Lesson 12.1.1 12-6 a: x 2 + y 2 = 9 b: 7 12-7. a: V = 1 3 π (32 )(10) = 30π 94.2 units 3 b: One method: BA = (21)(18) (12)(12) = 234 units 2, V = (234)(10) = 2340 units 3 12-8. Think of this as an anagram ( 4!2! 6! = 15 ). 12-9. a: 1.005 b: f (t) = 8500(1.005) t c: $11,465 12-10. a: x 2 + 8 2 = (x + 2) 2 ; x = 15 b: tan 1 ( 15 8 ) 28.1, 180 90 28.1 61.9 ; Sample tools: Trigonometry and the Triangle Sum Theorem 12-11. a: 124 b: 25π units 2 c: 12.3 units 12-12. C, by SAS Selected Answers 2014 CPM Educational Program. All rights reserved. 1
Lesson 12.1.2 (Day 1) 12-21. a: The slant height of the cone is 9.22 m, LA(cone) 6π(9.22) 173.78 m 2, and LA(cylinder) = 12π(11) = 132π 414.69 m 2, so total surface area is 173.78 + 414.69 = 588.47 m 2 b: V(cylinder) = 36π(11) = 396π 1244.07 m 3 and V(cone) = 1 3 (36π )(7) = 84π 263.89 m3, so total the volume is 1244.07 + 263.89 = 1507.96 m 3. 12-22. a: 36 b: b = c = 108, d = 72º 12-23. a: E(1, 3) and F(7, 3) ; AB = 9, DC = 3, EF = 6 ; EF seems to be the average of AB and CD. b: Yes; EF = 4, while AB = 6 and CD = 2 c: Sample response: The midsegment of a trapezoid is parallel to the bases and has a length that is the average of the lengths of the bases. 12-24. a: (2, 3), r = 5 b: ( 1, 3), r = 4 12-25. a: sin 27 = x 18, x 8.17 b: sin102 7 = sin 62 x, x 6.32 c: tan x = 6 4, x 56.31 12-26. a: Vertical angles are equal, 2x + 9 = 4x 2, x = 5.5 12-27. D b: The sum of the angles of a quadrilateral is 360, 116 + (3x + 8) + 32 + (2x 1) = 360, x = 41 c: When lines are parallel, same-side exterior angles are supplementary, so 7x 3 + 4x +12 = 180 and x = 15.55. Selected Answers 2014 CPM Educational Program. All rights reserved. 2
Lesson 12.1.2 (Day 2) 12-28. a: BA 77.25 m 2, LA = (8)(4)(16) = 512 m 2, total SA 2(77.25) + 512 666.51 m 2 b: Slant height = 12 ft, LA = 4( 1 2 )(10)(12) = 240 ft2, BA = (10)(10) = 100 ft 2, total SA = 240 + 100 = 340 ft 2 12-29. a: u = 1 b: x = 5, 8 3 c: k = 1 or 9 d: p = 4.5 or 1 12-30. a: Since the hypotenuse is 1, sinθ = y 1, and y = sinθ. Also, cosθ = 1 x, so x = cosθ. b: It must be 1 because of the Pythagorean Theorem. c: Yes, this appears to be true for all angles. 12-31. a: (x 4) 2 + (y 2) 2 = 9 b: Pick any two points, such as those that are parallel to the axes on the circle and then find the distance between the points using the Pythagorean Theorem. 12-32. a: 2 5 5 = 50 b: 2 4 3 = 24 12-33. a: k 2 = (8)(18) = 144, k = 12 b: k = 7 12-34. A Selected Answers 2014 CPM Educational Program. All rights reserved. 3
Lesson 12.1.3 12-40. Since 2πr = 40 feet, then r = 20 π V = 4 3 π ( 20 π )3 1080.8 ft 3 6.4 feet; SA 4π ( 20 π )2 509.3 square feet; 12-41. a: x = 6 b: x = 4 or 4 c: x = 4 d: x = 30 e: x = 3 or 3 f: x = 3 or 5 12-42. Answers vary. Typical cross-sections: regular hexagon, circle, rectangle, etc. 12-43. a: Yes; the graph includes the circle and all of the points inside the circle. b: No; the graph only includes the points outside the circle. The circle itself would be dashed. 12-44. This is an anagram of five Hs and five Ts. 10! 5!5! = 252. It is not a combination because order matters. 12-45. a: The point is not on the circle. This can be shown using the fact that all of the points on the circle are 3 units away from the origin and then finding the distance from the origin to the point (1, 5) with the Pythagorean Theorem, 1 2 + 5 2 3 2. 12-46. B b: x = 2 or 2 c: Possible answers will satisfy the equation x 2 + y 2 = 9. Selected Answers 2014 CPM Educational Program. All rights reserved. 4
Lesson 12.1.4 12-50. V = 820( 1 2 )3 = 102.5 cm 3 12-51. See solution graph at right. a: C: (0, 0); r = 4.5 b: C: (0, 0); r = 75 8.7 c: C: (3, 0); r = 1 d: C: (2, 1); r = 19 4.4 12-52. a: x = 33, y = 120 b: a 36.9, b = 4 c: z = 12, w = 5 d: x = 55 12-53. a: FG = 3.5 cm, BC = 14 cm b: 16(3) 4(3) = 36 cm 2 12-54. base radius = 14 in.; V = 1 3 (196π )(18) = 1176π 3695 in.3 SA = π rl = π (14)( 520) 1003 in. 2 12-55. a: 4 C 3 12C3 = 4 220 = 1.8% b: 5C 2 4 C 1 12C3 c: 5C 3 + 4 C 3 + 3 C 3 12C3 = 40 220 18.2% = 10+4+1 220 6.8% 12-56. B Selected Answers 2014 CPM Educational Program. All rights reserved. 5
Lesson 12.2.1 12-61. 65 ; One method: The base angles of ΔPSR must add up to 40 so that the sum of all three angles is 180. Then add the 40 and 35 of QPS and QRS, respectively, and the sum of the base angles of ΔPQR must be 115. Thus, m Q must be 180 115 = 65. 12-62. A 1, 459, 379.5 square feet 12-63. a: No; the triangle is equilateral, so all angles must be 60. b: Yes; 5 2 +12 2 = 13 2. 12-64. See graph at right. x-intercepts: (4, 0) and ( 4, 0), y-intercepts: (0, 2) and (0, 8) 12-65. a: This has one solution, because 4( 23 4 3) = 11. b: This has no real solution because x 2 must be positive or zero. c: This has two solutions because x = ± 6. d: This has no solution because the absolute value must be positive or zero. 12-66. See graph at right. 12-67. C 8 6 4 2-5 5-2 -4-6 -8 Selected Answers 2014 CPM Educational Program. All rights reserved. 6
Lesson 12.2.2 12-71. a: a = 120, b = 108, so a is greater. b: Not enough information is given since it is not known if the lines are parallel. c: Third side is approximately 8.9 units, so b is opposite the greater side and must be greater than a. d: a is three more than b, so a must be greater. e: a = 7 tan 23 2.97 and b = 2 cos 49 3.05, so b is greater than a. 12-72. x = 8 and y = 12.5 12-73. See graph at right. a: y-intercept: (0, 6); x-intercepts: (3, 0) and ( 1, 0) b: (1, 8) c: f (100) = 19, 594 and f ( 15) = 504 12-74. 49π sq. units 12-75. 2(5!)(3!) = 1440 12-76. SA = 76 units 2 ; V = 40 units 3 12-77. B Selected Answers 2014 CPM Educational Program. All rights reserved. 7
Lesson 12.2.3 12-85. Side length = 4, so height of triangle is 2 3. Thus, the y-coordinate of point C could be 2 ± 2 3 ; (5, 5.46) or (5, 1.46). 12-86. length of diagonal = 4sin54º 6.4721, or 32 32 cos108 6.4721 (using the Law of Cosines) height of shaded triangle = 6.4721 2 2 2 6.155 units, area of triangle = 1 2 (4)(6.155) 12.310 sq. units 12-87. The graph should include a circle with radius 5, center (0, 0), and a line with slope 1 and y-intercept (0, 1). (3, 4) and ( 4, 3). 12-88. Jamila s product does not equal zero. She cannot assume that if the product of two quantities is 8, then one of the quantities must be 8; Correct solution: x = 6 or 3 12-89. 14 12-90. a: 3C 1 10 C 3 13C4 12-91. C = 360 715 50.3%, or 4 P 1 3 P 1 10 P 3 13 P 4 b: 11 13C4 = 11 715 1.5% or 4 P 4 11 = 24 11 13P4 17160 1.5% = 4 3 10 9 8 13 12 11 10 = 50.3% 12-92. Base length 9.713 units, perimeter 30.97 units, area 62.84 square units 12-93. a: m PCQ = 46º, tan 46 = CP 5, CP = CQ 4.83, sin 46 = 5 CR, CR 6.95, so x 6.95 4.83 2.12 b: x 2 = 7 2 + 7 2 2(7)(7) cos102, x 10.88 12-94. a: It is a prism with dimensions 2 3 4 units. b: SA = 52 units 2 ; V = 24 units 3 c: SA = 52(3 2 ) = 468 units 2 ; V = (24)(3 3 ) = 648 units 3 12-95. a: y = 6 5 x + 4 b: y = 1 2 x 2 8 12-96. See graph at right. 12-97. A 6 4 2-5 5-2 -4-6 -8 Selected Answers 2014 CPM Educational Program. All rights reserved. 8
Lesson 12.2.4 12-101. a: 60 sq. units b: Side length 7.282 units, so area (7.282)(7) 50.97 sq. units. 12-102. a: It was 80 feet above ground because y = 80 when x = 0. b: 16(3) 2 + 64(3) + 80 = 128 feet; 16( 1 2 )2 + 64( 1 2 ) + 80 = 108 feet c: 16x 2 + 64x + 80 = 0 ; x = 5 seconds 12-103. They all are 30-60 - 90 triangles, so they are all similar to each other. In addition, the triangles in (a) and (c) are congruent because corresponding sides have equal length. 12-104. a: See views at right. b: 11 cubic units c: The volume of the new solid must be 5 3 = 125 times the original, so the increased volume must be 11(5 3 ) = 1375 cubic units. Front Right Top 12-105. a: C: ( 5, 0), r = 10 b: C: (3,1), r = 15 12-106. See answers in table below and graph at right. y x 4 3 2 1 0 1 2 3 4 y 6 5 4 3 2 3 4 5 6 x 12-107. B Selected Answers 2014 CPM Educational Program. All rights reserved. 9