is the sine of the angle also doubled? 40. a. Use a calculator to find cos 25° and cos 50°. b. If the measure of an angle is doubled, is the cosine of the angle also doubled? 41. As an angle increases in measure from 1° to 89°: a. Does the sine of the angle increase or decrease? b. Does the cosine of the angle increase or decrease? 322 Trigonometry of the Right Triangle In 42 and 43, complete each sentence by replacing? with a degree measure that makes the sentence true. 42. a. sin 70° cos? b. sin 23° cos? c. sin 38° cos? d. sin x° cos? Applying Skills 43. a. cos 50° sin? b. cos 17° sin? c. cos 82° sin? d. cos x° sin? 44. In ABC, mC 90, BC = 20, and BA 40. a. Find sin A. b. Find the measure of A. 45. In ABC, mC 90, AC 40, and AB 80. a. Find cos A. b. Find the measure of A. 46. In ABC, C is a right angle, AC 8, BC 15, and AB 17. Find: a. sin A e. the measure of A to the nearest degree f. the measure of B to the nearest degree b. cos A c. sin B d. cos B 47. In RST, mT 90, ST 11, RT 60, and RS 61. Find: a. sin R e. the measure of R to the nearest degree f. the measure of S to the nearest degree b. cos R c. sin S d. cos S 48. In ABC, C is a right angle, AC 1.0, BC 2.4, and AB 2.6. Find: d. cos B c. sin B a. sin A e. the measure of A to the nearest degree. f. the measure of B to the nearest degree. b. cos A 49. In rectangle ABCD, AB 3.5 and CB 1.2. Find: a. sin ABD e. the measure of ABD to the nearest degree. f. the measure of CBD to the nearest degree. b. cos AB |
D c. sin CBD d. cos CBD 50. In right triangle ABC, C is the right angle, BC 1, AC 3 " a. Using the given lengths, write the ratios for sin A and cos A. and AB 2. b. Use a calculator to find sin 30° and cos 30°. c. What differences, if any, exist between the answers to parts a and b? 51. In ABC, mC 90 and sin A cos A. Find mA. 8-5 APPLICATIONS OF THE SINE AND COSINE RATIOS Applications of the Sine and Cosine Ratios 323 Since the sine and cosine ratios each have the length of the hypotenuse of a right triangle as the second term of the ratio, we can use these ratios to solve problems in the following cases: 1. We know the length of one leg and the measure of one acute angle and want to find the length of the hypotenuse. 2. We know the length of the hypotenuse and the measure of one acute angle and want to find the length of a leg. 3. We know the lengths of the hypotenuse and one leg and want to find the measure of an acute angle. EXAMPLE 1 While flying a kite, Betty lets out 322 feet of string. When the string is secured to the ground, it makes an angle of 38.0° with the ground. To the nearest foot, what is the height of the kite above the ground? (Assume that the string is stretched so that it is straight.) Solution Let K be the position of the kite in the air, B be the point on the ground at which the end of the string is secured, and G be the point on the ground directly below the kite, as shown in the diagram. The height of the kite is the length of the perpendicular from the ground to the kite. Therefore, mG 90, mB 38.0, and the length of the string BK 322 feet. K x G 322 ft 38.0° B Let x KG, the height of the kite. We know the length of the hypotenuse and the measure of one acute angle and want to find the length of a leg, KG. opp In Method 1 below, since leg KG is opposite B, we can use sin B. hyp In Method 2 below, since leg KG is adjacent to K, we can use cos K adj hyp with m |
K 90 38.0 52.0. METHOD 1 METHOD 2 sin B sin 38.0° KG BK x 322 cos K cos 52.0° KG BK x 322 x 322 sin 38.0° x 198.242995 x 322 cos 52.0° x 198.242995 Write the ratio: Substitute the given values: Solve for x: Compute using a calculator: Answer The height of the kite to the nearest foot is 198 feet. 324 Trigonometry of the Right Triangle EXAMPLE 2 A wire reaches from the top of a pole to a stake in the ground 3.5 meters from the foot of the pole. The wire makes an angle of 65° with the ground. Find to the nearest tenth of a meter the length of the wire. Solution In BTS, B is a right angle, BS 3.5, mS 65, and mT 90 65 25. Let x ST, the length of the wire. Since we know the length of one leg and the measure of one acute angle and want to find the length of the hypotenuse, we can use either the sine or the cosine ratio. METHOD 1 BS ST cos S adj hyp 3.5 cos 65° x x cos 65° 3.5 3.5 x cos 65o x 8.281705541 METHOD 2 sin T opp hyp 5 BS ST 3.5 sin 25° x x sin 25° 3.5 3.5 x sin 25o x 8.281705541 T x 65° 3.5 m S B Answer The wire is 8.3 meters long to the nearest tenth of a meter. EXAMPLE 3 A ladder 25 feet long leans against a building and reaches a point 23.5 feet above the ground. Find to the nearest degree the angle that the ladder makes with the ground. Solution In right triangle ABC, AB, the length of the hypotenuse is 25 feet and BC, the side opposite A, is 23.5 feet. Since the problem involves A, and (the hypotenuse), the sine ratio is used. opp hyp 5 23.5 25 sin A AB BC (its opposite side), B 25.0 ft 23.5 ft ENTER: 2nd SIN1 23.5 25 ) ENTER A C DISPLAY Answer To the nearest degree, the measure of the angle is 70°. Applications of the Sine and Cosine Ratios 325 EXERCISES |
Writing About Mathematics 1. Brittany said that for all acute angles, A, (tan A)(cos A) sin A. Do you agree with Brittany? Explain why or why not. 2. Pearl said that as the measure of an acute angle increases from 1° to 89°, the sine of the angle increases and the cosine of the angle decreases. Therefore, cos A is the reciprocal of sin A. Do you agree with Pearl? Explain why or why not. In 3–11, find to the nearest centimeter the length of the side marked x. x 4. x 124 cm 65.0° 5. x 55.0° 1 4 3 c m 7. 10. x 38° 45 cm 5 c m 2 x 71° 8. 11. 15 cm x 55° 61° 32 cm 40.0° x 15. 12 ft x 18 ft In 12–15, find to the nearest degree the measure of the angle marked x. 12. 13. 10.5 ft x 8.0 ft Applying Skills 24 ft x 12 ft 14. 15 ft x 21 ft 16. A wooden beam 6.0 meters long leans against a wall and makes an angle of 71° with the ground. Find to the nearest tenth of a meter how high up the wall the beam reaches. 3. 6. 9. 22 c m 40° 3 1 c m 38° x x 45° 15 cm 326 Trigonometry of the Right Triangle 17. A boy flying a kite lets out 392 feet of string, which makes an angle of 52° with the ground. Assuming that the string is tied to the ground, find to the nearest foot how high the kite is above the ground. 18. A ladder that leans against a building makes an angle of 75° with the ground and reaches a point on the building 9.7 meters above the ground. Find to the nearest meter the length of the ladder. 19. From an airplane that is flying at an altitude of 3,500 feet, the angle of depression of an airport ground signal measures 27°. Find to the nearest hundred feet the distance between the airplane and the airport signal. 20. A 22-foot pole that is leaning against a wall reaches a point that is 18 feet above the ground. Find to the nearest degree the measure of the angle that the pole makes with the ground. 21. To reach the top of a hill that is 1.0 kilometer high, one must travel 8.0 kilometers up a straight road that leads |
to the top. Find to the nearest degree the measure of the angle that the road makes with the horizontal. 22. A 25-foot ladder leans against a building and makes an angle of 72° with the ground. Find to the nearest foot the distance between the foot of the ladder and the building. 23. A wire 2.4 meters in length is attached from the top of a post to a stake in the ground. The measure of the angle that the wire makes with the ground is 35°. Find to the nearest tenth of a meter the distance from the stake to the foot of the post. 24. An airplane rises at an angle of 14° with the ground. Find to the nearest hundred feet the distance the airplane has flown when it has covered a horizontal distance of 1,500 feet. 25. A kite string makes an angle of 43° with the ground. The string is staked to a point 104 meters from a point on the ground directly below the kite. Find to the nearest meter the length of the kite string, which is stretched taut. 26. The top of a 43-foot ladder touches a point on the wall that is 36 feet above the ground. Find to the nearest degree the measure of the angle that the ladder makes with the wall. 27. In a park, a slide 9.1 feet long is perpendicular to the ladder to the top of the slide. The distance from the foot of the ladder to the bottom of the slide is 10.1 feet. Find to the nearest degree the measure of the angle that the slide makes with the horizontal. 9.1 ft 10.1 ft Solving Problems Using Trigonometric Ratios 327 28. A playground has the shape of an isosceles trapezoid ABCD. The length of the shorter base, CD, is 185 feet. The length of each of the equal sides is 115 feet and mA 65.0. a. Find DE, the length of the altitude from D, to the nearest foot. b. Find AE, to the nearest tenth of a foot. c. Find AB, to the nearest foot. d. What is the area of the playground to the nearest hundred square feet? e. What is the perimeter of the playground? 29. What is the area of a rhombus, to the nearest ten square feet, if the measure of one side is 43.7 centimeters and the measure of one angle is 78.0°? 30. A roofer wants to reach the roof |
of a house that is 21 feet above the ground. The measure of the steepest angle that a ladder can make with the house when it is placed directly under the roof is 27°. Find the length of the shortest ladder that can be used to reach the roof, to the nearest foot. 8-6 SOLVING PROBLEMS USING TRIGONOMETRIC RATIOS When the conditions of a problem can be modeled by a right triangle for which the measures of one side and an acute angle or of two sides are known, the trigonometric ratios can be used to find the measure of another side or of an acute angle. Procedure To solve a problem by using trigonometric ratios: 1. Draw the right triangle described in the problem. 2. Label the sides and angles with the given values. 3. Assign a variable to represent the measure to be determined. 4. Select the appropriate trigonometric ratio. 5. Substitute in the trigonometric ratio, and solve the resulting equation. sin A Given ABC with mC 90: opp hyp 5 BC adj hyp 5 AC opp adj 5 BC AB 5 a c AB 5 b c AC 5 a b tan A cos A c b A B a C sin B opp hyp 5 AC adj hyp 5 BC adj 5 AC AB 5 b c AB 5 a c BC 5 b a cos B tan B opp 328 Trigonometry of the Right Triangle EXAMPLE 1 Given: In isosceles triangle ABC, AC CB 20 and mA mB 68. is an altitude. C CD Find: a. Length of altitude CD to the nearest tenth. b. Length of AB to the nearest tenth. Solution a. In right BDC, sin B CD CB Let x CD. 20 x 20 68° A D 68° y B sin 68° x 20 x 20 sin 68° 18.54367709 18.5 b. Since the altitude drawn to the base of an isosceles triangle bisects the base, AB 2DB. Therefore, find DB in BDC and double it to find AB. In right BDC, cos B = Let y DB. DB CB. cos 68° y 20 y 20 cos 68° 7.492131868 AB 2y 2(7.492131868) 14.9843736 Answers a. CD 18.5, to the nearest tenth. b. AB 15.0, to the nearest tenth. EXERCISES Writing About Mathematics 1. If |
the measures of two sides of a right triangle are given, it is possible to find the measures of the third side and of the acute angles. Explain how you would find these measures. 2. If the measures of the acute angles of a right triangle are given, is it possible to find the measures of the sides? Explain why or why not. Solving Problems Using Trigonometric Ratios 329 Developing Skills In 3–10: In each given right triangle, find to the nearest foot the length of the side marked x; or find to the nearest degree the measure of the angle marked x. Assume that each measure is given to the nearest foot or to the nearest degree. 3. 7. x 37° 22 ft x 18 ft 12 ft 4. 8. x 33 ft 60° 12 ft x 15 ft 5. x 41° 25 ft 9. 8.0 ft 24 ft x 6. 10. 35 ft 65° x x 1 ° 5 3 9 ft BD is the altitude to AC. Find BD to the nearest tenth. 11. In ABC, mA 42, AB 14, and 12. In ABC, AC > BC, mA 50, and AB 30. Find to the nearest tenth the length of the altitude from vertex C. 13. The legs of a right triangle measure 84 and 13. Find to the nearest degree the measure of the smallest angle of this triangle. 14. The length of hypotenuse number of degrees in B. AB of right triangle ABC is twice the length of leg BC. Find the 15. The longer side of a rectangle measures 10, and a diagonal makes an angle of 27° with this side. Find to the nearest integer the length of the shorter side. 16. In rectangle ABCD, diagonal degree the measure of CAB. AC measures 11 and side AB measures 7. Find to the nearest 17. In right triangle ABC, CD is the altitude to hypotenuse AB, AB 25, and AC 20. Find lengths AD, DB, and CD to the nearest integer and the measure of B to the nearest degree. 18. The lengths of the diagonals of a rhombus are 10 and 24. a. Find the perimeter of the rhombus. b. Find to the nearest degree the measure of the angle that the longer diagonal makes with a side of the rhombus. 19. The altitude to the hypotenuse of a right triangle ABC divides the hypotenuse into segments whose measures are 9 and 4. The measure of the |
altitude is 6. Find to the nearest degree the measure of the smaller acute angle of ABC. 330 Trigonometry of the Right Triangle 20. In ABC, AB = 30, mB = 42, mC 36, and AD is an altitude. a. Find to the nearest integer the length of AD. b. Using the result of part a, find to the nearest integer the length of DC. 21. Angle D in quadrilateral ABCD is a right angle, is perpendicular to AC and diagonal mB 35, and mDAC = 65. a. Find AC to the nearest integer. BC, BC = 20, b. Using the result of part a, find DC to the nearest integer. A D 65° C 36° C A D 3 0 42° B B 35° 20 22. The diagonals of a rectangle each measure 198 and intersect at an angle whose measure is 110°. Find to the nearest integer the length and width of the rectangle. Hint: The diagonals of a rectangle bisect each other. 23. In rhombus ABCD, the measure of diagonal AC is 80 and mBAC = 42. a. Find to the nearest integer the length of diagonal BD. b. Find to the nearest integer the length of a side of the rhombus. 24. In right triangle ABC, the length of hypotenuse AB is 100 and mA 18. a. Find AC and BC to the nearest integer. b. Show that the results of part a are approximately correct by using the relationship (AB)2 = (AC)2 + (BC)2. Applying Skills 25. Find to the nearest meter the height of a church spire that casts a shadow of 53.0 meters when the angle of elevation of the sun measures 68.0°. 26. From the top of a lighthouse 194 feet high, the angle of depression of a boat out at sea measures 34.0°. Find to the nearest foot the distance from the boat to the foot of the lighthouse. 27. A straight road to the top of a hill is 2,500 meters long and makes an angle of 12° with the horizontal. Find to the nearest ten meters the height of the hill. 28. A wire attached to the top of a pole reaches a stake in the ground 21 feet from the foot of the pole and makes an angle of 58° with the ground. Find to the nearest foot the length of the wire. 29. An airplane climbs at an |
angle of 11° with the ground. Find to the nearest hundred feet the distance the airplane has traveled when it has attained an altitude of 450 feet. 30. Find to the nearest degree the measure of the angle of elevation of the sun if a child 88 cen- timeters tall casts a shadow 180 centimeters long. 31. CD and AB represent cliffs on opposite sides of a river 125 meters wide. From B, the angle of elevation of D measures 20° and the angle of depression of C measures 25°. Find to the nearest meter: a. the height of the cliff represented by AB. b. the height of the cliff represented by CD. B A 20.0° 25.0° 125 m 32. Points A, B, and D are on level ground. CD represents the height of a building, BD = 86 feet, and mD 90. At B, the angle of elevation of the top of the building, CBD, measures 49°. At A, the angle of elevation of the top of the building, CAD, measures 26°. a. Find the height of the building, CD, to the nearest foot. b. Find AD to the nearest foot. 26° 49° A B 86 ft D Chapter Summary 331 D E C C CHAPTER SUMMARY The Pythagorean Theorem relates the lengths of the sides of a right triangle. If the lengths of the legs of a right triangle are a and b, and the length of the hypotenuse is c, then c2 a2 b2. The trigonometric functions associate the measure of each acute angle A with a number that is the ratio of the measures of two sides of a right triangle. The three most commonly used trigonometric functions are sine, cosine, and tangent. In the application of trigonometry to the right triangle, these ratios are defined as follows: sin A opp hyp cos A adj hyp tan A opp adj BC In right triangle ABC, with hypotenuse is opposite A and adjacent to B; is opposite B and adjacent to A. AC AB • • sin A cos A tan A BC AB AC AB BC AC BC cos B AB AC sin B AB tan B AC BC B C A 332 Trigonometry of the Right Triangle An angle of elevation, GDE in the diagram, is an angle between a horizontal line and a line of sight to an object at a higher elevation. An angle of depression, FED in the diagram, is an angle between a horizontal line and a |
line of sight to an object at a lower elevation. Angle of depression F D E G Angle of elevation VOCABULARY 8-1 Trigonometry • Direct measurement • Indirect measurement • Pythagorean Theorem • Pythagorean triple 8-2 Opposite side • Adjacent side • Similar • Tangent of an acute angle of a right triangle 8-3 Angle of elevation • Angle of depression 8-4 Sine of an acute angle of a right triangle • Cosine of an acute angle of a right triangle REVIEW EXERCISES 1. Talia’s calculator is not functioning properly and does not give the correct value when she uses the calculator are operating correctly. TAN key. Assume that all other keys of the a. Explain how Talia can find the measure of the leg ABC when BC 4.5 and mA 43. AC of right triangle b. Explain how Talia can use her calculator to find the tangent of any acute angle, given the measure of one side of a right triangle and the measure of an acute angle as in part a. 2. Jill made the following entry on her calculator: ENTER: 2nd SIN1 1.5 ) ENTER Explain why the calculator displayed an error message. In 3–8, refer to RST and express the value of each ratio as a fraction. 3. sin R 5. sin T 7. cos T 4. tan T 6. cos R 8. tan R 17 15 R T 8 S In 9–12: in each given triangle, find to the nearest centimeter the length of the side marked x. Assume that each given length is correct to the nearest centimeter. Review Exercises 333 40 c m 42° 9. 11. 50 cm x x 54° 18 cm 35° 10. x 12. 24° x 41 cm 13. If cos A sin 30° and 0° A 90°, what is the measure of A? 14. In right triangle ACB, mC 90, mA 66, and AC 100. Find BC to the nearest integer. 15. In right triangle ABC, mC 90, mB 28, and BC 30. Find AB to the nearest integer. 16. In ABC, mC 90, tan A 0.7, and AC 40. Find BC. 17. In ABC, mC 90, AB 30, and BC 15. What is the measure, in degrees, of A? 18. In ABC, mC 90, |
BC 5, and AC 9. Find to the nearest degree the measure of A. 19. Find to the nearest meter the height of a building if its shadow is 42 meters long when the angle of elevation of the sun measures 42°. 20. A 5-foot wire attached to the top of a tent pole reaches a stake in the ground 3 feet from the foot of the pole. Find to the nearest degree the measure of the angle made by the wire with the ground. 21. While flying a kite, Doris let out 425 feet of string. Assuming that the string is stretched taut and makes an angle of 48° with the ground, find to the nearest ten feet how high the kite is. 22. A rectangular field ABCD is crossed by a path from A to C. If mBAC 62 and BC 84 yards, find to the nearest yard: a. the width of the field, AB. b. the length of path, AC. 334 Trigonometry of the Right Triangle 23. Find the length of a leg of an isosceles right triangle if the length of the hypotenuse is 72. " 24. The measure of each of the base angles of an isosceles triangle is 15 degrees more than twice the measure of the vertex angle. a. Find the measure of each angle of the triangle. b. Find to the nearest tenth of a centimeter the measure of each of the equal sides of the triangle if the measure of the altitude to the base is 88.0 centimeters. c. Find to the nearest tenth of a centimeter the measure of the base of the triangle. d. Find the perimeter of the triangle. e. Find the area of the triangle. 25. ABCD is a rectangle with E a point on BC. AB 12, BE 5, and EC 9. a. Find the perimeter of triangle AED. b. Find the area of triangle AED. /CDE. c. Find the measure of d. Find the measure of BAE. e. Find the measure of AED. Exploration A regular polygon with n sides can be divided into n congruent isosceles triangles. a. Express, in terms of n, the measure of the vertex angle of one of the isosceles triangles. b. Express, in terms of n, the measure of a base angle of one of the isosceles triangles. c. Let s be the measure of a side of the regular polygon. Express |
, in terms of n and s, the measure of the altitude to the base of one of the isosceles triangles. d. Express the area of one of the isosceles triangles in terms of n and s. e. Write a formula for the area of a regular polygon in terms of the measure of a side, s, and the number of sides, n. CUMULATIVE REVIEW CHAPTERS 1–8 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following does not represent a real number when x 3? Cumulative Review 335 3 x (1) x 2 3 x 2. The coordinates of one point on the x-axis are (2) (3) x x (1) (1, 1) (2) (1, 1) (3) (1, 0) 3. The expression 0.2a2(10a3 2a) is equivalent to (1) 2a5 0.4a3 (2) 20a5 4a3 (3) 2a6 0.4a2 (4) 20a6 0.4a3 4. If 7 3, then x equals 1 4x 3 4x (1) 20 (2) 10 (3) 5 5. If sin A 0.3751, then, to the nearest degree, mA is (1) 21 (2) 22 (3) 68 6. Which of the following statements is false? (1) If a polygon is a square, then it is a parallelogram. (2) If a polygon is a square, then it is a rhombus. (3) If a polygon is a rectangle, then it is a parallelogram. (4) If a polygon is a rectangle, then it is a rhombus. (4) x x 2 3 (4) (0, 1) (4) 5 2 (4) 69 7. If the measures of two legs of a right triangle are 7.0 feet and 8.0 feet, then, to the nearest tenth of a foot, the length of the hypotenuse is (4) 48.9 (1) 10.6 (2) 15.0 (3) 41.2 8. The measure of the radius of a cylinder is 9.00 centimeters and the measure of its height is 24. |
00 centimeters. The surface area of the cylinder to the correct number of significant digits is (1) 1,610 square centimeters (2) 1,620 square centimeters (3) 1,860 square centimeters (4) 1,870 square centimeters 9. When 5b2 2b is subtracted from 8b the difference is (3) 5b2 10b (2) 5b2 6b (1) 6b 5b2 (4) 5b2 6b 10. When written in scientific notation, the fraction (1.2 3 1024) 3 (3.5 3 108) 8.4 3 1022 equals (1) 5.0 102 Part II (2) 5.0 101 (3) 5.0 105 (4) 5.0 106 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 336 Trigonometry of the Right Triangle 11. The vertices of pentagon ABCDE are A(2, 2), B(7, 2), C(7, 5), D(0, 5), E(2, 0). a. Draw pentagon ABCDE on graph paper. b. Find the area of the pentagon. 12. In ABC, mC = 90, mB = 30, value of x. Part III AC 5 6x2 2 4x, and AB 5 2x. Find the Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Plank Road and Holt Road are perpendicular to each other. At a point 1.3 miles before the intersection of Plank and Holt, State Street crosses Plank at an angle of 57°. How far from the intersection of Plank and Holt will State Street intersect Holt? Write your answer to the nearest tenth of a mile. 14. Benny, Carlos, and Danny each play a different sport and have different career plans. Each of the four statements given below is true. The boy who plays baseball plans to be an engineer. Benny wants to be a lawyer. Carlos plays soccer. The boy who plans to be a doctor does |
not play basketball. What are the career plans of each boy and what sport does he play? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Bart wants to plant a garden around the base of a tree. To determine the amount of topsoil he will need to enrich his garden, he measured the circumference of the tree and found it to be 9.5 feet. His garden will be 2.0 feet wide, in the form of a ring around the tree. Find to the nearest square foot the surface area of the garden Bart intends to plant. 16. Samantha had a snapshot that is 3.75 inches wide and 6.5 inches high. She cut a strip off of the top of the snapshot so that an enlargement will fit into a frame that measures 5 inches by 8 inches. What were the dimensions of the strip that she cut off of the original snapshot? GRAPHING LINEAR FUNCTIONS AND RELATIONS The Tiny Tot Day Care Center charges $200 a week for children who stay at the center between 8:00 A.M. and 5:00 P.M. If a child is not picked up by 5:00 P.M., the center charges an additional $4.00 per hour or any part of an hour. Mr. Shubin often has to work late and is unable to pick up his daughter on time. For Mr. Shubin, the weekly cost of day care is a function of time; that is, his total cost depends on the time he arrives at the center. This function for the daily cost of day care can be expressed in terms of an equation in two variables: y 30 4x.The variable x represents the total time, in hours, after 5:00 P.M., that a child remains at the center, and the variable y represents the cost, in dollars, of day care for the day. CHAPTER 9 CHAPTER TABLE OF CONTENTS 9-1 Sets, Relations, and Functions 9-2 Graphing Linear Functions Using Their Solutions 9-3 Graphing a Line Parallel to an Axis 9-4 The Slope of a Line 9-5 The Slopes of Parallel and Perpendicular Lines 9-6 The Intercepts of a Line 9-7 Graphing Linear |
Functions Using Their Slopes 9-8 Graphing Direct Variation 9-9 Graphing First-Degree Inequalities in Two Variables 9-10 Graphs Involving Absolute Value 9-11 Graphs Involving Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review 337 338 Graphing Linear Functions and Relations 9-1 SETS, RELATIONS, AND FUNCTIONS Set-Builder Notation In Chapters 1 and 2, we used roster form to describe sets. In roster form, the elements of a set are enclosed by braces and listed once. Repeated elements are not allowed. For example, {..., 3, 2, 1, 0, 1, 2, 3,...} is the set of integers, in roster form. A second way to specify a set is to use set-builder notation. Set-builder notation is a mathematically concise way of describing a set without listing the elements of the set. For instance, using set-builder notation, the set of counting numbers from 1 to 100 is: {x x is an integer and 1 x 100} This reads as “the set of all x such that x is an integer and x is at least 1 and at most 100.” The vertical bar “” represents the phrase “such that,” and the description to the right of the bar is the rule which defines the set. Here are some other examples of set-builder notation: 1. {x x is an integer and x 6 } {7, 8, 9, 10,...} the set of integers greater than 6 2. {x 1 x 3} any real number in the interval [1, 3] 3. {2n 1 n is a whole number} {2(0) 1, 2(1) 1, 2(2) 1, 2(3) 1,...} {1, 3, 5, 7,...} the set of odd whole numbers Frequently used with set-builder notation is the symbol, which means “is an element of.” This symbol is used to indicate that an element is a member of a set. The symbol means “is not an element of,” and is used to indicate that an element is not a member of a set. For instance, 2 {x 1 x 3} 2 {x 1 x 3} since 2 is between 1 and 3. since 2 |
is not between 1 and 3. EXAMPLE 1 List the elements of each set or indicate that the set is the empty set. a. A {x x of the set of natural numbers and x 0} c. C {x x 3 5} b. B {2n n of the set of whole numbers} d. D {m m is a multiple of 5 and m 25} Sets, Relations, and Functions 339 Solution a. Since there are no natural numbers b. B {2(0), 2(1), 2(2), 2(3),...} less than 0, A 5 {0, 2, 4, 6,... } c. C represents the solution set of the equation x 3 5. Therefore, C {2} d. D consists of the set of multiples of 5 that are less than 25. D {..., 45, 40, 35, 30} Answers a. A 5 b. B {0, 2, 4, 6,...} c. C {2} d. D {..., 45, 40, 35, 30} Relations That are Finite Sets There are many instances in which one set of information is related to another. For example, we may identify the persons of a group who are 17, 18, or 19 to determine who is old enough to vote. This information can be shown in a diagram such as the one at the right, or as a set of ordered pairs. 17 18 19 Debbie Kurt Kim Eddie {(17, Debbie), (17, Kurt), (18, Kim), (19, Eddie)} DEFINITION A relation is a set of ordered pairs. The domain of a relation is the set of all first elements of the ordered pairs. For example, in the relation shown above, the domain is {17, 18, 19}, the set of ages. The range of a relation is the set of all second elements of the ordered pairs. In the relation above, the range is {Debbie, Kurt, Kim, Eddie}, the set of people. Let us consider the relation “is greater than,” using the set {1, 2, 3, 4} as both the domain and the range. Let (x, y) be an ordered pair of the relation. Then the relation can be shown in any of the following ways. 1. A Rule The relation can be described by the inequality x y. 2. Set of Ordered P |
airs The relation can be listed in set notation as shown below: {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)} 340 Graphing Linear Functions and Relations 3. Table of Values The ordered pairs shown in 2 can be displayed in a table. Graph In the coordinate plane, the domain is a subset of the numbers on the x-axis and the range is a subset of the numbers on the y-axis. The points that correspond to the ordered pairs of numbers from the domain and range are shown. Of the 16 ordered pairs shown, only six are enclosed to indicate the relation x y Relations That are Infinite Sets Charita wants to enclose a rectangular garden. How much fencing will she need if the width of the garden is to be 5 feet? The answer to this question depends on the length of the garden. Therefore, we say that the length of the garden, x, and the amount of fencing, y, form a set of ordered pairs or a relation. The formula for the perimeter of a rectangle can be used to express y in terms of x: P 2l 2w y 2x 2(5) y 2x 10 Some possible values for the amount of fencing can also be shown in a table. As the length of the garden changes, the amount of fencing changes, as the following table indicates. x 1 3 4.5 7.2 9.1 2x 10 2(1) 10 2(3) 10 2(4.5) 10 2(7.2) 10 2(9.1) 10 y 12 16 19 24.4 28.2 (x, y) (1, 12) (3, 16) (4.5, 19) (7.2, 24.4) (9.1, 28.2) Sets, Relations, and Functions 341 Each solution to y 2x 10 is a pair of numbers. In writing each pair, we place the value of x first and the value of y second. For example, when x 1, y 12, (1, 12) is a solution. When x 3, y 16, (3, 16) is another solution. It is not possible to list all ordered pairs in the solution set of the equation because the solution set is infinite. However, it is possible to determine whether a given ordered pair is a member of the solution set. Replace x with the first element of the pair (the |
x-coordinate) and y with the second element of the pair (the y-coordinate). If the result is a true statement, the ordered pair is a solution of the equation. • (1.5, 13) {(x, y) y 2x 10} because 13 2(1.5) 10 is true. • (4, 14) {(x, y) y 2x 10} because 14 2(4) 10 is false. Since in the equation y 2x 10, x represents the length of a garden, only positive numbers are acceptable replacements for x. Therefore, the domain of this relation is the set of positive real numbers. For every positive number x, 2x 10 will also be a positive number. Therefore, the range of this relation is the set of positive real numbers. Although a solution of y 2x 10 is (–2, 6), it is not possible for 2 to be the length of a garden; (2, 6) is not a pair of the relation. When we choose a positive real number to be the value of x, there is one and only one value of y that makes the equation true. Therefore, y 2x 10 is a special kind of relation called a function. DEFINITION A function is a relation in which no two ordered pairs have the same first element. or A function is a relation in which every element of the domain is paired with one and only one element of the range. Since a function is a special kind of relation, the domain of a function is the set of all first elements of the ordered pairs of the function. Similarly, the range of a function is the set of all second elements of the ordered pairs of the function. The notation f(a) b or “f of a equals b,” signifies that the value of the function, f, at a is equal to b. The independent variable is the variable that represents the first element of an ordered pair. The domain of a function is the set of all values that the independent variable is allowed to take. The dependent variable is the variable that represents the second element of an ordered pair. The range of a function is the set of all values that the dependent variable is allowed to take. 342 Graphing Linear Functions and Relations EXAMPLE 2 Find five members of the solution set of the sentence 3x y 7. Solution How to Proceed (1) Transform the equation into an equivalent equation with y alone as one member: ( |
2) Choose any five values for x. Since no replacement set is given, any real numbers can be used: (3) For each selected value of x, determine y: 3x 3x y 7 3x y 3x 7 x 2 0 1 3 3 5 3x 7 3(2) 7 3(0) 7 A 1 1 7 23 3 B 3(3) 7 3(5) 7 y 13 7 6 2 8 Answer (–2, 13), (0, 7),, (3, –2), (5, –8) 1 3, 6 A B Note that many other solutions are also possible, and for every real number x, one and only one real number y will make the equation 3x y 7 true. Therefore, 3x y 7 defines a function. EXAMPLE 3 Determine whether each of the given ordered pairs is a solution of the inequality y 2x 4. a. (4, 0) b. (–1, 2) c. (0, 6) d. (0, 7) Solution a. y 2x 4? 0 2 2(4) $ 4 0 8 4 ✘ y 2x 4 c.? 6 2 2(0) $ 4 6 0 4 ✔ b. y 2x 4? 2 2 2(21) $ 4 d. 2 2 4 ✔ y 2x 4? 7 2 2(0) $ 4 7 0 4 ✔ Answers a. Not a solution b. A solution c. A solution d. A solution EXAMPLE 4 Note that for the inequality y 2x 4, the pairs (0, 6) and (0, 7) have the same first element. Therefore y 2x 4 is a relation but not a function. Sets, Relations, and Functions 343 The cost of renting a car for 1 day is $64.00 plus $0.25 per mile. Let x represent the number of miles the car was driven, and let y represent the rental cost, in dollars, for a day. a. Write an equation for the rental cost of the car in terms of the number of miles driven. b. Find the missing member of each of the following ordered pairs, which are elements of the solution set of the equation written in part a, and explain the meaning of the pair. (1) (155,?) (2) (?, 69) Solution a. Rental cost is $64.00 plus $0.25 times the number of miles. b. |
(1) For the pair (155,?), x 155 and y is to be determined. Then: y 64.00 0.25x y 64.00 0.25x 64.00 0.25(155) 64.00 38.75 102.75 When the car was driven 155 miles, the rental cost was $102.75. (2) For the pair (?, 69), y 69 and x is to be determined. Then: y 64 0.25x 69 64 0.25x 64 64 5 0.25x 0.25x 0.25 5 0.25 20 x When the rental cost was $69, the car was driven 20 miles. Answers a. y 64.00 0.25x b. (1) (155, 102.75) The car was driven 155 miles, and the cost was $102.75. (2) (20, 69) The car was driven 20 miles, and the cost was $69.00. 344 Graphing Linear Functions and Relations EXAMPLE 5 Determine whether or not each set is a relation. If it is a relation, determine whether or not it is a function. a. {x 0 x 10} b. {(x, y) y 4 x} c. {(a, b) a 2 and b is a real number} d. {(x, y) y x2} Solution a. The set is not a set of ordered pairs. It is not a relation. Answer b. The set is a relation and a function. It is a set of ordered pairs in which every first element is paired with one and only one second element. Answer c. The set is a relation that is not a function. The same first element is paired with every second element. Answer d. The set is a relation and a function. Every first element is paired with one and only one second element, its square. Answer EXAMPLE 6 Jane would like to construct a rectangular pool having an area of 102 square feet. If l represents the length of the pool and w its height, express the set of all ordered pairs, (l, w), that represent the dimensions of the pool using set-builder notation. Solution Since the area must be 102 square units, the variables l and w are related by the area formula: A lw. Therefore, the set of ordered pairs representing the possible dimensions of the pool is: {(l, w) 102 lw, where l and |
w are both positive} Answer Note: Since the length and the width cannot be zero or negative, the rule must specify that the dimensions, l and w, of the pool are both positive. EXERCISES Writing About Mathematics 1. A function is a set of ordered pairs in which no two different pairs have the same first element. In Example 4, we can say that the cost of renting a car is a function of the number of miles driven. Explain how this example illustrates the definition of a function. Sets, Relations, and Functions 345 2. Usually when a car rental company charges for the number of miles driven, the number of miles is expressed as a whole number. For example, if the car was driven 145.3 miles, the driver would be charged for 146 miles. In Example 4, what would be the domain of the function? (Recall that the domain is the set of numbers that can replace the variable x.) Developing Skills In 3–7, find the missing member in each ordered pair if the second member of the pair is twice the first member. 3. (3,?) 4. (0,?) 5. (2,?) 6. (?, 11) 7. (?, 8) In 8–12, find the missing member in each ordered pair if the first member of the pair is 4 more than the second member. 8. (?, 5) 9. 10. (?, 0) 11. 12. (8,?)?, 1 2 A B 91 4,? A B In 13–27, state whether each given ordered pair of numbers is a solution of the equation or inequality. 13. y 5x; (3, 15) 16. 3x 2y 0; (3, 2) 19. 3y 2x l; (4, 3) 1 1 22. 4x 3y 2; 4, 3 25. y 6x; (1, 2) B A 14. y 4x; (16, 4) 17. y 4x; (2, 10) 20. 2x 3y 9; (0, 3) 23. 3x y 4; (7, 1) 26. 3x 4y; (5, 2) 15. y 3x 1; (7, 22) 18. y 2x 3; (0, 2) 21. x y 8; (4, 5) 24. x 2y 15; (1, 7) 27. 5x 2y 19; (3, 2) |
In 28–31, state which sets of ordered pairs represent functions. If the set is not a function, explain why. 28. {(1, 2), (2, 3), (3, 4), (5, 6)} 30. {(5, 5), (5, 5), (6, 6), (6, 6)} 29. {(1, 2), (2, 1), (3, 4), (4, 3)} 31. {(81, 9), (81, 9), (25, 5), (25, 5)} In 32–35, find the range of each function when the domain is: a. A {x 10 x 13 and x of the set of integers} b. B {6x x of the set of whole numbers} 32. y 2x 3 33. y 5 x 6 34. y x 2 35. y x 1 Applying Skills In 36–39, in each case: a. Write an equation or an inequality that expresses the relationship between x and y. b. Find two ordered pairs in the solution set of the equation that you wrote in part a. c. Is the relation determined by the equation or inequality a function? 36. The cost, y, of renting a bicycle is $5.00 plus $2.50 times the number of hours, x, that the bicycle is used. Assume that x can be a fractional part of an hour. 37. In an isosceles triangle whose perimeter is 54 centimeters, the length of the base in centime- ters is x and the length of each leg in centimeters is y. 346 Graphing Linear Functions and Relations 38. Jules has $265 at the beginning of the month. What is the maximum amount, y, that he has left after 30 days if he spends at least x dollars a day? 39. At the beginning of the day, the water in the swimming pool was 2 feet deep. Throughout the day, water was added so that the depth increased by less than 0.4 foot per hour. What was the depth, y, of the water in the pool after being filled for x hours? 40. A fence to enclose a rectangular space along a river is to be constructed using 176 feet of fencing. The two sides perpendicular to the river have length x. a. Complete five more rows of the table shown on the right. b. Is the area, A, a function of x? If so, write the |
function and determine its domain. Length (x) Width Area (A) 1 2 176 2(1) 176 2(2) 1[176 2(1)] 174 2[176 2(2)] 344 41. (1) At the Riverside Amusement Park, rides are paid for with tokens purchased at a central booth. Some rides require two tokens, and others one token. Tomas bought 10 tokens and spent them all on x two-token rides and y one-token rides. On how many two-token and how many one-token rides did Tomas go? (2) Minnie used 10 feet of fence to enclose three sides of a rectangular pen whose fourth side was the side of a garage. She used x feet of fence for each side perpendicular to the garage and y feet of fence for the side parallel to the garage. What were the dimensions of the pen that Minnie built? a. Write an equation that can be used to solve both problems. b. Write the six solutions for problem (1). c. Which of the six solutions for problem (1) are not solutions for problem (2)? d. Write two solutions for problem (2) that are not solutions for problem (1). e. There are only six solutions for problem (1) but infinitely many solutions for problem (2). Explain why. 42. A cylindrical wooden base for a trophy has radius r and height h. The volume of the base will be 102 cubic inches. Using set-builder notation, describe the set of ordered pairs, (r, h), that represent the possible dimensions of the base. 9-2 GRAPHING LINEAR FUNCTIONS USING THEIR SOLUTIONS Think of two numbers that add up to 6. If x and y represent these numbers, then x y 6 is an equation showing that their sum is 6. If the replacement set for both x and y is the set of real numbers, we can find infinitely many solutions for the equation x y 6. Some of the solutions are shown in the following table.5 1.5 4 2 3 3 2.2 3.8 2 4 0 6 2 8 Graphing Linear Functions Using Their Solutions 347 Each ordered pair of numbers, such as (8, –2), locates a point in the coordinate plane. When all the points that have the coordinates associated with the number pairs in the table are located, the points appear to lie on a straight line, as shown at the right. In fact, if the replacement set |
for both x and y is the set of real numbers, the following is true: All of the points whose coordinates are solutions of x y 6 lie on this same straight line, and all of the points whose coordinates are not solutions of x y 6 do not lie on this line2 –1 –1 – This line, which is the set of all the points and only the points whose coordinates make the equation x y 6 true, is called the graph of x y 6. DEFINITION A first-degree equation in standard form is written as Ax By C where A, B, and C are real numbers, with A and B not both 0. We call an equation that can be written in the form Ax + By C a linear equation since its graph is a straight line that contains all points and only those points whose coordinates make the equation true. The replacement set for both variables is the set of real numbers unless otherwise indicated. When we graph a linear equation, we can determine the line by plotting two points whose coordinates satisfy that equation. However, we usually plot a third point as a check on the first two. If the third point lies on the line determined by the first two points, we have probably made no error. Procedure To graph a linear equation by means of its solutions: 1. Transform the equation into an equivalent equation that is solved for y in terms of x. 2. Find three solutions of the equation by choosing values for x and finding corresponding values for y. 3. In the coordinate plane, graph the ordered pairs of numbers found in Step 2. 4. Draw the line that passes through the points graphed in Step 3. 348 Graphing Linear Functions and Relations EXAMPLE 1 Does the point (2, 3) lie on the graph of x 2y 4? Solution The point (2, 3) lies on the graph of x 2y 4 if and only if it is a solu- tion of x 2y 4. x 2y 4 2 2 2(23) 5? 2 1 6 5? 4 4 8 4 ✘ Answer Since 8 4 is not true, the point (2, 3) does not lie on the line x 2y 4. EXAMPLE 2 What must be the value of d if (d, 4) lies on line 3x y 10? Solution The coordinates (d, 4) must satisfy 3x y 10. Let x d and y 4. Then: 3d 4 10 3d 6 d 2 Answer The value of d is |
2. EXAMPLE 3 a. Write the following verbal sentence as an equation: The sum of twice the x-coordinate of a point and the y-coordinate of that point is 4. b. Graph the equation written in part a. Solution a. Let x the x-coordinate of the point, and y the y-coordinate of the point. Then: 2x y 4 Answer b. How to Proceed (1) Transform the equation into an equivalent equation that has y alone as one member: 2x y 4 y 2x 4 Graphing Linear Functions Using Their Solutions 349 (2) Choose three values of x and find the corresponding values of y: (3) Plot the points that are associated with the three solutions: (4) Draw a line through the points that were plotted. Label the line with its equation: x 2x 4 0 2(0) 4 1 2(1) 4 2 2(23 –2 –1 –1 – To display the graph of a function on a calculator, follow the steps below. 1. Solve the equation for y and enter it as Y1. 2 X,T,,n ENTER: 4 DISPLAY: Y (-) ENTER. Use the standard viewing window to view the graph. The standard viewing window is a 20 by 20 graph centered at the origin. ENTER: ZOOM 6 ENTER DISPLAY: M e m o r y Zoom Tr i g 350 Graphing Linear Functions and Relations 3. Display the coordinates of points on the graph. ENTER: TRACE In the viewing window, the equation appears at the top left of the screen. The point at which the graph intersects the y-axis is marked with a star and the coordinates of that point appear at the bottom of the screen. Press the right and left arrow keys to move the star along the line to display other coordinates EXERCISES Writing About Mathematics 1. The points whose coordinates are (3, 1), (5, 1) and (7, 3) all lie on the same line. What could be the coordinates of another point that lies on that line? Explain how you found your answer. 2. Of the points (0, 5), (2, 4), (3, 3), and (6, 2), which one does not lie on the same line as the other three? Explain how you found your answer. Developing Skills In 3–5, state, in each case, whether the pair of values for x and y is a |
member of the solution set of the equation 2x y 6. 3. x 4, y 2 4. x 0, y 6 5. x 4, y 2 For 6–9, state in each case whether the point whose coordinates are given is on the graph of the given equation. 6. x y 7; (4, 3) 8. 3x 2y 8; (2, 1) 7. 2y x 7; (1, 3) 9. 2y 3x 5; (1, 4) In 10–13, find in each case the number or numbers that can replace k so that the resulting ordered number pair will be on the graph of the given equation. 10. x 2y 5; (k, 2) 12. x 3y 10; (13, k) 11. 3x 2y 22; (k, 5) 13. x y 0; (k, k) In 14–17, find in each case a value that can replace k so that the graph of the resulting equation will pass through the point whose coordinates are given. Graphing Linear Functions Using Their Solutions 351 14. x y k; (2, 5) 16. y 2x k; (2, 1) 15. x y k; (5, 3) 17. y x k; (5, 0) In 18–23, solve each equation for y in terms of x. 19. 4x y 6 22. 4x 2y 8 18. 3x y 1 21. 12x 3 2y 20. 2y 6x 23. 6x 3y 5 In 24–26: a. Find the missing values of the variable needed to complete each table. b. Plot the points described by the pairs of values in each completed table; then, draw a line through the points. 24. y 4x 25. y 3x 1 26. x 2y??? In 27–50, graph each equation. 27. y x 28. y 5x 29. y 3x 30. y x 1 2y 1 1 31. x 2y 3 35. y 3x 1 39. y x 0 43. 2x y 6 47. 3x 2y 4 51. a. Through points (0, 2) and (4, 0), draw a straight line. 32. x 36. y 2x 4 40. 3x y 12 44. 3x y 6 48. x 3y 9 33. y x 3 |
37. x y 8 41. x – 2y 0 45. x 3y 12 49. 2x y 4 34. y 2x 1 38. x y 5 42. y 3x –5 46. 2x 3y 6 50. 4x 3y b. Write the coordinates of two other points on the line drawn in part a. 52. a. Through points (2, 3) and (1, 3), draw a straight line. b. Does point (0, 1) lie on the line drawn in part a? Applying Skills In 53–60: a. Write an equation that can be used to represent each sentence. b. Graph the equation. 53. The length of a rectangle, y, is twice the width, x. 54. The distance to school, y, is 2 miles more than the distance to the library, x. 55. The cost of a loaf of bread, x, plus the cost of a pound of meat, y, is 6 dollars. 352 Graphing Linear Functions and Relations 56. The difference between Tim’s height, x, and Sarah’s height, y, is 1 foot. 57. The measure of each of three sides of the trapezoid is x, the measure of the remaining side is y, and the perimeter is 6. 58. The measure of each of the legs of an isosceles triangle is x, the measure of the base is y, and the perimeter is 9. 59. Bob’s age, y, is equal to five more than twice Alice’s present age, x. 60. The sum of the number of miles that Paul ran, x, and the number of miles that Sue ran, y, is 30 miles. 9-3 GRAPHING A LINE PARALLEL TO AN AXIS Lines Parallel to the x-Axis An equation such as y 2 can be graphed in the coordinate plane. Any pair of values whose y-coordinate is 2, no matter what the x-coordinate is, makes the equation y 2 true.Therefore, (3, 2), (2, 2), (1, 2), (0, 2), (1, 2), or any other pair (a, 2) for all values of a, are points on the graph of y 2. As shown at the right, the graph of y 2 is a horizontal line parallel to the x-axis and 2 units above it. The equation y 2 |
defines a function and can be displayed on a graphing calculator. If there are other equations entered in the before Y= menu, press entering the new equation. DISPLAY: ENTER: CLEAR 4 –3 –2 –1 –1 –2 –3 Y 2 GRAPH The equation of a line parallel to the x-axis and b units from the x-axis is y b. If b is positive, the line is above the x-axis, and if b is negative, the line is below the x-axis. Graphing a Line Parallel to an Axis 353 Lines Parallel to the y-Axis An equation such as x 3 can be graphed in the coordinate plane. Any pair of values whose x-coordinate is 3, no matter what the y-coordinate is, makes the equation x 3 true. Therefore, (3, 2), (–3, –1), (3, 0), (3, 1), (3, 2), or any other pair (3, b) for all values of b, are points on the graph of x 3. As shown at the right, the graph of x 3 is a vertical line parallel to the y-axis and 3 units to the left of it. y 3 2 1 O –4 –2 –1 1 2 3 –1 –2 –3 –4 3 – = x x The equation x 3 defines a relation that is not a function because it defines a set of ordered pairs but every ordered pair has the same first element. The method used to draw the graph of a function on a graphing calculator cannot be used to display this graph. The equation of a line parallel to the y-axis and a units from the y-axis is x a. If a is positive, the line is to the right of the y-axis, and if a is negative, the line is to the left of the x-axis. EXERCISES Writing About Mathematics 1. Mike said that the equation of the x-axis is x 0. Do you agree with Mike? Explain why or why not. 2. Does the line whose equation is y 4 intersect the x-axis? Determine the coordinates of the point of intersection if one exists or explain why a point of intersection does not exist. Developing Skills In 3–17, draw the graph of each equation. 3. x 6 8. y 4 x 5 1 2 13. 4. x 4 9. y 5 y 5 11 2 14. 5. x |
0 10. y 0 15. y 2.5 6. x 3 11. y 4 x 5 23 2 16. 18. Write an equation of the line that is parallel to the x-axis 7. x 5 12. y 7 17. y 3.5 a. 1 unit above the axis. b. 5 units above the axis. c. 4 units below the axis. d. 8 units below the axis. e. 2.5 units above the axis. f. 3.5 units below the axis. 354 Graphing Linear Functions and Relations 19. Write an equation of the line that is parallel to the y-axis a. 3 units to the right of the axis b. 10 units to the right of the axis c. 41 2 units to the left of the axis d. 6 units to the left of the axis. e. 2.5 units to the right of the axis. f. 5.2 units to the left of the axis. 20. Which statement is true about the graph of the equation y 6? (1) It is parallel to the y-axis (2) It is parallel to the x-axis. (3) It is not parallel to either axis. (4) It goes through the origin. 21. Which statement is not true about the graph of the equation x 5? (3) It is a function. (1) It goes through (5, 0). (2) It is a vertical line. (4) It is parallel to the y-axis. 22. Which statement is true about the graph of the equation y x? (1) It is parallel to the y-axis. (2) It is parallel to the x-axis. (3) It goes through (2, 2). (4) It goes through the origin. Applying Skills 23. The cost of admission to an amusement park on Family Night is $25.00 for a family of any size. a. What is the cost of admission for the Gauger family of six persons? b. Write an equation for the cost of admission, y, for a family of x persons. 24. The coordinates of the vertices of rectangle ABCD are A(2, 1), B(4, 1), C(4, 5), and D(2, 5). a. Draw ABCD on graph paper. b. Write the equations of the lines g AB, g BC, g CD, and g DA. c. Draw a horizontal line |
and a vertical line that separate ABCD into four congruent rectangles. d. Write the equations of the lines drawn in part c. 25. During the years 2004, 2005, and 2006, there were 5,432 AnyClothes stores. a. Write a linear equation that gives the number of stores, y. b. Predict the number of stores for the years 2007 and 2008. 26. During the years 1999 through 2006, SmallTown News’ circulation was 74,000 each year. a. Write a linear equation that gives the circulation, y. b. Predict the circulation of SmallTown News in the years 2007 and 2008. 9-4 THE SLOPE OF A LINE The Slope of a Line 355 Meaning of the Slope of a Line Easy Hill Tough Hill A 80 m B 20 m C D 80 m E 40 m F It is more difficult to hike up Tough Hill, shown above, than to hike up Easy Hill. Tough Hill rises 40 meters vertically over a horizontal distance of 80 meters, whereas Easy Hill rises only 20 meters vertically over the same horizontal distance of 80 meters. Therefore, Tough Hill is steeper than Easy Hill. To compare the steepness of roads, which lead up the two hills, we and compare their slopes. The slope of road is the ratio of the change in vertical distance, CB, to DE AB AB the change in horizontal distance, AC: slope of road AB 5 change in vertical distance, CB change in horizontal distance, AC 5 20 m 80 m 5 1 4 Also: slope of road DE 5 change in vertical distance, FE change in horizontal distance, DF 5 40 m 80 m 5 1 2 Note that the steeper hill has the larger slope. Finding the Slope of a Line To find the slope of the line determined by the two points (2, 3) and (5, 8), as shown at the right, we write the ratio of the difference in y-values to the difference in x-values as follows: slope 5 difference in y-values difference in x-values Suppose we change the order of the points (2, 3) and (5, 8) in performing the computation. We then have: slope 5 difference in y-values 23 5 5 difference in x-values 5 3 2 8 3 The result of both computations is the same. 2 2 5 5 25 5, 8) 5 = 3 – 8 (2, 3) 5 – 2 = 3 O –1 –1 1 2 3 4 5 6 7 |
x 356 Graphing Linear Functions and Relations When we compute the slope of a line that is determined by two points, it does not matter which point is considered as the first point and which the second. Also, when we find the slope of a line using two points on the line, it does not matter which two points on the line we use because all segments of a line have the same slope as the line. Procedure To find the slope of a line: 1. Select any two points on the line. 2. Find the vertical change, that is, the change in y-values by subtracting the y-coordinates in any order. 3. Find the horizontal change, that is, the change in x-values, by subtracting the x-coordinates in the same order as the y-coordinates. 4. Write the ratio of the vertical change to the horizontal change. We can use this procedure to find the, the lines shown in the g LM g RS and slopes of graph at the right. slope of g LM slope of g RS 5 5 or 2 5 5 6 – 2 vertical change horizontal change 2 5 2 3 – 1 5 4 1 vertical change horizontal change 4 – 1 5 22 24 2 (22) 3 5 22 3, 6) 4 O (1, 2) L –2 –1 1 2 3 4 5 6 x R Slope is a rate of change. It is a measure of the rate at which y changes compared to x. For g 2 LM whose slope is 2 or, y increases 2 units 1 when x increases 1 unit. When the second element of a rate is 1, the rate is a unit rate of change. If x is the independent variable and y is the dependent variable, then the slope is the rate of change in the dependent variable compared to the independent variable. –2 –3 –4 (4, –4) (1, –2) 2 – S 3 For g RS whose slope is, y decreases 2 units when x increases 3 units. We could also write this rate as to write the slope as a unit rate of change, that 22 3 2 3 1 is, y decreases of a unit when x increases 1 unit. 2 3 The Slope of a Line 357 In general: The slope, m, of a line that passes through any two points P1(x1, y1) and x2, is the ratio of the change or difference of the P2(x2, y2), where x1 y- |
values of these points to the change or difference of the corresponding x-values. Thus: slope of a line difference iny-values difference in x-values Therefore: slope of g P1P2 m 2 y y 1 2 x2 2 x1 x1, can The difference in x-values, x2 be represented by x, read as “delta x.” Similarly, the difference in y-values, y1, can be represented by y, read y2 as “delta y.” Therefore, we write: slope of a line m Dy Dx Positive Slopes g AB Examining from left to right and observing the path of a point, from C to D for example, we see that the line is rising. As the x-values increase, the y-values also increase. Between point C and point D, the change in y is 1, and the change in x is 2. Since both y and x are positive, the slope of must be positive. Thus: g AB slope of g AB m Dy Dx 5 1 2 y x2 – x1 1 y – 2 y O P1 (x1, y1) P2 (x2, y2) x x y A 1 –1 –1 B 2 1 D(3, 3) C(1, 2) O 1 Principle 1. As a point moves from left to right along a line that is rising, y increases as x increases and the slope of the line is positive. 358 Graphing Linear Functions and Relations Negative Slopes g EF from left to right and Now, examining observing the path of a point from C to D, we see that the line is falling. As the x-values increase, the y-values decrease. Between point C and point D, the change in y is 2, and the change in x is 3. Since y is negative and x is g EF positive, the slope of must be negative. slope of g EF m Dy Dx 5 22 3 5 22 3 y 1 –1 –1 E –2 O 1 C (2, 3) 3 D(5, 1) F x Principle 2. As a point moves from left to right along a line that is falling, y decreases as x increases and the slope of the line is negative. Zero Slope g GH g GH is parallel to the x-axis. We consider On the graph, a point moving along from left to right, for example from C to |
D. As the x-values increase, the y-values are unchanged. Between point C and point D, the change in y is 0 and the change in x is 3. Since y is 0 and x is 3, the slope of must be 0. Thus: g GH y 1 O –1 –1 G (–2, –2) C x H (1, –2) D slope of g GH m Dy Dx 5 0 3 5 0 Principle 3. If a line is parallel to the x-axis, its slope is 0. Note: The slope of the x-axis is also 0. No Slope g LM g LM is parallel to the y-axis. We consider On the graph, a point moving upward along, for example from C to D. The x-values are unchanged, but the y-values increase. Between point C and point D, the change in y is 3, and the change in x is 0. Since the slope is the change in y divided by the change in x and of has no defined a number cannot be divided by 0, slope. g LM g LM y M 1 O –1 –1 D(2, 1) x C(2, –2) L The Slope of a Line 359 slope of g MC m Dy Dx 5 1 2 (22) 0 5 undefined Principle 4. If a line is parallel to the y-axis, it has no defined slope. Note: The y-axis itself has no defined slope. EXAMPLE 1 Find the slope of the line that is determined by points (2, 4) and (4, 2). Solution Plot points (2, 4) and (4, 2). Let point (2, 4) be P1(x1, y1), and let point 2, y1 y 2 y 1 2 x2 2 x1 4, and y2 4, x2 2. y g slope of P1P2 (4, 2) be P2(x2, y2). Then, x1 Dy Dx 22 6 5 21 3 4 2 (22) Answer P1(–2, 4) –2 6 O 1 1 –1 –1 P2(4, 2) x EXAMPLE 2 3 Through point (2, 1), draw the line whose slope is. 2 Solution How to Proceed (1) Graph point A(2, 1): y (2) Note that since slope Dy Dx 5 3 2, |
when y changes by 3, then x changes by 2. Start at point and move 3 units upward and 2 units to the right to locate point B: A(2, 21) (3) Start at B and repeat these movements to locate point C: (4) Draw a line that passes through points A, B, and C: C 2 B 3 2 1 –1–1 3 O 1 A(2, –1) x 360 Graphing Linear Functions and Relations KEEP IN MIND A fundamental property of a straight line is that its slope is constant. Therefore, any two points on a line may be used to compute the slope of the line. Applications of Slope In real-world contexts, the slope is either a ratio or a rate. When both the x and y variables have the same unit of measure, slope represents a ratio since it will have no unit of measure. However, when the x and y variables have different units of measure, slope represents a rate since it will have a unit of measure. In both cases, the slope represents a constant ratio or rate of change. EXAMPLE 3 The following are slopes of lines representing the daily sales, y, over time, x, for various sales representatives during the course of a year: (1) m 20 (3) m 39 (2) m 45 (4) m 7 a. Interpret the meaning of the slopes if sales are given in thousands of dollars and time is given in months. b. Which sales representative had the greatest monthly increase in sales? c. Is it possible to determine which sales representative had the greatest total sales at the end of the year? Solution a. The slopes represent the increase in sales per month: (1) $20,000 increase each month (2) $45,000 increase each month (3) $39,000 increase each month (4) $7,000 decrease each month b. At $45,000, (2) has the greatest monthly increase in sales. c. The slopes give the increase in sales each month, not the total monthly sales. With the given information, it is not possible to determine which sales representative had the greatest total sales at the end of the year. EXERCISES Writing About Mathematics 1. Regina said that in Example 1, the answer would be the same if the formula for slope had been written as. Do you agree with Regina? Explain why or why not. Use y1 y2 2 x1 2 x2 to y1 y |
2 2 x1 2 x2 find the answer to Example 1 to justify your response. The Slope of a Line 361 2. Explain why any two points that have the same x-coordinate lie on a line that has no slope. Developing Skills In 3–8: a. Tell whether each line has a positive slope, a negative slope, a slope of zero, or no slope. b. Find the slope of each line that has a slope. 3. 5. 7. y 1 –1 1 x y 1 –1 –1 1 y 1 –1 –1 1 x x 4. 6. 8. y 1 –1 1 x y 1 –1 –1 1 y 1 –1 –1 1 x x 362 Graphing Linear Functions and Relations In 9–17, in each case: a. Plot both points, and draw the line that they determine. b. Find the slope of this line. 9. (0, 0) and (4, 4) 12. (1, 5) and (3, 9) 15. (5, 2) and (7, 8) 10. (0, 0) and (4, 8) 13. (7, 3) and (1, 1) 16. (4, 2) and (8, 2) 11. (0, 0) and (3, 6) 14. (2, 4) and (0, 2) 17. (1, 3) and (2, 3) In 18–29, in each case, draw a line with the given slope, m, through the given point. m 5 2 3 18. (0, 0); m 2 21. (4, 5); 24. (1, 5); m 1 25 27. (1, 0); m 4 Applying Skills 19. (1, 3); m 3 22. (3, 1); m 0 25. (2, 4); m 28. (0, 2); m 23 2 22 3 20. (2, 5); m 4 23. (3, 4); m 2 26. (2, 3); m 29. (2, 0); m 21 3 1 2 30. Points A(2, 4), B(8, 4), and C(5, 1) are the vertices of ABC. Find the slope of each side of ABC. 31. Points A(3, 2), B(9, 2), C(7, 4), and |
D(1, 4) are the vertices of a quadrilateral. a. Graph the points and draw quadrilateral ABCD. b. What type of quadrilateral does ABCD appear to be? c. Compute the slope of and the slope of d. What is true of the slope of and the slope of g BC g BC g AD. g AD? e. If two segments such as AD and BC, or two lines such as what appears to be true of their slopes? g AD and g BC, are parallel, f. Since AB and CD are parallel, what might be true of their slopes? g. Compute the slope of AB and the slope of CD. h. Is the slope of AB equal to the slope of CD? 32. A path over a hill rises 100 feet vertically in a horizontal distance of 200 feet and then descends 100 feet vertically in a horizontal distance of 150 feet. B 100' a. Find the slope of the path up the hill when walking from point A to point B. A 200' C 150' b. Find the slope of the path down the hill when walking from point B to point C. c. What is the unit rate of change from point A to point B? d. What is the unit rate of change from point B to point C? The Slopes of Parallel and Perpendicular Lines 363 33. The amount that a certain internet phone company charges for international phone calls is represented by the equation C $0.02t $1.00, where C represents the total cost of the call and t represents time in minutes. Explain the meaning of the slope in terms of the information provided. 34. The monthly utility bill for a certain school can be represented by the equation C $0.075g $100, where C is the monthly bill and g is the amount of gas used by the hundred cubic feet (CCF). Explain the meaning of the slope in terms of the information provided. 35. A road that has an 8% grade rises 8 feet vertically for every 100 feet horizontally. a. Find the slope of the road. b. Explain the meaning of the slope in terms of the information provided. c. If you travel a horizontal distance of 200 feet on this road, what will be the amount of vertical change in your position? 9-5 THE SLOPES OF PARALLEL AND PERPENDICULAR LINES Parallel Lines /2 /1 and In the diagram, lines are parallel. Right triangle |
1 has been drawn with one leg coinciding with the x-axis and with its hypotenuse coinciding with. Similarly, right triangle 2 has been drawn with one leg coinciding with the x-axis and with its hypotenuse coinciding with /1 /2. Since triangles 1 and 2 are right triangles, tan a oppa adja tan b oppb adjb y oppa a adja Triangle 1 oppb x b adjb Triangle 2,1,2 The x-axis is a transversal that intersects parallel lines sponding angles, a and b, are congruent. Therefore, /1 and /2. The corre- oppa adja oppb adjb Since oppa adja is the slope of /1 and oppb adj b two parallel lines have the same slope. is the slope of /2, we have shown that If the slope of <1 is m1, the slope of <2 is m2, and <1 <2, then m1 m2. 7 The following statement is also true: If the slope of <1 is m1, the slope of <2 is m2, and m1 m2, then <1 <2. 7 364 Graphing Linear Functions and Relations Perpendicular Lines y a C a b,2,1'B'C /2 /1 /1 and. The slope of /1 are perpenIn the diagram, lines dicular. Right triangle ABC has been drawn with one leg parallel to the x-axis, the other leg parallel to the y-axis, and the hypotenuse is since y coinciding with increases by a units as x increases by b units. Triangle ABC is then rotated 90° counterclockwise about A so that the hypotenuse of ABC coincides with /2. Notice that since each leg has been rotated by 90°, the two legs are still parallel to the axes. However, they are now parallel to different axes. Also, in the rotated triangle, the change in y is an increase of b units and the change in x is a decrease of a units, that is, a. Therefore, the slope b /2 /2 or of is the negative reciprocal of the 2a /1 slope of. We have thus shown that perpendicular lines have slopes that are negative reciprocals of each other.. In other words, the slope of 2b a is A B b b a x In general: If the slope of m1? m2 5 |
21. <1 is m1, the slope of <2 is m2, and <1'<2, then m1 2 1 m2 or The following statement is also true: If the slope of <1'<2. then <1 is m1, the slope of <2 is m2, and m1 2 1 m2 or m1? m2 5 21, EXAMPLE 1 The equation of g AB is y 2x 1. a. Find the coordinates of any two points on g AB. b. What is the slope of g AB? c. What is the slope of a line that is parallel to g AB? d. What is the slope of a line that is perpendicular to g AB? Solution a. If x 0, y 2(0) 1 1. One point is (0, 1). If x 1, y 2(1) 1 1. Another point is (1, 1). b. Slope of g AB 1 2 (21) 0 2 1 5 2 21 5 22. The Slopes of Parallel and Perpendicular Lines 365 c. The slope of a line parallel to g AB d. The slope of a line perpendicular to is equal to the slope of 2 1 g AB 22 5 1 2 is. g AB, –2. Answers a. (0, 1) and (1, 1) but many other answers are possible. b. 2 c. 2 d. 1 2 EXERCISES Writing About Mathematics 1. a. What is the slope of a line that is perpendicular to a line whose slope is 0? Explain your answer. b. What is the slope of a line that is perpendicular to a line that has no slope? Explain your answer. 2. What is the unit rate of change of y with respect to x if the slope of the graph of the equa- tion for y in terms of x is? 1 4 Developing Skills In 3–11: a. Write the coordinates of two points on each line whose equation is given. b. Use the coordinates of the points found in part a to find the slope of the line. c. What is the slope of a line that is parallel to each line whose equation is given? d. What is the slope of a line that is perpendicular to the line whose equation is given? 4. y x 2 7. x y 4 10. x 4 5. y 3x 7 8. 2x 3y 6 11. y 5 3. y |
2x 6 6. 3y x 9. x 2y 1 Applying Skills 12. A taxi driver charges $3.00 plus $0.75 per mile. a. What is the cost of a trip of 4 miles? b. What is the cost of a trip of 8 miles? c. Use the information from parts a and b to draw the graph of a linear function described by the taxi fares. d. Write an equation for y, the cost of a taxi ride, in terms of x, the length of the ride in miles. e. What is the change in the cost of a taxi ride when the length of the ride changes by 4 miles? f. What is the unit rate of change of the cost with respect to the length of the ride in dol- lars per mile? 366 Graphing Linear Functions and Relations 13. Mrs. Boyko is waiting for her oven to heat to the required temperature. When she first turns the oven on, the temperature registers 100°. She knows that the temperature will increase by 10° every 5 seconds. a. What is the temperature of the oven after 10 seconds? b. What is the temperature of the oven after 15 seconds? c. Use the information from parts a and b to draw the graph of the linear function described by the oven temperatures. d. Write an equation for y, the temperature of the oven, in terms of x, the number of sec- onds that it has been heating. e. What is the unit rate in degrees per second at which the oven heats? 9-6 THE INTERCEPTS OF A LINE The graph of any linear equation that is not parallel to an axis intersects both axes. The point at which the graph intersects the y-axis is a point whose x-coordinate is 0. The y-coordinate of this point is called the y-intercept of the equation. For example, for the equation 2x y 6, let x 0: 2x y 6 2(0 The graph of 2x y 6 intersects the y-axis at (0, 6) and the y-intercept is 6. The point at which the graph intersects the x-axis is a point whose y-coordinate is 0. The x-coordinate of this point is called the x-intercept of the equation. For example, for the equation 2x y 6, let y 0: 2x y 6 2x 0 6 2x 6 x 3 The |
graph of 2x y 6 intersects the x-axis at (3, 0) and the x-intercept is 3. Divide each term of the equation 2x y 6 by the constant term, 6 y 6 5 6 2x Note that x is divided by the x-intercept, 3, and y is divided by the y-intercept, –6. The Intercepts of a Line 367 In general, if a linear equation is written in the form y x b 5 1 a 1 y-intercept is b., the x-intercept is a and the y x b 5 1 a 1 a y-intercept Q x-intercept An equation of the form x a 1 equation. Two points on the graph of these two points to find the slope of the line. a 2 0 5 2b 0 2 b slope a 5 2b a y b 5 1 is called the intercept form of a linear y x b 5 1 a 1 are (a, 0) and (0, b). We can use Slope and y-Intercept Consider the equation 2x y 6 from another point of view. Solve the equation for y: 2x y 6 y 2x 6 y 2x 6 There are two numbers in the equation that has been solved for y: the coefficient of x, 2, and the constant term, 6. Each of these numbers gives us important information about the graph. The y-intercept of a graph is the y-coordinate of the point at which the graph intersects the y-axis. For every point on the y-axis, x 0. When x 0, y 2(0) 6 0 6 6 Therefore, –6 is the y-intercept of the graph of the equation y 2x 6 and (0, 6) is a point on its graph. The y-intercept is the constant term of the equation when the equation is solved for y. Another point on the graph of y 2x 6 is (1, 4). Use the two points, (0, 6) and (1, 4), to find the slope of the line. slope of y 2x 6 26 2 (24) 0 2 1 5 22 21 5 2 The slope is the coefficient of x when the equation is solved for y. For any equation in the form y mx b: 1. The coefficient of x is the slope of the line. 2. The constant term is the y-intercept. 368 Grap |
hing Linear Functions and Relations The equation we have been studying can be compared to a general equation of the same type: y 2x 3 Q slope a y-intercept y mx b Q slope a y-intercept The following statement is true for the general equation: If the equation of a line is written in the form y mx b, then the slope of the line is m and the y-intercept is b. An equation of the form y mx b is called the slope-intercept form of a linear equation. EXAMPLE 1 Find the intercepts of the graph of the equation 5x 2y 6. Solution To find the x intercept, let y 0: 5x 2y 6 5x 2(0) 6 5x 0 6 5x 6 x 6 5 To find the y intercept, let x 0: 5x 2y 6 5(0) 2y 6 0 2y 6 2y 6 y 3 6 Answer The x intercept is and the y intercept is 3. 5 EXAMPLE 2 Find the slope and the y-intercept of the line that is the graph of the equation 4x + 2y 10. Solution How to Proceed (1) Solve the given equation for y to obtain an equivalent equation of the form y mx b: (2) The coefficient of x is the slope: (3) The constant term is the y-intercept: 4x 2y 10 2y 4x 10 y 2x 5 slope 2 y-intercept 5 Answer slope 2, y-intercept 5 The Intercepts of a Line 369 EXERCISES Writing About Mathematics 1. Amanda said that the y-intercept of the line whose equation is 2y 3x 8 is 8 because the y-intercept is the constant term. Do you agree with Amanda? Explain why or why not. 2. Roberto said that the line whose equation is y x has no slope because x has no coefficient and no y-intercept because there is no constant term. Do you agree with Roberto? Explain why or why not. Developing Skills In 3–18, in each case, find the slope, y-intercept, and x-intercept of the line that is the graph of the equation. 3. y 3x 1 7. x 3 4. y 3 x 8. y 2 11. 2x y 5 12. 3y 6x 9 5. y 2x 22 9. y 3x 1 4 13. 2y 5 |
x 4 y x 3 5 1 2 1 17. 6. y x 14. 10. y 3x 7 2x 1 3 4 5 1 1 3y 4x 5 2 2y 5 1 15. 4x 3y 0 19. What do the graphs of the equations y 4x, y 4x 2, and y 4x 2 all have in 16. 2x 5y 10 18. common? 20. What do the lines that are the graphs of the equations y 2x + 1, y 3x + 1, and y 5 24x 1 1 all have in common? 21. If two lines are parallel, how are their slopes related? 22. What is true of two lines whose slopes are equal? In 23–26, state in each case whether the lines are parallel, perpendicular, or neither. 23. y 3x 2, y 3x 5 25. y 4x 8, 4y x 3 27. Which of the following statements is true of the graph of y 3x? 24. y 2x 6, y 2x 6 26. y 2x, x 2y (1) It is parallel to the x-axis. (2) It is parallel to the y-axis. (3) Its slope is 3. (4) It does not have a y-intercept. 28. Which of the following statements is true of the graph of y 8? (1) It is parallel to the x-axis. (2) It is parallel to the y-axis. (3) It has no slope. (4) It goes through the origin. 370 Graphing Linear Functions and Relations 9-7 GRAPHING LINEAR FUNCTIONS USING THEIR SLOPES The slope and any one point can be used to draw the graph of a linear function. One convenient point is the point of intersection with the y-axis, that is, the point whose x-coordinate is 0 and whose y-coordinate is the y-intercept. EXAMPLE 1 Draw the graph of 2x 3y 9 using the slope and the y-intercept. Solution How to Proceed (1) Transform the equation into the form y 5 mx 1 b : (2) Find the slope of the line (the coefficient of x): (3) Find the y-intercept of the line (the constant): (4) On the y-axis, graph point A, whose y-coordinate is the y-intercept: |
(5) Use the slope to find two more points on the line. Since Dy Dx 5 22 slope, when y changes 3 by 2, x changes by 3. Therefore, start at point A and move 2 units down and 3 units to the right to locate point B. Then start at point B and repeat the procedure to locate point C: (6) Draw the line that passes through the three points: This line is the graph of 2x 3y 9. Check To check a graph, select two or more points on the line drawn and substitute their coordinates in the original equation. For example, check this graph using points (4.5, 0), (3, 1), and (6, 1). 2x 3y 9 3y 2x 9 y 22 3 x 1 3 slope 22 3 y-intercept 3 y 1 O y 1 O A(0, 3) 1 A(0, 3) 1 B B x C x C Graphing Linear Functions Using Their Slopes 371 2x 3y 9 2x 3y 9 2x 3y 9 2(4.5) 1 3(0) 5? 9 1 0 5? 9 9 9 9 ✔ 2(3) 1 3(1) 5? 6 1 3 5? 9 9 9 9 ✔ 2(6) 1 3(21) 5? 12 2 3 5? 9 9 9 9 ✔ In this example, we used the slope and the y-intercept to draw the graph. We could use any point on the graph of the equation as a starting point. EXAMPLE 2 Draw the graph of 3x 2y 9 using the slope and any point. Solution How to Proceed (1) Solve the equation for y. The slope 3x 2 2y 5 9 is and the y-intercept is 3 2 29 2 : (2) Since is not a convenient 0, 2 9 2 A B point to graph, choose another point. Let x 1. A convenient point on the graph is A(1, 3): (3) Graph point A whose coordinates are (1,23) : (4) Use the slope to find two more points on the line. Since slope when y changes by 3, x changes by 2. Dy Dx 5 3 2, Therefore, start at point A and move 3 units up and 2 units to the right to locate point B. Then start at point B and repeat the procedure to locate point C: (5) Draw the line the passes |
through the three points. This is the graph of 3x 2y 9: 22y 5 23x 1 9 22 1 9 y 5 23x 22 2x 2 9 y 5 3 2 2(1 26 2 y 5 23 y 1 O C B x A(1, –3) Note that the graph intersects the x-axis at B(3, 0). The x-intercept is 3. 372 Graphing Linear Functions and Relations Translating, Reflecting, and Scaling Graphs of Linear Functions An alternative way of looking at the effects of changing the values of the y-intercept and slope is that of translating, reflecting, or scaling the graph of the linear function y x. For instance, the graph of y x 3 can be thought of as the graph of y x shifted 3 units up. Similarly, the graph of y x can be thought of the graph of y x reflected in the x-axis. On the other hand, the graph of y 4x can be thought of as the graph of stretched vertically by a factor is the graph of y x compressed vertically by a of 4, while the graph of 1 factor of. 4 y 5 1 4x = 4x Translation Rules for Linear Functions If c is positive: The graph of y x c is the graph of y x shifted c units up. The graph of y x c is the graph of y x shifted c units down. Reflection Rule for Linear Functions The graph of y x is the graph of y x reflected in the x-axis. Scaling Rules for Linear Functions When c 1, the graph of y cx is the graph of y x stretched vertically by a factor of c. When 0 c 1, the graph of y cx is the graph of y x compressed ver- tically by a factor of c. Graphing Linear Functions Using Their Slopes 373 EXAMPLE 3 In a–d, write an equation for the resulting function if the graph of y 5 x is: a. shifted 6 units down b. reflected in the x-axis and shifted 3 units up c. compressed vertically by a factor of 1 10 and shifted 1 unit up d. stretched vertically by a factor of 4 and shifted 1 unit down Solution a. y x 6 Answer b. First, reflect in the x-axis: Then, translate the resulting function 3 units up: c. First, compress vertically by a factor of 1 10 : Then, translate the resulting function 1 unit up: d. First, stretch vertically |
by a factor of 4: Then, translate the resulting function 1 unit down: y x y x 3 Answer y y 1 10x 1 10x 1 1 Answer y 4x y 4x 1 Answer EXERCISES Writing About Mathematics 1. Explain why (0, b) is always a point on the graph of y mx b. 2. Gunther said that in the first example in this section, since the slope of the line,, could have been written as, points on the line could have been located by moving up 2 units and to the left 3 units from the point (0, 3). Do you agree with Gunther? Explain why or why not. 2 23 22 3 3. Hypatia said that, for linear functions, vertical translations have the same effect as horizontal translations and that reflecting across the x-axis has the same effect as reflecting across the y-axis. Do you agree with Hypatia? Explain why or why not. Developing Skills In 4–15, graph each equation using the slope and the y-intercept. 5. y 2x 4. y 2x 3 6. y 2x 7. y 3x 2 374 Graphing Linear Functions and Relations 2 8. y 3x 1 2 12. y 2x 8 1 9. y 2x 2 1 13. 3x y 4 21 10. y 3x y x 3 5 1 2 1 14. 23 11. y 15. 4x 2y 4x 1 6 In 16–21, graph the equation using the slope and any point. 16. 3y 4x + 9 19. 2x 3y 1 17. 5x 2y 3 20. 3y x 5 18. 3x + 4y 7 21. 2x 3y 6 0 In 22–25, describe the translation, reflection, and/or scaling that must be applied to y x to obtain the graph of each given function. 23. y 25. y (x 2) 2 24. y 2x 1.5 22. y x 2 21 3x 2 2 Applying Skills 26. a. Draw the line through (2, 3) whose slope is 2. b. What appears to be the y-intercept of this line? c. Use the slope of the line and the answer to part b to write an equation of the line. d. Do the coordinates of point (2, 3) satisfy the equation written in part c? 2 27. a. Draw the line through |
(3, 5) whose slope is. 3 b. What appears to be the y-intercept of this line? c. Use the slope of the line and the answer to part b to write an equation of the line. d. Do the coordinates of point (3, 5) satisfy the equation written in part c? 28. a. Is (1, 1) a point on the graph of 3x 2y 1? b. What is the slope of 3x 2y 1? c. Draw the graph of 3x 2y 1 using point (1, 1) and the slope of the line. d. Why is it easier to use point (1, 1) rather than the y-intercept to draw this graph? 9-8 GRAPHING DIRECT VARIATION Recall that, when two variables represent quantities that are directly proportional or that vary directly, the ratio of the two variables is a constant. For example, when lemonade is made from a frozen concentrate, the amount of lemonade, y, that is obtained varies directly as the amount of concentrate, x, that is used. If we can make 3 cups of lemonade by using 1 cup of concentrate, the relationship can be expressed as y x 5 3 1 or y 3x Graphing Direct Variation 375 The constant of variation is 3. The equation y 3x can be represented by a line in the coordinate plane, as shown in the graph below. x 0 1 2 3 3x 3(0) 3(1) 3(2) 3(35 –5 x 5 Here, the replacement set for x and y is the set of positive numbers and 0. Thus, the graph includes only points in the first quadrant. In our example, if 0 cups of frozen concentrate are used, 0 cups of lemonade can be made. Thus, the ordered pair (0, 0) is a member of the solution set of y 3x. Using any two points from the table above, for example. (0, 0) and (1, 3), we can write: Dy Dx 5 3 2 0 1 2 0 5 3 Thus, we see that the slope of the line is also the constant of variation. An equation of a direct variation is a special case of an equation written in slopeand m the constant of variation, k. A direct variintercept form; that is, ation indicates that y is a multiple of x. Note that the graph of a direct variation is always a line |
through the origin. 1 5 3 b 5 0 The unit of measure for the lemonade and for the frozen concentrate is the same, namely, cups. Therefore, no unit of measure is associated with the ratio which, in this case, is or 3. or 6 cups 2 cups 5 3 1 lemonade concentrate 5 lemonade concentrate 5 9 cups 3 cups 5 3 1 There are many applications of direct variation in business and science, and it is important to recognize how the choice of unit can affect the constant of variation. For example, if a machine is used to pack boxes of cereal in cartons, the rate at which the machine works can be expressed in cartons per minute or in cartons per second. If the machine fills a carton every 6 seconds, it will fill 10 cartons in 1 minute. The rate can be expressed as: lemonade concentrate 5 or 3 1 3 cups 1 cup 5 3 1 1 carton 6 seconds 5 1 6 carton/second or 10 cartons 1 minute 10 cartons/minute Here each rate has been written as a unit rate. 376 Graphing Linear Functions and Relations As shown at right, each of these rates can be represented by a graph, where the rate, or is the constant of variation, slope of the line. The legend of the graph, that is, the units in which the rate is expressed, must be clearly stated if the graph is to be meaningful. y 10 10 Seconds 6 x 1 2 3 4 5 Minutes Cartons per second Cartons per minute EXAMPLE 1 The amount of flour needed to make a white sauce varies directly as the amount of milk used. To make a white sauce, a chef used 2 cups of flour and 8 cups of milk. Write an equation and draw the graph of the relationship. Solution Let x the number of cups of milk, and y the number of cups of flour. Then: y x 5 2 8 8y 2x y y 1 4 x 1 4(2) 1 4(4) 1 4(6) 2 8x 1 4x Milk EXERCISES Writing About Mathematics 1. A chef made the white sauce described in Example 1, measuring the flour and milk in ounces. Explain why the ratio of flour to milk should still be 1 : 4. 2. When Eduardo made white sauce, he used 1 cup of flour and 1 quart of milk. Is the ratio of flour to milk in Eduardo’s recipe the same as in the recipe used by the chef in the |
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