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Solutions Key 6 Polygons and Quadrilaterals ARE YOU READY? PAGE 377 19 CHAPTER Solutions Key 6 Polygons and Quadrilaterals ARE YOU READY? PAGE 377 19. F (counterexample: with side lengths 5, 6, 10); if a triangle is an acute triangle, then it is a scalene 1. F 2. B triangle; F (counterexample: any equilateral 3. A 4. D triangle). 5. E 6-1 PROPERTIES AND ATTRIBUTES OF 6. Use Sum Thm. POLYGONS, PAGES 382–388 x ° + 42° + 32° = 180° x ° = 180° - 42° - 32° CHECK IT OUT! x ° = 106° 1a. not a polygon b. polygon, nonagon 7. Use Sum Thm. x ° + 53° + 90° = 180° c. not a polygon x ° = 180° - 53° - 90° 2a. regular, convex b. irregular, concave x ° = 37° 3a. Think: Use Polygon ∠ Sum Thm. 8. Use Sum Thm. (n - 2)180° x ° + x ° + 32° = 180° (15 - 2)180° 2 x ° = 180° - 34° 2340° 2 x ° = 146° b. (10 - 2)180° = 1440° x ° = 73° m∠1 + m∠2 + … + m∠10 = 1440° 9. Use Sum Thm. 10m∠1 = 1440° 2 x ° + x ° + 57° = 180° m∠1 = 144° 3 x ° = 180° - 57° 4a. Think: Use Polygon Ext. ∠ Sum Thm. 3 x ° = 123° m∠1 + m∠2 + … + m∠12 = 360° x ° = 41° 12m∠1 = 360° 10. By Lin. Pair Thm., 11. By Alt. Ext. Thm., m∠1 = 30° m∠1 + 56 = 180 m∠2 = 101° b. 4r + 7r + 5r + 8r = 360 m∠1 = 124° By Lin. Pair Thm., 24r = 360 By Vert. Thm., m∠1 + m∠2 = 180 r = 15 m∠2 = 56° m∠1 + 101 = 180 By Corr. Post., m∠1 = 79° 5. By Polygon Ext. ∠ Sum Thm., sum of ext. ∠ m∠3 = m∠1 = 124° Since ⊥ m, m n measures is 360°. Think: There are 8 ext. , so By Alt. Int. Thm., → ⊥ n, divide sum by 8. 360 m∠4 = 56° m∠3 = m∠4 = 90°. m(ext. ∠) = _ ° = 45° 8 12. By Same-Side Int. Thm., 3x + 2x = 180 THINK AND DISCUSS 5x = 180 1. Possible answers: x = 36 Concave pentagon Convex pentagon By Lin. Pair Thm., m∠1 + 3(36) = 180 m∠1 + 108 = 180 m∠1 = 72° By Corr. Post., m∠2 = 3(36) = 108° A concave polygon seems to “cave in” or have a dent. A convex polygon does not have a dent. 13. 45°-45°-90° 14. 30°-60°-90° x = (11 √2 ) √2 14 = 2x 2. Since polygon is not regular, you cannot assume = 11(2) = 22 x = 7 that each of the ext. has the same measure. 3. 15. 45°-45°-90° 16. 30°-60°-90° )NTERIOR!NGLES %XTERIOR!NGLES x = 3 √2 x = 2(8) = 16 3UMOF!NGLE-EASURES n 17. T (Lin. Pair Thm.); if 2 are supp., then they form a n lin. pair; F (counterexample: any supp. but non-adj. /NE!NGLE-EASURE nn pair of ). 18. F (counterexample: a pair of with measure 30°); if 2 are rt , then they are ; T (Rt. ∠ Thm.). 125 Holt McDougal Geometry ge07_SOLKEY_C06_125-150.indd 125 12/22/09 4:30:06 PM EXERCISES 25. 9m(ext. ∠) = 360 GUIDED PRACTICE m(ext. ∠) = 40° 1. Possible answer: If a polygon is equil., all its sides 26. 5a + 4a + 10a + 3a + 8a = 360 are , but all its are not necessarily . For a 30a = 360 polygon to be regular, all its sides must be , and a = 12 all its must be . 27. 6m∠JKM = (6 - 2)180 28. 6m∠MKL = 360 6m JKM 720 m MKL 60 2. polygon, decagon 3. not a polygon ∠ = ∠ = ° m∠JKM = 120° 4. polygon, quadrilateral 5. not a polygon 29. x + x - 3 + 110 + 130 = (4 - 2)180 6. regular, convex 7. irregular, concave 2x + 237 = 360 8. irregular, convex 2x = 123 x = 61.5 9. Think: Use Polygon ∠ Sum Thm. (5 - 2)180° = 540° 30. 2(90) + 2x + 2(x + 22) = (6 - 2)180 3z + 4z + 5z + 3z + 5z = 540 4x + 224 = 720 20z = 540 4x = 496 z = 27 x = 124 m∠A = m∠D = 3(27) = 81° 31. 5x = 360° m∠B = 4(27) = 108° x = 72° m∠C = m∠E = 5(27) = 135° 32. m∠ = m(ext. ∠) 10. Think: Use Polygon ∠ Sum Thm. n(m∠) = n(m(ext. ∠)) (12 - 2)180° = 1800° (n - 2)180 = 360 m∠1 + m∠2 + … + m∠12 = 1800° 180n = 720 12m∠1 = 1800° n = 4 m∠1 = 150° 33. m∠ = 4m(ext. ∠) 11. Think: Use Polygon ∠ Sum Thm. n(m∠) = 4n(m(ext. ∠)) (n - 2)180° (n - 2)180 = 4(360) (20 - 2)180° 180n = 1800 3240° n = 10 12. Think: Use Polygon Ext. ∠ Sum Thm. 34. m(ext. ∠) = _1 m∠ 4y 2y 6y 4y 360 8 + + + = 8n(m(ext. ∠)) = n(m∠) 16y 360 = 8(360) = (n - 2)180 y 22.5 = 3240 = 180n 13. Think: Use Polygon Ext. ∠ Sum Thm. n = 18 m 1 m 2 … m 5 360 ∠ + ∠ + + ∠ = ° 35. (n - 2)180 = 540 36. (n - 2)180 = 900 5m 1 360 ∠ = ° n - 2 = 3 n - 2 = 5 m1 72 ∠ = ° n = 5 n = 7 14. pentagon pentagon heptagon 15. By Polygon ∠ Sum Thm., sum of ∠ measures is 37. (n - 2)180 = 1800 38. (n - 2)180 = 2520 (5 - 2)180 = 540°. Think: m∠Q = m∠S by def. of . n - 2 = 10 n - 2 = 14 m∠P + m∠Q + m∠R + m∠S + m∠T = 540 n = 12 n = 16 90 + m∠S + 90 + m∠S + 90 = 540 dodecagon 16-gon 2m∠S = 270 39. 360 = n(120) 40. 360 = n(72) m∠Q = m∠S = 135° n = 3 n = 5 PRACTICE AND PROBLEM SOLVING m ∠ = 180 - 120 = 60° m ∠ = 180 - 72 = 108° 16. polygon, hexagon 17. not a polygon 41. 360 = n(36) 42. 360 = n(24) 18. polygon, quadrilateral 19. irregular, concave n = 10 n = 15 m ∠ = 180 - 36 = 144° m ∠ = 180 - 24 = 156° 20. regular convex 21. irregular, convex 43. A; possible answer: this is not a plane figure, so it 22. 2n + 6n + 2n + 5n = (4 - 2)180 cannot be a polygon. 15n = 360 n = 24 m∠R = m∠T = 2(24) = 48° m∠S = 6(24) = 144° m∠V = 5(24) = 120° 23. 18m∠ = (18 - 2)180 24. (7 - 2)180 = 900° 18m∠ = 2880 m∠ = 160° 126 Holt McDougal Geometry ge07_SOLKEY_C06_125-150.indd 126 12/22/09 4:30:10 PM 44. C y CHALLENGE AND EXTEND x 56. ∠ measures are a, a + 4, …, a + 16, where a is a multiple of 4. B D a + a + 4 + … + a + 16 = (5 - 2)180 5a + 40 = 540 5a = 500 E A a = 100 ∠ measures are 100°, 104°, 108°, 112°, and 116°. Check students’ estimates; possible answer: −− −− −− −− 57. PQ ST, QR RS, and ∠Q ∠S. So by SAS, pentagon is not equiangular; m∠A = 100°; −− −− m∠B = 113°; m∠C = 113°; m∠D = 101°; PRQ SRT. By CPCTC, PR RT, so PRT m∠E = 113°; yes, pentagon is not equiangular. is isosc. By Isosc. Thm., ∠RTP ∠RPT, so m∠RTP = m∠RPT = z°. 45a. heptagon b. (7 - 2)180 = 900° By Sum Thm., c. m∠A + m∠B + m∠C + m∠D 2z + y = 180 (1) + m∠E + m∠F + m∠G = 900 By CPCTC and Isosc. Thm., 95 + 125 + m∠F + 130 ∠PRQ ∠SRT ∠ QPR ∠RTS + 130 + m∠F + m∠F = 900 m∠PRQ = m∠SRT = m∠QPR = m∠RTS = x° 3m∠F + 480 = 900 Since PQRST is reg., Subtr. (3) from (1): 3m∠F = 420 5m∠QRS = (5 - 2)180 z = 180 - 108 = 72° m∠F = 140° 5(2x + y) = 540 Subst: in (3): 46. Let n be number of sides and s (= 7.5) be side length. 2x + y = 108 (2) y + 72 = 108 P = ns 5m∠PTS = (5 - 2)180 y = 36° 45 = n(7.5) 5(y + z) = 540 Subst. in (2): n = 6 y + z = 108 (3) 2x + 36 = 108 Polygon is a (regular) hexagon. 2x = 72 x = 36° 47. 48. Possible answer: −− −− −− 58. KA EF LC. By Alt. Int. Thm., ∠BLC ext. ∠A and ∠CLD ext. ∠E 360 m∠ALC = m∠CLE = _ = 36° 10 m∠BLD = m∠BLC + m∠CLD = 72° 49. Possible answer: 50. Possible answer: 59. Yes, if you allow for ∠ measures greater than 180°. m∠A + m∠B + m∠C + m∠D + m∠E + m∠F = 720° C B 51. The figure has 6 sides, so it is a hexagon. The 6 A D sides are , so the hexagon is equilateral. The 6 are , so the hexagon is equiangular. Since the E hexagon is equilateral and equiangular, it is regular. F No diagonal contains pts. in the interior, so it is convex. SPIRAL REVIEW 2 2 52. As number of sides increases, isosc. formed by 60. x + 3x - 10 = 0 61. x - x - 12 = 0 each side become thinner, and dists. from any pt. (x + 5)(x - 2) = 0 (x - 4)(x + 3) = 0 on base of each triangle to its apex approach same x = -5 or x = 2 x = 4 or x = -3 value. For a circle, each pt. is the same dist. from 62. x 2 - 12x = -35 63. x + 4 > 4 center. So polygon begins to resemble a circle. x 2 - 12x + 35 = 0 x > 0 (x - 7)(x - 5) = 0 4 + 4 > x TEST PREP x = 7 or x = 5 8 > x 53. A 54. H 0 < x < 8 (16 -2)180 = 2520° 64. Check x + 6 > 12 and 65. Check x + 3 > 7 and ≠ 2880° 6 + 12 > x, since 6 < 12. 3 + 7 > x, since 3 < 7. 55. D x + 6 > 12 x + 3 > 7 49 + 107 + 2m∠D + m∠D = (4 - 2)180 x > 6 x > 4 3m∠D = 204 6 + 12 > x 3 + 7 > x m∠D = 68° 18 > x 10 > x m∠C = 2(68) = 136° 6 < x < 18 4 < x < 10 127 Holt McDougal Geometry ge07_SOLKEY_C06_125-150.indd 127 12/22/09 4:30:13 PM 66.
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