PYRAMIDS, CONES, AND SPHERES
11.2.1 – 11.2.3
Students have already worked with solids, calculating the volume and surface area of
prisms and cylinders, and investigating the relationship between the volumes of similar
solids. Now these skills are extended to determining the volumes and surface areas of
pyramids, cones, and spheres.
For additional information see the Math Notes boxes in Lessons 11.2.2, 11.2.3, and
12.1.1.
Example 1
The base of the pyramid at right is a regular hexagon. Using the
measurements provided, calculate the surface area and volume of
the pyramid.
14"
area)(height) . Calculate
The volume of any pyramid is V =
the surface area the same way as for all solids: calculate the area of
8"
each face and base, then add them all together. The lateral faces of
this pyramid are all congruent triangles. The base is a regular hexagon. Since the area of the
hexagon is needed for both the volume and the surface area, calculate it first.
1 (base
3
There are several ways to determine the area of a regular hexagon.
One way is to divide the hexagon into six congruent equilateral
triangles, each with a side of 8 inches. Calculate the area of one
triangle, then multiply by 6 to get the area of the hexagon. The value
of h, the height of the triangle, is needed first. Observe that the height
divides the equilateral triangle into two congruent 30°-60°-90°
triangles. To determine h, use the Pythagorean Theorem or the
pattern for a 30°-60°-90° triangle. Using either method h = 4 3 " .
Therefore the area of one equilateral triangle is shown at right.
The area of the hexagon is 6 ⋅16 3 = 96 3 ≈ 166.28
Then calculate the volume of the pyramid using
the formula as shown at right.
8
8
h
4
4
A = 12 bh
(
= 12 ⋅ ( 8 ) ⋅ 4 3
in 2 .
)
= 16 3 ≈ 27.71 in 2
V = 13 (base area)(height)
(
)
= 13 ⋅ 96 3 ⋅ (14 )
= 448 3 ≈ 776 in 3
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Core Connections Integrated II
Chapter 11
Next determine the area of one of the triangular faces.
These triangles are slanted, and the height of one of them
is called a slant height. The problem does not give us the
value of the slant height (labeled c at right), but it can
calculated based on the information already given.
c
a
b
A cross section of the pyramid at right shows a right triangle in its
interior. One leg is labeled a, another b, and the hypotenuse c. The
original picture indicates that a = 14". The length of b was
calculated previously: it is the height of one of the equilateral
triangles in the hexagonal base. Therefore, b = 4 3 " . To
calculate c, use the Pythagorean Theorem.
The base of one of the slanted triangles is 8", the length of the
side of the hexagon. Therefore the area of one slanted triangle is
8 61 ≈ 62.48 in 2 as shown below right.
a2 + b2 = c2
14 2 + (4 3)2 = c 2
196 + 48 = c 2
c 2 = 244
c = 244 = 2 61 ≈ 15.62 "
Since there are six of these triangles, the area of the lateral faces is
6(8 61) = 48 61 ≈ 374.89 in 2 .
A = 12 ⋅b ⋅ h
(
= 12 ⋅ ( 8 ) ⋅ 2 61
)
= 8 61 ≈ 62.48 in 2
Now determine the total surface area: 96 3 + 48 61 ≈ 541.17 in 2 .
Example 2
The cone at right has the measurements shown. What are the lateral
surface area and volume of the cone?
The volume of a cone is the same as the
volume of any pyramid:
V = 13 (Area of base)(height) .
The only difference is that the base is a
circle.
11 cm
V = 13 (base area)(height)
(π r2 ) h
= 13 ( π ⋅ 4 2 ) ⋅11
=
4 cm
1
3
= 13 ⋅ (176π ) = 1763 π
l
≈ 184.3 cm 3
Calculating the lateral surface area of a cone is a different matter.
Think of a cone as a child’s party hat. Then imagine cutting it
apart to make it lay flat. If you did, you would see that the cone is
really a sector of a circle––not the circle that makes up the base of
the cone, but a circle whose radius is the slant height of the cone,
labeled as l at right.
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4 cm
127
By using ratios the formula for the lateral surface area of the cone
can be derived as SA = π rl , where r is the radius of the base and
l is the slant height. In this problem, r is given, but l is not.
Determine the length of l by taking a cross section of the cone to
create a right triangle. The legs of the right triangle are 11 cm and
4 cm, and l is the hypotenuse. Use the Pythagorean Theorem to
calculate l ≈ 11.7 cm, as shown below right.
Now calculate the lateral surface area:
SA = π (4)(11.7) ≈ 147.1 cm 2
l
11 cm
4 cm
4 2 + 112 = l 2
l
11
16 + 121 = l 2
l 2 = 137
l = 137 ≈ 11.7 cm
4
Example 3
The sphere at right has a radius of 6 feet. Calculate the surface area
and the volume of the sphere.
Since spheres are related to circles, the formulas for the surface area
and volume have π in them. The surface area of a sphere with radius
r is 4π r 2 . Since the radius of the sphere is 6,
SA = 4π (6)2 = 144π ≈ 452.39 ft 2 . To calculate the volume of the
sphere, use the formula V = 43 π r 3 . Therefore,
π = 288π ≈ 904.78 ft 3 .
V = 43 π (6)3 = 4⋅216⋅
3
128
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6'
Core Connections Integrated II
Chapter 11
Problems
1.
The figure at right is a square-based pyramid. Calculate its
surface area and its volume.
9 cm
2.
Another pyramid, congruent to the one in the previous
problem, is glued to the bottom of the first pyramid, so that
their bases coincide. Calculate the surface area and volume of
the new solid.
3.
A regular pentagon has a side length of 10 inches. Calculate the area of the pentagon.
4.
The pentagon of the previous problem is the base of a right pyramid with a height of
18 inches. What is the surface area and volume of the pyramid?
5.
What is the total surface area and volume of the cone at right?
7 cm
5 ft
12 ft
6.
A cone fits perfectly inside a cylinder as shown. If the
volume of the cylinder is 81π cubic units, what is the
volume of the cone?
7.
A sphere has a radius of 12 cm. What are the surface
area and volume of the sphere?
Determine the volume of each solid.
8.
9.
8 in
5 in
10.
13 in
11 cm
10 in
12 in
8 cm
11.
1.0 m
2.6 m
Parent Guide with Extra Practice
10 in
12.
13.
11 m
16 in
16 in
18 m
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129
14.
15.
6.1 cm
6.2 cm
8.3 cm
16.
5 cm
2 cm
7 in
8 in
8 in
Determine the total surface area of the figures in the previous volume problems.
17. Problem 8
18.
Problem 9
21. Problem 12
22.
Problem 16
19.
Problem 10
20.
Problem 11
Use the given information to calculate the volume of the cone.
23. radius = 1.5 in
height = 4 in
24.
diameter = 6 cm
height = 5 cm
25.
base area = 25π
height = 3
26. base circum. = 12π
height = 10
27.
diameter = 12
slant height = 10
28.
lateral area = 12π
radius = 1.5
Use the given information to calculate the lateral area of the cone.
29. radius = 8 in
slant height = 1.75 in
30.
slant height = 10 cm
height = 8 cm
31.
base area = 25π
slant height = 6
32. radius = 8 cm
height = 15 cm
33.
volume = 100π
height = 5
34.
volume = 36π
radius = 3
Use the given information to calculate the volume of the sphere.
35. radius = 10 cm
36.
diameter = 10 cm
37.
circumference of
great circle = 12π
38. surface area = 256π
39.
circumference of
great circle = 20 cm
40.
surface area = 100
Use the given information to calculate the surface area of the sphere.
41. radius = 5 in
42.
diameter = 12 in
43.
circumference of
great circle = 14
44.
45.
circumference of
great circle = π
46.
volume =
130
volume = 250
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9π
2
Core Connections Integrated II
Chapter 11
Answers
1.
V = 147 cm3, SA ≈ 184 cm2
2.
V = 294 cm3, SA ≈ 270 cm2
3.
A ≈ 172 in2
4.
V ≈ 1032 in3, SA ≈ 654 in2
5.
V ≈ 314 ft3, SA = 90π ≈ 283 ft2
6.
27π cubic units
7.
SA = 576π ≈ 1090 cm2, V = 2304π ≈ 7238 cm3
8.
320 in3
9.
100 π ≈ 314.2 in3
10.
≈ 610 cm3
11.
≈ 2.5 m3
12.
512 in3
13.
≈ 514 m3
14.
≈ 52.3 cm3
15.
20 π
3
≈ 21 cm3
16.
≈ 149 in3
17.
≈ 229 in2
18.
90π ≈ 283 in2
19.
≈ 478 cm2
20.
3.6π ≈ 11.3 m2
21.
576 in2
22.
193.0 in2
23.
3π ≈ 9.4 in3
24.
15π ≈ 47 cm3
25.
25π ≈ 79 units3
26.
120π ≈ 377 units3
27.
96π ≈ 302 units3
28.
≈ 18.5 units3
29.
14π ≈ 43.98 in2
30.
60π ≈ 189 cm2
31.
30π ≈ 94 units2
32.
136π ≈ 427 cm2
33.
≈ 224 units2
34.
117 units2
37.
288π ≈ 905 units3
35.
4000π
3
≈ 4189 cm 3
36.
38.
2048π
3
≈ 2145 units 3
39.
≈ 135. cm3
40.
≈ 94 units3
500π
3
≈ 524 cm 3
41.
100π ≈ 314 in2
42.
144π ≈ 452 in2
43.
≈ 62 units2
44.
≈ 192 units2
45.
π ≈ 3 units2
46.
9π ≈ 28 units2
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131