12.1
Exploring Solids
Goals p Use properties of polyhedra.
p Use Euler’s Theorem.
VOCABULARY
Polyhedron A polyhedron is a solid that is bounded by
polygons that enclose a single region of space.
Face The faces of a polyhedron are polygons.
Edge An edge of a polyhedron is a line segment formed by
the intersection of two faces of the polyhedron.
Vertex A vertex of a polyhedron is a point where three or
more edges of the polyhedron meet.
Regular polyhedron A regular polyhedron is a polyhedron
whose faces are all congruent regular polygons.
Convex polyhedron A convex polyhedron is a polyhedron
such that any two points on its surface can be connected by
a line segment that lies entirely inside or on the polyhedron.
Cross section A cross section is the intersection of a plane
and a solid.
Platonic solids A Platonic solid is one of five regular
polyhedra: a regular tetrahedron, a cube, a regular
octahedron, a regular dodecahedron, and a regular
icosahedron. These solids are named after Plato, a Greek
mathematician and philosopher.
Tetrahedron A tetrahedron is a polyhedron with four faces.
Octahedron An octahedron is a polyhedron with eight faces.
Dodecahedron A dodecahedron is a polyhedron with twelve
faces.
Icosahedron An icosahedron is a polyhedron with twenty
faces.
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TYPES OF SOLIDS
Of the five solids below, the prism and pyramid are polyhedra.
The cylinder, sphere, and cone are not polyhedra.
Prism
Example 1
Pyramid
Cylinder
Sphere
Cone
Identifying Polyhedra
Decide whether the solid is a polyhedron. If so, count the number of
faces, vertices, and edges of the polyhedron.
a.
b.
c.
Solution
a. This is a polyhedron. It has 6 faces, 8 vertices, and
12 edges.
b. This is a polyhedron. It has 8 faces, 12 vertices, and
18 edges.
c. This is not a polyhedron. Some of the faces are not polygons.
Example 2
Classifying Polyhedra
Is the polyhedron convex? Is it regular?
a.
b.
convex, regular
c.
convex,
nonregular
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nonconvex,
nonregular
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THEOREM 12.1: EULER’S THEOREM
The number of faces (F), vertices (V ), and edges (E) of a
polyhedron are related by the formula F V E 2 .
Example 3
Using Euler’s Theorem
The solid has 10 faces: 8 trapezoids and
2 octagons. How many vertices does the
solid have?
On their own, 8 trapezoids and 2 octagons
have 8( 4 ) 2( 8 ) 48 sides. In the
solid, each side is shared by exactly two
polygons. So the number of edges is 24 .
Use Euler’s Theorem to find the number
of vertices.
FVE2
Write Euler’s Theorem.
10 V 24 2
Substitute.
V 16
Solve for V.
Answer The solid has 16 vertices.
Checkpoint Is the solid a polyhedron? If so, is it convex?
Is it regular?
1.
2.
Yes, convex;
nonregular
3.
No
Yes, nonconvex;
nonregular
4. Critical Thinking Is it possible for a polyhedron to have
16 faces, 34 vertices, and 50 edges? Explain.
No; From Euler’s Theorem, the number of faces (F),
vertices (V), and edges (E) of a polyhedron are related by
the formula F V E 2.
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12.2
Surface Area of
Prisms and Cylinders
Goal
p Find the surface area of a prism and of a cylinder.
VOCABULARY
Prism A prism is a polyhedron with two congruent faces,
called bases, that lie in parallel planes. The other faces,
called lateral faces, are parallelograms formed by
connecting the corresponding vertices of the bases. The
segments connecting these vertices are lateral edges.
Right prism In a right prism, each lateral edge is
perpendicular to both bases.
Oblique prisms Oblique prisms are prisms that have lateral
edges that are not perpendicular to the bases. The length of
the oblique lateral edges is the slant height of the prism.
Surface area of a polyhedron The surface area of a
polyhedron is the sum of the areas of its faces.
Lateral area of a polyhedron The lateral area of a polyhedron
is the sum of the areas of its lateral faces.
Net A net is a two-dimensional representation of all of the
faces of a polyhedron.
Cylinder A cylinder is a solid with congruent circular bases
that lie in parallel planes. The altitude, or height, of a
cylinder is the perpendicular distance between its bases. The
radius of the base is also called the radius of the cylinder.
Right cylinder A cylinder such that the segment joining the
centers of the bases is perpendicular to the bases
Lateral area of a cylinder The lateral area of a cylinder is the
area of its curved surface. The lateral area is equal to the
product of the circumference and the height, which is 2πrh.
Surface area of a cylinder The sum of the lateral area and the
areas of the two bases
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THEOREM 12.2: SURFACE AREA OF A RIGHT PRISM
The surface area S of a right prism can be found using the
formula S 2B Ph , where B is the area of a base, P is the
perimeter of a base, and h is the height.
Example 1
Using Theorem 12.2
Find the surface area of the right prism.
6 in.
Solution
Each base is an equilateral triangle with a side
length, s, of 6 inches. Using the formula for the
area of an equilateral triangle, the area of each
base is
6 in.
8 in.
6 in.
1
1
(s2) 3
( 6 2) 9 3
in.2
B 3
4
4
The perimeter of each base is P 18 in.
and the height is h 8 in.
6 in.
6 in.
6 in.
Answer So, the surface area is
S 2B Ph 2( 9 3
) 18 ( 8 ) ≈ 175 in.2
Checkpoint Find the surface area of the right prism.
1.
5m
4m
11 m
238 m2
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THEOREM 12.3: SURFACE AREA OF A RIGHT CYLINDER
The surface area S of a right cylinder is
B r 2
S 2B Ch 2πr 2 2πrh ,
C 2r
where B is the area of a base, C is the
circumference of a base, r is the radius
of a base, and h is the height.
Example 2
h
r
Finding the Surface Area of a Cylinder
Find the surface area of the right cylinder.
Solution
Each base has a radius of 4 meters,
and the cylinder has a height of 5 meters.
S 2πr2 2πrh
5m
4m
Formula for surface area of a cylinder
2π( 4 )2 2π( 4 )( 5 )
Substitute.
32 π 40 π
Simplify.
72 π
Add.
≈ 226.19
Use a calculator.
Answer The surface area is about 226 square meters.
Checkpoint Find the surface area of the right cylinder. Round
your result to two decimal places.
2.
9 ft
6 ft
565.49 ft2
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12.3
Surface Area of
Pyramids and Cones
Goal
p Find the surface area of a pyramid and of a cone.
VOCABULARY
Pyramid A pyramid is a polyhedron in which the base is a
polygon and the lateral faces are triangles with a common
vertex. The intersection of two lateral faces is a lateral edge.
The intersection of the base and a lateral face is a base
edge. The altitude, or height, of the pyramid is the
perpendicular distance between the base and the vertex.
Regular pyramid A regular pyramid has a regular polygon for
a base and its height meets the base at its center. The slant
height of a regular pyramid is the altitude of any lateral
face. A nonregular pyramid does not have a slant height.
Circular cone or cone A circular cone, or cone, has a circular
base and a vertex that is not in the same plane as the base.
The altitude, or height, is the perpendicular distance
between the vertex and the base.
Right cone In a right cone, the height meets the base at its
center and the slant height is the distance between the
vertex and a point on the base edge.
Lateral surface of a cone The lateral surface of a cone
consists of all segments that connect the vertex with points
on the base edge.
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Chapter 12 • Geometry Notetaking Guide
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Example 1
A regular
pyramid is
considered a regular
polyhedron only if all
its faces, including
the base, are
congruent. So, the
only pyramid that is
a regular polyhedron
is the regular
triangular pyramid,
or tetrahedron.
Finding the Area of a Lateral Face
Find the area of each lateral face
of the regular pyramid shown at
the right.
h 97 m
Solution
To find the slant height of the
pyramid, use the Pythagorean
Theorem.
s 90 m
1 2
(Slant height)2 h2 s
2
Write formula.
(Slant height)2 97 2 45 2
Substitute.
(Slant height)2 11,434
Simplify.
slant height
1
s
2
Slant height 11,43
4
Take the positive square root.
Slant height ≈ 106.93
Use a calculator.
Answer So, the area of each lateral face is
1
1
(base of lateral face)(slant height), or about ( 90 )( 106.93 ),
2
2
which is about 4812 square meters.
Checkpoint Complete the following exercise.
1. Find the area of a lateral face of
the regular pyramid. Round the
result to one decimal place.
h 11 in.
slant height
40.4 in.2
s 7 in.
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1
s
2
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THEOREM 12.4: SURFACE AREA OF A REGULAR PYRAMID
1
The surface area S of a regular pyramid is S B Pl , where
2
B is the area of the base, P is the perimeter of the base, and l is
the slant height.
Example 2
Finding the Surface Area of a Pyramid
To find the surface area of the regular
pyramid shown, start by finding the area of
the base.
Use the formula for the area of a regular
1
polygon, (apothem)(perimeter).
2
18 in.
8 in.
A diagram of the base is shown at the right.
After substituting, the area of the base is
1
(43
)(6 p 8 ), or 96 3
square inches.
2
4
3 in.
4
3 in.
8 in.
Now you can find the surface area, using
for the area of the base, B.
96 3
1
S B Pl
2
Write formula.
1
( 48 )( 18 )
96 3
2
Substitute.
432
96 3
Simplify.
≈ 598.3
Use a calculator.
Answer So, the surface area is about 598.3 square inches.
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THEOREM 12.5: SURFACE AREA OF A RIGHT CONE
The surface area S of a right cone is
S πr 2 πrl ,
l
where r is the radius of the base and l is the
slant height.
r
Finding the Surface Area of a Right Cone
Example 3
To find the surface area of the right cone
shown, use the formula for the surface area.
S πr2 πrl
5m
Write formula.
π( 3 )2 π( 3 )( 5 )
Substitute.
9 π 15 π
Simplify.
24 π
Add.
3m
Answer The surface area is 24 π square meters, or about
75.4 square meters.
Checkpoint Find the surface area of the solid. Round your
result to two decimal places.
2. Regular pyramid
3. Right cone
30 mm
10 ft
6 ft
20 mm
1600 mm2
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301.60 ft2
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12.4
Volume of Prisms and Cylinders
Goals p Use volume postulates.
p Find the volumes of prisms and cylinders.
VOCABULARY
Volume of a solid The volume of a solid is the number of
cubic units contained in the solid’s interior.
POSTULATE 27: VOLUME OF A CUBE
The volume of a cube is the cube of the length of its side, or
V s3 .
POSTULATE 28: VOLUME CONGRUENCE POSTULATE
If two polyhedra are congruent, then they have the same
volume .
POSTULATE 29: VOLUME ADDITION POSTULATE
The volume of a solid is the sum of the volumes of all its
nonoverlapping parts.
THEOREM 12.6: CAVALIERI’S PRINCIPLE
If two solids have the same height and the same cross-sectional
area at every level, then they have the same volume .
THEOREM 12.7: VOLUME OF A PRISM
The volume V of a prism is V Bh , where B is the area of a
base and h is the height.
THEOREM 12.8: VOLUME OF A CYLINDER
The volume V of a cylinder is V Bh πr 2h , where B is the
area of a base, h is the height, and r is the radius of a base.
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262
Example 1
Finding Volumes
Find the volume of the right prism and the right cylinder.
a.
3 ft
b.
1 ft
5m
6m
2 ft
Solution
1
3
a. The area B of the base is ( 1 )( 3 ), or ft 2. Use h 2 to
2
2
find the volume.
3
V Bh ( 2 ) 3 ft 3
2
b. The area B of the base is π p 5 2, or 25 π m 2. Use h 6 to
find the volume.
V Bh 25 π( 6 ) 150 π ≈ 471.24 m 3
Checkpoint Find the volume of the solid. Round your result to
two decimal places.
1. Right prism
2. Right cylinder
6 ft
5m
9 ft
4m
11 m
220 m3
Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
1017.88 ft3
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263
Using Volumes
Example 2
Use the measurements given to solve for x.
a. Cube,
V 90 ft3
b. Right cylinder,
V 1253 m3
xm
x ft
10 m
x ft
x ft
Solution
a. A side length of the cube is x feet.
V s3
90 x 3
4.48 ≈ x
Formula for volume of cube
Substitute.
Take the cube root.
Answer So, the height, width, and length of the cube are about
4.48 feet.
b. The area of the base is πx 2 square meters.
V Bh
Formula for volume of cylinder
1253 πx 2( 10 )
Substitute.
1253 10 πx 2
Rewrite.
1253
10 π
x2
39.88 ≈ x 2
6.32 ≈ x
Divide each side by 10 π.
Simplify.
Find the positive square root.
Answer So, the radius of the cylinder is about 6.32 meters.
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12.5
Volume of Pyramids and Cones
Goal
p Find the volume of pyramids and cones.
THEOREM 12.9: VOLUME OF A PYRAMID
1
The volume V of a pyramid is V Bh , where
3
B is the area of the base and h is the height.
h
B
THEOREM 12.10: VOLUME OF A CONE
1
1
The volume V of a cone is V Bh πr 2h ,
3
3
where B is the area of the base, h is the height, and
r is the radius of the base.
Example 1
h
B
r
Finding the Volume of a Pyramid
Find the volume of the pyramid with the
regular base.
10 in.
Solution
The base can be divided into six equilateral
triangles. Using the formula for the area of an
1
p s 2, the area of the
equilateral triangle, 3
4
base B can be found as follows:
1
1
6 p 3
p s2 6 p 3
p 4 2 24 3
in.2
4
4
Use Theorem 12.9 to find the volume of the pyramid.
1
1
V Bh ( 243
)( 10 ) 80 3
3
3
Answer The volume of the pyramid is 80 3
, or about
138.6 cubic inches.
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4 in.
4 in.
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265
Example 2
Finding the Volume of a Cone
Find the volume of each cone.
a. Right circular cone
b. Oblique circular cone
16.8 mm
12 ft
9.5 mm
4.5 ft
Solution
a. Use the formula for the volume of a cone.
1
V Bh
Formula for volume of cone
3
1
(πr2)h
3
Base area equals πr 2.
1
(π 9.5 2) 16.8
3
Substitute.
505.4 π
Simplify.
Answer The volume of the cone is 505.4 π, or about
1588 cubic millimeters.
b. Use the formula for the volume of a cone.
1
V Bh
Formula for volume of cone
3
1
(πr 2)h
3
Base area equals πr 2.
1
(π 4.5 2) 12
3
Substitute.
81 π
Simplify.
Answer The volume of the cone is 81 π, or about 254 cubic feet.
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266
Checkpoint Find the volume of the solid. Round your result to
two decimal places.
1. Pyramid with regular base
2. Right circular cone
15 cm
10 ft
6 ft
9 cm
1052.22 cm3
Example 3
301.60 ft3
Using the Volume of a Cone
Use the given measurements to solve for x.
8m
Solution
1
V (πr 2)h
3
Formula for volume
1
135 (πx 2)( 8 )
3
Substitute.
405 8 πx 2
Multiply each side by 3 .
16.11 ≈ x 2
Divide each side by 8π .
4.01 ≈ x
Find positive square root.
x
Volume 135 m3
Answer The radius of the cone is about 4.01 meters.
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12.6
Surface Area and
Volume of Spheres
Goals p Find the surface area of a sphere.
p Find the volume of a sphere.
VOCABULARY
Sphere A sphere is the locus of points in space that are a
given distance from a point. The point is called the center of
the sphere.
Radius of a sphere A radius of a sphere is a segment from
the center to a point on the sphere.
Chord of a sphere A chord of a sphere is a segment whose
endpoints are on the sphere.
Diameter of a sphere A diameter is a chord that contains the
center of the sphere.
Great circle A great circle is the intersection of a sphere and
a plane that contains the center of the sphere.
Hemisphere Half of a sphere, formed when a great circle
separates a sphere into two congruent halves
THEOREM 12.11: SURFACE AREA OF A SPHERE
The surface area S of a sphere with radius r is S 4πr 2 .
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Chapter 12 • Geometry Notetaking Guide
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Example 1
Finding the Surface Area of a Sphere
Find the surface area. When the radius doubles, does the surface
area double?
a.
b.
3 cm
6 cm
Solution
a. S 4πr 2 4π( 3 )2 36 π cm2
b. S 4πr 2 4π( 6 )2 144 π cm2
The surface area of the sphere in part (b) is four times
greater than the surface area of the sphere in part (a) because
36 π p 4 144 π.
Answer When the radius of a sphere doubles, the surface area
does not double .
Example 2
Using a Great Circle
The circumference of a great circle of a sphere is 7.4π feet. What is
the surface area of the sphere?
Solution
Begin by finding the radius of the sphere.
C 2πr
7.4π 2πr
3.7 r
Formula for circumference of circle
Substitute for C.
Divide each side by 2π.
Using a radius of 3.7 feet, the surface area is
S 4πr 2 4π( 3.7 )2 54.76 π ft2.
Answer The surface area of the sphere is 54.76 π ft2, or about
172 ft2.
THEOREM 12.12: VOLUME OF A SPHERE
4
The volume V of a sphere with radius r is V πr 3 .
3
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Chapter 12 • Geometry Notetaking Guide
269
Example 3
Finding the Volume of a Sphere
5 cm
What is the radius of a sphere made from
the cylinder of modeling clay shown?
Assume the sphere has the same volume
as the cylinder.
15 cm
Solution
To find the volume of the cylinder of
modeling clay, use the formula for the
volume of a cylinder.
Cylinder of
modeling clay
r
V πr 2h π( 5 )2( 15 ) 375 π cm3
To find the radius of the sphere, use the
formula for the volume of a sphere and
solve for r.
4
V πr 3
3
4
375 π πr 3
3
Formula for volume of sphere
Substitute for V.
1125 π 4πr 3
Multiply each side by 3 .
281.25 r 3
Divide each side by 4π .
6.55 ≈ r
Sphere made from cylinder
of modeling clay
Use a calculator to take the cube root.
Answer The radius of the sphere is about 6.55 centimeters.
Checkpoint Find the surface area and volume of the sphere.
Round your results to two decimal places.
1.
5 ft
314.16 ft2; 523.60 ft3
Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
2.
6.5 m
530.93 m2; 1150.35 m3
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12.7
Similar Solids
Goals p Find and use the scale factor of similar solids.
p Use similar solids to solve real-life problems.
VOCABULARY
Similar solids Two solids with equal ratios of corresponding
linear measures, such as heights or radii, are called similar
solids.
Identifying Similar Solids
Example 1
Decide whether the two solids are similar. If so, compare the
volumes of the solids.
a.
b.
7
6
8
14
3
5
3
5
3
5
10
6
Solution
a. The solids are not similar because the ratios of corresponding
linear measures are not equal, as shown.
5
lengths: 3
5
widths: 3
8
4
heights: 6
3
b. The solids are similar because the ratios of corresponding linear
measures are equal, as shown. The solids have a scale factor
of 2 : 1 .
10
2
6
2
lengths: widths: 5
1
3
1
14
2
heights: 7
1
The volume of the larger prism is V Bh 60 ( 14 ) 840 .
The volume of the smaller prism is V Bh 15 ( 7 ) 105 .
The ratio of side lengths is 2 : 1 and the ratio of volumes is
840 : 105 , or 8 : 1 .
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Chapter 12 • Geometry Notetaking Guide
271
Checkpoint Decide whether the two solids are similar.
1.
2.
5m
10 ft
15 m
5m
4 ft
5 ft
7.5 m
3 ft
not similar
similar
THEOREM 12.13: SIMILAR SOLIDS THEOREM
If two similar solids have a scale factor of a : b, then
corresponding areas have a ratio of a2 : b2 , and corresponding
volumes have a ratio of a3 : b3 .
Example 2
Using the Scale Factor of Similar Solids
Cylinders A and B are similar with a scale
factor of 2 : 5. Find the surface area and
volume of cylinder B given that the surface
area of cylinder A is 96π square feet and the
volume of cylinder A is 128π cubic feet.
B
A
Solution
Begin by using Theorem 12.13 to set up two proportions.
Surface area of A
a2
Surface area of B
b2
Volume of A
a3
Vo lume of B
b3
96π
4
Surface area of B
25
128π
8
Volume of B
125
Surface area of B 600π
Volume of B 2000π
Answer The surface area of cylinder B is 600π square feet and the
volume of cylinder B is 2000π cubic feet.
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Example 3
Finding the Scale Factor of Similar Solids
The two cones are similar. Find the
scale factor.
Solution
Find the ratio of the two volumes.
V 108 cm3
V 256 cm3
a3
108π
b3
256π
Write ratio of volumes.
a3
27
3
b
64
Simplify.
a
3
b
4
Find the cube root.
Answer The two cones have a scale factor of 3 : 4 .
Example 4
Comparing Similar Solids
Two punch bowls are similar with a scale factor of 2 : 3. The
amount of concentrate to be added is proportional to the volume.
How much concentrate does the smaller bowl require if the larger
bowl requires 48 ounces?
Solution
Using the scale factor, the ratio of the volume of the smaller punch
bowl to the larger punch bowl is
a3
b3
2 3
8
≈
27
3 3
1
.
3.4
The ratio of the volumes of the concentrates is about 1 : 3.4 . The
amount of concentrate for the smaller punch bowl can be found by
multiplying the amount of concentrate for the larger punch bowl by
1
3.4
as follows: 48
1
3.4
≈ 14.1 ounces.
Answer The smaller bowl requires about 14.1 ounces
of concentrate.
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