MCRBG-0901-SK.qxd 5-25-2001 11:17 AM Page 182 CHAPTER 9 Lesson 9.1 9.1 Practice and Applications (pp. 531–534) Think and Discuss (p. 525) 11. Q Q S 1. AB and ED are both to BD; in a plane, 2 lines to the same line are . 2. According to Theorem 4.8, if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, the two triangles are congruent. 13. 1. mL 180 30 60 90 JKL is a right triangle. 3. leg, altitude x T R 14. 4 x x 9 36 x2 1 3 x6 15. x 5 x 3 15 x2 x 15 17. SQR ~ TQS ~ TSR; RQ 18. GFE ~ HFG ~ HGE; EH 1 19. CBA ~ DBC ~ DCA L x 12 16 12 x x3 5 3 4. S 16. ZYX ~ WYZ ~ WZX; ZW altitude leg, altitude T 12x 400 y 1 K x 20 20 12 x 33 J hypotenuse R 12. SRQ ~ TSQ ~ TRS Skill Review (p. 526) 2. S 5. B 3x 3 5x x9 60° 3x 9 5x A 9 2x 20. GEF ~ HGF ~ HEG 30° C Two angles of JKL are congruent to two angles of ABC, therefore ABC JKL by the AA Sim. Post. 9 x 2 144 16x 20 x 20 25 25x 400 x 16 21. LKJ ~ MLJ ~ MKL x 32 32 15 Developing Concepts Activity (p. 527) 15x 1024 3. All of the triangles are similar. x 68.3 9.1 Guided Practice (p. 531) 1. geometric mean 4. JM 10. 5. JK 2. KML; JMK 6. LJ 7. KM 22. RSQ ~ TRQ ~ TSR 3. MK 8. KM 9. LK 72 97 72 DC DC 53.4 FD 53.4 FD 43.6 DF 48.3 40 x 40 32 32x 1600 x 50 23. CBA ~ DBC ~ DCA 4 x 4 x x2 16 x4 182 Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0901-SK.qxd 5-25-2001 11:17 AM Page 183 Chapter 9 continued 24. HGE ~ FHE ~ FGH xy h 5.5 32. 18 h 5.5 18 5.5 x2 126 x 126 314 11.2 25. x 9 3 x 26. x2 27 x 16 12 20 5 m7 7 5 27. 20x 192 7m 49 25 x 9.6 7m 74 x 33 m 10 28. 4 7 Z h 64.4 ft Not drawn to scale CD 1.5 2 2.5 CD 1.2 m AC 2 2 2.5 DB 1.5 1.5 2.5 AD 1.6 m DB 0.9 m 1 2 1.20.9 Area of DAC 121.61.2 0.96 m2 Area of CAB 1221.5 e c e d d 16 c 1 16 12 4 1 124 e 3 3 4 34. Given ABC is a right triangle and altitude CD is drawn e 33 e2 49 15 44 e 715 4 663 y y 24 32 24 z 40 y2 1600 2 3 1.5 m2 4 2 24 32 32 x 24x 1024 30. Y 1 5 2 ft 0.54 m2 1 4 x 42 W 5.5h 30.25 324 Area of DCB 16 14 14 c c 12 h 18 ft 33. DCB ~ DAC ~ CAB 16c 196 29. X wy xy wy zy x 18 7 x 24z 1280 z 53 y 40 1 3 8 x9 18 x9 x2 18x 81 144 35. From Ex. 34, CBD ~ ACD. Corresponding side x2 18x 63 0 lengths are in proportion, so x 21x 3 0 x 21 0 x 21 x30 x3 Solution: x 3 31. About 76 cm; ABC and ADC are congruent right triangles by the SSS Congruence Postulate, so AC is a perpendicular bisector of BD. By Geometric Mean Theorem 9.3, the altitude from D to hypotenuse AC divides AC into segments of lengths about 23.7 cm and 61.1 cm. By Geometric Mean Theorem 9.2, the length of the altitude to the hypotenuse of each right triangle is about 38 cm long, so the crossbar BD should be about 2 38, or 76 cm long. Copyright © McDougal Littell Inc. All rights reserved. to hypotenuse AB; DBC and DCA are right triangles by the definition of right triangles; CDB ACB because all right angles are congruent; B B by reflexive property for angles; therefore ACB ~ CDB by the AA Similarity Postulate; ADC ACB because all right angles are congruent; A A by the reflexive property for angles; therefore ACB ~ ADC by the AA Similarity Postulate; mACD mDCB 90 by the Angle Addition Postulate; mDCB mB 90 because the two acute angles in a right triangle are complementary; m∠ACD m∠B by the Transitive and Subtraction Properties of Equality, so ACD B by the def. of congruent ; CDA CDB because all right angles are congruent; so DCA ~ DBC by the AA Similarity Postulate. CD BD . CD AD 36. From Ex. 34, ABC ~ CBD and ABC ~ ACD. Corresponding side lengths are in proportion, so AB BC AB AC . and BC BD AC AD 37. Values of the ratios will vary, but will not be equal. The theorem says they are equal. 38. The ratios are equal. 39. The ratios are equal when the triangle is a right triangle but are not equal when the triangle is not a right triangle. Geometry Chapter 9 Worked-out Solution Key 183 MCRBG-0902-SK.qxd 5-25-2001 11:16 AM Page 184 Chapter 9 continued AB CB , , CB DB AB AC CB AB , . These proportions should be true: AC AD CB DB AC AB and . Now drag C to change the value of AC AD AB CB AB mC (so m∠C 90) and recalculate , , , and CB DB AC 40. Using the right triangle, calculate the values of CB AC AB . The values of the ratios will vary but AD CB DB AC AB and . AC AD Lesson 9.2 9.2 Guided Practice (p. 538) 1. Sample answer: 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 legs. 2. A, C 3. 22 12 x2 5 4. x2 82 102 x2 36 x2 5 x 41. D x6 no DC 12 42. 12 24 AD 24 6 AD 18 144 DC 24 yes 5. 42 x2 82 6. 52 d2 62 x2 48 d2 11 C x 43 3.3 ft no DC 6 43. Method 1 9.2 Practice and Applications (pp. 538–541) Measure the distance from the ground to the person’s eye level (DC) and the distance from the person to the BC AC building (AC). Use the proportion and solve AC DC for BC (the height of the building). One advantage of this method is you only need two measurements. One disadvantage is you need a friend to help. Measure the length of the building’s shadow (QS), the height of the pole (NP) and the length of the pole’s MP NP shadow (MP). Use the proportion and solve QS RS for RS (the height of the building). One advantage is it can be done by one person. One disadvantage is it must be done when the building and pole cast a shadow. 652 722 x2 8. 62 x2 92 4225 5184 x2 36 x2 81 7. 9409 45. 14 x2 78 46. d2 18 99 n ± 13 x2 64 d2 81 d ±9 x ±8 47. If the measure of one of the angles of a triangle is greater 1521 1 51. A in.2 1 2 12 50. A 74.5 31.5 cm2 10. 92 402 x2 7921 81 1600 x2 x2 6400 1681 x2 x2 x 80 yes 41 x yes 11. 72 x2 92 12. 22 32 x2 49 x2 81 4 9 x2 x2 32 13 x2 x 42 no 13 x no 2 13. 8 x 16 14. 202 x2 292 64 x2 256 400 x2 841 x2 192 x2 441 2 2 x 83 48. If the corresponding angles of two triangles are congru49. A 2 612 no 392 x2 892 than 90, then the triangle is obtuse; true. ent, then the two triangles are congruent; false. x 35 yes 9.1 Mixed Review (p. 534) 44. n2 169 x2 45 x2 97 x 9. Method 2 36 d 11 no x 21 yes 2 15. 14 14 x 16. 82 x2 162 196 196 x2 64 x2 256 2 135 2 392 62.5 m2 x2 142 x x2 192 x 83 no 184 Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0902-SK.qxd 5-25-2001 11:17 AM Page 185 Chapter 9 continued 17. 62 b2 102 36 b2 18. 32 b2 52 9 100 b2 b2 64 b4 42 62 x2 32 72 x2 16 36 x2 9 49 x2 52 x2 58 x2 21. b8 A 144 256 t2 81 144 t2 400 t2 225 t2 20 t 15 t s2 324 s2 900 182 302 20. 22. 122 400 r2 10,201 r2 9801 s 576 1225 s2 1369 26. 52 b2 142 81 b2 144 25 b2 196 b2 63 27. b5 d1 12 8 20 d2 5 5 10 A 122010 100 m2 b 319 A cm2 1 2 319 5 33. 42 b2 8.52 b2 72.25 16 56.25 b 7.5 m base a b 3 7.5 10.5 m 9.7 ft 2.5 ft 3 ft 36 inches 1521 c2 39 in. c b 7.2 a3m 10 ft 1296 225 c2 b2 51.75 a2 25 16 9 wall 362 152 c2 12.25 b2 64 28. a2 42 52 ladder 2 ft 6 in. 30 inches 3.52 b2 82 25.2 cm2 Distance from pitcher’s plate to home is 50 feet. The distance from second base to home is about 91.9 feet so the distance from second to the pitcher’s plate is 91.9 50 or about 41.9 feet. base of the ladder from the wall 10 is 4 or 2.5 feet. The ladder, if placed 2.5 feet from the wall, will reach 100 6.25 9.7 feet up the wall. 32.7 m2 A 1277.2 91.9 ft c 32. The minimum distance of the b2 171 b 37 35.7 b2 25 r 468 25. 92 b2 122 37 8450 c2 1012 r2 219,024 s 12 A 144 b2 169 354,025 r2 573,049 s2 144 1 2 9 31. 652 652 c2 5952 r2 7572 24. 16 30. 122 b2 132 r 99 s 24 352 s2 372 1 2 810 104 cm2 t2 r2 202 2 23. b2 100 36 64 58 x 92 10 cm 62 b2 102 t2 162 b 6 cm 11 4 7 213 x 10 cm 10 cm b2 16 b8 19. 25 12 8 4 122 29. 39 in. 39 in. 16 in. 94 in. 34. 300 ft 3600 in. 300 ft 1 in. 3601 in. 36002 h2 36012 h2 7201 h 84.9 in. 35. r 3 in.4 6 in.2 12 in.2 12 12 24 48 in. A 1210.54 21 m2 Copyright © McDougal Littell Inc. All rights reserved. Geometry Chapter 9 Worked-out Solution Key 185 MCRBG-0902-SK.qxd 5-25-2001 11:17 AM Page 186 Chapter 9 continued 36. 182 302 r2 49. 73 147 324 900 54. yes 2 r2 1224 r2 37. The area of the large square is a b2. Also, the area of the large square is the sum of the areas of the four congruent right triangles plus the area of the small square, or 412 a b c2. Thus, a b2 412 a b c2, and so a2 2ab b2 2ab c2. Subtracting 2ab from each side gives a2 b2 c2. 38. The area of the trapezoid is 2 a b2. Also, the area of 1 the trapezoid is equal to the sum of the areas of the two congruent right triangles plus the area of the isosceles triangle or 12 a b 12 a b 12 c2. Thus b2 a b 12c2, and so 2 2ab b2 2ab c2. Subtracting a 2ab from each side gives a2 b2 c2. 1 2 a 39. a. AB 82 42 b. AB 144 64 45 BD 208 100 9.8 ft No. The longest space in the room is the diagonal of the room which is only about 17.5 ft long. h2 The length of the diagonal of the base is l2 w2. The length of the diagonal of the box is 40. l 2 w 22 h2 l 2 w 2 h2. 2 2 The length of one side of the rhombus is 12a 12b or 12a2 b2. Multiplying the length of one side by 4 gives the perimeter of the rhombus, which is 412a2 b2 or 2a2 b2. 41. 56. Sample answer: slope of PQ 3 slope of RS; slope 3 of QR 8 slope of PS. Both pairs of opposites sides are parallel, so PQRS is a parallelogram by the definition of a parallelogram. Lesson 9.3 Activity 9.3 Investigating Sides and Angles of Triangles (p. 542) Construct Constructions may vary. Investigate Values in tables may vary. Conjecture AC2 BC2 < AB2 when mC > 90 9.3 Guided Practice (p. 545) 1. Sample answer: 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, then the triangle is a right triangle. 2. acute: c2 < 242 182 c < 30 right: c2 242 182 c 30 obtuse: c2 > 242 182 c > 30 3. C P 2a2 b2; a x, b 0.75x 4. D 5. D Since the two numbers are not equal, the triangles formed by the crossbars and the sides are not right triangles so the crossbars are not perpendicular. 40 x2 0.5625x2 40 1.5625x2 40 1.25x 9.3 Practice and Applications (pp. 546–548) 32 x 8. 972 ? 652 722 a 32 cm, b 0.7532 24 cm 2 45. 22 8 2 2 44. 14 14 2 46. 413 208 47. 549 1225 7921 7921 right 43. 9 9 2 9. 892 ? 802 392 9409 9409 9.2 Mixed Review (p. 541) 42. 6 6 6. A 7. No; the sum of 222 382 1928, while 452 2025. 80 2x2 0.75x2 186 5 of QR 4 slope of PS. Both pairs of opposite sides are parallel, so PQRS is a parallelogram by the definition of a parallelogram. AC2 BC2 > AB2 when mC < 90 17.5 ft yes 53. no 4. AC2 BC2 AB2 when mC 90 413 BD 80 16 c. d 52. no 11 Method 2 uses less ribbon. w2 51. no 55. Sample answer: slope of PQ 2 slope of RS; slope 35.0 in. r l2 50. no 2 48. 49 36 2 Geometry Chapter 9 Worked-out Solution Key 2 10. 23 right ? 20.8 10.5 2 2 11. 262 ? 12 52 529 < 542.89 26 26 not right right Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0903-SK.qxd 5-25-2001 11:16 AM Page 187 Chapter 9 continued 12. 33 2 ? 22 52 13. 435 2 ? 202 132 27 < 29; not right 14. 202 992 ? 1012 5 560 < 569; not right 15. 212 28 2 distance from B to A 3 42 7 62 2 ? 35 400 9801 ? 10,201 441 784 ? 1225 10,201 10,201; right 1225 1225; right 16. 102 172 ? 262 30. distance from B to C 3 02 7 32 52 distance from A to C 4 02 6 32 5 17. not a triangle 52 52 52 2 100 289 ? 676 25 25 50; therefore ABC is a right triangle by the Converse of the Pythagorean Theorem. 389 < 676; obtuse 18. 42 67 2 ? 92 24. 2 2 ? 7 31. Computing slopes is easier because it involves two calcu- 16 67 ? 81 13 36 ? 49 83 > 81; acute 49 49; right 20. 162 302 ? 342 22. 19. 13 6 2 21. 102 112 ? 142 256 900 ? 1156 100 121 ? 196 1156 1156; right 221 > 196; acute 42 52 ? 52 lations, not three. Computing slopes also does not involve square roots. 23. 172 144 2 2 ? 145 41 > 25; acute 21,025 21,025; right ? 502 25. 5 52 ? PR 3 62 4 22 Q(5, 0) 2 289 20,736 ? 21,025 492 45 P(3, 4) 16 25 ? 25 102 PQ 3 52 4 02 y 32. 35 6 x R(6, 2) RQ 6 52 2 02 55 45 2 35 2 5.52 ? 55 2 100 2401 ? 2500 5 25 ? 30.25 80 45 125 2501 > 2500; acute 30 < 30.25; obtuse The triangle is a right triangle. 26. Rectangle; the quadrilateral has two pairs of congruent opposite sides. Each triangle formed by either diagonal is a right triangle because in each case 142 82 265 2; 260 260. Therefore, the quadrilateral has four right angles. The quadrilateral is a rectangle. PQ 1 42 2 12 y 33. 26 P(1, 2) Q(4, 1) PR 1 02 2 12 1 10 x 2 R(0, 1) RQ 0 42 1 12 27. Square; the diagonals bisect each other, so the quadri- lateral is a parallelogram. The diagonals are congruent, so the parallelogram is a rectangle. 12 12 2 2, so the diagonals intersect at right angles to form perpendicular lines; thus, the parallelogram is also a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square. 28. Rhombus; the diagonals bisect each other so the quadri- lateral is a parallelogram. so the diagonals intersect at right angles to form perpendicular lines so the parallelogram is a rhombus. 32 42 52, 63 3 ; 40 4 73 4 slope of BC ; 30 3 3 4 Since 1, AC BC , so ABC is a right 4 3 angle. Therefore, ABC is a right triangle by the definition of a right triangle. 29. slope of AC Copyright © McDougal Littell Inc. All rights reserved. 17 10 2 17 2 ? 26 2 27 > 26 The triangle is an acute triangle. 34. Since 22 32 < 42, ABC is obtuse and ABC is obtuse. 1 and ABC are a linear pair and are therefore supplementary. By the definition of supplementary angles, mABC m1 180. Since ABC is obtuse, mABC > 90. Therefore, m1 < 90. 1 is an acute angle by definition of an acute angle. 35. Since 10 2 22 < 42, ABC is obtuse and C is obtuse. By the Triangle Sum Theorem, mA mABC mC 180. C is obtuse, so mC > 90. It follows that mABC < 90. Vertical angles are congruent, so mABC m1. By substitution, m1 < 90. By the definition of an acute angle, 1 is acute. Geometry Chapter 9 Worked-out Solution Key 187 MCRBG-0903-SK.qxd 5-25-2001 11:16 AM Page 188 Chapter 9 continued 36. If a, b, and c are a Pythagorean triple, then a2 b2 c2. Let k represent a positive integer. Multiplying both sides of the equation by k2 gives the equation k2a2 b2 k2c2, or k2a2 k2b2 k2c2 by the Distributive Property. So ka2 kb2 kc2 by a property of powers. Since k 0, ka, kb, and kc represent the side lengths of a right triangle by the Converse of the Pythagorean Theorem. 37. A, C, D 38. rectangle 1192 1202 39. 1692 ? 66492 ? 48002 46012 ✓ 28,561 28,561 44,209,201 44,209,201 Statements x2 1. a2 Reasons b2 1. Pythagorean Theorem 2. c2 a2 b2 2. Given 3. c2 x2 3. Substitution prop. of equality 4. c x 4. A property of square roots 5. mN mR 5. Converse of Hinge Thm. 6. N is a right angle. 6. Given, def. of rt. angle, def. of obtuse angle, subst. prop. of equality ✓ 18,5412 ? 13,5002 12,7092 7. LMN is a right ✓ 343,768,681 343,768,681 43. 7. Def. of right triangle (N is the largest .) triangle. 2 2 2 40. 714 ? 599 403 44. 77 36 2 2 822 402 ? 912 2 ? 85 509,796 < 521,210 so the is acute. 5629 1296 ? 7225 6724 1600 ? 8281 Cincinnati is not directly north of Tallahassee. It is northwest of Tallahassee. 6925 < 7225 8324 > 8281 ABC is obtuse DEF is acute A is obtuse D is acute 41. Reasons 1. Pythagorean Theorem A 2. Given 45. mA > 90, so mB mC < 90 3. Substitution property of equality mD < 90, so mE mF > 90 5. Converse of the Hinge Theorem 6. Given, def. of right angle, def. of acute angle, and substitution property of equality 7. Def. of acute triangle (C is the largest angle of ABC .) 42. Given: In ABC, c2 > a2 b2 Prove: ABC is an obtuse triangle. Plan for Proof: Draw right triangle PQR with side lengths a, b, and hypotenuse x. Compare lengths c and x. Q C a B P a 1. Given 2. NPMQ 2. Def. of altitude 3. ∠MPN and ∠QPN are 3. Def. of perpendicular right angles. 4. MPN and QPN are 4. Def. of right triangle 5. MN2 s2 t2, 5. Pythagorean Theorem 6. MN2 NQ2 6. Addition and s2 t2 r2 t2 s2 2t2 r2 R Reasons 1. x2 a2 b2 1. Pythagorean Theorem 2. c2 > a2 b2 2. Given 3. c2 > x2 3. Substitution prop. of equality 4. c > x 4. A property of square roots 5. mC > mP 5. Converse of Hinge Thm. 6. C is an obtuse angle. 6. Given, def. of rt. angle, def. of obtuse angle, subst. prop. of equality triangle. Reasons 1. NP is an altitude. NQ2 r2 t2 Statements 7. ABC is an obtuse Statements x b b 46. right triangles. A c B 7. Def. of obtuse triangle (C is the largest of ABC .) Substitution properties of equality 7. t is the geometric mean of 7. Given r and s. 8. r t t s 8. Def. of geometric mean 9. t2 rs 9. Cross product prop. 10. MN NQ 2 2 10. Substitution prop. of s2 2rs r2 r s2 11. r s MQ equality 11. Given (diagram) 12. MN NQ MQ 2 2 2 12. Substitution prop. of equality 13. MQN is a right triangle. 13. Converse of the Pythagorean Thm. 188 Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0903-SK.qxd 5-25-2001 11:16 AM Page 189 Chapter 9 continued 9.3 Mixed Review (p. 549) 47. 22 48. 6 Conjecture 2 44 211 3. The length of the hypotenuse is the product of the length of one side and 2. 8 48 43 6 84 221 15 6 90 310 49. 14 50. 3 311 51. 11 11 12 53. 18 54. 24 8 Exploring the Concept 4. Triangles may vary. 4 45 52. 5 5 6. triangle with side length 4 cm: side lengths: 2 cm, 4 cm, 23 cm 12 42 22 32 2 triangle with side length 6 cm: side lengths: 3 cm, 6 cm, 33 cm 8 46 26 26 6 3 triangle with side length 8 cm: side lengths: 4 cm, 8 cm, 43 cm 55. an enlargement with center C and scale factor 3 5 56. reduction with center C and scale factor 57. 5x x 36 7 4 2y y 11 4x 36 Conjecture 7. y 11 33 43 longer leg 23 : 3; 3; 3 shorter leg 2 3 4 x9 ratio of hypotenuse:longer leg:shorter leg 2:3:1 Quiz 1 (p. 549) 1. CDB ~BDA ~ CBA 3. 9 15 15 AC 9.4 Guided Practice (p. 554) BD 9 15 20 1. Right triangles with angle measures 454590 and 306090 180 15BD AC 25 2. According to the AA Similarity Postulate, since two BD 12 5. 3 x 7 2 2. BD 4. 9AC 225 2 hypotenuse 4 6 8 : 2; 2; 2 shorter leg 2 3 4 angles of one triangle are congruent to two angles of the other triangle, the two triangles are similar. 6. x 12 18 2 2 x 210 2 2 x 65 7. x2 62 182 3. true 4. false 8. true 9. x 42 7. true 10. a 2; b 23 9 2k 8. 2192 ? 1682 1402 9 47,961 > 47,824 No; the square of the longest side is larger than the sum of the squares of the smaller sides. Lesson 9.4 2 k 92 kh 2 9.4 Practice and Applications (pp. 554–556) Activity 9.4 Developing Concepts (p. 550) 12. x 5; y 52 13. a 123; b 24 14. e 22 Exploring the Concept 15. d 1. Triangles may vary. 2. side length 3 cm: 6. true hk 11. x 122 5. false 2 8 d 42 32 32 c2 c 32 cm side length 4 cm: 42 42 c2 c 42 cm c 42 17. q 162; r 16 19. f 3 8 f side length 5 cm: 52 52 c2 c 52 cm h Copyright © McDougal Littell Inc. All rights reserved. 16. c 5; d 53 18. m 12; p 63 20. n 6 83 3 163 3 Geometry Chapter 9 Worked-out Solution Key 189 MCRBG-0903-SK.qxd 5-25-2001 11:16 AM Page 190 Chapter 9 continued 2.52 x2 52 21. 39. Stage 1: x2 x2 12 6.25 x2 25 5 cm 5 cm 2x2 1 x2 18.75 x 1 x 2 2 x 4.3 cm 2.5 cm 5 cm 92 92 x2 x 9 in. Stage 3: x2 x2 9 in. 81 81 x2 162 x2 1 4 x 1 2 2x2 676 x x2 338 x 18.4 in. Stage 4: x2 x2 27.7 ft2 1 8 x2 1 16 x 1 4 of the stage. 31.2 41. ft2 26. A 5 23 17.3 m2 1 1 16 simplify. 28 22 x2 1 22 1 2n 2 1 1 8 40. The pattern of the lengths is x 1 2 43 8 1 2 6 63 2 2x2 x 26 in. 12 1 4 23. x2 x2 262 25. A x2 2x2 9 in. 12.7 in. x 1 2 9 in. s 9 in. 12 2x2 1 x 2 22. 4s 36 24. A Stage 2: x2 x2 , where n the number Substitute 8 for n into the formula 1 2n and 27. A 6 2 4 23 41.6 ft2 9.4 Mixed Review (p. 557) 28. x 3 cm 1.7 cm 42. Let x length of the third side; 14 9 > x; x 9 > 14; 1 30. y 23 3.5 cm; x 2 31. r 2 s 3 29. x 1.42 2.0 cm 2 4 cm 5 cm < x < 23 cm t2 v 6 43. Q1, 2 u 5 w 7 46. B0, 10 I used the Pythagorean theorem in each right triangle, working from left to right. 32. Going from left to right: triangle 1 33. Going from left to right: triangle 3 Theorem, x2 x2 DE2; 2x2 DE2; DE 2x2 2 x by a property of square roots. Thus the hypotenuse is 2 times as long as a leg. → 36. Construct CD on BC so that CD BC a. Then ADC ABC by the SAS Cong. Post. B D and BAC CAD, because they are corresponding parts of congruent triangles. Therefore mD 60 and mCAD 30. mBAD mBAC mCAD 30 30 60. BAD is equiangular so it is also equilateral. Since it is equilateral, AB 2a. If BC a and AB 2a, then AC 2a2 a2 3 a. The side lengths are in the following ratio: hypotenuse:longer leg: shorter leg 2a:3 a:a. Therefore, in a 306090 triangle, the hypotenuse is twice as long as the shorter leg and the longer leg is 3 times as long as the shorter leg. 37. C 190 44. P8, 3 45. A4, 5 47. AA Similarity Postulate 48. SAS Similarity Theorem 49. SSS Similarity Theorem Math & History 1. area of triangles: 4 34. n 1 35. Let DF x. Then EF x. By the Pythagorean 2 12 ab 2ab area of square: b a2 b2 2ab a2 2. 2ab b2 2ab a2 c2 a2 b2 c2 Lesson 9.5 9.5 Guided Practice (p. 562) 1. sin A BC AB cos A AC AB tan A BC AC 2. The value of a trigonometric ratio depends only on the measure of the acute angle, not on the particular right triangle used to compute the value. 3. 4 5 4. 3 5 5. 4 3 6. 3 5 7. 4 5 8. 3 4 38. A; 6 63 12 28.4 cm Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0904-SK.qxd 5-25-2001 11:38 AM Page 191 Chapter 9 continued 9. sin 25 7 d 29. sin 23 30. cos 36 9.5 Practice and Applications (pp. 562–565) 45 0.8491 53 28 0.5283 53 cos R 45 tan R 1.6071 28 28 sin S 0.5283 53 45 cos S 0.8491 53 28 tan S 0.6222 45 6 0.6 10 11. sin B 6 tan B 0.75 8 6 0.6 10 cos A 3 12. sin X 13 0.8321 3 1.5 2 tan X tan D tan J sin Y 13 2 0.5547 0.5547 13 24 0.96 25 7 0.2917 24 sin F 24 0.96 25 2 0.8944 2 2 1 cos H 15. sin J 2 cos X cos D 5 tan G 8 1.3333 6 tan A 7 0.28 25 0.8321 7 cos F 0.28 25 14. sin G 8 sin A 0.8 10 2 0.6667 3 13 13. sin D 8 0.8 10 cos B tan Y 3 cos Y 2 5 5 34 cos K 24 tan F 3.4286 7 1 cos G tan H 0.8575 cos J sin K 0.4472 5 1 sin H 0.8944 5 1.6667 3 5 0.4472 1 0.5 2 3 34 0.5145 3 0.5145 34 3 5 0.8575 tan K 0.6 34 5 16. 0.7431 17. 0.9744 18. 6.3138 19. 0.4540 20. 0.3420 21. 0.0349 22. 0.9781 23. 0.8090 24. 0.4245 25. 0.4540 26. 0.8290 27. 2.2460 28. tan 37 6 y y 8.0 Copyright © McDougal Littell Inc. All rights reserved. cos 23 t 13.3 d 16.6, or about 17 ft 10. sin R t 34 sin 37 6 x s 31.3 4 r tan 36 r 4.9 31. sin 65 s 4 s 2.9 t 8 cos 65 t 7.3 32. sin 70 s 34 u 8 u 3.4 9 v tan 70 v 9.6 9 w w 3.3 6 33. sin 22 x tan 22 6 y y 14.9 x 16.0 1 2 34. A 2 22 22 4 cm 35. A 1 126.9 41.6 m2 2 1 2 36. A 116 37. tan 13 3 34.9 m2 h 58.2 38. tan 42 h 13.4 m d 40 d 36.0 m 39. vertical drop, x 5500 5018 482 ft sin 20 482 d 1409.3 ft 40. tan 55 30 26 x 30 x 26 sin 45 x 16.4 in. d 500 41. sin 45 d 714.1 m 42. tan 20 x 8 x 2.9 ft 43. sin A a c cos A b c tan A a b sin B b c cos B a c tan B b a 44. The tangent of one acute angle of a right triangle is the reciprocal of the tangent of the other acute angle. The sine of one acute angle of a right triangle is the same as the cosine of the other acute angle and the cosine of one acute angle of a right triangle is the same as the sine of the other acute angle. x 10.0 Geometry Chapter 9 Worked-out Solution Key 191 MCRBG-0904-SK.qxd 5-25-2001 11:38 AM Page 192 Chapter 9 continued 45. Procedures may vary. One method is to reason that since 52. Statements the tangent ratio is equal to the ratio of the lengths of the legs, the tangent is equal to 1 when the legs are equal in length, that is, when the triangle is a 454590 triangle. Tan A > 1 when mA > 45, and tan A < 1 when mA < 45, since increasing the measure of A increases the length of the opposite leg and decreasing the measure of A decreases the length of the opposite leg. 1. ABC is a right b c x 55 x sin 30 18 9 sin 55 BC x9 47. Reasons 55. tan 53 x 60 2 9.5 Mixed Review (p. 566) 2 2 cos 45 sin 45°2 cos 45°2 50. sin 60 3 2 3 2 enlargement; 56. Q 51. sin 13 0.2250 2 Q 2 2 3 2 10 P 5 P 4 8 6 3 2 QR 10 PR 8 R R 2 2 1 4 4 57. MNP ~ NQP ~ MQN 1 2 2 2 3 1 2 2 3 1 1 4 4 cos 13 0.9744 sin 13°2 cos 13°2 0.22502 0.97442 7 QP 15 7 18.27 NP NP 3.27 49 15QP NP2 59.7429 QP 3.27 58. x2 952 1932 Geometry x2 502 652 x2 2500 4225 28,224 x2 1725 x2 x 168 yes Chapter 9 Worked-out Solution Key NP 7.73 59. x2 9025 37,249 1 192 scale factor 6 22 22 cos 60 sin 60°2 cos 60°2 y 33.26 h 46 ft 1 3 1 4 4 2 y 60 xyh 3 1 sin 30° cos 30° 2 49. sin 45 tan 29 79.62 33.26 h 5. Substitution property of equality 2 54. C x 79.62 3. Division property of equality 2 equality 8 ;D CD 2. Pythagorean Theorem 2 dividing and simplifying fractions 5. Transitive property of 53. sin 25 1. Given 4. Subst. prop of equality and sin A cos A 5. tan A C BC 11.0 cos 30 a 3. Def. of sine and cosine c a sin A c a 4. cos A b b c 30 1 2 2. Def. of tangent 3. cos A ; sin A B 48. sin 30 a b 2. tan A trigonometric ratios. A 1. Given triangle with side lengths a, b, and hypotenuse c. 46. ABC is not a right triangle, so you cannot use the 18 Reasons x 569 no Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0904-SK.qxd 5-25-2001 11:38 AM Page 193 Chapter 9 continued 42.92 702 x2 60. Lesson 9.6 1840.41 4900 x2 Activity 9.6 Developing Concepts (p. 567) 6740.41 x2 1. B 82.1 x no 5 cm 3 cm Quiz 2 (p. 566) x 23 m 1. 4m A 3.5 m 4m cos A x 42 in. 4 in. 1. To solve a right triangle is to find the measures of all angles and the lengths of all sides of the triangle. 4 in. 2. true 4. mA 35.0 3. false 6. mA 64.2 4 in. 1 h 1.53 in. A 31.53 2 3. 3 in. 3.9 in.2 3 in. h sin A 10 x x 15.6 x 18 x 8.5 x 20 x 9.3 7. tan 11 950 d d 4887.3 ft Copyright © McDougal Littell Inc. All rights reserved. 56 65 sin B mA 59.5 4225 c2 33 65 mB 30.5 65 c sin D 9. 912 d2 1092 60 109 sin E mD 33.4 d2 3600 1.5 in. 3 in. 5. mA 79.5 7. mA 84.3 8. 332 562 c2 6. tan 25 3 0.75 4 9.6 Guided Practice (p. 570) 5.7 in. x 5. cos 62 tan A 4. The values are approximately equal. 4m 4. sin 40 4 0.8 5 3. sin1 0.6 36.9 cos1 0.8 36.9 tan1 0.75 36.9 2m 4 in. 3 0.6 5 2. sin A x 2. C 4 cm 91 109 mE 56.6 d 60 tan 40 10 y y 11.9 sin 62 y 18 y 15.9 cos 25 20 y y 22.1 10. sin 60 y 4 cos 60 y 3.5 x 4 m∠ X 30° x2 9.6 Practice and Applications (pp. 570–572) 55 73 mQ 48.9 12. sin Q 11. 482 552 QS2 5329 QS2 73 QS 13. sin S 48 73 mS 41.1 14. mA 26.6 15. mA 45 17. mA 20.5 18. mA 81.4 20. mA 65.6 21. mA 6.3 16. mA 30 19. mA 50.2 Geometry Chapter 9 Worked-out Solution Key 193 MCRBG-0904-SK.qxd 5-25-2001 11:38 AM Page 194 Chapter 9 continued 22. 202 212 AB2 841 AB2 sin B 20 29 sin A 21 29 mB 43.6 mA 46.4 7 sin E 98 7 sin D 98 mE 45 mD 45 32. tan 51 98 DE2 40 GH 2 sin H 40 mG 71.6 mH 18.4 mF 39 33. sin 34 4 m tan 34 8 9.2 sin L ML2 20.64 mL 60.4 cos K 5.9 mL 56 8 9.2 mK 29.6 69 36 35. tan1B 1.9167 36. 692 362 AB2 ML 4.5 26. 42 NQ2 13.62 NQ2 168.96 4 cos P 13.6 mN 17.1 mP 72.9 37. 38. tan x 4855 17,625 x 15.4 sin S TS2 120.25 6 12.5 mS 28.7 cos R 6 12.5 39. 2402 172 x2 57,311 x2 sin y TS 11.0 p 4.5 p 4.0 sin A 0.4626 x 239.4 in. or 19 ft 11 in. mR 61.3 cos 26 36 BC AB 6057 sin A 77.8 in. AB NQ 13.0 27. 62 TS2 12.52 69 36 mB 62.4 6057 AB2 4 sin N 13.6 4 mL 90 34 34. tan B 25. 82 ML2 9.22 q 4.5 q 2.0 e 4.8 m 7.2 6 sin G 40 6.3 GH 28. sin 26 3 e mF 90 51 9.9 DE 24. 22 62 GH2 cos 51 d 3.7 29 AB 23. 72 72 DE2 d 3 17 240 y 4.1 40. 55.2 in. 8.33° 54.6 in. 8 in. mP 90 26 41. Answers may vary. mP 64 29. sin 20 s 12 cos 20 t 12 t 11.3 s 4.1 cos 52 8.5 z z 13.8 x 10.9 mY 90 52 mY 38 31. cos 56 5 c x 32.5° tan x mT 70 x 8.5 7 11 43. tan x 8.25 9 x 42.5° 44. Sample answer: riser length: 6 in.; tread length: 12 in. mT 90 20 30. tan 52 42. tan x tan 56 c 8.9 a 5 6 12 x 26.6° 45. Sample answer: The riser-to-tread ratio affects the safety of the stairway in several ways. First, the deeper the tread the more of a person’s foot can fit on the step. This makes a person less likely to fall. Also, the smaller the angle of inclination the less steep the stairway. This makes the stairs less tiring to climb, and therefore, safer. a 7.4 mB 90 56 34 194 Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0905-SK.qxd 5-25-2001 11:38 AM Page 195 Chapter 9 continued 46. Draw an altitude, CD , from C to AB , and let CD h. In h h . In rt. BCD, sin B . Thus, b a h b sin A and h a sin B. By the substitution prop. of equality, b sin A a sin B. Dividing both a b sides by sin A sin B gives , or sin B sin A a b . sin A sin B 8. tan x rt. ACD, sin A x 51.3 north of east 9. 4, 5 4, 2 0, 3 9.7 Practice and Applications (pp. 576–579) \ \ \ 12. 47. 3, 2 48. 2, 2 49. 1, 3 50. 1, 0 51. 1, 2 52. 3, 1 y 2, 7 Q PQ 2 02 7 02 \ 53 7.3 6x 150 7y 784 x 25 y 112 3 g 10 42 56. 1 14. y x 1 P 84 7 18 k 10g 126 3, 5 Q PQ 5 22 1 62 \ 7k 1512 g 12.6 57. 13. 7 49 54. 16 y x 5 53. 30 6 55. 3, 6 JK 2 12 2 42 35 6.7 4, 4 EF 0 42 1 52 42 5.7 10. 4, 1 RS 1 52 1 22 17 4.1 11. 9.6 Mixed Review (p. 572) 5 4 34 k 216 m 7 2 1 4 8 t 11 58. x 1 88 4t m 14 5.8 P 1 t 22 59. not a triangle 15. 60. 2282 ? 2202 602 61. 8.52 ? 7.72 3.62 51,984 < 52,000 72.25 72.25 acute 62. 2632 ? 802 63. 1132 ? 1122 69,169 > 68,900 12,769 12,769 obtuse right PQ 3 72 2 62 \ Q right 2502 10, 4 y P 152 64. not a triangle 116 10.8 2 x 2 16. 6, 4 y 1 Lesson 9.7 1 x PQ 4 22 3 72 \ 52 P 7.2 9.7 Guided Practice (p. 576) 1. The magnitude of a vector is the distance from its initial point to its terminal point. The direction of a vector is the angle it makes with a horizontal line. \ \ \ Q \ 2. AB : 2, 2; PQ : 3, 3; MN : 0, 3; UV : 0, 2 \ \ \ \ 3. UV is parallel to MN ; AB is parallel to PQ 4. 2, 2 4, 2 PQ 5 12 4 22 25 4.5 2, 5 MN 3 12 4 12 29 5.4 \ 5. 4, 5 AB 4 02 5 02 41 6.4 \ 6. \ 7. Copyright © McDougal Littell Inc. All rights reserved. Geometry Chapter 9 Worked-out Solution Key 195 MCRBG-0905-SK.qxd 5-25-2001 11:38 AM Page 196 Chapter 9 continued 17. 23. 10, 10 and 50, 50 6, 4 y 1 LM 10 502 10 502 \ P x 1 PQ 5 12 0 42 \ 52 57 miles per hour 7.2 tan x Q 40 40 x 45 north of west 24. 0, 0 and 50, 40 OP 0 502 0 402 \ 18. 8, 2 y PQ 6 2 \ P Q 2 64 miles per hour 3 1 2 tan x 68 1 x 38.7 south of west 8.2 x 1 40 50 \ \ \ \ 25. EF , CD , and AB \ \ 26. EF and CD \ \ 27. EF and CD \ 28. GH and JK 29. yes; no 19. 1, 4 y P 1 1 x 30. Round 2; the vectors have the same magnitude and PQ 6 52 0 42 \ 17 4.1 Q \ opposite directions. In Round 1, team A won; since CA has a greater magnitude than CB , CA represents a greater force applied. \ \ \ 31. u : 4, 1 y \ v : 2, 4 \ v 20. u 1 3, 0 y \ u v 6, 5 uv x 1 \ P PQ 0 32 5 52 Q 9 \ 32. u : 6, 2 y \ 3 v : 5, 3 5 1 \ \ u v 1, 5 v x 1 uv u 1 x 21. 0, 20 and 60, 30 ST 60 02 30 202 \ \ 33. 61 miles per hour u : 2, 4 y uv 1 \ v : 3, 6 x 1 10 tan x 60 \ \ u v 5, 2 v u x 9.5 north of east 22. 0, 0 and 40, 50 AB 40 02 50 02 \ \ 34. u : 2, 3 y 64 miles per hour tan x 50 40 x 51.3 south of east 196 Geometry Chapter 9 Worked-out Solution Key \ v : 1, 6 u \ 1 \ u v 3, 3 v uv 1 x Copyright © McDougal Littell Inc. All rights reserved. MCRBG-0905-SK.qxd 5-25-2001 11:38 AM Page 197 Chapter 9 continued 35. 4, 11 36. 8, 7 38. 2, 3 37. 10, 10 39. 4, 4 40. 0, 0 \ 41. u : 0, 120 \ of u and the directions are the same. When k < 0, the magnitude of v is k times the magnitude of u and the direction of v is opposite the direction of u . Justifications may vary. \ \ \ v : 40, 0 42. \ 47. When k > 0, the magnitude of v is k times the magnitude 48. a. UP \ \ N s 2 W u 2 v u E s S \ b. s 10 02 2 02 10.2 mih tan x v 10 W E 10 2 10 x 11.3 north of east c. Answers may vary. DOWN \ 49. AB : 54, 24 \ \ 43. s 20 602 140 202 126.5 mih the speed at which the skydiver is falling, taking into account the breeze. 44. tan x \ \ AB BC 18, 60 \ 50. CA : 18, 60 18, 60 18, 60 0, 0 120 40 BC 362 362 50.9 ft CA 182 602 62.6 ft \ 51. AB 542 242 59.1 ft x 71.6 45. BC : 36, 36 \ The new velocity is 30, 120. UP \ Total distance 59.1 50.9 62.6 172.6 ft 52. The answer to Ex. 50 is a vector which gives the final position of the bumper car, while the answer to Ex. 51 is a number which gives the total distance traveled by the bumper car. 53. Since D and E are right angles and all right angles 10 W E 10 DOWN \ 46. JK 10 \ Sample answer: AB 3, 1 The component form must give the same magnitude as JK . AB 10. \ \ are congruent, D E . Since ABC is equilateral, AB BC. DE AC, so DBA BAC and EBC BCA by the Alternate Interior Angles Theorem. An equilateral triangle is also equiangular, so mBAC mBCA 60. By the definition of congruent angles and the substitution property of equality, DBA EBC . ADB CEB by the AAS Congruence Theorem. Corresponding parts of congruent triangles are congruent, so DB EB. By the definition of midpoint, B is the midpoint of DE. 54. x 45 55. x 120 56. x 30 y 90 y 30 2 2 x 1 x 2x 1 57. 2 2 58. x 7 x 14x 49 y 60 2 2 59. x 11 x 22x 121 60. 7 x2 49 14x x2 Copyright © McDougal Littell Inc. All rights reserved. Geometry Chapter 9 Worked-out Solution Key 197 MCRBG-0905-SK.qxd 5-25-2001 11:38 AM Page 198 Chapter 9 continued Quiz 3 (p. 580) 1. sin 25 \ 9. b 46 cos 25 b 19.4 Q 5.8 a 41.7 12 z 1 12 y \ 10. y 12 z 17.0 x 1 P tan 45 y mY 45 3. tan 40 PQ 3 02 4 12 \ a 46 mA 90 25 65 2. sin 45 PQ : 3, 5 y PQ : 7, 11 P PQ 2 52 6 52 \ 4 m 16 cos 40 m 13.4 13.0 16 q 2 q 20.9 x Q mN 90 40 50 4. sin 75 p 8 cos 75 p 7.7 q 8 11. T 6 7.6 2 f 7.6 62 2 2 2 S f 4.7 mF 90 52.1 37.9 3 12.4 32 2 12.42 mK 14.0 12. 4, 2 13. 2, 4 15. 2, 1 16. 6, 13 144.76 y 5 y 9 6 9 6 1. x \ PQ : 5, 1 PQ 3 2 \ P 2 36 9x 4 3 2 5.1 Q 2. 1 x 1 \ 8. PQ : 6, 5 y PQ 2 4 \ P 2 2 3. 2 3 7.8 2 17. 0, 3 9.1 Similar Right Triangles mL 90 14.0 76.0 y 14. 2, 8 Chapter 9 Review (pp. 582–584) 2 12.0 7. x f 2 21.76 mG 52.1 6. sin K 8 3 x 69.4 north of east q 2.1 mQ 90 75 15 5. sin G tan x y x 2 y2 45 x4 y 35 25 x x 9 x2 225 y 9 16 y y2 144 x 15 y 12 36 x 27 36 27x 1296 48 27 y y 21 z 21 27 z z2 567 z 97 x 48 23.8 Q 9.2 The Pythagorean Theorem 4. 122 162 t2 5. 82 s2 122 400 t2 s2 80 s 45 8.9 20 t yes 198 Geometry Chapter 9 Worked-out Solution Key no Copyright © McDougal Littell Inc. All rights reserved. MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 199 Chapter 9 continued 6. r2 162 342 7. 42 62 t2 r2 900 52 t2 r 30 18. sin B 42 0.6285 9 cos B 7 0.7778 9 tan B 42 0.8081 7 sin A 7 0.7778 9 cos A 42 0.6285 9 tan A 7 1.2374 42 t 213 7.2 yes no 9.3 The Converse of the Pythagorean Theorem 8. 102 ? 62 72 9. 412 ? 402 92 100 > 85 1681 1681 obtuse right 9.6 Solving Right Triangles 10. not a triangle 11. 9 2 x2 80 ? 3 45 2 2 sin Z mX 48.2 8 12 mZ 41.8 8.9 acute d 20 20. sin 50 9.4 Special Right Triangles 21. 6 32 13. leg 2 P 432 14. shorter leg 8 15 tan R 15 17 mT 61.9 \ 22. PQ : 1, 4 y PQ 1 16 17 4.1 \ P 1 x 2 Q 9.5 Trigonometric Ratios cos J 60 0.9836 61 \ 23. 11 0.1803 61 tan L 60 5.4545 11 17. sin P 35 0.9459 37 cos P 35 0.3243 37 35 tan P 2.9167 12 12 0.3243 sin N 37 35 cos N 0.9459 27 12 tan N 0.3429 35 PQ : 12, 5 y PQ 144 25 13 \ 60 0.9836 sin L 61 cos L Copyright © McDougal Littell Inc. All rights reserved. sin T 9.7 Vectors 1 18 93 813 140.3 cm2 2 11 tan J 0.1833 60 289 s2 40 17 s 15. altitude 93 cm 11 0.1803 61 mF 90 50 82 152 s2 18 in.2 1 12 6 in.; longer leg 63 in. 2 f 20 f 12.9 mR 28.1 A 32 2 122 in. cos 50 d 15.3 12. hypotenuse 232 6 16. sin J 8 12 x 45 81 < 89 A cos X 19. 82 x2 122 P 2 2 x Q \ 24. PQ : 3, 2 y PQ 9 4 13 3.6 \ 2 P Q 1 x Geometry Chapter 9 Worked-out Solution Key 199 MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 200 Chapter 9 continued \ \ \ 25. u v 14, 9 14. y u v 196 81 277 16.6 \ LM 16 9 5 \ tan x \ 9 14 M x 32.7 north of east 1. E 2. A 3. C 4. D x 5. B 15. sin 40 quadrilateral has two pairs of consecutive congruent sides, but opposite sides are not congruent. 8. PQ 25 4 29 ? sin 35 b2 12,544 PR 4 36 210 b 112 29 2 52 sin 50 10 AB AB 13.1 16. mBCA 90 35 55 9. 152 b2 1132 QR 9 16 5 CD 10 CD 6.4 7. WXYZ is a kite. The diagonals are perpendicular and the 3 4 x 36.9 south of east 1 6. DBA; DAC 210 2 tan x 1 Chapter 9 Chapter Test (p. 585) BC 40 tan 35 BC 22.9 17. 2, 8 22.9 DE DE 32.7 18. 4, 1 19. 2, 3 Chapter 9 Standardized Test (pp. 586–587) 40 < 54 acute 5 x1 20 x1 1. 10. 1 A d1d2 2 6 in. 3 in. 3 3 in. x 12 100 x2 2x 99 0 1 663 2 6 in. 6 in. x 11x 9 0 183 30 60 6 in. x 11 31.2 in.2 KL 11. sin 30 9 JL cos 30 9 KL 4.5 JL 7.8 12 DF tan 25 DF 28.4 4. B QR 25 4 6 mP 48.2 C D 5. D P 482 322 in. 2x 256 2 x 128 82 in. 12 DE 7. tan 67 DE 25.7 cos P C 6. x2 x2 162 x 8 D cos 67 x 18.8 mF 90 25 65 13. 42 QR2 62 154 in.2 3. P 2125 214 50.4 in. mK 90 30 60 12. sin 25 x9 2. A 1114 side length 6 in. sin R 4 6 Geometry Chapter 9 Worked-out Solution Key 8 y y 20.5 E 8. tan x mR 41.8 4.5 200 LM : 4, 3 L 12 9 x 53.1 A 9. sin A 8 13 mA 38.0 B Copyright © McDougal Littell Inc. All rights reserved. MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 201 Chapter 9 continued 10. 2 x 6 y 4 11 x8 5. BD is the median from point B, AD CD, BD BD, y7 B \ 11. AB 8 12 3 92 15 12. 24 2 x2 D 252 6. Sample answer: Given quadrilateral ABCD x7 p 7 24 25 56 13. sin x 7 25 sin y x 16.3 14. A 25 and it is given that AB CB. Thus, ABD CBD by the SSS Congruence Postulate. Also, ABD CBD since corresponding parts of congruent triangles are congruent. By the definition of an angle bisector, BD bisects ABC. 24 25 y 73.7 12 2724 216 square units 1 15. BD 103 3 30 BC 302 42.4 FG 15 GC 450 21.2 DC 30 AF 103 15 32.3 where mA 37, mB 143, and mC 37. Since there are 360 in a quadrilateral, mD 360° 37° 143° 37° 143. A and C are opposite angles, and B and D are opposite angles. A C and B D. Since both pairs of opposite angles are congruent, quadrilateral ABCD is a parallelogram. 7. Yes; clockwise and counterclockwise rotational symmetry of 120. 8. 26 4y 90 26 x 10 90 4y 64 x 54 16. mABC 75; mFEA 60; mBGF 135 FE 17. tan 30 32.3 32.3 cos 30 AE FE 18.6 AE 37.3 1 18. A 3032.3 15 709.5 square units 2 5 b 19. a. tan 30 b. tan 40 b 8.7 cm b 6.0 cm 5 d. tan 60 b 5 c. tan 50 b b 2.9 cm b 4.2 cm e. tan 70 5 b 5 b 20. As the sun rises, the value of b decreases. 5 5.25 x 43.6 y 324 5 180 8x 192 y 67 180 x 24 y 113 10. z 55; x 90; y 65 11. m 8 4 62 5 1 6 3 slope of perpendicular bisector midpoint: 3 4 52 1, 6 2 2 2, 2 3 6 y2 x 4 4 12. AB 16 9 5 approaches the horizon. AC 36 64 10 BC 4 121 Chapter 9 Cumulative Practice (pp. 588–589) 1. No; if two planes intersect, then their intersection is a line. The three points must be collinear, so they cannot be the vertices of a triangle. 3. never y 3x 5 180 8x 12 180 3 7 y x 4 2 22. The value of the expression increases as the sun 2. always 9. 3x 5 5x 7 180 3 y 2 x 2 4 b 1.8 cm 21. tan x y 16 125 55 scalene right triangle 13. A1, 2, B3, 5, C5, 6 4. always Copyright © McDougal Littell Inc. All rights reserved. Geometry Chapter 9 Worked-out Solution Key 201 MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 202 Chapter 9 continued 14. A2, 1 y C 24. B5, 3 C6, 5 25. x 1 23 2 22 ZX2 ZP2 12 16 ZX2 ZP 23 A 1 2 ZP 6 ZP 23 XY A B C B 4 ZX 3 23 26. 102 242 XY2 12 3XY 676 XY2 4 XY 26 XY 27. 192 ? 152 122 15. A3, 6, B7, 9, C9, 2 12 5 16. x 2 7x 24 4 x4 5 3 x3 7 26 3 29. Let x the measure of the smaller acute angle. 2y 21 19. No; in ABCD, the ratio of the length to width is 8:6 or sin x 8 17 tan x 8 15 30. sin 57 4:3. In APQD, the ratio of length to width is 4:6, or 2:3. Since these ratios are not equal, the rectangles are not similar. cos x ST 20 8 12 2 21. Yes; the ratios 9 , 12 , and 18 all equal 3 , so the triangles are ST 16.8 in. x 7 9x 8 31. TR 10.9 in. 8x 63 7x u v 4 256 16.1 v tan x \ \ 2 16 u x 2 32. Construct a circle inscribed in the triangle by bisecting two angles of the triangle. The point at which the bisectors intersect is the center of the circle. Construct a segment from the center of the circle perpendicular to a side of the triangle. The length of this segment is the radius of the desired circle. 33. 1 4 has endpoints 2, 3 3 13 13 and 4, 3; its slope is 3 . The image with scale 2 6 1 factor has endpoints 3, 2 and 6, 4.5; its slope is 2 13 . The two image segments are parallel. 6 23. The image with scale factor uv x 7.1 south of east segment 2 4.8 cm x 4.2 \ u v 2, 16 y 2 segment 1 4.2 cm 15x 63 TR 20 mS 90 57 33 similar by the SSS Similarity Theorem. 22. 15 17 cos 57 \ DE is a midsegment of ABC. By the Midsegment Theorem, DE AC. Since D E the lines containing these segments are parallel, A C BDE BAC and BED BCA, by the Corresponding Angles Postulate. Since two angles of BDE are congruent to two angles of BAC, the two triangles are similar by the AA Similarity Postulate. B 6 3 3 acute 1 y 10 2 20. 82 22 3 43 361 < 369 7 y 18. 9 y3 9y 7y 21 3 x 17. 7 8 24 5x 28. 376 470 16 x 34. P 3 3 5 5 16 3 16 36 w 376x 7520 x 20 gallons 16w 108 w 6.75 in. 36 13.5 2 11.25 in. 35. d 127.5 140 2 2 2 d2 35,856.25 mi d 189.4 mi 202 Geometry Chapter 9 Worked-out Solution Key Copyright © McDougal Littell Inc. All rights reserved. MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 203 Chapter 9 continued Project: Investigating Fractals (pp. 590–591) Investigation 1. stage 1: 4 3 2. stage 3: 64 27 ; stage 2: 16 9 4 256 ; stage 4: 81 ; the length at stage 1 is 3 times the length at stage 0, and similarly the length at stage 2 is 4 3 times the length at stage 1. 3. At each stage the length would be 4 3 times the previous stage, so the length gets increasingly large. 4. Yes; the graph is a curve that increases sharply as n, the number of stages, increases Stages of a Koch Snowflake 5. Stage 0 6. Stage 1 Stage, n Perimeter, P Stage 2 0 3 1 2 3 4 4 16 3 64 9 256 27 7. P 3 3 4 n Copyright © McDougal Littell Inc. All rights reserved. Geometry Chapter 9 Worked-out Solution Key 203