Practice: surface area and volume
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Determine the surface area of this regular tetrahedron to the nearest square centimetre.
5.0 cm
/
5.8 cm
a. 116 cm2
/
b. 44 cm2
c. 58 cm2
d. 29 cm2
____
2. The lateral area of a cone is 198.6 cm2. The diameter of the cone is 10.2 cm. Determine the height of the cone
to the nearest tenth of a centimetre.
a. 12.4 cm
b. 8.0 cm
c. 8.8 cm
d. 11.3 cm
____
3. The slant height of a right square pyramid is 17 ft. and the side length of the base is 13 ft. Determine its lateral
area to the nearest square foot.
a. 111 square feet
b. 408 square feet
c. 884 square feet
d. 442 square feet
____
4. A regular tetrahedron has edge length 19.0 m and a slant height of 16.5 m. Calculate the surface area of the
tetrahedron to the nearest square metre.
a. 470 m2
b. 627 m2
c. 157 m2
d. 1254 m2
____
5. A right cone has a height of 16 in. and a base diameter of 8 in. Determine the lateral area of the cone to the
nearest square inch.
a. 201 square inches
c. 207 square inches
b. 258 square inches
d. 225 square inches
____
6. In 2008, the Queen Sesheshet Pyramid was discovered in Egypt. Archeologists determined that the original
height of this right square pyramid was about 14 m and the original base side length was about 22 m.
Determine its original lateral area to the nearest square metre.
a. 1267 m2
b. 783 m2
c. 196 m2
d. 616 m2
____
7. A right rectangular pyramid has base dimensions 8 ft. by 6 ft. and a height of 12 ft. Calculate the surface area
of the pyramid to the nearest square foot.
a. 216 square feet
b. 159 square feet
c. 271 square feet
d. 223 square feet
____
8. A right pyramid has a square base with side length 12 m and a height of 7 m. Calculate the surface area of the
pyramid to the nearest square metre.
a. 365 m2
b. 443 m2
c. 664 m2
d. 312 m2
____
9. The surface area of a right cone is 400.2 m2. The radius of the cone is 6.0 m. Determine the height of the cone
to the nearest metre.
a. 13 m
b. 14 m
c. 16 m
d. 15 m
____ 10. A right cone has a height of 13 cm and a base diameter of 17 cm. Determine the surface area of the cone to
the nearest square centimetre.
a. 1057 cm2
b. 642 cm2
c. 574 cm2
d. 415 cm2
____ 11. Calculate the slant height, s, of this right square pyramid to the nearest tenth of a centimetre.
SA = 129.5 cm 2
s
|
|
4.5 cm
a. 6.1 cm
b. 16.6 cm
c. 12.1 cm
d. 11.9 cm
____ 12. Calculate the edge length, l,of this regular tetrahedron to the nearest tenth of a metre.
4.6 m
SA = 48.9 m 2
/
/
l
a. 10.6 m
b. 5.3 m
c. 7.1 m
d. 6.5 m
____ 13. Calculate the volume of this right square pyramid to the nearest cubic foot.
|
7 ft.
|
5 ft.
a. 54 cubic feet
b. 163 cubic feet
c. 62 cubic feet
d. 58 cubic feet
____ 14. Calculate the volume of this right rectangular pyramid to the nearest cubic inch.
9 in.
3 in.
8 in.
a. 64 cubic inches
b. 72 cubic inches
c. 216 cubic inches
d. 78 cubic inches
____ 15. Calculate the volume of this right cone to the nearest tenth of a cubic metre.
5.0 m
3.0 m
a. 47.1 m3
b. 49.3 m3
c. 141.4 m3
d. 55.0 m3
____ 16. A regular tetrahedron has base area 146.4 m2 and height 10.7 m. Determine its volume to the nearest cubic
metre.
a. 522 m3
b. 1566 m3
c. 586 m3
d. 3133 m3
____ 17. A right rectangular prism with base dimensions 7.8 m by 5.1 m has a volume of 110.1 m3. Determine the
height of the prism to the nearest tenth of a metre.
a. 8.3 m
b. 1.2 m
c. 5.5 m
d. 2.8 m
____ 18. A right rectangular pyramid has base dimensions 9 ft. by 5 ft., and a height of 12 ft. Determine its volume to
the nearest cubic foot.
a. 237 cubic feet
b. 192 cubic feet
c. 180 cubic feet
d. 184 cubic feet
____ 19. A right cone has a height of 8 cm and a volume of 250 cm3. Determine the radius of the base of the cone to
the nearest centimetre.
a. 5 cm
b. 11 cm
c. 3 cm
d. 17 cm
____ 20. A right cone has slant height 15 in. and base diameter 12 in. Determine its volume to the nearest cubic inch.
a. 543 cubic inches b. 518 cubic inches c. 1555 cubic inches d. 396 cubic inches
____ 21. A right square pyramid has a base side length of 11 ft. and a slant height of 19 ft. Calculate the volume of the
pyramid to the nearest cubic foot.
a. 2201 cubic feet
b. 539 cubic feet
c. 766 cubic feet
d. 734 cubic feet
____ 22. This regular tetrahedron has a height of 4.7 cm. Calculate its volume to the nearest cubic centimetre.
5.0 cm
5.8 cm
a. 23 cm3
b. 68 cm3
c. 58 cm3
d. 45 cm3
____ 23. This right square pyramid has a volume of 254.7 cm3. Calculate the side length of its base, x, to the nearest
tenth of a centimetre.
14.8 cm
x
a. 5.9 cm
b. 4.3 cm
c. 7.2 cm
d. 4.1 cm
____ 24. A right cylindrical can has a volume of 263.1 cm3. What is the volume of a right cone with the same base and
the same height, to the nearest tenth of a centimetre?
a. 131.6 cm
b. 89.7 cm
c. 91.7 cm
d. 87.7 cm
____ 25. The volume of this right cone is 14.7 mm3. Calculate its height, h, to the nearest tenth of a millimetre.
3.7 mm
h
a. 4.1 mm
b. 2.8 mm
c. 1.4 mm
d. 1.0 mm
____ 26. A right square pyramid has a base side length of 4.5 cm and a slant height of 6.1 cm. Determine the volume of
the pyramid to the nearest cubic centimetre.
a. 13 cm3
b. 38 cm3
c. 43 cm3
d. 71 cm3
____ 27. The radius of a volleyball is approximately 11 cm. Determine the surface area of a volleyball to the nearest
square centimetre.
a. 1521 cm2
b. 380 cm2
c. 6082 cm2
d. 5575 cm2
____ 28. The surface area of a tennis ball is approximately 23 square inches. What is the diameter of the tennis ball to
the nearest inch?
a. 3 in.
b. 4 in.
c. 6 in.
d. 1 in.
____ 29. Mars approximates a sphere with radius 2100 mi. What is the approximate volume of Mars?
a. 3.9  1010 mi.3
b. 5.5  107 mi.3
c. 6.8  1011 mi.3
d. 3.1  1011 mi.3
____ 30. A hemisphere has radius 11.6 cm. What is the surface area of the hemisphere to the nearest tenth of a square
centimetre?
a. 3269.1 cm2
b. 918.4 cm2
c. 845.5 cm2
d. 1268.2 cm2
____ 31. A hemisphere has radius 11.4 cm. What is the volume of the hemisphere to the nearest tenth of a cubic
centimetre?
a. 6205.9 cm3
b. 3102.9 cm3
c. 1224.8 cm3
d. 1633.1 cm3
____ 32. A ten-pin bowling ball has a radius of approximately
nearest square inch.
a. 57 square inches
b. 908 square inches
c. 227 square inches
d. 322 square inches
____ 33. A ten-pin bowling ball has a radius of approximately
cubic inch.
a. 5642 cubic inches
b. 2572 cubic inches
in. Determine the surface area of the ball to the
in. Determine the volume of the ball to the nearest
c. 322 cubic inches
d. 227 cubic inches
____ 34. A sphere has a surface area of 6.4 m2. What is the diameter of the sphere to the nearest tenth of a metre?
a. 2.3 m
b. 1.4 m
c. 2.0 m
d. 0.7 m
____ 35. A sphere has a surface area of 10.1 m2. What is the radius of the sphere to the nearest tenth of a metre?
a. 3.7 m
b. 4.8 m
c. 1.8 m
d. 0.9 m
____ 36. The circumference of a beach ball is 55 cm. Determine its volume to the nearest cubic centimetre.
a. 963 cm3
b. 2810 cm3
c. 22 476 cm3
d. 307 cm3
____ 37. The circumference of a medicine ball is 28 in. Determine its surface area to the nearest square inch.
a. 371 square inches
c. 998 square inches
b. 111 square inches
d. 250 square inches
____ 38. A stadium has a roof that approximates a hemisphere with circumference 2500 ft. Determine the surface area
of the roof to the nearest square foot.
a. 1 492 078 square feet
c. 994 718 square feet
b. 3 978 874 square feet
d. 131 928 625 square feet
____ 39. A flat basketball is inflated using a hand pump. The pump inflates the ball at a rate of
230 cm3 per pump, to a diameter of 23.5 cm. How many pumps are required to inflate the ball?
a. 27 pumps
b. 30 pumps
c. 29 pumps
d. 28 pumps
____ 40. A china bowl approximates a hemisphere with diameter 27.0 cm. What is the capacity of the bowl to the
nearest tenth of a litre? (1000 cm3 = 1L)
a. 5.2 L
b. 10.3 L
c. 0.4 L
d. 2.6 L
____ 41. A china bowl approximates a hemisphere with diameter 30 cm. One cup is 250 mL. How many cups are
required to completely fill the bowl?
a. 30 cups
b. 28 cups
c. 9 cups
d. 29 cups
Short Answer
42. A right cone has a slant height of 14 in. and a base diameter of 10 in. Determine the surface area of the cone
to the nearest square inch.
43. A regular tetrahedron with edge length 12.7 mm has a surface area of 229.0 mm2. Determine the slant height
of the tetrahedron to the nearest millimetre.
44. A right square pyramid has a height of 15 cm and a slant height of 17 cm. Determine the side length of the
base of the pyramid to the nearest centimetre.
45. A regular tetrahedron has an edge length of 9.0 m and a slant height of 7.8 m. Calculate the surface area of the
tetrahedron to the nearest tenth of a square metre.
46. In 2008, the Queen Sesheshet Pyramid was discovered in Egypt. Archeologists determined that the original
height of this right square pyramid was about 14 m and the original base side length was about 22 m.
Determine its original volume to the nearest cubic metre.
47. A regular tetrahedron has base area 98.9 m2 and height 8.6 m. Determine its volume to the nearest tenth of a
cubic metre.
48. A right rectangular pyramid has base dimensions 11 cm by 7 cm and height 9 cm. Determine the volume of
the pyramid to the nearest cubic centimetre.
49. A right cone has a diameter of 17.1 cm and a height of 11.3 cm. Determine the volume of the cone to the
nearest tenth of a cubic centimetre.
50. A right pyramid has a base that is a regular hexagon with side length 2.0 cm. The pyramid has a height of 5.3
cm and a base area of 10.4 cm2. Calculate the volume of the pyramid to the nearest tenth of a cubic
centimetre.
51. A right cone has a volume of 871 cubic inches. The radius of the cone is 8 in. Determine the height of the
cone to the nearest inch.
52. Determine the surface area of this sphere to the nearest square centimetre. Determine its volume to the nearest
cubic centimetre.
19 cm
53. Determine the surface area of this sphere to the nearest square inch. Determine its volume to the nearest cubic
inch.
11 in.
54. A hemisphere has radius 7 ft. Determine the surface area of the hemisphere to the nearest square foot.
55. A hemisphere has radius 12 m. Determine the volume of the hemisphere to the nearest tenth of a cubic metre.
56. A spherical balloon has a surface area of 88 cm2. What is the diameter of the balloon to the nearest tenth of a
centimetre?
57. A spherical globe has diameter 41.3 cm. What is the volume of the globe to the nearest tenth of a centimetre?
Problem
58. A right pyramid with a base that is a regular hexagon has a slant height of 5.0 m. The base area is 10.4 m2 and
the side length of the base is 2.0 m. Calculate the surface area of the pyramid to the nearest tenth of a square
metre.
59. Three wooden blocks need to be painted. The first block is a right rectangular pyramid with base dimensions
1.5 cm by 2.5 cm and a height of 2.0 cm . The second block is a right square pyramid with a base length of
2.8 cm and a height of 2.0 cm. The third block is a right cone with a height of 2.0 cm and a base diameter of
3.6 cm. Which block requires the most paint? Which block requires the least paint? Sketch diagrams to help
explain your answer.
60. Nicole has this right cone, which has lateral area 414.5 cm2. She needs a cone with height at least 15.5 cm for
a craft project. Is this cone tall enough? Justify your answer.
h
7.0 cm
61. A right rectangular pyramid has base dimensions 6 cm by 4 cm and a height of 8 cm. Determine its lateral
area to the nearest square centimetre.
62. A right square pyramid has a height of 7.5 m and a base perimeter of 36 m. Calculate the surface area of the
pyramid to the nearest square metre.
63. A right square pyramid has base perimeter 62.4 m and height 6.4 m. Calculate the volume of the pyramid to
the nearest cubic metre.
64. A right rectangular pyramid has base dimensions 23.2 cm by 17.0 cm and volume 1552.4 cm3. Calculate the
height of the pyramid to the nearest tenth of a centimetre.
65. A right cylinder has base radius 22.9 cm and height 17.1 cm. Determine the volume of a right cone with the
same base and the same height, to the nearest tenth of a cubic centimetre.
66. Francis has three empty containers: a right rectangular prism, a right square pyramid, and a right cone. Each
container has height 2.0 cm. The prism has base dimensions 1.5 cm by 2.5 cm. The pyramid has base side
length 3.4 cm. The cone has base diameter 3.8 cm. Determine the volume of each container to the nearest
tenth of a cubic centimetre. Which container has the least volume? Which container has the greatest volume?
Explain your answer.
67. Determine the volume of a right prism that has the same base and the same height as the right square pyramid
below. Give your answer to the nearest tenth of a cubic metre. Explain your answer.
12.9 m
4.5 m
68. A right cone has a base diameter of 8 in. and a volume of 259 cubic inches. Determine the slant height of the
cone to the nearest inch.
69. A candle approximates a sphere with circumference 21 cm. The surface of the candle is to be covered with
glitter.
a) Determine the radius of the candle to the nearest centimetre.
b) Determine the surface area of the candle to the nearest square centimetre.
70. A hemisphere has radius 23.3 m.
a) Determine the surface area of the hemisphere to the nearest tenth of a square metre.
b) Determine the volume of the hemisphere to the nearest tenth of a cubic metre.
71. A sphere has a diameter of 24 in. A hemisphere has a radius of 18 in. Franco says the surface area of the
sphere is greater than the surface area of the hemisphere. Do you agree? Justify your answer.
72. A baby’s rattle contains a plastic ball inside a spherical case. The diameter of the plastic ball is 2 cm and the
diameter of the case is 7 cm.
7 cm
2 cm
a) Calculate the volume of the spherical case, to the nearest cubic centimetre.
b) Calculate the volume of the plastic ball, to the nearest cubic centimetre.
c) Calculate the volume of air in the rattle, to the nearest cubic centimetre.
73. A pail of ice cream is cylindrical, with diameter 10 in. and height 12 in. A scoop makes a sphere of ice cream
with diameter 2 in. How many full scoops of ice cream can be made from this pail?
Practice: surface area and volume
Answer Section
MULTIPLE CHOICE
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14. ANS:
REF:
TOP:
C
PTS: 1
DIF: Easy
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Easy
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
C
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Difficult
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Moderate
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
C
PTS: 1
DIF: Difficult
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
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31. ANS:
A
PTS: 1
DIF: Easy
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Easy
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
C
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Moderate
1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Easy
1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
A
PTS: 1
DIF: Easy
1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
D
PTS: 1
DIF: Easy
1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
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1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
C
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
C
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
B
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
D
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
B
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
D
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
A
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
B
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
A
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
D
PTS: 1
DIF:
1.6 Surface Area and Volume of a Sphere
Measurement
KEY:
LOC: 10.M3
Procedural Knowledge
Easy
LOC: 10.M3
Procedural Knowledge
Easy
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Difficult
LOC: 10.M3
Procedural Knowledge
Moderate
LOC: 10.M3
Procedural Knowledge
Difficult
LOC: 10.M3
Procedural Knowledge
SHORT ANSWER
42. ANS:
298 square inches
PTS: 1
LOC: 10.M3
43. ANS:
9 mm
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
44. ANS:
16 cm
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
45. ANS:
140.4 m2
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
46. ANS:
2259 m3
DIF: Easy
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
47. ANS:
283.5 m3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
48. ANS:
231 cm3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
49. ANS:
865.0 cm3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
50. ANS:
18.4 cm3
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
51. ANS:
13 in.
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
52. ANS:
SA = 1134 cm2
V = 3591 cm3
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
53. ANS:
SA = 380 square inches
V = 697 cubic inches
PTS: 1
LOC: 10.M3
54. ANS:
462 square feet
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
55. ANS:
3619.1 m3
DIF: Easy
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
56. ANS:
5.3 cm
DIF: Easy
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
57. ANS:
36 884.9 cm3
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
PROBLEM
58. ANS:
SA = (slant height)(perimeter of base) + (base area)
SA = ( )(5.0)(6  2.0) + 10.4
SA = ( )(5.0)(12.0) + 10.4
SA = 30.0 + 10.4
SA = 40.4
The surface area of the pyramid is 40.4 m2.
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
59. ANS:
Surface area of right rectangular pyramid:
Sketch the pyramid and label its vertices.
In EFH, FH is
the length of BC, so FH is
E
2.0 cm
0.75 cm.
EF is the height of the pyramid, which is 2.0 cm.
Use the Pythagorean Theorem in right EFH.
A
D
B
G
F
H
1.5 cm
C
2.5 cm
Area, A, of EDC is:
A = (2.5)(
A = 1.25(
)
)
Since EDC and EAB are congruent, the area of EAB is 1.25(
In EFG, FG is
).
the length of DC, so FG is 1.25 cm.
Use the Pythagorean Theorem in right EFG.
Area, A, of EBC is:
A = (1.5)(
A = 0.75(
)
)
Since EBC and EAD are congruent, the area of EAD is 0.75(
).
Area, B, of the base of the pyramid is:
B = (1.5)(2.5)
B = 3.75
Each of two triangles has area 1.25(
), and each of the other two triangles has area 0.75(
Surface area, SA, of the right rectangular pyramid is:
The surface area of the right rectangular pyramid is approximately 12.6 cm2.
).
Surface area of right square pyramid:
Sketch the pyramid and label its vertices.
the length of BC, so FH is
2.0 cm
1.4 cm.
Use the Pythagorean Theorem in right EFH
to find the slant height, s.
s2 = EF2 + FH2
s2 = 2.02 + 1.42
s2 = 4.0 + 1.96
s2 = 5.96
s=
s
A
|
|
|
|
B
G
2.8 cm
F
|
Surface area, SA, of the right square pyramid
is:
D
| |
|
In EFH, FH is
E
C
H
2.8 cm
SA = ( )s(perimeter of base) + (base area)
SA = ( )(
)(2.8  4) + (2.8  2.8)
SA = ( )(
)(11.2) + 7.84
SA = 21.5113...
The surface area of the right square pyramid
is approximately 21.5 cm2.
Surface area of right cone:
Sketch a diagram.
In ABC, BC is
A
the diameter of the cone,
so BC is 1.8 cm.
Use the Pythagorean Theorem to find the
slant height, s.
s2 = AC2 + BC2
s2 = 2.02 + 1.82
s2 = 4.0 + 3.24
s2 = 7.24
s=
2.0 cm
s
C
3.6 cm
B
Surface area, SA, of the right cone is:
The surface area of the right cone is
approximately 25.4 cm2.
Sothe block that is a right cone requires the most paint and the block that is a right rectangular pyramid
requires the least paint.
PTS: 1
DIF: Difficult
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
60. ANS:
Use the formula for lateral area, AL, of the cone and solve for s.
AL =
To determine the height of the cone, use the Pythagorean Theorem in right ABC.
A
h
s
The height of the cone is approximately
17.5 cm. The cone is tall enough for Nicole’s
craft project.
B
7.0 cm
C
PTS: 1
DIF: Difficult
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
61. ANS:
The lateral area is the area of the triangular faces of the pyramid.
Sketch the pyramid.
In UVW, VW is
the length of TS, so VW is
U
3 cm.
UV is the height of the pyramid, which is 8 cm.
Use the Pythagorean Theorem in right UVW.
UW2 = UV2 + VW2
UW2 = 82 + 32
UW2 = 64 + 9
UW2 = 73
UW =
Area, A, of URS is:
A = (4)(
A = 2(
8 cm
Q
T
)
R
W
V
S
X
4 cm
6 cm
)
Since URS and UQT are congruent, the area of UQT is 2(
In UVX, VX is
).
the length of RS, so VX is 2 cm.
Use the Pythagorean Theorem in right UVX.
UX2 = UV2 + VX2
UX2 = 82 + 22
UX2 = 64 + 4
UX2 = 68
UX =
Area, A, of UST is:
A = (6)(
A = 3(
)
)
Since UST and URQ are congruent, the area of URQ is 3(
).
Each of two triangles has area 2(
), and each of the other two triangles has area 3(
).
The lateral area, AL, of the right rectangular pyramid is:
The lateral area of the right rectangular pyramid is approximately 84 cm2.
PTS: 1
DIF: Difficult
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
62. ANS:
Since the perimeter of the square base is 36 m, its side length is:
Visualize cutting the pyramid vertically in half. Sketch a diagram.
9m
Use the Pythagorean Theorem in right ACD.
s2 = 7.52 + 4.52
s2 = 56.25 + 20.25
s2 = 76.5
s=
Use the formula for the surface area of a right
pyramid with a regular polygon base:
A
7.5 m
s
B
4.5 m
s
C 4.5 m
D
SA = s(perimeter of base) + (base area)
SA = (
)(36) + (81)
SA = 238.4357...
The surface area of the pyramid is approximately 238 m2.
PTS: 1
LOC: 10.M3
63. ANS:
DIF: Difficult
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
TOP: Measurement
KEY: Problem-Solving Skills
The perimeter of the square base is 62.4 m. So, the side length of the base is:
Use the formula for the volume of a right rectangular pyramid.
lwh
The volume of the pyramid is approximately 519 m3.
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
64. ANS:
Use the formula for the volume of a right rectangular pyramid.
The height of the pyramid is approximately 11.8 cm.
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
65. ANS:
Use the formula for the volume of a right cylinder.
The volume of a right cone is
the volume of a right cylinder with the same base and the same height.
The volume of the right cone is approximately 9390.7 cm3.
PTS: 1
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
66. ANS:
Right rectangular prism:
Use the formula for the volume of a right rectangular prism.
The volume of the prism is 7.5 cm3.
Right square pyramid:
Use the formula for the volume of a right rectangular pyramid.
The volume of the pyramid is approximately 7.7 cm3.
Right cone:
The radius, r, of the base of the cone is
the diameter.
Use the formula for the volume of a right cone.
The volume of the cone is approximately 7.6 cm3.
Since
greatest volume.
the right rectangular prism has the least volume and the right square pyramid has the
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
67. ANS:
Calculate the height of the pyramid.
Let h metres represent the height.
In right ABC, BC is
the side length of the base, so BC = 2.25 m.
Use the Pythagorean Theorem in right ABC to calculate h.
A
h
12.9 m
B
4.5 m
The height is
m.
Use the formula for the volume of a right
rectangular pyramid.
C
Volume = lwh
The volume of a right prism is 3 times the volume of a right pyramid with the same base and the same height.
The volume of the right prism is approximately 257.2 m3.
PTS: 1
DIF: Difficult
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
68. ANS:
The radius, r, of the base of the cone is
Use the formula for the volume of a cone.
the diameter.
Use the Pythagorean Theorem to calculate the slant height, s.
s
h
r
The slant height is approximately 16 in.
PTS: 1
DIF: Difficult
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
69. ANS:
a) Use the circumference, C, to determine the radius, r.
The radius of the candle is approximately 3 cm.
b) Use the formula for the surface area of a sphere.
The surface area of the candle is approximately 140 cm2.
PTS: 1
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
70. ANS:
a) SA of a hemisphere = SA of one-half a sphere + area of a circle
The surface area of the hemisphere is approximately 5116.6 m2.
b) Volume of a hemisphere = volume of one-half a sphere
The volume of the hemisphere is approximately 26 492.7 m3.
PTS: 1
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
71. ANS:
Use the formula for the surface area of a sphere.
The radius, r, is:
The surface area of the sphere is approximately 1810 square inches.
Use the formula for the surface area of a hemisphere.
The surface area of the hemisphere is approximately 3054 square inches.
Franco is not correct. The surface area of the hemisphere is greater than the surface area of the sphere.
PTS: 1
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
72. ANS:
a) Use the formula for the volume of a sphere.
The radius, r, is:
The volume of the spherical case is approximately 180 cm3.
b) Use the formula for the volume of a sphere.
The radius, r, is:
The volume of the plastic ball is approximately 4 cm3.
c) The volume of air in the rattle is: 179.5943... cm3 – 4.1887... cm3 = 175.4055... cm3
The volume of air in the rattle is approximately 175 cm3.
PTS: 1
DIF: Difficult
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Problem-Solving Skills
73. ANS:
Volume of ice cream in the pail:
Use the formula for the volume of a cylinder.
The radius, r, is:
Volume of ice cream in a scoop:
Use the formula for the volume of a sphere.
The radius, r, is:
Number of scoops of ice cream:
The number of full scoops of ice cream that can be made from this pail is 225.
PTS: 1
LOC: 10.M3
DIF: Difficult
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Problem-Solving Skills