Selected Answers 1

Lesson 11.1.1 11-8. a: Yes; the sum of the interior angles only depends on the number of sides of the polygon. b: No; only regular hexagons have a guarantee of having an angle with the measure 120°. c: Yes; that is the definition of hexagon. 11-9. a: 7!= 5040 b: 1⋅5!⋅1= 120 11-10. Lateral surface area = (circumference of the base)(height) = 8π ⋅15 = 120π ≈ 377 cm2 11-11. a: 2x = 180 −106 , x = 37° b: x + 67 = 180 , x = 113 , 5y + 3y −16 = 180 , y = 24.5

� c: sin 739 = sin 57

x = sin 50y , x ≈ 7.9 , y ≈ 7.2

d: 4x − 2 + 2(8x − 9) = 180 , x = 10 11-12. a: A = 144 square units, P = 84 units b: A = 16 square units, P = 28 units 11-13. a: A sphere. b: A cylinder with a cone on top and bottom. c: A double cone: two cones attach at the vertices. 11-14. B 11-15. a: 4 b: 6 c: 4 11-16. n!

(n−4)! 11-17. (10)(12)h = 840 , so h = 840 ÷120 = 7 mm 11-18. a: b is larger, even though we are not told that b is a central angle. b: The missing angle = 180�− 62�− 70�= 48� and since the angle opposite side a is

bigger, a must be larger than b. c: a = 9 3 ≈15.6 units2 and b = 16 units2, so b is larger than a. 11-19. a: x2 + y2 = r2 b: sinθ = y

r ; y c: cosθ = xr ; x

11-20. If the circle’s center is C and if the midpoint of AB is D, then ΔADC is a 30°-60°-90°

triangle. Then the radius, AC , is 10 units long and the area of the circle is 100π ≈ 314.16 units .

11-21. C

Selected Answers 2

Lesson 11.1.2 11-25. 12 inches 11-26. a: 23P3 = 10, 626 b: 23P3 = 10, 626 c: 1⋅22 ⋅22 = 484 d: 44 ⋅22 ⋅21= 1848 11-27. The areas are all equal because the triangles have the same base and height. 11-28. a: x = 12.9 b: x = 2

c: x = 0 c: x = 239 ≈ 2.56

11-29. PA = PA (Reflexive Property), m∠PBA = m∠PCA = 90º (tangents are ⊥ to radii drawn to

the point of tangency), PB = PC (radii of a circle must be equal), so ΔPAB ≅ ΔPAC (HL). Therefore AB = AC ( ≅ Vs → ≅ parts ).

11-30. a: 10 b: tanθ = 8

6 , θ ≈ 53.13� 11-31. ±6, 18, ±54 11-32. D

Lesson 11.1.2 (Day 1) 11-39. a: Yes. One way is to cut off a corner so the cross-section is a triangle. b: A tetrahedron (also called a triangle-based pyramid). 11-40. V = 13 (6

2 )(4) = 48 units3; slant height = 32 + 42 = 5; SA = 4(12 ·6 ·5)+ 62 = 96 units2

11-41. a: 900C12 ≈ 5.48 ⋅1026 b: 899C11 ≈ 7.30 ⋅1024 c: ≈1.3% 11-42. a: 8 faces, 12 edges, and 6 vertices b: A square 11-43. a: x = 12, y = 7.5 b: 43 c: 48 square units 11-44. A = 16π square units; C = 8π units 11-45. B

Selected Answers 3

Lesson 11.1.2 (Day 2) 11-46. region B, since area ≈17.46 units2 and perimeter ≈ 24.3 units 11-47. a: π (1.5)2(4.5) ≈ 31.8 in.3

b: Volume of the pot is π (7)2(10) ≈1539.4 in.3 . Therefore, Aimee would need 1539.4 ÷ 31.8 = 48.4 or 49 cans of soup to fill the pot.

c: 2π (1.5)(4.5) ≈ 42.4 in.3 11-48. m∠ECB = m∠EAD (given), AE ≅ CE (definition of midpoint), ∠DEA = ∠BEC

(vertical angles are congruent), ΔAED ≅ ΔCEB (AAS≅), so AD ≅ CB ( ≅ Vs → ≅ parts ).

11-49. a: 6P4 = 6 ⋅5 ⋅4 ⋅3= 360 b: 1⋅5 ⋅4 ⋅3+ 5 ⋅1⋅4 ⋅3+ ⋅5 ⋅4 ⋅1⋅3+ 5 ⋅4 ⋅3⋅1= 4 ⋅ 5P4 = 240 11-50. a: 3711000 +

2501000 −

1521000 =

4691000 = 46.9%

b: 1− 4691000 =

5311000 = 53.1%

c: 1− 2501000 =

7501000 = 75%

d: P(A given under 20) = 152152+54+44 = 60.8%

11-51. a: 18x +174° = 540° , x = 20 13 ° b: x2 + (x +17)2 = 252 , x = 7 c: 2(x + 4°) = 134° , x = 63° d: x + 9° + 3x +15° = 180° , x = 39° 11-52. E

Selected Answers 4

Lesson 11.1.4 11-58. a: BA = 1

2 (7)(24) = 84 in.2 , V = (84)(12) = 1008 in.3 , SA = (2)(84)+ (12)(7 + 24 + 25) = 840 in.2

b: BA = 25π 2 , V = 13 (25π )(12) = 100π m3 , lateral SA = π (5)(13) = 65π m2 ,

total SA = 25π + 65π = 90π ≈ 282.7 m2 11-59. a: 8C3 = 56 b: There are 6 choices left for the third filling. c: 7

C28C3

= 2156 = 37.5%

11-60. Only one right triangle can be built: 3-4-5; 3⋅2⋅16⋅6⋅6 =

6216 . You cannot use combinations

because each combination is not equally likely. For example, rolling a combination of 1, 1, 2 on the dice (where order does not matter) is three times more likely than 1, 1, 1.

11-61. a: Vertically through the vertex of the cone. b: A circle c: A sphere 11-62. a: ΔABC : FED (AA~) b: ΔABC ~ ΔMKL (SSS~) c: Not similar because the zoom factors for corresponding sides are not equal. 11-63. See graph at right. 11-64. a: Corresponding angles, ≅ b: Alternate interior angles, ≅ c: Straight angle, supplementary d: Alternate exterior angles, neither because lines intersect. 11-65. A

Selected Answers 5

Lesson 11.1.5 11-72. r = 4 cm, SA = 4π (42 ) = 64π ≈ 201.1 cm2 , V = 4

3 π (4)3 = 2563 π ≈ 268.1 cm3

11-73. 8C3 ⋅ 10C3 = 6720 11-74. a: 12C3 + 12C3 = 715

b: If raspberry and custard are known filling, then there are two fewer fillings to choose from, so 10

C2 +10C1715 = 55

715 ≈ 7.7% . 11-75. small cone: 25 =

r6 , r = 2.4"; V = 1

3 π (2.4)2(2) ≈12.06 in.3 ; large cone: V = 1

3 π (6)2(5) ≈188.50 in.3 ; new volume = 188.50 − 2(12.06) ≈164.37 in.3 11-76. a: An icosahedron has 20 faces, so the surface area is (20)(45) = 900 mm2 . b: Since it has 12 faces, 180 ÷12 = 9 cm2 . c: The area of each face is 1

2 (6)(3 3) = 9 3 ≈15.6 in.2 , so total SA = 4(15.6) ≈ 62.4 in.2 . 11-77. ≈ 29° 11-78. 72π square units 11-79. a: ( 43 , 10) b: (−1, −9)

Selected Answers 6

Lesson 11.2.1 11-84. a: Earth’s circumference: 8000π = 25,132.7 miles. Therefore the distance to the moon

is 230, 000 ÷ 25,132.7 ≈ 9.5 times greater. b: cos 89.85° = 238,900

x , x ≈ 91, 253,182.4 miles 11-85. C ≈102.6 , so mAB

≈ 32360 (102.6) ≈ 9.1 mm

11-86. a: 1

3 (92 )(12) = 324 cm3 b: 12 cm3 11-87. a: 52C2 = 1326

b: 16C21326 = 120

1326 ≈ 9.0% or using permutations 16P2

52 P2= 16⋅1552⋅51 ≈ 9.0%

c: 16C252C2

d: 12C2

52C2= 661326 ≈ 5.0% or using permutations 12

P252 P2

= 12⋅1152⋅51 ≈ 5.0%

e: 4C1 ⋅ 12C11326 = 48

1326 ≈ 3.6% or using permutations, 2 ⋅ 4⋅1252⋅51 ≈ 3.6% 11-88. a: 108177 ≈ 61% b: 4488 = 50% 11-89. Answers vary. Sample responses: The rings in a tree, the ripples created when a stone is

tossed in a pond, the rings of a dartboard, etc. 11-90. D

Selected Answers 7

Lesson 11.2.2 11-96. radius of slice = 3

π ≈ 0.977 in., x ≈ 2.52 − 0.9772 ≈ 2.3 inches 11-97. The surface area of the moon ≈ 4π (1080)2 ≈14, 657, 414.7 which makes it larger

than Africa and smaller than Asia. 11-98. 4C1⋅6C1⋅2C1

12C3= 1255 ≈ 21.8% , or permutations and the Fundamental Principle of

Counting could be used: there are 3P3 = 6 ways to arrange the colors, and 12P3 = 12 ⋅11⋅10 ways to arrange twelve pens, so, 6 ⋅ 6⋅4⋅2

12⋅11⋅10 = 21.8% . 11-99. Central angle = 36°, distance from center to midpoint of side = 30.777 units,

A = 12 (20)(307.77) ≈ 3077.7 square units

11-100. V = 324 −12 = 312 cm3 11-101. ΔART : ΔPIT by AA~ 11-102. a: V = 2100 units3 , SA = 1007.34 units2 b: V = 1000π cm3 , SA = 1100π

3 + 240 ≈1391.92 cm2 c: V = 60 in.3 , SA = 144 in.2 11-103. D

Selected Answers 8

Lesson 11.2.3 11-110. a: x = 270 b: x = 132° , y ≈15.7 c: 3(x + 2) = 6x , x = 2 11-111. a: a = 44°, b = 28°, c = 56° b: Some students may just report that c is smaller. But some may notice that the

vertical angles (72°) are each the average of the two arcs they intercept. This will be revisited in problem 11-120.

11-112. Since P(academic) � P(Arts) = P(academic and Arts) the events are independent.

There is no association between winning an academic award and a Fine Arts award. Another possible method is P(Arts given academic) = P(arts), or that P(academic given Arts) = P(academic).

11-113. Base area = 36 units2, slant height = 109 ≈10.44 units,

lateral SA = 12 109 ≈125.3 units2, total SA ≈ 36 +125.3≈161.3 units2 11-114. a: ≈ 436,000 miles b: The sun’s radius is almost double the distance between the Earth and the moon.

That means that if the sun were placed next to the Earth, its center would be farther away than the moon!

c: 1,295,029 11-115. V (16)(16)(16) = 4096 units3; SA = (6)(16)(160 = 1536 units2 11-116. C 11-117. a: x = 117°, y = 88 ≈ 9.4 b: r = 5

sin 25° ≈11.8 , z = 310° c: 9(9 + a) = 8(21) , a = 9 23 11-118. V (prism) = (34)(84)(99) = 282, 744 units3;

V (cylinder) = π (38)2(71) ≈ 322, 088.6 units3, so the cylinder has more volume. 11-119. 6C3 + 2 ⋅6 C2 + 6C1 = 20 + 2(15) + 6 = 56 11-120. 21° + x = 2(62°) , x = 103° 11-121. f (x) = 4(

32 )x

11-122. The solution is shown with dashed lines in the diagram above. 11-123. D