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CHAPTER 9
Geometry, Concepts and Skills 137Chapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter Opener
Chapter Readiness Quiz (p. 472)
1. A; 52 � 122 � c2
25 � 144 � c2
169 � c2
�1�6�9� � �c2�13 � c
2. G; A � �12
�bh
� �12
�(12)(5)
� 30 units2
3. C; A � πr2
� π(6)2
� 113 in.2
Lesson 9.1
9.1 Checkpoint (p. 475)
1. The solid is a polyhedron with triangular bases, so it is atriangular prism.
2. A cone has a curved surface, so it is not a polyhedron.
3. The solid is a polyhedron with a rectangular base, so it isa rectangular pyramid.
4.
9.1 Guided Practice (p. 476)
1. C 2. A 3. B 4. true 5. false
6. true 7. false 8. true 9. true
10. The solid is a polyhedron with rectangular bases, so it isa rectangular prism.
11. A cylinder has a curved surface, so it is not a polyhedron.
12. The solid is a polyhedron with a pentagonal base, so it isa pentagonal pyramid.
13. hexagonal prism; 8 faces and 18 edges;
congruent faces: ABCDEF � UVWXYZ and AFZU �FEYZ � EDXY � DCWX � CBVW � BAUV;
congruent edges: AB**** � BC**** � CD**** � DE**** � EF**** �FA**** � UV**& � VW**& � WX**& � XY**& � YZ**** � ZU**** andAU**** � BV**** � CW**& � DX**** � EY**** � FZ**&
14. triangular pyramid; 4 faces and 6 edges;
congruent faces: T PQR �T PQS;
congruent edges: QR**** � QS**& and PR**** � PS****
15. cube or square prism; 6 faces and 12 edges;
congruent faces: JKLM � TUVW � JMWT � MLVW �LKUV � KJTU;
congruent edges: JK**** � KL**** � LM**&� MJ**&� TU**** � UV**&�VW**&� WT**&� JT**** � KU**&� LV**** � MW**&&*
16.
6 faces and 12 edges
9.1 Practice and Applications (pp. 477–480)
17. A cone has a curved surface, so it is not a polyhedron.
18. The solid is a polyhedron with triangular bases, so it is atriangular prism.
19. The solid is a polyhedron with a rectangular base, so it isa rectangular pyramid.
20. False; a pyramid only has one base.
21. True; prisms have two bases.
22. True; the bases of a prism are congruent polygons.
23. False; a cone only has one base.
24. False; a sphere has a rounded surface, so it is not a polyhedron.
25. F 26. D 27. A 28. E 29. B 30. C
31. triangular prism
32. Sample answer: a pyramid with a square base has onesquare face and four triangular faces, while a triangularprism has two congruent triangular faces and three rectangular faces.
33. The solid is a polyhedron with a rectangular base, so it isa rectangular pyramid.
34. The solid is a polyhedron with pentagonal bases, so it is apentagonal prism.
35. The solid has a curved surface, so it is not a polyhedron.
36. rectangular pyramid; 5 faces and 8 edges;
congruent faces: T ABE �T ACD and T ABC �T AED;
congruent edges: BE**** � CD**** , BC**** � ED**** , and AB**** � AC**** �AD**** � AE**** .
37. triangular prism; 5 faces and 9 edges;
congruent faces: T JKL �T FGH;
congruent edges: FH**** � JL**** , FG**** � JK****, GH**** � KL**** , andFJ**** � GK**&� HL**&
Chapter 9 continued
38. pentagonal pyramid; 6 faces and 10 edges;
congruent faces: TNPT �TNTS �TNSR � TNRQ �TNQP;
congruent edges: NP**** � NT**** � NS**** � NR**** � NQ**& andPT**** � TS**** � SR**** � RQ**** � QP****
39. True; solids are three-dimensional shapes.
40. False; cylinders, cones, and spheres are not polyhedrabecause they have curved surfaces.
41. False; if a prism is not rectangular, then the bases are thecongruent faces that are not rectangular.
42. True; all plane surfaces of a prism are faces.
43. 44.
45. 46. Sample answer:
47. Sample answer: 48. Sample answer:
49. Sample answer: 50. Sample answer:
51. Sample answer:
52. F � V � E � 2 53. F � V � E � 2
6 � V � 12 � 2 8 � V � 12 � 2
6 � V � 14 8 � V � 14
V � 8 V � 6
54. F � V � E � 2 55. F � V � E � 2
12 � V � 30 � 2 20 � V � 30 � 2
12 � V � 32 20 � V � 32
V � 20 V � 12
56. F � V � E � 2 57. F � V � E � 2
5 � 6 � E � 2 F � 7 � 12 � 2
11 � E � 2 F � 7 � 14
9 � E F � 7
58. F � V � E � 2
8 � 12 � E � 2
20 � E � 2
18 � E
9.1 Standardized Test Practice (p. 480)
59. D 60. F 61. C
9.1 Mixed Review (p. 480)
62. A � s263. A � bh
� 92 � (7)(4)
� 81 cm2 � 28 m2
64. A � bh
60 � 6b
10 in. � b
65. A � s2
169 � s2
�1�6�9� � s
13 ft � s
66. C � πd � π(12) � 38 ft;
A � πr2 � π(6)2 � π(36) � 113 ft2
67. C � 2πr � 2π(5) � 10π � 31 cm;
A � πr2 � π(5)2 � π(25) � 79 cm2
68. C � 2πr � 2π(14) � 28π � 88 yd;
A � πr2 � π(14)2 � π(196) � 616 yd2
9.1 Algebra Skills (p. 480)
69. 92 � (12 � 39) � 92 � 51 � 41
70. 8 � 4 � 3 � 5 � 8 � 12 � 5 � 20 � 5 � 15
71. (7 � 5) � 14 � 2 � 14 � 28
72. 10 � (5 � 2)2 � 8 � 10 � 32 � 8� 10 � 9 � 8� 1 � 8� 9
73. 14 � 42 � 26 � 14 � 16 � 26 � 30 � 26 � 4
74. 3(10 � 3)2 � 3(7)2 � 3(49) � 147
75. l � w � h � 3 � 2 � 5 � 30
76. 2l � 2w � 2h � 2(3) � 2(2) � 2(5) � 6 � 4 � 10 � 20
77. 2πh � 2π(5) � 10π
78. πw2h � π(2)2(5) � π(4)(5) � 20π
138 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
79. πl2 � π(3)2 � 9π
80. 2l � 2w � 2(3) � 2(2) � 6 � 4 � 10
Lesson 9.2
9.2 Activity (pp. 481–482)
Step 4. rectangular prism
Step 5.
Step 6. A � Area of rectangle A
� l � w
� 7 � 3 � 21 square units;
P � Perimeter of rectangle A
� 2l � 2w
� 2(7) � 2(3)
� 14 � 6
� 20 units;
h � Height of rectangles B, C, D, and E � 5 units
1. 2A � Ph � 2(21) � 20(5) � 42 � 100 � 142
2. They are equal.
3. Surface Area � 2A � Ph, where A is the area of the base,P is the perimeter of the base, and h is the height of theprism.
4. Sample answer:
Surface Area � 2A � Ph
� 2(20) � 18(2)
� 40 � 36
� 76 units2
F
B
A
C ED
9.2 Checkpoint (pp. 485–486)
1. B � 2 � 6 � 12
P � 2(2) � 2(6) � 4 � 12 � 16
h � 3
S � 2B � Ph
� 2(12) � 16(3)
� 24 � 48
� 72 in.2
2. B � 5 � 8 � 40
P � 2(5) � 2(8) � 10 � 16 � 26
h � 6
S � 2B � Ph
� 2(40) � 26(6)
� 80 � 156
� 236 ft2
3. B � �12
� � 8 � 6 � 4 � 6 � 24
P � 8 � 6 � 10 � 24
h � 4
S � 2B � Ph
� 2(24) � 24(4)
� 48 � 96
� 144 cm2
4. S � 2πr2 � 2πrh 5. S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(5) � 2π(6)2 � 2π(6)(10)
� 18π � 30π � 72π � 120π
� 48π � 192π
� 151 in.2 � 603 ft2
6. L � 2πrh � 2π(1)(2) � 4π � 13 m2
9.2 Guided Practice (p. 487)
1. true 2. false 3. true 4. true
5. B � 6 � 4 � 24
P � 2(6) � 2(4) � 12 � 8 � 20
h � 10
S � 2B � Ph
� 2(24) � 20(10)
� 48 � 200
� 248 in.2
6. S � 2πr2 � 2πrh
� 2π(4)2 � 2π(4)(6)
� 32π � 48π
� 80π
� 251 ft2
Geometry, Concepts and Skills 139Chapter 9 Worked-Out Solution Key
Rectangle Length Width Area
A 7 units 3 units 21 units2
B 7 units 5 units 35 units2
C 3 units 5 units 15 units2
D 7 units 5 units 35 units2
E 3 units 5 units 15 units2
F 7 units 3 units 21 units2
Total Area 142 units2
Chapter 9 continued
7. B � �12
� � 3 � 4 � 6
P � 3 � 4 � 5 � 12
h � 2
S � 2B � Ph
� 2(6) � 12(2)
� 12 � 24
� 36 m2
9.2 Practice and Applications (pp. 487–490)
8. 5 in.
9. A � 3 � 2 � 6 in.2
10. P � 2(3) � 2(2) � 6 � 4 � 10 in.
11. 5 m
12. A � πr2 � π(4)2 � 16π � 50 m2
13. C � 2πr � 2π(4) � 8π � 25 m
14. 15 ft
15. A � �12
� � 6 � 8 � 24 ft2
16. P � 6 � 8 � 10 � 24 ft
17. B � 10 � 9 � 90
P � 2(10) � 2(9) � 20 � 18 � 38
h � 6
S � 2B � Ph
� 2(90) � 38(6)
� 180 � 228
� 408 ft2
18. B � 4 � 4 � 16
P � 2(4) � 2(4) � 8 � 8 � 16
h � 4
S � 2B � Ph
� 2(16) � 16(4)
� 32 � 64
� 96 in.2
19. B � �12
� � 6 � 8 � 24
P � 6 � 8 � 10 � 24
h � 7
S � 2B � Ph
� 2(24) � 24(7)
� 48 � 168
� 216 m2
20. No; doubling the height does not double the surface area.
Prism A: B � 3 � 3 � 9
P � 2(3) � 2(3) � 6 � 6 � 12
h � 2
S � 2B � Ph
� 2(9) � 12(2)
� 18 � 24
� 42 m2
Prism B: B � 3 � 3 � 9
P � 2(3) � 2(3) � 6 � 6 � 12
h � 1
S � 2B � Ph
� 2(9) � 12(1)
� 18 � 12
� 30 m2
21. rectangular prism 22. cylinder
23. triangular prism
24. S � 2πr2 � 2πrh 25. S � 2πr2 � 2πrh
� 2π(6)2 � 2π(6)(11) � 2π(3)2 � 2π(3)(13)
� 72π � 132π � 18π � 78π
� 204π � 96π
� 641 ft2 � 302 m2
26. S � 2πr2 � 2πrh 27. S � 2πr2 � 2πrh
� 2π(8)2 � 2π(8)(8) � 2π(5)2 � 2π(5)(7)
� 128π � 128π � 50π � 70π
� 256π � 120π
� 804 cm2 � 377 in.2
28. B � 16 � 16 � 256
P � 2(16) � 2(16) � 32 � 32 � 64
h � 16
S � 2B � Ph
� 2(256) � 64(16)
� 512 � 1024
� 1536 mm2
29. B � 3
P � 3(2.7) � 8.1
h � 18
S � 2B � Ph
� 2(3) � 8.1(18)
� 6 � 145.8
� 152 in.2
140 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
30. S � 2πr2 � 2πrh
� 2π(1.5)2 � 2π(1.5)(1)
� 4.5π � 3π
� 7.5π
� 24 in.2
31. B � 260
P � 6(10) � 60
h � 11
S � 2B � Ph
� 2(260) � 60(11)
� 520 � 660
� 1180 cm2
32. The formula Juanita used to find the surface area is incorrect. The radius of the cylinder is 6 inches, not 12 inches. Also, 122 � 24. The correct solution is as follows.
S � 2πr2 � 2πrh
� 2π(6)2 � 2π(6)(10)
� 72π � 120π
� 192π
� 603 in.2
33. 34.
35.
36.
B � 6 � 3 � 18
P � 2(6) � 2(3) � 12 � 6 � 18
h � 10
S � 2B � Ph
� 2(18) � 18(10)
� 36 � 180
� 216 ft2
6 ft
10 ft
3 ft
37.
S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(7)
� 18π � 42π
� 60π
� 188 m2
38.
B � �12
� � 6 � 8 � 24
P � 6 � 8 � 10 � 24
h � 5
S � 2B � Ph
� 2(24) � 24(5)
� 48 � 120
� 168 in.2
39. P � 2(2) � 2(6) � 4 � 12 � 16
h � 7
L � Ph � 16(7) � 112 m2
40. L � 2πrh
� 2π(3)(8)
� 48π
� 151 yd2
41. P � 6 � 8 � 10 � 24
h � 5
L � Ph � 24(5) � 120 ft2
42. L � 2πrh 43. L � 2πrh
� 2π(2)(90) � 2π(1)(4)
� 360π � 8π
� 1131 cm2 � 25 in.2
44. L � 2πrh
� 2π(60)(1.2)
� 144π
� 452 mm2
45. P � 64(4) � 256
h � 415
L � Ph � 256(415) � 106,240 m2
5 in.8 in.
6 in.
3 m
7 m
Geometry, Concepts and Skills 141Chapter 9 Worked-Out Solution Key
Chapter 9 continued
46. B � 642 � 4096
S � B � Ph
� 4096 � 106,240
� 110,336 m2
9.2 Standardized Test Practice (p. 490)
47. B; S � 2πr2 � 2πrh
� 2π(5)2 � 2π(5)(15)
� 50π � 150π
� 200π
� 628 cm2
48. H; B � 5 � 2 � 10
P � 2(5) � 2(2) � 14
S � 2B � Ph
104 � 2(10) � 14x
104 � 20 � 14x
84 � 14x
6 ft � x
9.2 Mixed Review (p. 490)
49. A � s2 � �12
�bh 50. A � bh � �12
�bh
� 62 � �12
�(11)(6) � 10(4) � �12
�(10)(5)
� 36 � 33 � 40 � 25
� 69 ft2 � 65 cm2
51. base of triangle: 52. A � bh
a2 � b2 � c2 70 � 5b
52 � b2 � 132 14 ft � b
25 � b2 � 169
b2 � 144
b � 12;
A � bh � �12
�bh
� 3(13) � �12
�(12)5
� 39 � 30
� 69 in.2
9.2 Algebra Skills (p. 490)
53. �6x
� � �25
� 54. �110� � �
4x
�
2x � 6(5) x � 10(4)
2x � 30 x � 40
x � 15
55. �35
� � �3x5� 56. �
47
� � �1x6�
5x � 3(35) 4x � 7(16)
5x � 105 4x � 112
x � 21 x � 28
57. �3x3� � �
1113� 58. �
3x2� � �
58
�
11x � 33(13) 8x � 32(5)
11x � 429 8x � 160
x � 39 x � 20
Lesson 9.3
9.3 Checkpoint (pp. 493–494)
1. B � 7 � 7 � 49
P � 7 � 7 � 7 � 7 � 28
l � 9
S � B � �12
�Pl
� 49 � �12
�(28)(9)
� 175 in.2
2. B � 35.1
P � 9 � 9 � 9 � 27
l� 12
S � B � �12
�Pl
� 35.1 � �12
�(27)(12)
� 197.1 cm2
3. B � 10 � 10 � 100
P � 10 � 10 � 10 � 10 � 40
l2 � 52 � 122
� 25 � 144
� 169
l � �1�6�9� � 13
S � B � �12
�Pl
� 100 � �12
�(40)(13)
� 360 ft2
4. S � πr2 � πrl 5. S � πr2 � πrl
� π(10)2 � π(10)(12) � π(4)2 � π(4)(8)
� 100π � 120π � 16π � 32π
� 220π � 48π
� 691 in.2 � 151 ft2
142 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
6. l2 � r2 � h2
� 32 � 42
� 9 � 16
� 25
l� �2�5� � 5;
S � πr2 � πrl
� π(3)2 � π(3)5
� 9π � 15π
� 24π
� 75 cm2
9.3 Guided Practice (p. 495)
1. The red line segment is the height of the pyramid.
2. The blue line segment is the slant height of the pyramid.
3. The height of the lateral faces of the pyramid is the slant height.
4. The height of a pyramid is the perpendicular distancebetween the vertex and base.
5.
l2 � 52 � 122
� 25 � 144
� 169
l � �1�6�9�
� 13
The slant height is 13 in.
6. B � 3.9
P � 3 � 3 � 3 � 9
l � 5
S � B � �12
�Pl
� 3.9 � �12
�(9)(5)
� 26 m2
7. S � πr2 � πrl
� π(2)2 � π(2)(5)
� 4π � 10π
� 14π
� 44 in.2
10 in.
12 in.
10 in.5 in.
8. B � 6 � 6 � 36
P � 6 � 6 � 6 � 6 � 24
l � 10
S � B � �12
�Pl
� 36 � �12
�(24)(10)
� 156 ft2
9.3 Practice and Applications (pp. 495–499)
9. height 10. height 11. slant height
12. l2 � 62 � 8213. l2 � 92 � 122
� 36 � 64 � 81 � 144
� 100 � 225
l � �1�0�0� l� �2�2�5�
� 10 m � 15 mm
14. l2 � 82 � 152
� 64 � 225
� 289
l � �2�8�9�
� 17 in.
15. B � 5 � 5 � 25 16. B � 21.2
P � 5 � 5 � 5 � 5 � 20 P � 7 � 7 � 7 � 21
l � 8 l � 11
S � B � �12
�Pl S � B � �12
�Pl
� 25 � �12
�(20)(8) � 21.2 � �12
�(21)(11)
� 105 m2 � 136.7 in.2
17. B � 24 � 24 � 576
P � 24 � 24 � 24 � 24 � 96
l2 � 122 � 152
� 144 � 225
� 369
l � �3�6�9�
� 19.21
S � B � �12
�Pl
� 576 � �12
�(96)(19.21)
� 1498 cm2
Geometry, Concepts and Skills 143Chapter 9 Worked-Out Solution Key
Chapter 9 continued
18. B � 1.7
P � 2 � 2 � 2 � 6
l � 4
S � B � �12
�Pl
� 1.7 � �12
�(6)(4)
� 13.7 yd2
19. B � 32 � 32 � 1024
P � 32 � 32 � 32 � 32 � 128
l2 � 162 � 122
� 256 � 144
� 400
l � �4�0�0�
� 20
S � B � �12
�Pl
� 1024 � �12
�(128)(20)
� 2304 mm2
20. B � 14 � 14 � 196
P � 14 � 14 � 14 � 14 � 56
l2 � 72 � 242
� 49 � 576
� 625
l � �6�2�5�
� 25
S � B � �12
�Pl
� 196 � �12
�(56)(25)
� 896 cm2
21. The slant height of a pyramid is the hypotenuse of a righttriangle formed by the height of the pyramid and half thelength of the base. Since the hypotenuse of a right triangleis always the longest side of the triangle, the slant heightof a pyramid is always greater than the height.
22. Jamie substituted the height of the pyramid into the formula where the slant height belongs;
P � 4(40) � 160
l2 � 202 � 152
� 400 � 225
� 625
l � �6�2�5� � 25
S � 402 � �12
�(160)(25)
� 1600 � 2000
� 3600 m2
23. S � πr2 � πrl 24. S � πr2 � πrl
� π(9)2 � π(9)(22) � π(4)2 � π(4)(10)
� 81π � 198π � 16π � 40π
� 279π � 56π
� 877 m2 � 176 ft2
25. S � πr2 � πrl 26. S � πr2 � πrl
� π(7)2 � π(7)(25) � π(6)2 � π(6)(10)
� 49π � 175π � 36π � 60π
� 224π � 96π
� 704 m2 � 302 cm2
27. l2 � 92 � 40228. l2 � 102 � 242
� 81 � 1600 � 100 � 576
� 1681 � 676
l � �1�6�8�1� l � �6�7�6�
� 41 � 26
S � πr2 � πrl S � πr2 � πrl
� π(9)2 � π(9)(41) � π(10)2 � π(10)(26)
� 81π � 369π � 100π � 260π
� 450π � 360π
� 1414 yd2 � 1131 in.2
29. L � �12
�Pl
� �12
�(28)(14)
� 196 cm2
30. L � πrl 31. L � πrl
� π(4.3)(22.3) � π(4)(14)
� 301.25 in.2 � 176 in.2
32.
B � 12 � 12 � 144
P � 4(12) � 48
l2 � 62 � 82 � 36 � 64 � 100
l� �1�0�0� � 10
S � B � �12
�Pl
� 144 � �12
�(48)(10)
� 384 m2
12 m12 m
8 m
144 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
33.
B � 27.7
P � 3(8) � 24
l � 13
S � B � �12
�Pl
� 27.7 � �12
�(24)(13)
� 184 ft2
34.
S � πr2 � πrl
� π(3)2 � π(3)(7)
� 9π � 21π
� 30π
� 94 yd2
35.
l2 � 102 � 142 � 100 � 196 � 296
l � �2�9�6� � 17.2
S � πr2 � πrl
� π(10)2 � π(10)(17.2)
� 100π � 172π
� 272π
� 855 in.2
36. square pyramid;
B � 6 � 6 � 36
P � 4(6) � 24
l � 4
S � B � �12
�Pl
� 36 � �12
�(24)(4)
� 84 ft2
10 in.
14 in.
6 yd
7 yd
13 ft 8 ft
8 ft8 ft
B �27.7 ft2
37. cone; 38. L � πrl
S � πr2 � πrl � π(6)(10)
� π(2)2 � π(2)(6) � 60π
� 4π � 12π � 188 in.2
� 16π
� 50 cm2
39. l2 � 32 � 42 � 9 � 16 � 25
l� �2�5� � 5
L � πrl
� π(3)(5)
� 15π
� 47 in.2
40. 60π � 15π � 45π � 141.4
Approximately 141 square inches of material is needed tomake the Elizabethan collar.
41. l2 � 142 � 182 � 196 � 324 � 520
l� �5�2�0� � 23 cm
42. P � 4(28) � 112
l � 23
L � �12
�Pl
� �12
�(112)(23)
� 1288 cm2
43. Lateral area of one pyramid:
P � 4(12) � 48
l2 � 62 � 8.82 � 36 � 77.44 � 113.44
l � �1�1�3�.4�4� � 10.65
L � �12
�Pl
� �12
�(48)(10.65)
� 255.6 square units
Surface area of solid � 2 � L
� 2 � 255.6
� 511 square units
44. l2 � 1.52 � 32 � 2.25 � 9 � 11.25
l� �1�1�.2�5� � 3.35
S � B � Ph � �12
�Pl
� (3 � 3) � (4 � 3)(6) � �12
�(4 � 3)(3.35)
� (9) � (12)(6) � �12
�(12)(3.35)
� 9 � 72 � 20.1
� 101 square units
Geometry, Concepts and Skills 145Chapter 9 Worked-Out Solution Key
Chapter 9 continued
45. l2 � 32 � 42 � 9 � 16 � 25
l � �2�5� � 5
S � πr2 � 2πrh � πrl
� π(3)2 � 2π(3)(10) � π(3)(5)
� 9π � 60π � 15π
� 84π
� 264 square units
9.3 Standardized Test Practice (p. 498)
46. a. Cone A: Cone B:
S � πr2 � πrl S � πr2 � πrl
� π(3)2 � π(3)(6) � π(3)2 � π(3)(8)
� 9π � 18π � 9π � 24π
� 27π � 33π
� 85 ft2; � 104 ft2;
Cone C:
S � πr2 � πrl
� π(3)2 � π(3)(10)
� 9π � 30π
� 39π
� 123 ft2
b. The radius stayed the same.
c. Cone D: Cone E:
S � πr2 � πrl S � πr2 � πrl
� π(2)2 � π(2)(8) � π(4)2 � π(4)(8)
� 4π � 16π � 16π � 32π
� 20π � 48π
� 63 ft2; � 151 ft2;
Cone F:
S � πr2 � πrl
� π(6)2 � π(6)(8)
� 36π � 48π
� 84π
� 264 ft2
d. The slant height stayed the same.
e. The radius has a greater influence on surface area. Theradius is used in computing both the base area and the lateral surface area, and is squared when computing thebase area, while the slant height affects only the lateral surface area.
9.3 Mixed Review (p. 499)
47. 3x2 � 3(6)248. x2 � 6 � (4)2 � 6
� 3(36) � 16 � 6
� 108 � 22
49. 2x2 � 3 � 2(3)2 � 3 50. 4x2 � 10 � 4(2)2 � 10
� 2(9) � 3 � 4(4) � 10
� 18 � 3 � 16 � 10
� 15 � 6
51. x2 � 4x � (6)2 � 4(6)
� 36 � 24
� 60
52. 3x2 � 5x � 3(5)2 � 5(5)
� 3(25) � 25
� 75 � 25
� 100
53. �18
x0
� � �130600
� 54. �
11x4
� � �36600
�
360x � 180 � 100 360x � 60 � 114
360x � 18,000 360x � 6840
x � 50 m2 x � 19 ft2
55. �25
x8
� � �316200
�
360x � 258 � 120
360x � 30,960
x � 86 cm2
9.3 Algebra Skills (p. 499)
56. �2x � 3 � 6x � (�2 � 6)x � 3
� 4x � 3
57. 4y � 5 � 4y � 1 � (4 � 4)y � (5 � 1)
� 4
58. 4 � (2x � 1) � x � 4 � 2x � 1 � x
� (�2 � 1)x � (4 � 1)
� �x � 5
59. 7x � 2 � (3x � 2) � 7x � 2 � 3x � 2
� (7 � 3)x � (�2 � 2)
� 4x � 4
60. x � 9x � 6x � (1 � 9 � 6)x
� �2x
61. 10c � (5 � 3c) � 10c � 5 � 3c
� (10 � 3)c � 5
� 13c � 5
Quiz 1 (p. 499)
1. Triangular base; triangular prism; yes, this is a polyhedron with 5 faces.
2. Circular base; cone;no, this is not a polyhedron.
3. Rectangular base; rectangular pyramid;yes, this is a polyhedron with 5 faces.
146 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
4. S � 2πr2 � 2πrh
� 2π(2)2 � 2π(2)(9)
� 8π � 36π
� 44π
� 138 ft2
5. B � 3 � 7 � 21
P � 2(3) � 2(7) � 20
h � 10
S � 2B � Ph
� 2(21) � (20)(10)
� 42 � 200
� 242 in.2
6. l2 � 62 � 82 � 36 � 64 � 100
l � �1�0�0� � 10
S � πr2 � πrl
� π(6)2 � π(6)(10)
� 36π � 60π
� 96π
� 302 m2
7. B � �12
�(12)(5) � 30
P � 13 � 12 � 5 � 30
h � 8
S � 2B � Ph
� 2(30) � (30)(8)
� 300 in.2
8. B � 10 � 10 � 100
P � 10 � 10 � 10 � 10 � 40
l � 9
S � B � �12
�Pl
� 100 � �12
�(40)(9)
� 280 m2
9. S � 2πr2 � 2πrh
� 2π(5)2 � 2π(5)(7)
� 50π � 70π
� 120π
� 377 cm2
Lesson 9.4
9.4 Checkpoint (pp. 501–502)
1. V � Bh 2. V � Bh
� (9 � 4) � 6 � (5 � 5) � 5
� 36 � 6 � 25 � 5
� 216 ft3 � 125 cm3
3. V � Bh
� ��12
� � 7 � 7� � 10
� 24.5 � 10
� 245 in.3
4. V � πr2h 5. V � πr2h
� π(2)2(3) � π(1)2(5)
� 12π � 5π
� 38 ft3 � 16 in.3
6. V � πr2h
� π(2)2(10)
� 40π
� 126 m3
9.4 Guided Practice (p. 503)
1. volume 2. surface area
3. surface area 4. volume
5. V � Bh � (63.6)(12) � 763.2 cm3
6. V � Bh
� ��12
� � 8 � 6� � 10
� (24) � 10
� 240 cm3
7. V � Bh � (23.4)(12) � 280.8 cm3
9.4 Practice and Applications (pp. 503–507)
8. 30 unit cubes; the base is 5 units by 3 units. So, 3 � 5, or15 unit cubes are needed to cover the bottom layer. Thereare 2 layers. So, the total number of cubes is 2 � 15, or 30.
9. 48 unit cubes; the base is 4 units by 4 units. So, 4 � 4, or16 unit cubes are needed to cover the bottom layer. Thereare 3 layers. So, the total number of cubes is 3 � 16, or 48.
10. 24 unit cubes; the base is 3 units by 2 units. So, 3 � 2, or 6 unit cubes are needed to cover the bottom layer. Thereare 4 layers. So, the total number of cubes is 6 � 4, or 24.
11. V � Bh 12. V � Bh
� (5 � 5) � 4 � (2 � 3) � 6
� 25 � 4 � 6 � 6
� 100 in.3 � 36 cm3
Geometry, Concepts and Skills 147Chapter 9 Worked-Out Solution Key
Chapter 9 continued
13. V � Bh
� ��12
� � 4 � 9� � 12
� 18 � 12
� 216 m3
14. 15.
V � Bh V � Bh
� (3 � 3) � 3 � (7 � 7) � 7
� 27 m3 � 49 � 7
� 343 ft3
16.
V � Bh
� (10 � 10) � 10
� 1000 cm3
17. 18.
V � Bh
� (4 � 4) � 7 V � Bh
� 16 � 7 � (3 � 6) � 8
� 112 m3 � 18 � 8
� 144 ft3
19. V � Bh � Bh
� (2 � 3) � 8 � (5 � 3) � 2
� 6 � 8 � 15 � 2
� 48 � 30
� 78 ft3
20. V � Bh � Bh
� (2 � 1) � 4 � (6 � 1) � 2
� 2 � 4 � 6 � 2
� 8 � 12
� 20 in.3
21. V � Bh � Bh
� (7 � 10) � 4 � (7 � 2) � 5
� 70 � 4 � 14 � 5
� 280 � 70
� 350 m3
3 ft
8 ft
6 ft4 m
7 m
4 m
10 cm10 cm
10 cm
7 ft7 ft
7 ft
3 m3 m
3 m
22. smaller box: V � Bh
� (8 � 2) � 10
� 16 � 10
� 160 in.3
larger box: V � Bh
� (10 � 4) � 16
� 40 � 16
� 640 in.3
23. 640 160 � 4 boxes
24. The larger box gives you the most cereal for your money.Four times the volume of the smaller box would cost $8,not $6.
25. V � Bh
� (800 � 80) � 22
� 64,000 � 22
� 1,408,000 ft3
26. 1,408,000 ft3 � �71.5
fgt3al
� � 10,560,000 gal
27. V � πr2h 28. V � πr2h
� π(4)2(9) � π(6)2(3)
� 144π � 108π
� 452 in.3 � 339 m3
29. V � πr2h 30. V � πr2h
� π(12)2(15) � π(10)2(4)
� 2160π � 400π
� 6786 m3 � 1257 ft3
31. V � πr2h 32. V � πr2h
� π(12)2(4) � π(7.5)2(3)
� 576π � 168.75π
� 1810 ft3 � 530 ft3
33. The pool in Exercise 32 has the smallest volume, so itwould require the least amount of water to fill it.
34. horizontal line: V � πr2h
� π(3)2(5)
� 45π
� 141 in.3
vertical line: V � πr2h
� π(5)2(3)
� 75π
� 236 in.3
35. The solid on the right has a greater volume. The solidwith the vertical line of rotation has a volume that isalmost twice the volume of the solid with the horizontalline of rotation.
148 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
36. V � πr2h
� π(20)2(23)
� 9200π
� 28,903 ft3
37. 28,903 ft3 � �71.5
fgt3al
� � 216,773 gal
38. V � Bh
� (20 � 10) � 11
� 200 � 11
� 2200 in.3
2200 in.3 � �23
11gianl.3
� � 9.5 gal
9 or 10 fish
39. small carton: V � πr2h
� π(5)2(10)
� 250π
� 785 cm3
jumbo carton: V � πr2h
� π(10)2(20)
� 2000π
� 6283 cm3
40. No; the jumbo carton contains about 8 times as much ice cream as the regular carton (6283 785 � 8).
41. V � Bh � Bh
� (6 � 8) � 2 � (1 � 2) � 2
� 48 � 2 � 2 � 2
� 96 �4
� 92 in.3
42. V � πr2h � πr2h
� π(8)2(10) � π(3)2(10)
� 640π � 90π
� 550π
� 1728 ft3
43. V � Bh � Bh
� (4 � 4) � 4 � (1 � 1) � 4
� 16 � 4 � 1 � 4
� 64 �4
� 60 m3
44. V � Bh
� (3 � x) � 4
� 12x
45. V � Bh
� ��12
� � x � 2x� � 7
� x2 � 7
� 7x2
46. V � Bh
� (3x � 5x) � x
� 15x2 � x
� 15x3
47. V � πr2h 48. V � πr2h
100.48 � π(2)2h 1536.6 � πr2(10)
100.48 � 4πh 1536.6 � 31.42r2
100.48 � 12.57h 48.91 � r2
8 in. � h 7 ft � r
49. Let x � width of prism
V � Bh
54 � (2x � x) � x
54 � 2x2 � x
54 � 2x3
27 � x3
3 in. � x
width � 3 in., length � 3 � 2 � 6 in., height � 3 in.
9.4 Standardized Test Practice (p. 507)
50. C; V � πr2h 51. H; V � Bh
� π(10)2(5) 168 � (7 � 3) � h
� 500π 168 � 21h
� 1570 in.3 8 ft � h
9.4 Mixed Review (p. 507)
52. 72 � a2 � 12253. 82 � b2 � 92
49 � a2 � 144 64 � b2 � 81
a2 � 95 b2 � 17
a � 9.7 b � 4.1
54. a2 � 142 � 182
a2 � 196 � 324
a2 � 128
a � 11.3
55. S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(7)
� 18π � 42π
� 60π
� 188 ft2
Geometry, Concepts and Skills 149Chapter 9 Worked-Out Solution Key
Chapter 9 continued
56. B � 2 � 9 � 18
P � 2(2) � 2(9) � 4 � 18 � 22
h � 8
S � 2B � Ph
� 2(18) � 22(8)
� 36 � 176
� 212 m2
57. S � πr2 � πrl
� π(5)2 � π(5)(12)
� 25π � 60π
� 85π
� 267 yd2
9.4 Algebra Skills (p. 507)
58. x � 7 � 0 59. m � 1 � �12
x � 7 � 7 � 0 � 7 m � 1 � 1 � �12 � 1
x � 7 m � �11
60. 10 � c � �3
10 � c � 10 � �3 � 10
c � �13
61. �34
�b � 24
�43
� � �34
�b � �43
� � 24
b � 32
62. �14d � 2
��
�
1144d
� � ��
214�
d � ��17
�
63. 6n � 102
�66n� � �
1062
�
n � 17
Lesson 9.5
9.5 Activity (pp. 508–509)
1. The base areas are the same.
2. The heights are the same.
3. The rectangular prism has the greater volume; the pyramidfits inside the prism with extra room around the pyramid.
4. 3 times; the volume of the prism is 3 times the volume ofthe pyramid.
5. Volume of the prism � 3 times the volume of the pyramid;
Volume of the pyramid � �13
� times the volume of the prism
6. Volume of a pyramid: V � �13
�Bh
9.5 Checkpoint (pp. 511–512)
1. V � �13
�Bh 2. V � �13
�Bh
� �13
�(6 � 6) � 7 � �13
�(8 � 9) � 5
� 84 in.3 � 120 ft3
3. V � �13
�Bh
� �13
���12
� � 12 � 9� � 10
� 180 cm3
4. V � �13
�πr2h
� �13
�π(5)2(9)
� 75π
� 236 in.3
5. V � �13
�πr2h
� �13
�π(5)2(7)
� �1735
�π
� 183 ft3
6. 102 � h2 � 262
100 � h2 � 676
h2 � 576
h � 24
V � �13
�πr2h
� �13
�π(10)2(24)
� 800π
� 2513 m3
7. V � �13
�πr2h
� �13
�π(4)2(6)
� 32π
� 101 in.3
8. 82 � h2 � 172
64 � h2 � 289
h2 � 225
h � 15
V � �13
�πr2h
� �13
�π(8)2(15)
� 320π
� 1005 ft3
150 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
9.5 Guided Practice (p. 513)
1. D 2. C 3. B 4. A
5. V � �13
�Bh
� �13
�(3 � 5) � 7
� 35 m3
6. V � �13
�πr2h
� �13
�π(9)2(10)
� 270π
� 848 in.3
7. 52 � h2 � 132
25 � h2 � 169
h2 � 144
h � 12
V � �13
�Bh
� �13
�(10 � 10) � 12
� 400 ft3
8. 92 � h2 � 152
81 � h2 � 225
h2 � 144
h � 12
V � �13
�πr2h
� �13
�π(9)2(12)
� 324π in.3
9.5 Practice and Applications (pp. 513–516)
9. B � 5 � 6 � 30 ft2
10. B � �12
� � 7 � 4 � 14 in.2
11. B � π(8)2 � 64π � 201 cm2
12. V � �13
�Bh 13. V � �13
�Bh
� �13
�(3 � 3) � 5 � �13
�(8 � 12) � 7
� 15 ft3 � 224 in.3
14. V � �13
�Bh 15. V � �13
�Bh
� �13
���12
� � 11� 9� � 12 � �13
�(48)(5)
� 198 yd3 � 80 ft3
16. V � �13
�Bh
� �13
�(4 � 4) � 3
� 16 in.3
17. V � �13
�πr2h 18. V � �13
�πr2h
� �13
�π(3)2(10) � �13
�π(12)2(7)
� 30π � 336π
� 94 yd3 � 1056 cm3
19. V � �13
�πr2h 20. V � �13
�Bh
� �13
�π(9)2(8) � �13
�(16)(3)
� 216π � 16 in.3
� 679 m3
21. V � �13
�πr2h 22. V � �13
�Bh
� �13
�π(5)2(8) � �13
�(46,535)(144)
� �2030
�π � 2,233,680 m3
� 209 cm3
23. V � �13
�Bh
� �13
�(6 � 6) � 8
� 96 ft3
V � �13
�Bh
� �13
�(6 � 6) � 16
� 192 ft3
192 96 � 2
The volume is doubled.
24. V � �13
�Bh
� �13
�(12 � 12) � 8
� 384 ft3
384 96 � 4
The volume is multiplied by 4.
25. The student used the slant height instead of the height of the pyramid.
52 � h2 � 132
25 � h2 � 169
h2 � 144
h � 12
V � �13
�Bh
� �13
�(10 � 10) � 12
� 400 in.3
Geometry, Concepts and Skills 151Chapter 9 Worked-Out Solution Key
Chapter 9 continued
26. The student used the diameter instead of the radius of the base.
V � �13
�πr2h
� �13
�π(4)2(6)
� 32π
� 100.53 ft3
27. V � �13
�Bh
20 � �13
�(15)h
20 � 5h
4 in. � h
28. V � �13
�πr2h 29. V � �13
�Bh
8π � �13
�πr2(6) 120 � �13
�s2(10)
8π � 2πr2 36 � s2
4 � r2 6 m � s
2 ft � r
30. 82 � r2 � 10231. 82 � h2 � 172
64 � r2 � 100 64 � h2 � 289
r2 � 36 h2 � 225
r � 6 h � 15
V � �13
�πr2h V � �13
�Bh
� �13
�π(6)2(8) � �13
�(16 � 16) � 15
� 96π � 1280 m3
� 302 ft3
32. 72 � h2 � 252
49 � h2 � 625
h2 � 576
h � 24
V � �13
�πr2h
� �13
�π(7)2(24)
� 392π
� 1232 in.3
33. 34.
62 � h2 � 102 42 � h2 � 82
36 � h2 � 100 16 � h2 � 64
h2 � 64 h2 � 48
h � 8 h � 4�3�
V � �13
�πr2h V � �13
�πr2h
� �13
�π(6)2(8) � �13
�π(4)2(4�3�)
� 96π � 36.95π
� 302 cm3 � 116 m3
35.
32 � h2 � 52
9 � h2 � 25
h2 � 16
h � 4
V � �13
�Bh
� �13
�(6 � 6) � 4
� 48 ft3
36. small container:
V � �13
�πr2h
� �13
�π(3)2(6)
� 18π
� 57 in.3
large container:
V � πr2h
� π(3)2(6)
� 54π
� 170 in.3
37. 170 57 � 3 containers
38. The large container gives you more popcorn for yourmoney; Sample answer: You need to buy three smallcontainers for $6 to equal the amount of popcorn in thelarge container which costs $4.
6 ft6 ft
5 ft
8 m
4 m6 cm
10 cm
152 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
39. V � πr2h � �13
�πr2h
� π(2.5)2(7.5) � �13
�π(2.5)2(4)
� 46.875π � �235�π
� 173.4 in.3
40. 5 cups � �14
1.4cu
inp.3
� � 72 in.3
41. Yes; the feeder holds about 173.4 cubic inches and only72 cubic inches are needed for five days.
42. V � �13
�πr2h
� �13
�π(3)2(1.83)
� 5.49π
� 17.25 mi3
43. V � �13
�πr2h
� �13
�π(0.4)2(0.25)
� �715�π
� 0.04 mi3
44. V � 17.25 � 0.042 � 17.208 mi3
The volume of Mount St. Helens today is about 17.21cubic miles.
9.5 Standardized Test Practice (p. 516)
45. A; V � �13
�Bh
96 � �13
�B(9)
96 � 3B
32 � B
4 � 8 � B
9.5 Mixed Review (p. 516)
46. C � πd � π(6) � 19 m;
A � πr2 � π(3)2 � 9π � 28 m2
47. C � 2πr � 2π(14) � 28π � 88 in.;
A � πr2 � π(14)2 � 196π � 616 in.2
48. C � πd � π(12) � 38 cm;
A � πr2 � π(6)2 � 36π � 113 cm2
49. B � 5 � 8 � 40
P � 2(5) � 2(8) � 10 � 16 � 26
h � 7
S � 2B � Ph
� 2(40) � 26(7)
� 80 � 182
� 262 in.2
50. S � 2πr2 � 2πrh
� 2π(5)2 � 2π(5)(9)
� 50π � 90π
� 140π
� 440 ft2
51. 122 � 92 � l2
144 � 81 � l2
225 � l2
15 � l
S � 2πr2 � πrl
� π(9)2 � π(9)(15)
� 81π � 135π
� 216π
� 679 in.2
9.5 Algebra Skills (p. 516)
52. 53.
The slope is positive The slope is positive because the line is rising. because the line is rising.
54. 55.
The slope is zero because The slope is positive the line is horizontal. because the line is rising.
56. 57.
The slope is undefined The slope is negative because the line is vertical. because the line is falling.
y
2
2
x
(�3, 5)
(4, �1)
y
2
2
x(�4, �1)
(�4, 7)
y
1
1
x
(0, 4)
(�2, �2)
y
2
6
x
(6, 2)(�3, 2)
y
4
2
x
(5, 5)
(1, �2)
y
3
1
x
(4, 3)
(0, 0)
Geometry, Concepts and Skills 153Chapter 9 Worked-Out Solution Key
Chapter 9 continued
Lesson 9.6
9.6 Checkpoint (pp. 518–519)
1. S � 4πr22. S � 4πr2
� 4π(4)2 � 4π(6)2
� 64π � 144π
� 201 in.2 � 452 ft2
3. S � 4πr2
� 4π(7)2
� 196π
� 616 in.2
4. V � �43
�πr3
� �43
�π(4)3
� �2536
�π
� 268 in.3
5. V � �12
���43
�πr3� 6. V � �43
�πr3
� �12
���43
�π(3)3� � �43
�π(9)3
� 18π � 972π
� 57 cm3 � 3054 ft3
9.6 Guided Practice (p. 519)
1. A sphere can be divided into two hemispheres, so ahemisphere is half a sphere.
2. S � 4πr23. S � 4πr2
� 4π(1)2 � 4π(3)2
� 4π � 36π
� 13 in.2 � 113 ft2
4. S � 4πr2
� 4π(9)2
� 324π
� 1018 m2
5. V � �43
�πr3
� �43
�π(3)3
� 36π
� 113 ft3
6. V � �43
�πr37. V � �
12
���43
�πr3�� �
43
�π(11)3 � �12
���43
�π(10)3�� �
53324�π � �
40600�π
� 5575 cm3 � 2094 yd3
9.6 Practice and Applications (pp. 520–523)
8. S � 4πr29. S � 4πr2
� 4π(7)2 � 4π(16)2
� 196π � 1024π
� 616 cm2 � 3217 in.2
10. S � 4πr2
� 4π(15)2
� 900π
� 2827 m2
11. Bob wrote V instead of S for surface area, used the wrongformula, used the diameter rather than the radius, andwrote the answer in cubic units rather than square units;
S � 4πr2
� 4π(5)2
� 100π
� 314 mm2
12. S � 4πr213. S � 4πr2
� 4π(4.3)2 � 4π(3.3)2
� 73.96π � 43.56π
� 232 in.2 � 137 cm2
14. S � 4πr215. S � 4πr2
� 4π(10.9)2 � 4π(0.85)2
� 475.24π � 2.89π
� 1493 cm2 � 9 in.2
16. S � 4πr217. S � 4πr2
� 4π(4.75)2 � 4π(4.8)2
� 90.25π � 92.16π
� 284 in.2 � 290 cm2
18. No; the surface area of the sphere is 4π(3)2 � 36π and ifyou double the radius, the surface area is 4π(6)2 � 144π.So, the surface area will quadruple if you double theradius (144π 36π � 4).
19. S � 4πr220. S � 4πr2
� 4π(3960)2 � 4π(1080)2
� 62,726,400π � 4,665,600π
� 197,000,000 mi2 � 14,700,000 mi2
21. �11947,,700000,,000000
mm
ii2
2�� 13.4
The surface area of Earth is approximately 13.4 timeslarger than the surface area of its moon.
22. (0.70)(197,000,000 mi2) � 138,000,000 mi2
23. V � �43
�πr324. V � �
43
�πr3
� �43
�π(8)3 � �43
�π(4)3
� �20
348�π � �
2536
�π
� 2145 m3 � 268 ft3
154 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
25. V � �43
�πr326. V � �
43
�πr3
� �43
�π(10)3 � �43
�π(11)3
� �40
300�π � �
53324�π
� 4189 cm3 � 5575 yd3
27. V � �43
�πr328. V � �
43
�πr3
� �43
�π(7)3 � �43
�π(3.5)3
� �13
372�π � �
1731.5�π
� 1437 ft3 � 180 in.3
29–31.
29. When the radius is doubled, the surface area is quadru-pled. When the radius is tripled, the surface area is ninetimes greater. When the radius is quadrupled, the surfacearea is 16 times greater.
30. The surface area is changed by a greater amount than the radius because the radius is squared in the formulaS � 4πr2.
31. Sample answer: If the radius of a sphere doubled, thevolume would be 23, or 8 times greater. If the radius of a sphere is tripled, the volume would be 33, or 27 timesgreater.
32.
a. When the radius is doubled, the volume is 23, or 8times greater. When the radius is tripled, the volume is33, or 27 times greater. When the radius is quadrupled,the volume is 43, or 64 times greater.
b. The volume is changed by a greater amount than the radius because the radius is cubed in the formula
V � �43
�πr3.
33. S � 4πr2
� 4π(43.5)2
� 7569π
� 23,779 ft2
34. V � �43
�πr3
� �43
�π(43.5)3
� 109,750.5π
� 344,791 ft3
35. V � s3 � 953 � 857,375 ft3
36. S � 2B � Ph, but excluding the ground and roof (thebases) leaves S � Ph.
P � 95 � 95 � 95 � 95 � 380
h � 95
S � Ph � (380)(95) � 36,100 ft2
37. V � �12
���43
�πr3� 38. V � �12
���43
�πr3�� �
12
���43
�π(7)3� � �12
���43
�π(15)3�� �
13672�π � �
13,6500�π
� 718 cm3 � 7069 m3
39. V � �12
���43
�πr3�� �
12
���43
�π(9)3�� 486π
� 1527 in.3
40. V � πr2h � �12
���43
�πr3�� π(10)2(18) � �
12
���43
�π(10)3�� 1800π � �
20300�π
� �74
300�π
� 7749 cm3
41. V � �13
�πr2h � �12
���43
�πr3�� �
13
�π(6)2(12) � �12
���43
�π(6)3�� 144π � 144π
� 288π
� 905 ft3
Geometry, Concepts and Skills 155Chapter 9 Worked-Out Solution Key
Comparing Spheres
A B C
1 Radius, Surface area, Surface area of new spherer 4π r2 Surface area of original sphere
2 3 113.1 1
3 6 452.4 4
4 9 1017.9 9
5 12 1809.6 16
Comparing Spheres
A B C
1 Radius, Volume, Volume of new spherer �43�π r3 Volume of original sphere
2 3 113.1 1
3 6 904.8 8
4 9 3053.6 27
5 12 7238.2 64
Chapter 9 continued
42. V � πr2h � �12
���43
�πr3�� π(52)(9) � ��
12
���43
�π(5)3�� 225π � �
2530
�π
� �4235
�π
� 445 in.3
43. V � �12
���43
�πr3�� �
12
���43
�π(25.3)3�� �
64,7767.108�π
� 33,917 ft3
44. S � �12
�(4πr2) 45. S � �12
�(4πr2)
� �12
�(4π(25.3)2) � �12
�(4π(24)2)
� 1208.18π � 1152π
� 4022 ft2 � 3619 ft2
9.6 Standardized Test Practice (p. 522)
46. A; S � 4πr2
� 4π(16)2
� 1024π
� 3217 in.2
9.6 Mixed Review (p. 523)
47. l2 � 32 � 122 � 9 � 144 � 153
l � �1�5�3� � 12.4
S � πr2 � πrl
� π(3)2 � π(3)(12.4)
� 9π � 37.2π
� 46.2π
� 145 m2
48. B � 4 � 4 � 16
P � 4 � 4 � 4 � 4 � 16
l � 3
S � B � �12
�Pl
� 16 � �12
�(16)(3)
� 40 ft2
49. S � 2πr2 � 2πrh
� 2π(9)2 � 2π(9)(9)
� 162π � 162π
� 324π
� 1018 cm2
50. �6� � 2.4 51. �1�8� � 4.2 52. �7�7� � 8.8
53. �4�0�0� � 20 54. �2�5�6� � 16 55. �9�9� � 9.9
56. �4�0� � 6.3 57. �1�2�0� � 11.0
9.6 Algebra Skills (p. 523)
58. P � 2w � 2l
� 2(6) � 2(11)
� 12 � 22
� 34 ft
59. A � s2
100 � s2
�1�0�0� � s
10 � s
P � 4s
� 4(10)
� 40 in.
60. A � lw 61. P � 4s
40 � 8w 44 � 4s
5 m � w 11 yd � s
Quiz 2 (p. 523)
1. V � Bh
� (9 � 4) � 4
� (36) � 4
� 144 in.3
2. V � Bh
� ��12
� � 6 � 10� � 7
� 210 ft3
3. V � πr2h
� π(5)2(7)
� 175π
� 550 m3
4. V � �13
�Bh
� �13
�(10 � 10) � 12
� 400 ft3
156 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
5. V � �43
�πr3
� �43
�π(8)3
� �20
348�π
� 2145 m3
6. h � �5�2��� 3�2�� �2�5� �� 9�
� �1�6�
� 4
V � �13
�πr2h
� �13
�π(3)2(4)
� 12π
� 38 in.3
7. 8.
V � πr2h S � 4πr2
� π(4)2(4) � 4π(9)2
� 64π � 324π
� 201 in.3 � 1018 cm2
Chapter 9 Summary and Review (pp. 524–527)
1. Polyhedra are named by the shape of their base(s).
2. The surface area of a polyhedron is the sum of the areasof its faces.
3. A prism is a polyhedron with two congruent faces that liein parallel planes.
4. The lateral faces of a prism are the faces of the prismthat are not bases.
5. The volume of a solid is the number of cubic units con-tained in its interior.
6. The solid is a polyhedron with pentagonal bases, so it isa pentagonal prism.
7. A sphere has a curved surface, so it is not a polyhedron.
8. The solid is a polyhedron with a hexagonal base, so it isa hexagonal pyramid.
9. S � 2B � Ph
� 2(3 � 5) � (2 � 3 � 2 � 5)(3)
� 2(15) � (16)(3)
� 30 � 48
� 78 ft2
9 cm
4 in.
4 in.
10. S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(6)
� 18π � 36π
� 54π
� 170 in.2
11. S � 2B � Ph
� 2��12
� � 8 � 6� � (6 � 8 � 10)(9)
� 2(24) � (24)(9)
� 48 � 216
� 264 m2
12. S � B � �12
�Pl
� (5 � 5) � �12
�(4 � 5)(7)
� 25 � �12
�(20)(7)
� 25 � 70
� 95 m2
13. S � πr2 � πrl
� π(8)2 � π(8)(11)
� 64π � 88π
� 152π
� 478 in.2
14. l2 � 32 � 42 � 25
l� �2�5� � 5
S � B � �12
�Pl
� (6 � 6) � �12
�(4 � 6)(5)
� 36 � �12
�(24)(5)
� 36 � 60
� 96 ft2
15. l2 � 62 � 82 � 100
l� �1�0�0� � 10
S � πr2 � πrl
� π(6)2 � π(6)(10)
� 36π � 60π
� 96π
� 302 cm2
16. S � B � �12
�Pl
� 6.9 � �12
�(3 � 4)(6)
� 6.9 � 36
� 43 ft2
Geometry, Concepts and Skills 157Chapter 9 Worked-Out Solution Key
Chapter 9 continued
17. l2 � 122 � 52 � 169
l � �1�6�9� � 13
S � πr2 � πrl
� π(5)2 � π(5)(13)
� 25π � 65π
� 90π
� 283 ft2
18. V � Bh
� (6 � 8)(3)
� (48)(3)
� 144 in.3
19. V � Bh
� ��12
� � 4 � 5�(3)
� (10)(3)
� 30 m3
20. V � πr2h 21. V � πr2h
� π(3)2(9) � π(9)2(8)
� 81π � 648π
� 254 cm3 � 2036 ft3
22. V � Bh 23. V � πr2h
� (4 � 4)(6) � π(10)2(14)
� (16)(6) � 1400π
� 96 yd3 � 4398 in.3
24. V � �13
�Bh
� �13
�(3 � 5)(6)
� 30 m3
25. V � �13
�πr2h
� �13
�π(5)2(12)
� 100π
� 314 in.3
26. h2 � 92 � 152
h2 � 81 � 225
h2 � 144
h � 12
V � �13
�πr2h
� �13
�π(9)2(12)
� 324π
� 1018 cm3
27. S � 4πr228. S � 4πr2
� 4π(6.5)2 � 4π(18)2
� 169π � 1296π
� 531 m2; � 4072 cm2;
V � �43
�πr3 V � �43
�πr3
� �43
�π(6.5)3 � �43
�π(18)3
� 1150 m3 � 7776π
� 24,429 cm3
29. S � 4πr2
� 4π(3.85)2
� 59.29π
� 186 ft2;
V � �43
�πr3
� �43
�π(3.85)3
� 239 ft3
Chapter 9 Chapter Test (p. 528)
1. The solid is a polyhedron with hexagonal bases, so it is ahexagonal prism.
2. The solid is a polyhedron with a square base, so it is asquare pyramid.
3. A cylinder has a curved surface, so it is not a polyhedron.
4. The solid is a polyhedron with triangular bases, so it is atriangular prism.
5. S � 2B � Ph
� 2(4 � 9) � (2 � 4 � 2 � 9)(5)
� 2(36) � (26)(5)
� 72 � 130
� 202 ft2
6. S � 2B � Ph 7. S � 2πr2 � 2πrh
� 2(3 � 3) � (4 � 3)(3) � 2π(4)2 � 2π(4)(5)
� 2(9) � (12)(3) � 32π � 40π
� 18 � 36 � 72π
� 54 in.2 � 226 cm2
8. S � 2πr2 � 2πrh
� 2π(5)2 � 2π(5)(7)
� 50π � 70π
� 120π
� 377 m2
158 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
9. l2 � 82 � 62 � 64 � 36 � 100
l � �1�0�0� � 10
S � πr2 � πrl
� π(8)2 � π(8)(10)
� 64π � 80π
� 144π
� 452 in.2
10. S � B � �12
�Pl
� (6 � 6) � �12
�(4 � 6)(5)
� 36 � 60
� 96 yd2
11. cylinder;
S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(10)
� 18π � 60π
� 78π
� 245 ft2;
V � πr2h
� π(3)2(10)
� 90π
� 283 ft3
12. V � Bh
� ��12
� � 4 � 8�(7)
� (16)(7)
� 112 yd3
13. V � Bh
� (6 � 2)(3)
� (12)(3)
� 36 in.3
14. V � �13
�Bh
� �13
�(6 � 6)(9)
� �13
�(36)(9)
� 108 m3
15. V � πr2h
� π(2)2(5)
� 20π
� 63 ft3
16. V � �43
�πr3
� �43
�π(9)3
� 972π
� 3054 ft3
17. V � �13
�πr2h
� �13
�π(2)2(5)
� �230�π
� 21 cm3
18. The volume is quadrupled.
V � π(2r)2h
� 4πr2h
19. The volume is doubled.
V � πr2h(2h)
� 2πr2h
20. The volume is multiplied by 8.
V � �43
�π(2r)3
� 8��43
��πr3
21. V � Bh
� (14 � 6)(8)
� (84)(8)
� 672 in.3
22. B � 14 � 6 � 84
P � 2(14) � 2(6) � 40
h � 10
There is only one base of the aquarium, so
S � B � Ph
� 84 � (40)(10)
� 484 in.2
Chapter 9 Standardized Test (p. 529)
1. B
2. G; S � 2πr2 � 2πrh 3. C;
� 2π(2)2 � 2π(2)(7) S � 4πr2
� 8π � 28π � 4π(4)2
� 36π � 64π
� 113 in.2 � 201 cm2
Geometry, Concepts and Skills 159Chapter 9 Worked-Out Solution Key
Chapter 9 continued
4. G; B � 12 � 12 � 144
h2 � 102 � 62 � 64
h � �6�4� � 8
V � �13
�Bh
� �13
�(144)(8)
� 384 ft3
5. C; When s � 3:
S � 2B � Ph
� 2(3 � 3) � (4 � 3)(3)
� 2(9) � (12)(3)
� 18 � 36
� 54 ft2
When s � 6:
S � 2B � Ph
� 2(6 � 6) � (4 � 6)(6)
� 2(36) � (24)(6)
� 72 � 144
� 216 ft2
216 54 � 4
6. J; V � �13
�πr2h
48π � �13
�π(6)2h
48π � 12πh
�4182ππ
� � h
4 in. � h
7. B; V � Bh � πr2h
� (6 � 6)(8) � π(1)2(6)
� (36)(8) � 6π
� 288 � 6π
� 269 ft3
8. V � Bh � �13
�Bh
� (6 � 6)(4) � �13
�(6 � 6)(4)
� (36)(4) � �13
�(36)(4)
� 144 � 48
� 192 cm3
9. S � B � Ph � �12
�Pl
� (6 � 6) � (4 � 6)(4) � �12
�(4 � 6)(5)
� 36 � (24)(4) � �12
�(24)(5)
� 36 � 96 � 60
� 192 cm2
10. The volume of the prism section is 3 times the volume of the pyramid section. Sample answer: The prism and the pyramid have the same base and height; the formulas for the volumes of a prism and a pyramid tell us that the volume of the prism is 3 times the volume of the pyramid.
Chapter 9 Algebra Review (p. 531)
1. ��15�� � 0.4 2. �
�72�� � 4.9 3. �
�31�0�� � 0.9
4. ��
91�7�� � 2.2 5. �
�10
7�� � 3.8 6. �
��1�5�1�
� � 0.7
7. ���1�3�9�
� � 2.5 8. �47��2�5�1�
� � 1.2
9. 6 � ��66�� � 6 � 2.4 � 3.6
10. 5 � ��23�� � 5 � 1.2 � 6.2
11. ��24
59� � �
��
2�4�
5�9�
� � �57
� 12. ��694� � �
��6�9�4�
� � �38
�
13. ��182
11� � �
��1�8�2�1�1�
� � �191� 14. ��
316� � �
��3�1�6�
� � �16
�
15. ��41
96� � �
��
4�1�
9�6�
� � �74
� 16. ��176� � �
��1�7�6�
� � ��47��
17. ��188
1� � �
��
1�8�
8�1�
� � ��
�9�8���1�2�
� � ��9�
�8��
1��2�
� � �3�
92�
� � ��32��
18. ��18
00� � �
��1�8�0�0�� � �
��
4�1�0�
��0�2�
� � ��
�4�1��
0��0�2�
� � �21�02�
� � ��52��
19. ��144
070
� � ���
1�4�
4�0�
7�0�
� � ��
�4�4�9�0���0�3�
� � ��4�
�9�4�0��
0��3�
� � �72�03�
�
20. ��112
484
� � ���
1�1�2�4�
8�4�
� � ��
�6�1�4�4���4�2�
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4��2�
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Chapters 1–9 Cumulative Practice (pp. 532–533)
1.
aABC is acute because its measure is less than 90.
2. By the Linear Pair Postulate,
ma1 � 126 � 180
ma1 � 54.
By the Vertical Angles Theorem,
ma2 � ma1 � 54 and ma3 � 126.
3. Perpendicular lines form 4 right angles, so ma1 � ma2 � 90.
x
A (4, 6)B(�1, 3)
C (5, �1)8�2
8
160 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 9 continued
4. By the Vertical Angles Theorem, ma3 � 118. By TheCorresponding Angles Theorem, ma4 � ma3 � 118.
5. By the Linear Pair Postulate, ma5 � 180 � 81 � 99.By the Alternate Interior Angles Theorem, ma6 � 81.By the Alternate Exterior Angles Theorem,ma7 � ma5 � 99.
6. maA � maB � maC � 180
70 � x � 54 � 180
124 � x � 180
x � 56
7. All three angles of T ABC are acute, so T ABC is acute.
8. maA � maB � maC, so BC � AC � AB. The sides of the triangle from longest to shortest are BC**** , AC**** ,and AB**** .
9. Yes; T ABC �T DCB by the SSS Congruence Postulatebecause AB**** � DC**** (Given), AC**** � DB**** (Given), andBC**** � BC**** (Reflexive Property of Congruence).
10. Yes; T JLK �T NLM by the AAS Congruence Theorembecause aK � aM (Given), aKLJ � aMLN (VerticalAngles Theorem), and JL**** � NL**** (Given).
11. Yes; T PQR �T SRT by the HL Congruence Theorembecause aQ � aR (Corresponding Angles Theorem)and aQ is a right angle, so aR is a right angle. In righttriangles PQR and SRT the hypotenuses PR**** and ST**** arecongruent and the legs QR**& and RT**** are congruent(Given).
12. Two acute angles are sometimes complementary.
13. The sides of a rhombus are always congruent.
14. A rectangle always has exactly two lines of symmetry.
15. Two squares are sometimes congruent.
16. UR � �12
�(PQ � TS)
x � �12
�(7 � 13)
� �12
�(20)
� 10
17. AD � BC 18. JK � ML
3x � 1 � 8 2x � 5 � 4x � 5
3x � 9 5 � 2x � 5
x � 3, 10 � 2x
maA � maC 5 � x,
70 � y, maJ � 90
maA � maB � 180 y � 90
70 � z � 180
z � 110
19. AA Similarity Postulate; aABC � aADE and aACB � aAEDby the Corresponding Angles Theorem.
20. �DD
BA� � �
EE
AC�
21. �BA
DB� � �
AC
CE�
�2x
� � �10
4� 4�
�2x
� � �46
�
4x � 12
x � 3
22. The polygon has 5 sides so it is a pentagon.
23. Convex; Sample answer: None of the extended sidespass through the interior.
24. (n � 2) � 180 � (5 � 2) � 180
� 3 � 180
� 540
The sum of the exterior angles is 360 by the PolygonExterior Angles Theorem.
25. (n � 2) � 180 � (8 � 2) � 180
� 6 � 180
� 1080
1080 8 � 135
26. Square; Sample answer: The sum of the measures of oneinterior angle in each of 2 octagons and in one of the yel-low figures is 360.
135 � 135 � x � 360
270 � x � 360
x � 90
So, each interior angle of the yellow figure must measure90. Since the sides of the yellow figure are congruent,the yellow figure is a square.
27. A � s2 � 142 � 196 m2
28. A � �12
�bh � �12
�(8)(17) � 68 ft2
29. A � bh
� (14)(8)
� 112 cm2
30. A � �12
�h(b1 � b2)
� �12
�(4)(4 � 9)
� �12
�(4)(13)
� 26 in.2
Geometry, Concepts and Skills 161Chapter 9 Worked-Out Solution Key
Chapter 9 continued
31. S � 2B � Ph
� 2(4 � 10) � (2 � 4 � 2 � 10)(8)
� 2(40) � (28)(8)
� 80 � 224
� 304 m2;
V � Bh
� (4 � 10)(8)
� (40)(8)
� 320 m3
32. l2 � 122 � 52 � 169
l � �1�6�9� � 13
S � B � �12
�Pl
� (10 � 10) � �12
�(4 � 10)(13)
� 100 � �12
�(40)(13)
� 100 � 260
� 360 ft2;
V � �13
�Bh
� �13
�(10 � 10)(12)
� 400 ft3
33. S � 2πr2 � 2πrh
� 2π(3)2 � 2π(3)(9)
� 18π � 54π
� 72π
� 226 cm2;
V � πr2h
� π(3)2(9)
� 81π
� 254 cm3
34. S � πr2 � πrl
� π(12)2 � π(12)(20)
� 144π � 240π
� 384π
� 1206 cm2;
h2 � 122 � 202
h2 � 144 � 400
h2 � 256
h � 16
V � �13
�πr2h
� �13
�π(12)2(16)
� 768π
� 2413 cm3
35. S � 4πr2
� 4π(16)2
� 1024π
� 3217 in.2
36. V � �43
�πr2h
� �43
�π(16)3
� 17,157 in.3
�12
�V � �12
�(17,157) � 8579 in.3
162 Geometry, Concepts and SkillsChapter 9 Worked-Out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
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