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LESSON

10.1 The Geometry of Solids

In this lesson you will
● Learn about polyhedrons, including prisms and pyramids
● Learn about solids with curved surfaces, including cylinders, cones,
and spheres
In this chapter you will study three-dimensional solid figures. Lesson 10.1
introduces various types of three-dimensional solids and the terminology
associated with them. Read the lesson. Then review what you read by
completing the statements and answering the questions below.
1. A polyhedron is a solid formed by ________________ that enclose a single
region of space.
2. A segment where two faces of a polyhedron intersect is called
a(n) ________________.
3. A polyhedron with six faces is called a(n) ________________.
4. A tetrahedron has ________________ faces.
5. If each face of a polyhedron is enclosed by a regular polygon, and each face
is congruent to the other faces, and the faces meet each vertex in exactly the
same way, then the polyhedron is called a(n) ________________.
6. A(n) ________________ is a polyhedron with two faces, called bases, that
are congruent, parallel polygons.
7. The faces of a prism that are not bases are called ________________.
8. What is the difference between a right prism and an oblique prism?
9. What type of solid is shown below?

10. How many bases does a pyramid have?

11. The point that all the lateral faces of a pyramid have in common is the
_______________ of the pyramid.
12. What is the difference between an altitude of a pyramid and the height of
a pyramid?
13. What type of solid is shown at right?

(continued)

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Lesson 10.1 • The Geometry of Solids (continued)

14. Name three types of solids with curved surfaces.

15. A(n) ________________ is the set of all points in space at a given distance
from a given point.
16. A circle that encloses the base of a hemisphere is called
a(n) ________________.
17. Give an example of a real object that is shaped like a cylinder. Explain how
you know it is a cylinder.
18. Tell which cylinder below is an oblique cylinder and which is a right
cylinder. For each cylinder, draw and label the axis and the altitude.

19. The base of a cone is a(n) ________________.

20. If the line segment connecting the vertex of a cone with the center of the
base is perpendicular to the base, then the cone is a(n) ________________.

After you have finished, check your answers below.

1. polygons 13. square pyramid

2. edge 14. possible answers: cylinder, cone, sphere,
3. hexahedron hemisphere
4. four 15. sphere
5. regular polygon 16. great circle
6. prism 17. The example will vary. Reason: It has two
parallel bases and both bases are circles.
7. lateral faces
18. The figure on the left is a right cylinder, and
8. In a right prism all lateral faces are rectangles
the figure on the right is an oblique cylinder.
and are perpendicular to the base. In an
See page 523 of the book for labeled pictures
oblique prism they are not.
of both.
9. pentagonal prism (it may be oblique or right)
19. circle
10. one
20. right cone
11. vertex
12. The altitude is a segment from the vertex of
the pyramid to the plane of the base that is
perpendicular to the plane of the base. The
height is the length of the altitude.

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CONDENSED
LESSON

10.2 Volume of Prisms and Cylinders

In this lesson you will
● Discover the volume formula for right rectangular prisms
● Extend the volume formula to right prisms and right cylinders
● Extend the volume formula to oblique prisms and oblique cylinders
Volume is the measure of the amount of space contained in a solid. You use cubic units
to measure volume: cubic inches
in3, cubic feet
ft3, cubic yards
yd3, cubic
centimeters
cm3, cubic meters
m3, and so on. The volume of an object is the
number of unit cubes that completely fill the space within the object.

1 1

1 1

Length: 1 unit Volume: 1 cubic unit Volume: 20 cubic units

Investigation: The Volume Formula for Prisms and Cylinders

Read Step 1 of the investigation in your book.
Notice that the number of cubes in the base layer is the number of square units
in the area of the base and that the number of layers is the height of the prism.
So, you can use the area of the base and the height to find the volume of a right
rectangular prism.

Rectangular Prism Volume Conjecture If B is the area of the base of a C-86a

right rectangular prism and H is the height of the solid, then the formula for
the volume is V
________________.

You can find the volume of any other right prism or cylinder
the same way—by multiplying the area of the base by the
height. For example, to find the volume of this cylinder, find
the area of the circular base (the number of cubes in the
base layer) and multiply by the height.
You can extend the Rectangular Prism Volume Conjecture
to all right prisms and right cylinders. Complete the
conjecture below.

Right Prism-Cylinder Volume Conjecture If B is the area of the base of a C-86b

right prism (or cylinder) and H is the height of the solid, then the formula
for the volume is V
________________.

(continued)

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Lesson 10.2 • Volume of Prisms and Cylinders (continued)

What about the volume of an oblique prism or cylinder? Page 532 of your book
shows that you can model an oblique rectangular prism with a slanted stack
of paper. You can then “straighten” the stack to form a right rectangular prism
with the same volume. The right prism has the same base area and the same
height as the oblique prism. So, you find the volume of the oblique prism by
multiplying the area of its base by its height. A similar argument works for
other oblique prisms and cylinders.
Now you can extend the Right Prism-Cylinder Volume Conjecture to oblique
prisms and cylinders.

Oblique Prism-Cylinder Volume Conjecture The volume of an oblique C-86c

prism (or cylinder) is the same as the volume of a right prism (or cylinder)
that has the same ____________ and the same ________________.

Be careful when calculating the volume of an oblique prism or cylinder. Because

the lateral edges are not perpendicular to the bases, the height of the prism or
cylinder is not the length of a lateral edge.
You can combine the last three conjectures into one conjecture.

Prism-Cylinder Volume Conjecture The volume of a prism or a cylinder is C-86

the ________________ multiplied by the ________________.

Examples A and B in your book show you how to find the volumes of a 8 cm
trapezoidal prism and an oblique cylinder. Read both examples. Try to find
the volumes yourself before reading the solutions. Then read the example below.
14 cm
EXAMPLE The solid at right is a right cylinder with a 135° slice removed.
Find the volume of the solid. Round your answer to the
nearest cm.
135
225 5
 Solution The base is




360 , or 8 , of a circle with radius 8 cm. The whole circle has
area 64
cm2, so the base has area
58
(64
), or 40
cm2. Now use the
volume formula.

V
BH The volume formula.

40
(14) Substitute the area of the base and the height.

560
Multiply.

 1759 Use the
key on your calculator to get an approximate answer.

The volume is about 1759 cm3.

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LESSON

10.3 Volume of Pyramids and Cones

In this lesson you will
● Discover the volume formula for pyramids and cones
There is a simple relationship between the volumes of prisms and pyramids with
congruent bases and the same height, and between cylinders and cones with
congruent bases and the same height.

Investigation: The Volume Formula for Pyramids and Cones

If you have the materials listed at the beginning of the investigation, follow the
steps in your book before continuing. The text below summarizes the results of
the investigation.
Suppose you start with a prism and a pyramid with congruent bases and the
same height. If you fill the pyramid with sand or water and then pour the
contents into the prism three times, you will exactly fill the prism. In other
words, the volume of the pyramid is 13 the volume of the prism.

You will get the same result no matter what shape the bases have, as long as the
base of the pyramid is congruent to the base of the prism and the heights are the
same.
If you repeat the experiment with a cone and a cylinder with congruent bases and
the same height, you will get the same result. That is, you can empty the contents
of the cone into the cylinder exactly three times.
The results can be summarized in a conjecture.

Pyramid-Cone Volume Conjecture If B is the area of the base of a C-87

pyramid or a cone and H is the height of the solid, then the formula for
the volume is V
13BH.

Example A in your book shows how to find the volume of a regular hexagonal
pyramid. In the example, you need to use the 30°-60°-90° Triangle Conjecture to
find the apothem of the base. Example B involves the volume of a cone. Work
through both examples. Then read the example on the next page.
(continued)

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Lesson 10.3 • Volume of Pyramids and Cones (continued)

EXAMPLE Find the volume of this triangular pyramid.

12 cm

10 cm
45°

 Solution The base is an isosceles right triangle. To find the area of the base, you
need to know the lengths of the legs. Let l be the length of a leg, and use
the Isosceles Right Triangle Conjecture.

l
2
10 The length of the hypotenuse is the length of a leg times 2.
10
l
 Solve for l.
 2
10
The length of each leg is  . Now find the area of the triangle.
2 
1
A
2bh Area formula for triangles.



1 10

2  

10
2 2
Substitute the known values.

25 Multiply.

So, the base has area 25 cm2. Now find the volume of the pyramid.
1
V
3BH Volume formula for pyramids and cones.

1

3(25)(12) Substitute the known values.

100 Multiply.

The volume of the pyramid is 100 cm3.

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LESSON

10.4 Volume Problems

In this lesson you will
● Use the volume formulas you have learned to solve problems
You have learned volume formulas for prisms, cylinders, pyramids, and cones.
In this lesson you will use these formulas to solve problems.
In Example A in your book, the volume of a right triangular prism is given and
you must find the height. In Example B, the volume of a sector of a right cylinder
is given and you must find the radius of the base. Try to solve the problems
yourself before reading the solutions. Below are some more examples.

EXAMPLE A A swimming pool is in the shape of the prism

shown at right. How many gallons of water can 30 ft
the pool hold? (A cubic foot of water is about
7.5 gallons.) 14 ft

6 ft
 Solution First, find the volume of the pool in cubic feet. The
16 ft
pool is in the shape of a trapezoidal prism. The
trapezoid has bases of length 6 feet and 14 feet and
a height of 30 feet. The height of the prism is 16 feet.

V
BH Volume formula for prisms.

1

2(30)(6  14)
16 Use the formula 12h
b1  b2 for the area of a trapezoid.

4800 Solve.

The pool has volume 4800 ft3. A cubic foot is about 7.5 gallons, so the pool holds
4800(7.5), or 36,000 gallons of water.

EXAMPLE B A sealed rectangular container 5 cm by 14 cm by 20 cm

is sitting on its smallest face. It is filled with water up to 5 cm
5 cm from the top. How many centimeters from the
20 cm
bottom will the water level reach if the container is
placed on its largest face? 15 cm

5 cm
 Solution The smallest face is a 5-by-14-centimeter rectangle. When
14 cm
the prism is resting on its smallest face, the water is in the
shape of a rectangular prism with base area 70 cm2 and
height 15 cm. So, the volume of the water is 1050 cm3.
(continued)

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Lesson 10.4 • Volume Problems (continued)

If the container is placed on its largest face, the volume will still be 1050 cm3, but
the base area and height will change. The area of the new base will be 14(20), or
280 cm2. You can use the volume formula to find the height.

V
BH Volume formula for prisms.

1050
280H Substitute the known values.

3.75
H Solve for H.

The height of the water will be 3.75 cm. So, the water level will be 3.75 cm from
the bottom of the container.

EXAMPLE C Find the volume of a rectangular prism with dimensions that are twice those of
another rectangular prism with volume 120 cm3.

 Solution For the rectangular prism with volume 120 cm3, let the dimensions of the
rectangular base be x and y and the height be z. The volume of this prism
is xyz, so xyz
120.
The dimensions of the base of the other prism are 2x and 2y, and the height
is 2z. Let V be the volume of this prism. Then
V
BH Volume formula for prisms.

(2x)(2y)(2z) Substitute the known values.

8xyz Multiply.

8(120) Substitute the known value.

960 Multiply.

The volume of the larger prism is 960 cm3.

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LESSON

10.5 Displacement and Density

In this lesson you will
● Learn how the idea of displacement can be used to find the volume of
an object
● Learn how to calculate the density of an object
To find the volumes of geometric solids, such as prisms and cones, you can use a
volume formula. But what if you want to find the volume of an irregularly
shaped object like a rock? As Example A in your book illustrates, you can
submerge the object in a regularly shaped container filled with water. The volume
of water that is displaced will be the same as the volume of the object. Read
Example A in your book. Then read the example below.

EXAMPLE A When Tom puts a rock into a cylindrical container with diameter 7 cm,
the water level rises 3 cm. What is the volume of the rock to the nearest tenth
of a cubic centimeter?

 Solution The “slice” of water that is displaced is a cylinder

with diameter 7 cm and height 3 cm.
Use the volume formula to find the volume of the
displaced water.
3 cm
V
BH

(3.5)2
3
 115.5

The volume of the rock is about 115.5 cm3, the same 7 cm

as the volume of the displaced water.

An important property of a material is its density. Density is the mass of matter

in a given volume. It is calculated by dividing mass by volume.
mass
density
 
volume
The table on page 551 of your book gives the densities of ten metals.
In Example B, the mass of a clump of metal and information about the
amount of water it displaces are used to identify the type of metal. Read
this example carefully. Then read the following example.
(continued)

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Lesson 10.5 • Displacement and Density (continued)

EXAMPLE B Chemist Preethi Bhatt is given a clump of metal and is told that it is sodium.
She finds that the metal has a mass of 184.3 g. She places it into a nonreactive
liquid in a cylindrical beaker with base diameter 10 cm. If the metal is indeed
sodium, how high should the liquid level rise?

 Solution The table on page 551 of your book indicates that the density of sodium is
0.97 g/cm3. Use the density formula to find what the volume of the clump
of metal should be if it is sodium.
mass
density
 
volume Density formula.
184.3
0.97
 
volume Substitute the known information.
volume
0.97
184.3 Multiply both sides by volume.
184.3
volume



0.97 Divide both sides by 0.97.

volume
190 Simplify.

So, if the metal is sodium, it should displace 190 cm3 of water. Use the volume
formula to find the height of the liquid that should be displaced.
V
BH Volume formula for cylinders.

190
(52
)H Substitute the known information. (The base is a circle with radius 5 cm.)

190
25
H Multiply.

2.4  H Solve for H.

The liquid should rise about 2.4 cm.

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LESSON

10.6 Volume of a Sphere

In this lesson you will
● Discover the volume formula for a sphere
You can find the volume of a sphere by comparing it to the volume of a cylinder.
In the investigation you will see how.

Investigation: The Formula for the Volume of a Sphere

If you have the materials listed at the beginning of the investigation, follow the
steps in your book before continuing. The text below summarizes the results of
the investigation.
Suppose you have a hemisphere and a cylinder. The radius of the cylinder r
equals the radius of the hemisphere, and the height of the cylinder is twice
the radius. Note that the cylinder is the smallest one that would enclose a
sphere made from two of the hemispheres. r
2r
The volume of the cylinder is
r 2(2r), or 2
r 3.
If you fill the hemisphere with sand or water and empty the contents into
the cylinder, the cylinder will be 13 full.
If you fill the hemisphere again and empty the contents into the cylinder, the
cylinder will be 23 full. So, the volume of the sphere (two hemispheres) is equal
to 23 the volume of the cylinder.
The volume of the cylinder is 2
r 3. So, the volume of the sphere is 23(2
r 3),
or 43
r 3.
The results can be summarized as a conjecture.

Sphere Volume Conjecture The volume of a sphere with radius r is given C-88
by the formula V
43
r 3.

Read Example A in your book, which involves finding the percentage of plaster
cut away when the largest possible sphere is carved from a cube. The solution
involves four steps:
Step 1 Find the volume of the sphere.
Step 2 Find the volume of the cube.
Step 3 Subtract the volume of the sphere from the volume of the cube to find
the volume of the plaster cut away.
Step 4 Divide the volume cut away by the volume of the original cube and
convert the answer to a percent to find the percentage cut away.
Read Example B in your book. Solve the problem yourself and then check your
work by reading the given solution.
The following two examples give you more practice working with the volume of a
sphere. (continued)

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Lesson 10.6 • Volume of a Sphere (continued)

EXAMPLE A Find the volume of this solid.

24 cm

60°

 Solution The solid is a hemisphere with a 60° sector cut away. First, find the volume of the
entire hemisphere. Because the formula for the volume of a sphere is V
43
r 3,
the formula for the volume of a hemisphere is V
23
r 3.
2
V
3
r 3 Volume formula for a hemisphere.

2

3
(24)3 The radius is 24 cm.

9216
Simplify.

The volume of the entire hemisphere is 9216
cm3. A 60° sector has been cut
300 5
away, so the fraction of the hemisphere that remains is




360 , or 6 . So, the volume
of the solid is
56
(9216
)
7680
cm3, or about 24,127 cm3.

EXAMPLE B A marble is submerged in water in a rectangular prism with a 2 cm-by-2 cm base.

The water in the prism rises 0.9 cm when the marble is submerged. What is the
diameter of the marble?

 Solution First find the volume of the water displaced by the marble.

V  BH Volume formula for a prism.

 (2)(2)(0.9) Substitute the known information.

 3.6 Simplify.

So, the volume of the displaced water, and thus of the marble, is 3.6 cm3.
Next use the volume of the marble to find its radius. Substitute 3.6 for V in the
formula for the volume of a sphere and solve for r.
4
V 
3

r 3 Volume formula for a sphere.

4
3.6 
3

r 3 Substitute the known information.

3
(3.6)
4
  r3 Multiply both sides by 3


and divide by
.

4

0.86  r 3 Simplify.

r 3
  0.95
0.86 Take the cube root of both sides.

The radius of the marble is about 0.95 cm, so the diameter is about 1.9 cm.

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LESSON

10.7 Surface Area of a Sphere

In this lesson you will
● Discover the formula for the surface area of a sphere
You can use the formula for the volume of a sphere, V
43
r 3, to find the
formula for the surface area of a sphere.

Investigation: The Formula for the Surface Area of a Sphere

Imagine a sphere’s surface divided into tiny shapes that are nearly flat. The surface
area of the sphere is equal to the sum of the areas of these “near polygons.” If you
imagine radii connecting each of the vertices of the “near polygons” to the center
of the sphere, you are mentally dividing the volume of the sphere into many
“near pyramids.” Each of the “near polygons” is a base for one of the pyramids,
and the radius, r, of the sphere is the height of the pyramid. The volume, V, of
the sphere is the sum of the volumes of all the pyramids.

B1

Step 1 Imagine that the surface of a sphere is divided into 1000 “near polygons”
with areas B1, B2, B3, . . . , B1000. The surface area, S, of the sphere is the sum of
the areas of these “near polygons”:

S
B1  B2  B3  · · ·  B1000

Step 2 The pyramid with base area B1 has volume 13
B1r, the pyramid with base
area B2 has volume 13
B2r, and so on. The volume of the sphere, V, is the sum of
these volumes:
1 1 1 1
V
3
B1r  3
B2r  3
B3r  · · ·  3
B1000r

Factor 13r from each term on the right side of the equation:
1
V
3r
B1  B2  B3  · · ·  B1000

Step 3 Because V
43
r 3, you can substitute 43
r 3 for V:

4 1

r 3
r
B1  B2  B3  · · ·  B1000
3 3
Now substitute S for B1  B2  B3  · · ·  B1000:
4 1

r 3
rS
3 3
(continued)

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Lesson 10.7 • Surface Area of a Sphere (continued)

Step 4 Solve the last equation for the surface area, S.

4 1

r 3
rS The equation from Step 3.
3 3
4
r 3
rS Multiply both sides by 3.

4
r 2
S Divide both sides by r.

You now have a formula for finding the surface area of a sphere in terms of the
radius. You can state the result as a conjecture.

Sphere Surface Area Conjecture The surface area, S, of a sphere with C-89
radius r is given by the formula S
4
r 2.

The example in your book shows how to find the surface area of a sphere if you
know its volume. Try to find the surface area on your own before reading the
solution. Then solve the problem in the example below.

EXAMPLE The base of this hemisphere has circumference 32
cm. Find the surface area
of the hemisphere (including the base).

 Solution Because the base of the hemisphere has circumference 32
cm, the radius is
16 cm.
The area of the base of the hemisphere is
(16)2, or 256
cm2.
The area of the curved surface is half the surface area of a sphere with
radius 16 cm.
1
S
2
4
r 2
1

2
4
(16)2

512

So, the total surface area is 256
 512

768
cm2, or about 2413 cm2.

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