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Lesson 12.1.1 12-6 a: x2 + y2 = 9 b: 7 12-7. a: V = 1

3 π (32 )(10) = 30π ≈ 94.2 units3

b: One method: BA = (21)(18) – (12)(12) = 234 units2, V = (234)(10) = 2340 units3 12-8. Think of this as an anagram ( 6!

4!2! = 15 ). 12-9. a: 1.005

b: f (t) = 8500(1.005)t

c: ≈ $11,465 12-10. a: x2 + 82 = (x + 2)2 ; x = 15

b: tan−1( 815 ) ≈ 28.1° , 180°− 90°− 28.1° ≈ 61.9° ; Sample tools: Trigonometry and the Triangle Sum Theorem

12-11. a: 124° b: 25π units2 c: ≈ 12.3 units 12-12. C, by SAS ≅

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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 m2, and

LA(cylinder) = 12π(11) = 132π ≈ 414.69 m2, so total surface area is ≈ 173.78 + 414.69 = 588.47 m2

b: V(cylinder) = 36π(11) = 396π ≈ 1244.07 m3 and V(cone) = 13 (36π )(7) = 84π ≈ 263.89m3, so total the volume is ≈ 1244.07 + 263.89 = 1507.96 m3.

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 = 64 , x ≈ 56.31°

12-26. a: Vertical angles are equal, 2x + 9° = 4x − 2° , x = 5.5 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°. 12-27. D

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Lesson 12.1.2 (Day 2) 12-28. a: BA ≈ 77.25 m2, LA = (8)(4)(16) = 512 m2, total SA ≈ 2(77.25) + 512 ≈ 666.51 m2

b: Slant height = 12 ft, LA = 4(12 )(10)(12) = 240 ft2, BA = (10)(10) = 100 ft2, total SA = 240 + 100 = 340 ft2

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 = x1 , 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: k2 = (8)(18) = 144 , k = 12 b: k = 7 12-34. A

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Lesson 12.1.3 12-40. Since 2πr = 40 feet, then r = 20π ≈ 6.4 feet; SA ≈ 4π (20π )

2 ≈ 509.3 square feet; V = 43 π (

20π )

3 ≈1080.8 ft3 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, 12 + 52 ≠ 32 .

b: x = –2 or 2 c: Possible answers will satisfy the equation x2 + y2 = 9 . 12-46. B

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Lesson 12.1.4 12-50. V = 820(12 )

3 = 102.5 cm3 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 cm2 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: 4C3

12C3= 4220 = 1.8%

b: 5C2⋅4C112C3

= 40220 ≈18.2%

c: 5C3+4C3+3C312C3

= 10+4+1220 ≈ 6.8%

12-56. B

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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; 52 +122 = 132 . 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(234 − 3) = 11 .

b: This has no real solution because x2 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

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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 units2; V = 40 units3 12-77. B

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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 = 6.47212 − 22 ≈ 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: 3C1⋅10C313C4

= 360715 ≈ 50.3% , or 4P1 3

P1 ⋅ 10 P313P4

= 4 ⋅ 3⋅10⋅9⋅813⋅12⋅11⋅10 = 50.3%

b: 1113C4

= 11715 ≈1.5% or 4P4 11

13P4= 24 ⋅ 11

17160 ≈1.5% 12-91. C 12-92. Base length ≈ 9.713 units, perimeter ≈ 30.97 units, area ≈ 62.84 square units 12-93. a: m∠PCQ = 46º, tan 46° = 5

CP , CP = CQ ≈ 4.83, sin 46° = 5CR , CR ≈ 6.95,

so x ≈ 6.95 – 4.83 ≈ 2.12

b: x2 = 72 + 72 − 2(7)(7) cos102° , x ≈10.88 12-94. a: It is a prism with dimensions 2 × 3 × 4 units. b: SA = 52 units2; V = 24 units3

c: SA = 52(32 ) = 468 units2; V = (24)(33) = 648 units3

12-95. a: y = − 6

5 x + 4 b: y =12 x − 2

12-96. See graph at right. 12-97. A

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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(12 )2 + 64(12 )+ 80 = 108 feet

c: −16x2 + 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

53 = 125 times the original, so the increased volume must be 11(53) = 1375 cubic units.

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.

x – 4 –3 –2 –1 0 1 2 3 4 y 6 5 4 3 2 3 4 5 6

12-107. B

Front Right Top

x

y