# Ratios and Proportions GEOMETRY LESSON 8 1 For

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Ratios and Proportions GEOMETRY LESSON 8 -1 (For help, go to Skills Handbook, page 718 and Lesson 5 -1. ) Simplify each ratio. 2 8 6 10 1. 4 2. 12 3. 8 4. 10 5. 20: 30 6. 8 to 2 7. 2 to 8 8. 12 : 9 9. Draw a triangle. Then draw its three midsegments to form a smaller triangle. How do the lengths of the sides of the smaller triangle compare to the lengths of the sides of the larger triangle? 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 Solutions 2 ÷ 2 1 1. Divide the numerator and denominator by 2. = 4 ÷ 2 2 8 ÷ 4 2 6 ÷ 2 3 2. Divide the numerator and denominator by 4. = 12 ÷ 4 3 3. Divide the numerator and denominator by 2. = 8 ÷ 2 4 10 ÷ 10 4. Divide the numerator and denominator by 10. = 1 10 ÷ 10 20 5. Rewrite 20: 30 as . Divide the numerator and denominator by 10. 30 20 ÷ 10 = 2 30 ÷ 10 3 8 6. Rewrite 8 to 2 as . Divide the numerator and denominator by 2. 2 8 ÷ 2 4 = or 4 to 1. 2 ÷ 2 1 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 Solutions (continued) 2 7. Rewrite 2 to 8 as . Divide the numerator and denominator by 2. 8 2 ÷ 2 1 = or 1 to 4. 8 ÷ 2 4 12 8. Rewrite 12 : 9 as . Divide the numerator and denominator by 3. 12 ÷ 3 4 = or 4 : 3. 9 ÷ 3 3 9 1 9. The lengths of the sides of the smaller triangle are the lengths of the 2 sides of the larger triangle. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 A scale model of a car is 4 in. long. The actual car is 15 ft long. What is the ratio of the length of the model to the length of the car? Write both measurements in the same units. 15 ft = 15 X 12 in. = 180 in. length of model 4 in. 4 1 = 15 ft. = = = 180 in. 180 45 length of car The ratio of the length of the scale model to the length of the car is 1 : 45. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 Complete: If a 12 b ? = , then =. 4 b 12 ? a 12 4 b 4 b = 4 b Multiply each side by 4 b. ab = 48 Cross-Product Property ab 48 = 12 a Divide each side by 12 a. b 4 = 12 a Simplify. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 Solve each proportion. 2 n x + 1 a. 5 = 35 x b. 3 = 2 5 n = 2(35) Cross-Product Property 3 x = 2(x + 1) Cross-Product Property 5 n = 70 Simplify. 3 x = 2 x + 2 Use the Distributive Property. Divide each side by 5. x = 2 Subtract 2 x from each side. n = 14 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 1 2 Two cities are 3 in. apart on a map with the scale 1 in. = 50 mi. Find the actual distance. Let d represent the actual distance. Map distance (in. ) = 1 in. actual distance (mi. ) 50 mi. 1 3 2 d = 1 50 Substitute. 1) d = 50(3 Use the Cross-Product Property. d = 175 Simplify. 2 The cities are actually 175 mi. apart. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 Pages 418 -421 Exercises 3 1. 1 : 1000 7. a 15. 6. 875 2. 1 : 370 8. 4 a 16. 7. 2 3– 11. Answers may vary. Samples are given. 9. 4 17. 7. 2 3 7 11. b + 4 18. 7. 5 19. 14 4 3 12. 4 20. 7 5. b 13. 1 6. b 14. 4 7 10. 3. 3 b 4. 4 a 4 2 3 21. 125 mi 22. about 75 mi 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 23. about 135 mi 24. about 67. 5 mi 32. 30 40. 12 18 33. b 2 19 41. 7 25. 18 in. by 22. 5 in. 26. 13 : 6 34. Check students’ work. 35. 6 42. 3 43. 16 cm 27. 5 : 4 36. 32 44. 0. 348 in. 28. 4 : 3 37. 16. 5 45. 9; 18 5 48 38. 6 30. 7 39. 4 46. 9; 12 47. 8; 21 29. 3 31. 9 4 8 -1 25

Ratios and Proportions GEOMETRY LESSON 8 -1 48– 51. Answers may vary. Samples are given. 48. 52. Elaine did not convert units, and thought the ratios equaled 1. 53. b or a 50. d c 54. c or a d b 55. c + 2 d 51. d 49. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 a c 56. = (Given); ad = bc (Cross-Product Prop. ); b d bc ad bc = ad (Symm. Prop. of =); = ac ac b d (Div. Prop. of =); = (Simplify) a c 57. = (Given); ad = bc (Cross-Prod. Prop. ); b d ad bc a b = (Div. Prop. of =); = (Simplify) cd cd c d a c a 61. x = 3; y = 21 62. C 63. G 64. D 65. H c 58. = (Given); + 1 = + 1 (Add. Prop. of =); b d a b c d a + b c + d + = + (Subst. ); = (Simplify) b b d d b d 2. 75 23. 2 66. [2] a. = 16 x 59. x = 5; y = 24 (OR equivalent proportion) 60. x = 3; y = 15 b. 135 km 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 66. (continued) [1] incorrect proportion OR incorrect distance 70. 72. trapezoid 67. 60% 71. rhombus, parallelogram 73. I and III 68. 25% 69. rectangle, parallelogram 8 -1 74. II and III 75. a. If an is not acute, then it does not measure between 0 and 90.

Ratios and Proportions GEOMETRY LESSON 8 -1 75. (continued) b. If an does not measure between 0 and 90, then it is not acute. 76. a. If two lines are not ||, then they are not coplanar. s 77. a. If two are not compl. , then the s are not both acute. s b. If two are not both acute, then they are not compl. b. If two lines are not coplanar, then they are not ||. 8 -1

Ratios and Proportions GEOMETRY LESSON 8 -1 1. A scale model of a boat is 9 in. long. The boat’s actual length is 60 ft. Find the ratio of the length of the scale model to the length of the boat. 1 : 80 10 15 2. Solve the proportion = . 12 8 x 3. A map uses the scale 1 cm = 20 mi. A county is 90 mi wide. How wide 1 is the county on the map? 4 cm 2 x 7 If = , complete each of the following. y 11 4. = ? x ? 11 7 5. 7 y = ? 11 x y x+y ? 6. = y 11 18 8 -1

Similar Polygons GEOMETRY LESSON 8 -2 (For help, go to Lessons 4 -1 and 8 -1. ) Solve each proportion. 1. ABC HIJ. Name three pairs of congruent sides. 3 x 2. = 4 8 2 8 3. = x 24 x 1 4. = 9 8 -2 3 10 2 5. = 25 x

Similar Polygons GEOMETRY LESSON 8 -2 Solutions 1. Choose pairs of letters in the same position on each part of the congruence statement ABC HIJ: AB HI, AC HJ, BC IJ 3 x 3 8 3 2 2. Multiply both sides of = by 8: x = = = 6 4 8 4 1 1 1 8 2 1 3. Simplify in = : = ; use the Cross Product Property: 24 x 3 (1)x = (2)(3); simplify: x = 6 x 1 1 9 9 4. Multiply both sides of = by 9: x = = = 3 9 3 3 10 10 2 2 2 5. Simplify in = : = ; use the Cross Product Property: 25 25 x (2)x = (5)(2); simplify: 2 x = 10; divide by 2: x = 5 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 ABC ~ XYZ Complete each statement. a. m B = ? b. BC = ? YZ XZ Two polygons are similar if (1) corresponding angles are congruent and (2) corresponding sides are proportional. a. B Y and m Y = 78, so m B = 78 because congruent angles have the same measure. BC AC b. Because AC corresponds to XZ, . = YZ 8 -2 XZ

Similar Polygons GEOMETRY LESSON 8 -2 Determine whether the parallelograms are similar. Explain. Check that the corresponding sides are proportional. AB 2 = JK 4 BC 1 = KL 2 CD 2 = LM 4 DA 1 = MJ 2 Corresponding sides of the two parallelograms are proportional. Check that corresponding angles are congruent. / B corresponds to K, but m B = m K, so corresponding angles are not congruent. Although corresponding sides are proportional, the parallelograms are not similar because the corresponding angles are not congruent. 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 If ABC ~ YXZ, find the value of x. Because ABC ~ YXZ, you can write and solve a proportion. AC BC = YZ XZ Corresponding sides are proportional. x = 12 40 30 Substitute. 12 x = 40 Solve for x. 30 x = 16 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 A painting is 24 in. wide by 36 in. long. The length of a postcard reduction of the painting is 6 in. How wide is the postcard? The postcard and the painting are similar rectangles, so you can write a proportion. Let x represent the width of the postcard. Postcard width = postcard length Painting width painting length x 6 = 24 36 6 x = 24 36 Corresponding sides are proportional. Substitute. Solve for x. x = 4 The postcard is 4 in. wide. 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 The dimensions of a rectangular tabletop are in the Golden Ratio. The shorter side is 40 in. Find the longer side. Let represent the longer side of the tabletop. 40 = 1. 1618 Write a proportion using the Golden Ratio. = 64. 72 Cross-Product Property 1 The table is about 65 in. long. 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 Pages 425 -429 Exercises 9. yes; KLMJ ~ PQNO; 3 5 10. yes; ABCD ~ FGHE; 4 1. JHY 2. R 5 3. JXY 4. HY 5. JT s 11. No; corr. are not . 7 12. yes; ABC ~ FED; 19. 70 mm 13. x = 4; y = 3 21. 2 : 3 22. 3 : 2 5 14. x = 20; y = 17. 5; z = 7. 5 6. HY 20 36 7. no; 30 =/ 52 8. yes; QRST ~ XWZY; 3 4 17. 6. 6 in. by 11 in. 18. 3. 6 in. by 6 in. 15. x = 16; y = 4. 5; z = 7. 5 16. x = 6; y = 8; z = 10 8 -2 20. 54 in. by 87. 37 in. 23. 50 24. 50

Similar Polygons GEOMETRY LESSON 8 -2 25. 70 2 26. 3 27. 7. 5 m 28. 5. 6 m s 29. Yes; corr. and sides are . 30. equal sign, similarity symbol; Answers may vary. Sample: figures are similar with = areas. 31. 20 ft 32. x = 60, y = 25 33. 2. 6 cm 34. 3 : 4 35. 3 : 1 s 40. sides of 2 cm; of 60° and 120° s 41. sides of 2 cm; of 60° and 120° s 42. sides of 3. 2 cm; of 60° and 120° s 43. sides of 0. 8 cm; of 60° and 120° s 44. sides of 1 cm; of 60° and 120° s 45. sides of 3 cm; of 60° and 120° 36. 2 : 1 37. 1 : 2 38. 4 : 3 39. 2 : 3 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 46. 16. 2 in. 50. Answers may vary. Sample: 47. 6. 2 in. 48. No; corr. sides are not in proportion. 49. Yes; explanations may vary. Sample: The ratios of radii, diameters, and circumferences of 2 circles are =. 51. a. (1) Corr. sides of ~ polygons are proport. (2) Subst. (3) Cross-Product Prop. (4) Subtr. Prop. 52. a. 21, 34, 55, 89, 144, 233, 377 b. 1. 6; 1. 625; 1. 6154; 1. 6190; 1. 6176; 1. 6182; 1. 6180; 1. 6181; 1. 6180 c. The ratios get closer to the golden ratio. 53. D b. Length cannot be negative. 54. C c. 1. 6180 55. A 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 56. [2] a. 12 cm b. m KLM = 38 [1] incorrect length OR incorrect angle measure 61. no; only one pair of || sides 62. Yes; both pairs of sides are ||. 63. CEA, FED, BCD 57. 7 y 58. 7 9 59. y + 9 9 64. BD, FA 65. 8 66. 69 60. Yes; the diagonals bisect each other. 8 -2

Similar Polygons GEOMETRY LESSON 8 -2 Use the trapezoids below for Exercises 1– 3. DFHN ~ BMLP. Complete each statement. 1. m H = ? 74 2. x = ? 3. m D = 49 ? 99 4. A 4 -in. by 6 -in. drawing is enlarged to fit on a poster that measures 20 in. by 24 in. What are the dimensions of the largest drawing possible? 16 in. by 24 in. 5. A rectangle with a perimeter 20 cm has a side 4 cm long. A rectangle with perimeter 40 cm has a side 8 cm long. Determine whether the rectangles are similar. If they are, give the similarity ratio. If they are not, explain. yes; 1 : 2 6. The longer side of the golden rectangle is 20 ft. Find the length of the shorter side, rounded to the nearest tenth. 12. 4 ft 8 -2

Proving Triangles Similar GEOMETRY LESSON 8 -3 (For help, go to Lessons 4 -2 and 4 -3. ) Name the postulate or theorem you can use to prove the triangles congruent. 1. 2. 3. 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 Solutions 1. All corresponding sides are marked congruent, so Side-Side, or SSS postulate 2. Two sides and an included angle of one are congruent to two sides and an included angle of the other, so Side-Angle-Side, or SAS Postulate 3. Two angles and an included side of one are congruent to two angles and an included side of the other, so Angle-Side-Angle, or ASA Postulate 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 MX AB. Explain why the triangles are similar. Write a similarity statement. Because MX AB, AXM and BXK are both right angles, so AXM BXK. A B because their measures are equal. AMX ~ BKX by the Angle-Angle Similarity Postulate (AA ~ Postulate). 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 Explain why the triangles must be similar. Write a similarity statement. YVZ WVX because they are vertical angles. VY 12 1 VZ 18 1 = = and = = , so corresponding sides are proportional. VW 24 2 VX 36 2 Therefore, YVZ ~ WVX by the Side-Angle-Side Similarity Theorem (SAS Similarity Theorem). 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 ABCD is a parallelogram. Find WY. Because ABCD is a parallelogram, AB || DC. XAW ZYW and AXW YZW because parallel lines cut by a transversal form congruent alternate interior angles. Therefore, AWX ~ YWZ by the AA ~ Postulate. Use the properties of similar triangles to find WY. WY WZ = WA WX 10 WY = 4 5 10 WY = 5 4 Corresponding sides of ~ triangles are proportional. Substitute. Solve for WY. WY = 12. 5 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 Joan places a mirror 24 ft from the base of a tree. When she stands 3 ft from the mirror, she can see the top of the tree reflected in it. If her eyes are 5 ft above the ground, how tall is the tree? Draw the situation described by the example. TR represents the height of the tree, point M represents the mirror, and point J represents Joan’s eyes. Both Joan and the tree are perpendicular to the ground, so m JOM = m TRM, and therefore JOM TRM. The light reflects off a mirror at the same angle at which it hits the mirror, so JMO TMR. Use similar triangles to find the height of the tree. 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 (continued) JOM ~ TRM AA ~ Postulate RM TR = OM JO TR = 24 3 5 24 TR = 5 3 5 Corresponding sides of ~ triangles are proportional. Substitute. Solve for TR. TR = 40 The tree is 40 ft tall. 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 Pages 435 -438 Exercises 1. Yes; ABC ~ FED; SSS ~ Thm. 2. No; more info. is needed. 2 3. Ex. 1: (for ABC to 3 FED); Ex. 2: Not possible; s the aren’t necessarily similar. 4. yes; FHG ~ KHJ; AA ~ Post. 6 10 5. No; . =/ 4 3 20 25 6. No; . =/ 45 55 7. Yes; APJ ~ ABC; SSS ~ Thm. or SAS ~ Thm. 8. Yes; NMP ~ NQR; SAS ~ Thm. 32 45 9. No; . =/ 22 30 10. AA ~ Post. ; 7. 5 8 -3 11. AA ~ Post. ; 2. 5 12. AA ~ Post. ; 12 5 6 13. AA ~ Post. ; 12 14. AA ~ Post. ; 8 15. AA , Post. ; 15 16. SAS ~ Thm. ; 12 m 17. AA ~ Post. ; 220 yd 18. AA ~ Post. ; 15 ft 9 in.

Proving Triangles Similar GEOMETRY LESSON 8 -3 19. AA ~ Post. ; 90 ft 20. Answers may vary. Sample: She can measure her shadow s and use ~ to find the length of the shadow of the proposed building. 21. 151 m 22. a. trapezoid b. RSZ ~ TWZ; AA ~ Post. s 23. a. No; the corr. may not be . b. Yes; every isosc. rt. is a 45°-45° 90° . Therefore, by AA ~ Thm. they are all ~. 27. No; there is only one of each . 28. 45 ft 24. Yes; GMK ~ SMP; SAS ~ Thm. 31. 2 : 1 32. 12 : 7 25. Yes; AWV ~ AST; SAS ~ Thm. 26. Yes; XYZ ~ MNK; SSS ~ Thm. 8 -3 29. Check students’ work. 30. 3 : 2 33. 4 : 3 34. 3 : 1

Proving Triangles Similar GEOMETRY LESSON 8 -3 35. 3 : 2 36. 3 : 2 37. 2 : 1 38. 3 : 1 39. 6 : 1 40. Check students’ work. Draw ABC. Construct A R. Construct RS such that RS = 3 AB, and RT such that RT = 3 AC. Connect points S and T. 41. a. 98 m; 98 m b. 420 m 2; 420 m 2 c. s No; the given are a counterexample to this conjecture, since the sides are not in proportion. 42. 1. 1 || 2, EF AF, BC AF (Given) 2. EFD and BCA s are right . (Def. of ) 8 -3 42. (continued) 3. EFD BCA s (All rt. are . ) 4. BAC EDF (If || lines, then s corr. are . ) 5. ABC ~ DEF (AA ~) BC EF 6. = AC DF (Def. of similar)

Proving Triangles Similar GEOMETRY LESSON 8 -3 BC EF 45. C 43. 1. = , EF AF, 43. (continued) AC DF 6. 1 || 2, BC AF (Given) s (If corr. are , 46. I then || lines. ) 2. ACB and DFE s 47. [2] a. All corr. of s are rt. . s 44. 1. RT • TQ = MT • TS similar are . (Def. of ) (Given) Q X; subtract m V 3. ACB DFE MT RT and m L from 2. = (Prop. of s (All rt. are . ) TQ TS 180 to get m Q Proportions) = m X. 4. ABC ~ DEF (SAS ~) 3. RTM STQ b. 52 s (Vert. are . ) 5. BAC EDF [1] incorrect (Def. of similar) 4. RTM ~ STQ explanation OR (SAS ~ Thm. ) incorrect measure 8 -3

Proving Triangles Similar GEOMETRY LESSON 8 -3 48. [4] a. ABC ~ ADE; AA , Post. 6 h b. = ; 18 ft 10 30 [3] appropriate methods and appropriate diagram but incorrect solution 48. (continued) [2] correct diagram and similarity statement OR correct proportion [1] correct height, no work shown 49. E 50. P 51. Y 52. ZY 8 -3 53. EZ 54. YZ 55. x-values may vary. Sample: W(–b, c); Z( –b, –c) 56. x-values may vary. Sample: W(–b, c); Z( –a, 0) 57. 6 < x < 24

Proving Triangles Similar GEOMETRY LESSON 8 -3 Are the triangles similar? If so, write a similarity statement and name the postulate or theorem you used. If not, explain. The congruent angle is not included between 1. 2. the proportional sides, so you cannot conclude that the triangles are similar. ABX ~ CDX by SAS ~ Theorem 3. 4. Find the value of x. 16 MRT ~ XRW by AA ~ Postulate 5. When a 6 ft tall man casts a shadow 18 ft long, a nearby tree casts a shadow 93 ft long. How tall is the tree? 31 ft 8 -3

Similarity in Right Triangles GEOMETRY LESSON 8 -4 (For help, go to Lesson 8 -1 and page 355. ) Solve each proportion. x 8 18 24 4 x 1. = 5. = 10 5 2 3 x 7 3 9 2. = 6. = m 8 15 4 18 x w 20 3. = 7. = 2 9 51 x 17 13 9 27 4. = 8. = 6 a 9. Draw a right triangle. Label the triangle ABC with right angle C. Draw the altitude to the hypotenuse. Label the altitude CD. Name the two smaller right triangles that are formed. 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 Solutions x 18 18 1. Multiply both sides of = by 8; x = = = 6 8 24 24 1 3 2 x 2 7 14 2 2. Multiply both sides of = by 7; x = = or 4 3 7 3 1 3 3 15 18 3. = ; (15)x = 18(4) (Cross Products); 4 x 72 24 4 Simplify: 15 x = 72; divide by 15: x = = or 4 15 5 5 51 17 4. = ; (51)13 = 17(x) (Cross Products); x 13 Simplify: 663 = 17 x; divide by 17: x = 39 4 5 4 4 x 5. Multiply both sides of = by 5; x = = = 2 10 5 10 8 -4 1 2

Similarity in Right Triangles GEOMETRY LESSON 8 -4 Solutions (continued) 3 9 6. = ; (3)8 = 9(m) (Cross Products); 8 m 24 9 8 3 3 w 20 20 2 40 4 7. Multiply both sides of = by 2; w = = or 4 2 9 9 1 9 9 Simplify: 24 = 9 m; divide by 9: m = = or 2 2 9 6 27 a 3 2 27 a 8. Simplify in = ; 3(a) = 27(2) (Cross Products); Simplify: 3 a = 54; divide by 3: a = 18 9. The altitude divides ABC into two smaller right triangles, ADC and CDB. 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 Find the geometric mean of 3 and 12. 3 = x x 12 Write a proportion. x 2 = 36 Cross-Product Property x = 36 Find the positive square root. x = 6 The geometric mean of 3 and 12 is 6. 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 Solve for x and y. Use Corollary 1 of Theorem 8 -3 (The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse) to find x. 2 = 6 6 x 2 x = 36 Write a proportion. Cross-Product Property x = 18 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 (continued) Use Corollary 2 of Theorem 8 -3 (The altitude to the hypotenuse of a right triangle intersects it so that the length of each leg is the geometric mean of the length of its adjacent segment of the hypotenuse and the length of the entire hypotenuse) to find y. x y = y 2 + x y 18 = 2 + 18 y Write a proportion. Substitute 18 for x. y 2 = 360 Cross-Product Property. y = 360 Find the positive square root. y = 6 10 Write in the simplest radical form. 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 At a golf course, Maria drove her ball 192 yd straight toward the cup. Her brother Gabriel drove his ball straight 240 yd, but not toward the cup. The diagram shows the results. Find x and y, their remaining distances from the cup. Use Corollary 2 of Theorem 8 -3 to solve for x. x + 192 240 = 240 192(x +192) = 2402 192 x + 36, 864 = 57, 600 192 x = 20, 736 x = 108 8 -4 Write a proportion. Use the Cross-Product Property. Use the Distributive Property. Solve for x.

Similarity in Right Triangles GEOMETRY LESSON 8 -4 (continued) Use Corollary 2 of Theorem 8 -3 to solve for y. y x + 192 = x y y 108 + 192 = 108 y y 300 = 108 y y 2 = 32, 400 y = 32, 400 Write a proportion. Substitute. Simplify. Use the Cross-Product Property. Find the positive square root. y = 180 Maria’s ball is 108 yd from the cup, and Gabriel’s ball is 180 yd from the cup. 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 Pages 435 -438 Exercises 1. 6 9. s 17. 10 2. 2 10 3. 4 3 10. r 11. c 18. 6 3 19. 12 4. 12 5. 14 2 12. a; a 13. h 6. 25 7. 6 6 14. b 15. 9 20. 60 21. a. 18 mi b. 24 mi 22. KNL; JNK 8. 3 7 16. 20 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 23. a. 4 cm b. 24. a. 26. 4 3 27. 14 28. 4 c. Answers may vary. Sample: Draw a 10 -cm segment. 2 cm from one endpoint, construct a of length 4 cm. Connect to form a . b. They are =. Explanations may vary. Sample: The altitude and hyp. segments are sides of s two isosc. . 25. (10, 6), (– 2, 6) 8 -4 29. 14 30. 1 31. 2. 5 32. 10 33. 121

Similarity in Right Triangles GEOMETRY LESSON 8 -4 34. x = 12; y = 3 7; z = 4 7 35. x = 12 5; y = 12; z = 6 5 36. x = 4; y = 2 13; z = 3 13 37. 12 2 38. a. Given b. Corollary 2 of Thm. 8 -3 38. (continued) c. Cross-Product Prop. d. Addition Prop. of = e. Dist. Prop. f. Segment Add. Post. g. Subst. 39. about 6. 5 m 40. 5 3 cm 8 -4 12 9 41. h = 5, a = , h 1 = , 5 5 h 2 = 16 5 42. 1 = 2 13, 2 = 3 13, h = 13, a = 6 43. 1 = 2 = 6 2, h = 12, h 2 = 6 20 3 25 3 h 1 = 3, h 2 = 16 3 60 45. 1 = 5, a = , 13 25 144 h 1 = , h 2 = 13 13 44. 2 = , h = ,

Similarity in Right Triangles GEOMETRY LESSON 8 -4 16 4 7 46. 2 = , h = , 3 3 49. a. 52. 4. 5 12 53. 5 7 a = 7, h 2 = 3 47. 1 = 8 5, 2 = 4 5, Given: rt. ABC with alt. CD; h 1 = 4, h 2 = 20 Prove: AC • BC = AB • CD 48. 1 = 6, h = 12, a = 3 3, h 2 = 9 b. Yes; AC • BC = 2 X area ABC and AB • CD = 2 X area ABC. 54. 60 13 55. 120 17 56. 420 29 57. D 50. 3 58. G 59. C 51. 4 60. H 8 -4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 61. [2] a. Solve for x by making a proport. from similar rt. . s The proportion x 9 is = . Now 13 x cross multiply and solve for x. b. 3 13 [1] incorrect proportion OR incorrect x value 62. a. RNM ~ PNQ b. AA ~ Post. 63. a. PRQ ~ ACB b. SSS ~ Thm. 64. not similar 65. 7. 5 66. 10 67. x = 5; y = 8 8 -4 68. x = 6; y = 9 69. x = 3; y = 4

Similarity in Right Triangles GEOMETRY LESSON 8 -4 1. Find the geometric mean of 32 and 2. 8 2. Find the geometric mean of 6 and 20. 2 30 3. Solve for x. 4. Solve for x and y. 5 3 x = 9, y = 16 5. The roof of a house forms a right angle, with each side of the roof measuring 28 ft in length. Find the width and the height of the roof. width = 28 2 ft, height = 14 2 ft 8 -4

Proportions in Triangles GEOMETRY LESSON 8 -5 (For help, go to Lesson 8 -2. ) Find the value of x. The triangles in each are similar. 1. 2. 3. 4. 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 Solutions 30 + 15 84 3 1. Simplify in = : = ; 84(1) = (3)x; 15 x 15 divide by 3: x = 28 cm x 1 x x + 8 2. Write a proportion: = ; 4(x + 8) = 22 x; Simplify: 4 x + 32 = 22 x: 4 22 32 16 7 Subtract 4 x: 32 = 18 x; divide by 18; x = = or 1 mm 18 9 9 7 7 + x 3. Write a proportion: = ; 7(12) = 5(7 + x); Simplify: 84 = 35 + 5 x; 5 12 49 Subtract 35: 49 = 5 x; divide by 5: x = or 9. 8 in. 5 6 6 7. 5 2 7. 5 4. Simplify in = : = ; (2)x = 7. 5(3); 6 + 3 x 3 x Simplify: 2 x = 22. 5; divide by 2: x = 11. 25 ft 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 Find y. CN CM = NA MB Side-Splitter Theorem 12 = 12 y 6 Substitute. 10 y = 72 Cross-Product Property y = 7. 2 Solve for y. 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 The segments joining the sides of trapezoid RSTU are parallel to its bases. Find x and y. Because RSTU is a trapezoid, RS || UT. 6 5 = x 12. 5 Corollary to the Side-Splitter Theorem 5 x = 75 x = 15 Cross-Product Property Solve for x. x 12. 5 = 9 y 15 12. 5 = 9 y Corollary to the Side-Splitter Theorem Substitute 15 for x. 15 y = 112. 5 Cross-Products Property y = 7. 5 Solve for y. x = 15 and y = 7. 5. 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 Find the value of x. IG IK = GH HK Triangle-Angle-Bisector Theorem 24 x = 40 30 Substitute. 24 (30) = x 40 Solve for x. x = 18 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 Pages 448 -452 Exercises 1. 7. 5 9. 3 17. KS 2. 8 3. 5. 2 10. 9. 6 11. 6 18. SQ 19. JP 4. d 5. c 12. 4. 8 13. 35 20. KP 21. KM 6. b 7. d 14. 3. 6 40 15. 7 22. PM 23. JP 8. 7. 5 16. 12 24. LW 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 25. 559 ft 26. 671 ft 27. 3. 8 cm and 9. 2 cm 28. Answers may vary. Sample: 9 cm and 13. 5 cm 29. x = 18 m; y = 12 m 30. a. b. isosceles; - Bisector Thm. 31. 20 32. 2. 5 33. 9 34. a. b. c. 36. 4. 5 cm or 12. 5 cm 37. 6 38. 2. 5 AB BC WX XY AB WX = BC XY 35. Measure AC, CE, and BD. Use the Side. Splitter Thm. Write the AC AB proport. = and CE solve for AB. 8 -5 BD 39. 19. 5 40. h = 10. 0, h 1 = 4. 3, h 2 = 5. 7 41. h = 13. 0, 1 = 5. 3, 2 = 11. 9 42. 1 = 5. 0, h 1 = 3. 8, h 2 = 9. 2

Proportions in Triangles GEOMETRY LESSON 8 -5 43. 2 = 7. 1, h 2 = 5. 0, h = 10. 0 44. h = 17. 0, h 1 = 11. 1, h 2 = 5. 9 45. h 2 = 4, 1 = 5. 7, 2 = 5. 7 46. The ratio of the legs of a 30°-60°-90° is 3. Let the rt. bisector divide the hyp. into segments x and y. Then by the - - Bis. Thm. , x = 3, so x = 3 y. y 47. 1. Given 2. Prop. of Proportions 3. Segment Add. Post. 4. Reflexive Prop. of 5. SAS ~ Thm. s s 6. Corr. of ~ are . s 7. If corr. are , s lines are ||. 8 -5 6 9 48. Yes; since = , 10 15 the segments are || by the Converse of the Side-Splitter Thm. 28 24 =/ 49. No; . 12 10 15 20 50. Yes; since = , 12 16 the segments are || by the Converse of the Side-Splitter Thm.

Proportions in Triangles GEOMETRY LESSON 8 -5 51. a. A midsegment of a connects the midpts. of 2 opp. sides. b. 51. (continued) 1. ABCD (Given) 2. AE || BF and ED || FC (Def. of ) 3. AD BC (Opp. sides of are . ) Given: ABCD with EF connecting the midpts. of AD and BC Prove: AB || EF; EF || CD 4. E and F are midpts. of AD and BC. (Given) 1 5. AE = ED = AD; 2 1 BF = FC = BC 2 (Def. of midpt. ) 8 -5 51. (continued) 6. AE = BF, ED = FC (Subst. ) 7. ABFE and EFCD s are (If one pair of opp. sides of a quad. is and ||, it is a . ) 8. AB || EF and EF || CD (Opp. sides of a are ||. )

Proportions in Triangles GEOMETRY LESSON 8 -5 51. (continued) c. 52. D 55. [4] a. 53. F n + 1 20 54. [2] = ; 28 35 35 n + 35 = 560; Given: ABCD with midsegment EF Prove: EF bisects AC and BD. Since AB || EF || DC by part (b), and EF bisects AD, by the Side-Splitter Thm. , EF bisects AC and BD. n = 15 [1] correct proportion solved incorrectly b. x = 6 4 5 6 y 5 x = 4. 8 cm = 4 y = 7. 5 cm 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 55. (continued) [3] correct drawings and proportions with one computational error [2] one possibility drawn and done correctly OR two correct drawings [1] one correct drawing OR one correct proportion 65. RT = SV = 48 66. RT = SV = 6. 5 56. m 57. m 58. c 59. h 60. x = 24; y = 12 3 61. x = 9; y = 9 3 62. x = 15; y = 30 63. RT = SV = 38 64. RT = SV = 37 8 -5

Proportions in Triangles GEOMETRY LESSON 8 -5 Solve for x in each diagram. 1. 2. 3. 11. 25 8 24 Use the diagram for Exercises 4– 6. 4. Find x. 9. 6 5. Find y. 5 6. Explain how you know that AM is not the angle bisector of BAC. AB BM If it were, would equal = 1. But AC > AB. AC MC 8 -5

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 (For help, go to Lesson 1 -7. ) Find the perimeter and area of each figure. 1. 2. 3. Find the perimeter and area of each rectangle with the given base and height. 4. b = 1 cm, h = 3 cm 5. b = 2 cm, h = 6 cm 6. b = 3 cm, h = 9 cm 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 Solutions 1. All sides are congruent, so each side measures 7 in. The perimeter is the sum of all the sides. 4 7 = 28 in. ; the area is the square of a side s. A = s 2 = 72 = 49 in. 2 2. Two sides measure 4 m and two sides measure 8 m. The perimeter is the sum of all sides: 2 4 + 2 8 = 2(4 + 8) = 2(12) = 24 m; the area is the product of the base b and the height h: A = bh = (8)(4) = 32 m 2 3. Use the Pythagorean Theorem: a 2 + b 2 = c 2 (6)2 + (8)2 = c 2 36 + 64 = c 2 = 100 c = 10 cm; the perimeter is the sum of all the sides: 10 + 8 + 6 = 24 cm; the area is half the product of the base b 1 1 2 and the height h; A = bh = (8)(6) = 24 cm 2 2 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 Solutions (continued) 4. The perimeter is the sum of the sides: 1 + 3 + 1 + 3 = 8 cm; the area is the product of the base b and the height h: A = bh = (1)(3) = 3 cm 2 5. The perimeter is the sum of the sides: 2 + 6 + 2 + 6 = 16 cm; the area is the product of the base b and height h: A = bh = (2)(6) = 12 cm 2 6. The perimeter is the sum of the sides: 3 + 9 + 3 + 9 = 24 cm; the area is the product of the base b and height h: A = bh = (3)(9) = 27 cm 2 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 The triangles below are similar. Find the ratio (larger to smaller) of their perimeters and of their areas. The shortest side of the triangle to the left has length 4, and the shortest side of the triangle to the right has length 5. 5 From larger to smaller, the similarity ratio is . 4 By the Perimeters and Areas of Similar Figures Theorem, the ratio of the 5 25 5 perimeters is also , and the ratio of the areas is , or . 2 2 4 4 8 -6 16

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 The ratio of the lengths of the corresponding sides of two 8 2. Find regular octagons is . The area of the larger octagon is 320 ft 3 the area of the smaller octagon. All regular octagons are similar. Because the ratio of the lengths of the corresponding sides of the regular 8 82 64 octagons is , the ratio of their areas is , or . 2 3 64 320 = 9 A 64 A = 2880 A = 45 3 9 Write a proportion. Use the Cross-Product Property. Divide each side by 64. The area of the smaller octagon is 45 ft 2. 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 Benita plants the same crop in two rectangular fields. Each 1 dimension of the larger field is 3 times the dimension of the smaller 2 field. Seeding the smaller field costs $8. How much money does seeding the larger field cost? The similarity ratio of the fields is 3. 5 : 1, so the ratio of the areas of the fields is (3. 5)2 : (1)2, or 12. 25 : 1. Because seeding the smaller field costs $8, seeding 12. 25 times as much land costs 12. 25($8). Seeding the larger field costs $98. 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 The areas of two similar pentagons are 32 in. 2 and 72 in. 2 What is their similarity ratio? What is the ratio of their perimeters? Find the similarity ratio a : b. a 2 32 b 2 = 72 The ratio of the areas is a 2 : b 2. a 2 16 = 2 b 36 Simplify. a 4 2 = = b 6 3 Take the square root. The similarity ratio is 2 : 3. By the Perimeters and Areas of Similar Figures Theorem, the ratio of the perimeters is also 2 : 3. 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 Pages 456 -459 Exercises 1. 1 : 2; 1 : 4 9. $384 17. 3 : 1; 9 : 1 2. 4 : 3; 16 : 9 3. 2 : 3; 4 : 9 10. $47. 20 11. 1 : 2; 1 : 2 18. 2 : 5; 4 : 25 19. 2 : 3; 4 : 9 4. 3 : 5; 9 : 25 5. 24 in. 2 12. 5 : 2; 5 : 2 13. 7 : 3; 7 : 3 20. 7 : 4; 49 : 16 21. 6 : 1; 36 : 1 6. 56 m 2 7. 59 ft 2 14. 3 : 4; 3 : 4 15. 4 : 1; 4 : 1 8. 439 m 2 16. 1 : 10; 1 : 10 22. 800 cm 2 23. While the ratio of lengths is 2 : 1, the ratio of areas is 4 : 1. 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 24. 0. 3 cm 2 31. x = 8 cm, y = 12 cm 25. 252 m 2 26. x = 2 cm, y = 3 cm 27. x = 2 2 cm, y = 3 2 cm 28. x = 4 cm, y = 6 cm 8 3 29. x = cm, 3 y = 4 3 cm 30. x = 4 2 cm, y = 6 2 cm 1 32. 2 in. by 12 in. ; 4 3 in. by 16 in. 33. a. Check students’ work. b. Check students’ work. c. Estimates may vary. Sample: 205 m 2 8 -6 34. Ratio of small to large is 1 : 2. 35. a. 5 2 b. 36. a. b. 37. a. b. 25 4 8 3 64 9 2 1 4 1

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 38. Answers may vary. Sample: The proposed playground is more than adequate. The number of students has approx. doubled. The proposed playground would be four times larger than the original. 39. Answers may vary. Sample: a. b. 114 mm; 475 mm 2 c. 456 yd; 7600 yd 2 40. a. 6 3 cm 2 b. 54 3 cm 2; 13. 5 3 cm 2; 96 3 cm 2 8 -6 41. Always; ~ rectangles with = perimeters have a similarity ratio of 1, so they are . 42. Sometimes; a 1 -by-8 rect. and 2 -by-4 rect. have the same areas, but are not ~. 43. Never; if they were , both measures would be the same. If they were ~, but not , their areas would not be =.

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 44. Sometimes; if they are , they have = areas and are ~. 45. 225 46. 675 47. 54 52. 50 cm 2 53. 690 units 2 57. y = –x – 2; 54. 480 units 2 55. 44. 4 units 2 56. y = 3 x – 4 48. 17303 49. 162. 3 50. x = 45; y = 107. 5 1 3 51. 5 cm; 12 cm 8 -6 3 4 58. y = x;

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 59. y = – 4 x + 5; 60– 61. Eq. forms may vary. Samples are given. 60. y = – 3 x + 1 8 61. y + 1 = x 3 8 -6

Perimeters and Areas of Similar Figures GEOMETRY LESSON 8 -6 1. For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas. 4 16 perimeters: ; areas: 9 81 2. The triangles are similar. The area of the larger triangle is 48 ft 2. Find the area of the smaller triangle. 27 ft 2 3. The similarity ratio of two regular octagons is 5 : 9. The area of the smaller octagon is 100 in. 2 Find the area of the larger octagon. 324 in. 2 4. The areas of two equilateral triangles are 27 yd 2 and 75 yd 2. Find their similarity ratio and the ratio of their perimeters. 3 : 5; 3 : 5 5. Mulch to cover an 8 -ft by 16 -ft rectangular garden costs $48. At the same rate, what would be the cost of mulch to cover a 12 -ft by 24 -ft rectangular garden? $108 8 -6

Similarity GEOMETRY CHAPTER 8 Page 464 1. 16 2. 4. 2 3. 2 4. x = 42; y = 138; z = 9 5. 4 6. x = 63; y = 8 7. PRQ ~ TWV; SSS ~ Thm. 8. No; corr. sides are not in prop. 9. ABC ~ FDE; AA ~ Thm. 10. AD= AB DC BC 11. Answers may vary. AD CD Sample: = 12. 6 m CD DB 15. 6 16. 6 2 17. Check students’ work. 18. 16 19. 10 20. 5 5 11 21. 10 13. 5 cm 14. 5 6 8 -A 2 3

Similarity GEOMETRY CHAPTER 8 22. Answers may vary. Sample: To measure the height of a tree on a sunny day, measure the length of the shadow it casts. Then measure the length of the shadow that you cast. Since you already know your height, you can use ~ to s write and solve a proportion. 23. 49 : 64 24. 9 : 4 8 -A

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