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4.5 Volume of a Sphere 379
4
379
LEARNING GOAL KEY TERMS
sphere
radius of a sphere
diameter of a sphere
great circle of a sphere
hemisphere
annulus
In this lesson, you will:
Derive the formula for the volume of
a sphere.
Archimedes of Syracuse, Sicily, who lived from 287 BC to 212 BC, was an ancient
Greek mathematician, physicist, and engineer. Archimedes discovered formulas
for computing volumes of spheres, cylinders, and cones.
Archimedes has been honored in many ways for his contributions. He has appeared
on postage stamps in East Germany, Greece, Italy, Nicaragua, San Marino, and Spain.
His portrait appears on the Fields Medal for outstanding achievement in mathematics.
You can even say that his honors are out of this world. There is a crater on the moon
named Archimedes, a mountain range on the moon named the Montes Archimedes,
and an asteroid named 3600 Archimedes!
Spheres à la ArchimedesVolume of a Sphere
4.5
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380 Chapter 4 Three-Dimensional Figures
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What operation is used to remove the base of the smaller cone from the cross
section of the cylinder?
When rewriting the expression, what is common to both terms?
What do the variables b and r represent?
Problem 1
The problem begins with
de!nitions of sphere, radius of
a sphere, diameter of a sphere,
cross section, great circle of
a sphere and hemisphere.
Students answer questions to
determine the area of a cross
section of the annulus formed
by a cone placed inside
a cylinder.
Grouping
Ask students to read the
introduction and de!nitions.
Discuss as a class.
Have students complete
Questions 1 through 5 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 through 5
What polygon is the base of
the small cone?
Which formula is used to
determine the area of the
base of the small cone?
What polygon is the base of
the large cone?
Which formula is used to
determine the area of the
base of the large cone?
What is the difference
between the two
area formulas?
If the area of the base of the
smaller cone is removed
from the cross section of
the cylinder, how would you
describe the shape of
this region?
PROBLEM 1 Starting with Circles . . . and Cones
Recall that a circle is the set of all points in two dimensions that are equidistant from the
center of the circle. A sphere can be thought of as a three-dimensional circle.
great circle
hemisphere
diameter
radius
center
A sphere is the set of all points in three dimensions that are equidistant from a given point
called the center.
The radius of a sphere is a line segment drawn from the center of the sphere to a point on
the sphere.
The diameter of a sphere is a line segment drawn between two points on the sphere
passing through the center.
A great circle of a sphere is a cross section of a sphere when a plane passes through the
center of the sphere.
A hemisphere is half of a sphere bounded by a great circle.
You have shown that stacking,
translating, and rotating plane figures can help you think about the volumes of three-dimensional figures.
Use this knowledge to build a formula for the volume of
any sphere.
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4.5 Volume of a Sphere 381
4
The cone shown on the left has a height and a radius equal to b. The height and the
radius form two legs of a right triangle inside the cone. The hypotenuse lies along the side
of the cone.
The cone shown on the right is an enlargement of the !rst cone. It also has a height that is
equal to its radius, r. The smaller cone is shown inside the larger cone.
b
b
r
r
1. Write an expression to describe the area of:
a. the base of the smaller cone.
pb2
b. the base of the larger cone.
pr2
2. Place both cones inside a cylinder with the same radius and height as the larger cone.
r
r
Then, make a horizontal cross section through the cylinder just at the base of the
smaller cone.
r
r
What is the area of this cross section? Explain your reasoning.
The cross section is a circle with a radius of r. So, the area of the cross section is pr2.
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382 Chapter 4 Three-Dimensional Figures
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3. Amy makes the following statement about the horizontal
cross section of a cylinder.
Amy
It doesn’t matter where you make the cross section in a cylinder. The cross section will always be a circle with an area of π 3 radius2.
Explain why Amy is correct.
You can think of a cylinder as a translation of a circle. That circle’s radius doesn’t
change over the translation, because a translation preserves congruency. So, no
matter where the cross section is on a cylinder, it will have the same radius as every
other cross section. That means it will also have the same area as every other cross
section.
The image on the left shows the cross section of the cylinder, including the base of the cone.
The image on the right shows the cross section of the cylinder with the base of the cone
removed. The area bound between the two concentric circles shown on the right is called
the annulus.
annulus
4. Calculate the area of the annulus of the cylinder. Explain your reasoning.
The area of the circular cross section is pr2. The area of the base of the smaller cone
is pb2. So, the area of the annulus is pr2 2 pb2.
5. Show how you can use the Distributive Property to rewrite your expression from
Question 4.
p(r2 2 b2)
Now that you have a formula that describes the area of the annulus of the cylinder, let’s
compare this with a formula that describes the area of a cross section of a hemisphere with
the same height and radius as the cylinder.
It dœsn’t matter how you slice it!
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4.5 Volume of a Sphere 383
4
Guiding Questions for Share Phase, Questions 1 through 4
How does the shape of the cross section in the hemisphere compare to the
shape of the cross section in the cylinder?
Where is the center point of the hemisphere located?
What do you know about the distance from the center point of the hemisphere
to any point on the hemisphere?
What do the variables b and r represent?
Problem 2
Using a hemisphere and
cylinder with a cone with the
same radius and equal heights,
students write an algebraic
expression representing the
area of the circular cross
section of the hemisphere. They
conclude that this area is equal
to the area of the cross section
in the previous problem.
?
PROBLEM 2 And Now Hemispheres
The diagram shows a hemisphere with the same radius and height of the cylinder from
Problem 1.
The diagram also shows that there is a cross section in the hemisphere at the same height,
b, as that in the cylinder.
r
r
r
b
r
r
r2 2 b2
1. Describe the shape of the cross section shown in the hemisphere.
The cross section is a circle.
2. Analyze the hemisphere. Write expressions for the side lengths of the right triangle in
the diagram. Label the diagram with the measurements.
The length of the shortest leg is b, and the length of the hypotenuse is r, because
a sphere or hemisphere is defined by all the points that are equidistant from a
center point. Using the Pythagorean Theorem, I can determine that the horizontal
side length is √_______
r 2 2 b 2 .
3. Lacy says that the length of the horizontal side has a measure of r. Is Lacy correct?
Explain your reasoning.
Lacy is not correct. The hypotenuse of a right triangle is the longest side of that
triangle. The horizontal length could be close to the hypotenuse, but it can’t be as
long as the hypotenuse, which has a length of r.
4. Write an expression to describe the area of the cross section in the hemisphere.
Explain your reasoning.
The cross section has a radius of √_______
r2 2 b2 . So, the area of the cross section is
p 3 radius2, or p(r2 2 b2).
To better follow and
read Problem 2,
students should make
their own copy of the
cylinder and hemisphere
diagram. They should
then add the cross
section diagram from
Problem 1 underneath
the cylinder, and draw
a corresponding cross
section diagram for the
sphere. They should
carefully label the
radiuses. Check their
labels for understanding.
Grouping
Ask students to read the
introduction. Discuss as
a class.
Have students complete
Questions 1 through 9 with a
partner. Then have students
share their responses as
a class.
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384 Chapter 4 Three-Dimensional Figures
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Guiding Questions for Share Phase, Questions 5 through 9
Is it possible for two sides of
the right triangle to be equal
in length? Why or why not?
How is the Pythagorean
Theorem used to write an
expression representing the
horizontal side length of the
right triangle in the diagram
of the hemisphere?
Does the expression
describing the area of
the cross section in the
hemisphere look familiar?
Does every cross section
of the hemisphere and
the cylinder, with the cone
removed, have the
same area?
What is the volume formula
of a cone?
What is the volume formula
of a cylinder?
Is the volume of the
hemisphere equal to the
volume of the cylinder with
equal height and radius
minus the volume of the cone
with equal height and radius?
Considering the
hemisphere’s height is equal
to its radius, how can the
formula be rewritten?
How is the volume formula
for a hemisphere used to
write the volume formula for
a sphere?
?
5. Compare the area of the cross section of the hemisphere to the area of the annulus of
the cylinder. What do you notice?
The areas of both cross sections at each height are equal to each other. They are
both equal to p( r 2 2 b 2 ).
6. Stacy says that the volume of the cylinder and the volume of the hemisphere are not
the same. But, if you remove the volume of the cone from the volume of the cylinder,
then the resulting volume would be the same as the volume of the hemisphere.
Is Stacy correct? Explain why or why not.
Stacy is correct. The area of any annulus of the cylinder is equal to the area of the
circular cross section minus the area of the base of the cone. This difference is
equal to the area of any horizontal cross section of the hemisphere. So, the volume
of the cylinder, with the volume of the cone removed, is equal to the volume of the
hemisphere with the same radius and height.
7. Write the formula for the volume of a sphere. Show your work and explain
your reasoning.
The volume of the hemisphere is equal to the volume of the cylinder minus the
volume of the cone:
Volume of hemisphere: p r 2 h 2 ( 1 __ 3 ) p r 2 h 5 ( 3 __
3 ) p r 2 h 2 ( 1 __
3 ) p r 2 h
5 ( 2 __ 3 ) p r 2 h
Because a hemisphere’s height is equal to its radius, its volume can be written as
( 2 __ 3 ) p r 2 3 r, or ( 2 __
3 ) p r 3 .
The volume of the sphere is twice the volume of the hemisphere, so the volume of
the sphere is
2 3 ( 2 __ 3 ) p r 3 , or ( 4 __
3 ) p r 3 .
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4.5 Volume of a Sphere 385
4
8. The sphere shown has an approximate volume of 268.08 cubic inches.
Apply the formula for the volume of a sphere to determine the radius.
a. Substitute the known values into the formula.
4 __
3 pr3 ¯ 268.08
b. Solve the equation for the radius.
4 __
3 pr3 ¯ 268.08
pr3 ¯ 201.06
r3 ¯ 64
r ¯ 4
9. The radius of the Earth is approximately 3960 miles. The Sun’s radius is approximately
432,450 miles. Assuming that both the Earth and the Sun are spheres, how many Earths
could !t in the Sun? Explain your reasoning.
Approximately 1,302,333 Earths would fit in the Sun.
Volume(Earth)
5 4 __
3 p(39603)
¯ 260,120,252,602 cubic miles
Volume(Sun)
5 4 __
3 p(432,4503)
¯ 338,763,267,878,015,100 cubic miles
Volume
(Sun) ___________
Volume(Earth)
¯ 1,302,333
Be prepared to share your solutions and methods.
Use a calculator to
help you solve this problem.
r
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386 Chapter 4 Three-Dimensional Figures
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Check for Students’ Understanding
1. For over seven years, John Bain has spent his life creating the Worlds Largest Rubber Band Ball.
The ball is completely made of rubber bands. Each rubber band was individually stretched around
the ball creating a giant rubber band ball. The weight of the ball is over 3,120 pounds, the
circumference is 15.1 feet, the cost of the materials was approximately $25,000 and the number of
rubber bands was 850,000.
Calculate the volume of the giant rubber band ball. Use 3.14 for pi.
C 5 2pr
15.1 5 2pr
r < 2.4
V 5 4 __ 3 p(2.4)3
V 5 4 __ 3 p(13.824)
V < 57.88 ft3
The volume of the rubber band ball is approximately 57.88 ft3.
2. The world’s largest twine ball is in Darwin, Minnesota. It weighs 17,400 pounds and was created by
Francis A. Johnson. He began this pursuit in March of 1950. He spent four hours a day, every day
wrapping the ball. At some point, the ball had to be lifted with a crane to continue proper wrapping.
It took Francis 39 years to complete. Upon completion, it was moved to a circular open air shed on
his front lawn for all to view.
If the volume of the world’s largest twine ball is 7234.56 cubic feet, determine the radius.
Use 3.14 for pi.
7234.56 5 4 __ 3
pr3
5425.92 5 pr3
1728 5 r3
3 Ï····· 1728 5 r
12 5 r
The radius of the world’s largest ball of twine is 12 feet.
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387A
ESSENTIAL IDEAS
The lateral surface area of a three-dimensional !gure is the sum of the areas of its lateral faces.
The total surface area of a three-dimensional !gure is the sum of the areas of its bases and lateral faces.
The formula for the total surface area of a right prism can change depending on which faces are considered the bases.
The formula for the area of a regular
polygon is A 5 1 __ 2 Pa, where P represents
perimeter and a represents the length of the apothem.
A sphere has a lateral face and bases each with an area of 0, so a sphere’s lateral surface area is equal to its total surface area.
TEXAS ESSENTIAL KNOWLEDGE
AND SKILLS FOR MATHEMATICS
(11) Two-dimensional and three-dimensional
!gures. The student uses the process skills
in the application of formulas to determine
measures of two- and three-dimensional !gures.
The student is expected to:
(A) apply the formula for the area of regular
polygons to solve problems using
appropriate units of measure
(C) apply the formulas for the total and lateral
surface area of three-dimensional !gures,
including prisms, pyramids, cones,
cylinders, spheres, and composite !gures,
to solve problems using appropriate units
of measure
Surface AreaTotal and Lateral Surface Area
4.6
lateral face
lateral surface area
total surface area
apothem
KEY TERMS
In this lesson, you will:
Apply formulas for the total surface area of prisms, pyramids, cones, cylinders, and spheres to solve problems.
Apply the formulas for the lateral surface area of prisms, pyramids, cones, and spheres to solve problems.
Apply the formula for the area of regular polygons to solve problems.
LEARNING GOALS
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387B Chapter 4 Three-Dimensional Figures
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Overview
Students apply the formulas for the total and lateral surface areas of three-dimensional !gures
to solve problems.
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4.6 Total and Lateral Surface Area 387C
4
Warm Up
1. The right rectangular prism shown has a length of 8 feet, a width of 2 feet, and a height of 3 feet.
8 ft
2 ft
3 ft
Determine the total surface area of the prism.
The surface area of the prism is 92 square feet.
SA 5 2(8)(2) 1 2(8)(3) 1 2(2)(3)
5 32 1 48 1 12
5 92
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387D Chapter 4 Three-Dimensional Figures
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4.6 Total and Lateral Surface Area 387
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387
LEARNING GOALS
I n football, a lateral pass happens when a player passes the ball to the side or
backward, instead of forward—in a direction parallel to the goal line or away from
the goal line.
The word lateral, meaning “side,” shows up a lot in math and in other contexts. An
equilateral triangle is a triangle with equal-length sides. Bilateral talks are discussions
that take place often between two opposing sides of a conflict.
KEY TERMS
lateral face
lateral surface area
total surface area
apothem
In this lesson you will:
Apply formulas for the total surface area of
prisms, pyramids, cones, cylinders, and
spheres to solve problems.
Apply the formulas for the lateral surface
area of prisms, pyramids, cones, cylinders,
and spheres to solve problems.
Apply the formula for the area of regular
polygons to solve problems.
4.6Surface AreaTotal and Lateral Surface Area
Discuss with students
the meaning of the word
lateral. Ask students to
think of other words that
use the same Latin stem
latus, meaning side. In
this case, students could
think of lateral surface
area as the area of the
sides (or walls) of a room.
This means that the base
needs to be well de!ned,
and orientation is
important. Have students
consider rectangular
prisms if the base has
not been de!ned and
how they would decide
what makes up the lateral
surface area.
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388 Chapter 4 Three-Dimensional Figures
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Problem 1
Students sketch nets of right
prisms, right pyramids, and
right cylinders in order to
determine formulas for their
lateral and total surface areas.
These formulas are then applied
to solve problems.
Grouping
Discuss the de!nitions as a
class. Have students complete
Questions 1 and 2 with a
partner. Then have students
share their responses
as a class.
Guiding Questions for Share Phase, Questions 1 and 2
Is a cube a
rectangular prism?
How can you write the
formula for the total and
lateral surface area
of a cube?
PROBLEM 1 Cylinders, Prisms, and Pyramids
In this lesson, you will identify and apply surface area formulas to solve problems. Let’s start
by reviewing some of these formulas.
A lateral face of a three-dimensional !gure is a face that is not a base. The lateral surface
area of a three-dimensional !gure is the sum of the areas of its lateral faces. The total
surface area of a three-dimensional !gure is the sum of the areas of its bases and
lateral faces.
1. What units are used to describe lateral and total surface area? Explain.
Both lateral and total surface area are described in square units, because area is
described in square units.
2. Consider the right rectangular prism shown. Its bases are shaded.
h
Ow
a. Sketch the bases and lateral faces of the prism. Include the dimensions of each.
w
O
w
O
w
h
w
h h
O
h
O
Bases:
Lateral faces:
b. Determine the area of each face.
Bases: ℓw, ℓw
Lateral faces: wh, wh, ℓh, ℓh
c. Use your sketch to write the formulas for the total
surface area and lateral surface area of the right
rectangular prism. Explain your reasoning.
The total surface area is the area of all of the
faces of the prism:
SA 5 2ℓw 1 2wh 1 2ℓh.
The lateral surface area is the area of all of the
faces of the prism except the bases:
SA 5 2wh 1 2ℓh.
For right rectangular prisms, you can
call any pairs of opposite faces “bases.”
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4.6 Total and Lateral Surface Area 389
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Grouping
Discuss Question 3 as a class.
Have students complete
Questions 4 and 5 with a
partner. Then have students
share their responses
as a class.
Guiding Question for Share Phase, Questions 4 and 5
Do you prefer to use the
formula 2B 1 L or the
more speci!c formula for
the surface area of a
prism? Why?
? 3. David says that the lateral surface area of a right
rectangular prism can change, depending on what the
bases of the prism are. He calculates 3 different lateral
surface areas, L, for the prism shown.
L1 5 36 ft2 L
2 5 60 ft2 L
3 5 48 ft2
Is David correct? Explain your reasoning.
David is correct.
There are 3 different formulas for the lateral surface area
of a right rectangular prism, depending on which faces are considered bases:
2wh 1 2ℓh, 2ℓw 1 2ℓh, and 2ℓw 1 2wh.
The possible bases are the top and bottom faces, the front and back faces, or the two
side faces.
When applied to the prism shown, each of these formulas gives a different lateral
surface area for the prism.
4. Determine the lateral surface area and total
surface area of the right prism shown.
The bases of the prism are shaded.
a. Identify the length, width, and height of
the prism.
Let ℓ 5 length (22 cm), w 5 width (4 cm), and h 5 height (4 cm).
b. Apply the formula to determine the lateral surface area of the prism.
The lateral surface area of the prism is 352 square centimeters.
Lateral surface area 5 2ℓw 1 2ℓh
5 2(22)(4) 1 2(22)(4)
5 352
c. Apply the formula to determine the total surface area of the prism.
The total surface area of the prism is 384 square centimeters.
Total surface area 5 2ℓw 1 2ℓh 1 2wh
5 2(22)(4) 1 2(22)(4) 1 2(4)(4)
5 384
3 feet
6 feet2 feet
22 centimeters4 centimeters
4 centimeters
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390 Chapter 4 Three-Dimensional Figures
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Grouping
Have students complete
Question 6 with a partner.
Then have students share their
responses as a class.
Guiding Question for Share Phase, Question 6
How can you use the formula
for the perimeter—2l 1 2w—
to show that Michael
is correct?
5. Explain why Vicki is correct.
The net of any prism includes 2 congruent
bases and a number of lateral faces. The
total area of these faces represents the
total surface area of the prism.
6. Consider the right rectangular pyramid shown.
a. Sketch the bases and lateral faces of the pyramid. Include
the dimensions.
w
O
Base: Lateral Faces:
O
s
O
s
w
s
w
s
b. Determine the area of each face.
Area of base: ℓw
Area of lateral faces: 1 __ 2
ℓs, 1 __ 2 ℓs, 1 __
2 ws, 1 __
2 ws
c. Use your sketch to write the formulas for the total surface area and lateral surface
area of the pyramid. Explain your reasoning.
Let ℓ 5 length of base, w 5 width of base, and s 5 slant height.
The total surface area is the area of all of the faces of the pyramid:
SA 5 ℓw 1 2( 1
__ 2
sw) 1 2( 1 __
2 ℓs)
5 ℓw 1 sw 1 ℓs
The lateral surface area is the area of all of the faces of the pyramid except
the bases: SA 5 sw 1 ℓs.
O
w
s
Vicki
One formula for the total surface
area of any prism can be written
as 2B + L, where B represents the
area of each base and L represents
the lateral surface area.
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4.6 Total and Lateral Surface Area 391
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Grouping
Have students complete
Question 7 with a partner.
Then have students share their
responses as a class.
Guiding Questions for Share Phase, Question 7
What units are used to
describe lateral and total
surface area?
How can you check that your
answers are reasonable?
d. Use your sketch and the formula you
determined to explain why Michael is
correct.
Answers will vary.
I determined that the formula for the
lateral surface area of a rectangular
pyramid is sw 1 ℓs.
This is equivalent to 1 __
2 s(2ℓ 1 2w), or 1 __
2 Ps,
where P, or 2ℓ 1 2w, is the perimeter of
the base.
For the total surface area, I added the
area of the base, B to the lateral surface
area, 1 __ 2 Ps.
1 __ 2 Ps 1 B
7. Determine the lateral surface area and total surface area of the right
rectangular pyramid shown.
a. Apply the formula to determine the lateral surface area of the
pyramid.
The lateral surface area of the pyramid is 32 square inches.
Lateral surface area 5 1 __
2 (4)(2 3 5 1 2 3 3)
5 32
b. Apply the formula to determine the total surface area of the pyramid.
The total surface area of the pyramid is 47 square inches.
Total surface area 5 1 __
2 (4)(2 3 5 1 2 3 3) 1 (5 3 3)
5 47
Michael
Lateral surface area of a right rectangular pyramid 5 1 _ 2 Ps.
Total surface area of a
right rectangular pyramid 5 1 _ 2 Ps + B.
P 5 perimeter of base, s 5 slant height, and B 5 area of base.
3 in.
4 in.
5 in.
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392 Chapter 4 Three-Dimensional Figures
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Grouping
Have students complete
Question 8 with a partner.
Then have students share their
responses as a class.
Guiding Questions for Share Phase, Question 8
What is the formula for the
area of a circle?
What shape is the lateral face
of the cylinder?
How is the circumference of
the base related to the lateral
face of the cylinder?
Grouping
Have students complete
Question 9 with a partner.
Then have students share their
responses as a class.
Guiding Questions for Share Phase, Question 9
What is the radius of the
paint roller?
Which measure—lateral
surface area or total surface
area—would matter when
painting with the paint roller?
How can you check the
reasonableness of
your answers?
8. Consider the right cylinder shown.
a. Sketch the bases and lateral faces of the cylinder. Include the
dimensions.
Bases: Lateral Face:
r r
2pr
h
b. Determine the area of each face.
Area of bases: pr2, pr2
Area of lateral face: 2prh
c. Use your sketch to write the formulas for the total
surface area and lateral surface area of the cylinder.
Explain your reasoning.
Let h 5 height of cylinder and r 5 radius of cylinder.
The total surface area is the area of all of the faces
of the cylinder: Total SA 5 2pr2 + 2prh
The lateral surface area is the area of all
of the faces of the cylinder except the
bases: Lateral SA 5 2prh.
9. A cylindrical paint roller has a diameter of 2.5 inches and a length of 10 inches.
a. Apply the formula to determine the lateral surface area of the paint roller.
The lateral surface area is 2p(1.25)(10), or approximately 78.54 square inches.
b. Apply the formula to determine the total surface area of the paint roller.
The total surface area is 2p(1.25)(10) 1 2p(1.25)2, or approximately
88.36 square inches.
r
h
Recall that the
width of the lateral face of a cylinder is equal to the
circumference of the base.
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4.6 Total and Lateral Surface Area 393
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Problem 2
Students investigate and apply
surface area formulas for solid
!gures with regular polygons
as bases.
Grouping
Discuss the formula for the
area of a regular polygon and
the de!nition of apothem as a
class. Have students complete
Questions 1 and 2 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 and 2
Which segment length in
the diagram represents
the apothem?
Would you need the apothem
to determine the lateral
surface area of the pyramid?
Why or why not?
How can you use formulas
you have learned previously
to verify the lateral and
total surface area of the
right hexagonal prism?
PROBLEM 2 A Regular Problem
You have learned previously that the formula for the area
of a regular polygon—a polygon with all congruent
sides—is A 5 1 __
2 Pa, where a represents the length
of the apothem and P represents the perimeter of
the polygon.
You can apply this formula to solve problems involving
surface area.
1. Consider the right hexagonal pyramid shown.
Its base is a regular hexagon.
14 ft
6 ft
8 ft
a. What formula is used to determine the total surface area of the pyramid?
The formula 1
__ 2 Ps 1 B describes the total surface area of the pyramid, where P
represents the perimeter of the base, s represents the slant height, and B
represents the area of the base.
b. Apply the formula for the area of a regular polygon to determine the area of the base
of the hexagonal pyramid. Show your work.
The area of the base of the hexagonal pyramid is 144 square feet.
B 5 1 __
2 (6 ft)(8 ft 3 6)
5 1 __
2 (288 ft2)
5 144
c. Determine the total surface area of the hexagonal pyramid.
The total surface area of the pyramid is 480 square feet.
The area of the base, B, is 144 square feet.
The perimeter of the base, P, is 8 ft 3 6, or 48 feet.
The slant height is 14 feet.
1 __
2 (48 ft)(14 ft) 1 144 ft2 5 480 ft2
Recall that
the apothem is the length of a line segment from the center of the polygon to
the midpoint of a side.
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Problem 3
Students investigate formulas
for the total and lateral
surface areas of cones and
spheres. Students conclude by
summarizing the surface area
formulas they have learned in
this lesson.
Grouping
Discuss the net of a cone as a
class. Have students complete
Questions 1 and 2 with a
partner. Then have students
share their responses as a
class. Discuss Question 3
as a class.
2. The right prism shown has shaded bases that are regular
polygons. Apply the formulas you know to determine the total
and lateral surface area of the prism.
Total surface area 5 2920 cm2
Lateral surface area 5 2720 cm2
The formula for the total surface area of a prism can be
written as 2B 1 L.
I can substitute 1
__ 2 Pa, the area of a regular polygon, for B in
this formula because each base is a regular pentagon:
2( 1
__ 2
Pa) 1 L.
P 5 8 cm 3 5 5 40 cm
a 5 5 cm
L 5 5(68 cm 3 8 cm) 5 2720 cm2
The total surface area of the pentagonal prism is 2( 1 __
2 3 40 3 5) 1 2720,
or 2920 square centimeters.
The lateral surface area of the pentagonal prism is equal to L, or
2720 square centimeters.
PROBLEM 3 Cones and Spheres
You can also apply the formulas for the lateral and total surface areas of cones and spheres
to solve problems.
A right cone is made up of two faces—a circular base and a wedge-shaped lateral face.
r
r
s
s
2pr
Base: Lateral Face:
The area of the circular base is given by pr2.
The area of the lateral face is given by 1 __
2 (2pr)(s), or prs, where s is the slant height of the cone.
8 cm
5 cm
68 cm
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4.6 Total and Lateral Surface Area 395
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Guiding Questions for Share Phase, Questions 1 and 2
Describe how the total and
lateral surface area of a cone
is similar to the surface areas
of other solid !gures.
How can you check the
reasonableness of
your answers?
4 feet
10 feet
6.4 feet
?
1. Write the formulas for the lateral surface area and total surface area of a right cone.
The lateral surface area of a cone is given by 1 __
2 (2pr)(s), or prs.
I add the area of the circular base, pr2, to determine the total surface area,
which is pr2 1 prs.
2. Determine the lateral and total surface area of the cone. Round to
the nearest hundredth.
a. Determine the slant height, s, and the radius, r, of the cone.
s 5 6.4 ft
r 5 5 ft
b. Apply the formula to determine the lateral surface area of the cone.
The lateral surface area of the cone is approximately 100.53 square feet.
p(5)(6.4) 5 32p
¯ 100.53
c. Apply the formula to determine the total surface area of the cone.
The total surface area of the cone is approximately 100.53 square feet.
p(5)2 1 p(5)(6.4) 5 25p 1 32p
5 57p
¯ 179.07
3. Benjamin argued that he could increase the total surface area of a cone without
increasing the radius of the base.
Is Benjamin correct? Use the lateral and total surface area formulas to explain your
reasoning.
Benjamin is correct. He can increase the total surface area by increasing the slant
height of the cone.
Given the formula for total surface area of a cone, pr2 1 prs, increasing the slant
height increases the lateral surface area, prs, and total surface area but does not
change the area of the circular base, pr2.
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Grouping
Discuss the surface area of a
sphere and complete Question
4 as a class. Have students
complete Question 5 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Question 5
Which surface area formulas
involve slant height?
What would the slant height
of a right cone be if its lateral
surface area were equal to
the lateral surface area of a
right cylinder?
Describe the height of a
right cylinder when its lateral
surface area is equal to the
surface area of a sphere.
You can think of a sphere as a solid !gure with
bases that are points. Each of these points has an
area of 0, so the total surface area of a sphere is
equal to its lateral surface area.
total surface area 5 lateral surface area
5 4pr2
4. Apply the formula to determine the total and
lateral surface area of the sphere shown.
The surface area of the sphere is approximately 339.79 square inches.
4p(5.2)2 ¯ 339.79
5. Complete the table to record the formulas for the lateral surface area and total surface
area of the !gures you studied in this lesson. Identify what the variables in your formulas
represent.
Surface Area Formulas
Figure Lateral Surface Area Total Surface Area
Right Rectangular Prism
2ℓh 1 2wh
ℓ 5 length
w 5 width
h 5 height
2ℓw 1 2ℓh 1 2wh, or 2B 1 L
ℓ 5 length
b 5 width
c 5 height
B 5 area of base
L 5 lateral surface area
Right Rectangular Pyramid
1 __
2 Ps
P 5 perimeter of base
s 5 slant height
1
__ 2 Ps 1 B, or B 1 L
P 5 perimeter of base
s 5 slant height
B 5 area of base
L 5 lateral surface area
Right Cylinder
2prh
r 5 radius of cylinder
h 5 height of cylinder
2pr2 1 2prh, or 2B 1 L
r 5 radius of cylinder
h 5 height of cylinder
B 5 area of base
L 5 lateral surface area
Right Cone
prs
r 5 radius of cone
s 5 slant height
pr2 1 prs, or B 1 L
r 5 radius of cone
s 5 slant height
B 5 area of base
L 5 lateral surface area
Sphere
4pr2
r 5 radius of sphere
4pr2
r 5 radius of sphere
base area 5 0 in.2
base area 5 0 in.2
r 5 5.2 in.
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4.6 Total and Lateral Surface Area 397
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Problem 4
Students apply the formulas
for total and lateral surface
area that they have learned
in this lesson to solve
application problems.
Grouping
Have students complete
Questions 1 through 5 with a
partner. Then have students
share their responses
as a class.
Guiding Questions for Share Phase, Questions 1 through 5
How can you tell whether a
problem is asking for total or
lateral surface area?
How can you check the
reasonableness of
your answers?
PROBLEM 4 Show What You Know
1. A new umbrella design was created in the shape of a hemisphere, or half-sphere,
with a special plastic coating on the material to better repel water. The diameter of
the umbrella is about 1 yard. Because the umbrella is still in its beginning stages, the
manufacturer only produces 200 of them to be sold in select markets. How much of
the specially coated material must be produced for the manufacture of these umbrellas?
Each umbrella is half of a sphere. Calculate the surface area of a sphere with diameter
of 1 yard, and divide by 2 to get the amount of plastic coating needed for each
umbrella. SA 5 4pr2 < 4(3.14)(0.52) < 3.14. So, the surface area of each umbrella
is 3.14 _____ 2 < 1.57 square yards. Because there are 200 umbrellas, 200(1.57), or
approximately 314 square yards of material must be produced.
2. A tunnel through a mountain is in the shape of half of a cylinder. The entrance to the
tunnel is 40 feet wide, as shown. The tunnel is 800 feet long.
40 ft
The inside of the tunnel is lined with cement, which includes the arc of the tunnel and
the road. What is the surface area of the inside of the tunnel and the road that will be
lined with cement?
The surface area of the inside of the tunnel is approximately 82,240 square feet.
S 5 1 __ 2 (2prh) 1 Iw 5 prh 1 Iw
5 p(20)(800) 1 (40)(800)
5 16,000p 1 32,000
< 82,240 ft2
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3. A store sells square pyramid-shaped scented candles. The dimensions of two of the
candles are shown below.
6 cm6 cm
16 cm
Candle A
8 cm 8 cm
9 cm
Candle B
a. Calculate the slant height of each candle.
The slant height of candle A is √____
265 , or approximately 16.28 centimeters.
The slant height of candle B is √___
97 , or approximately 9.85 centimeters.
<2 5 32 1 162 <2 5 42 1 92
<2 5 9 1 256 <2 5 16 1 81
<2 5 265 <2 5 97
< 5 √____
265 < 16.28 < 5 √___
97 < 9.85
b. Calculate the total and lateral surface area of each candle.
The total surface area of candle A is approximately 231.35 square centimeters.
The total surface area of candle B is approximately 221.58 square centimeters.
The lateral surface area of candle A is 12( √____
265 ), or approximately
195.35 square centimeters.
The lateral surface area of candle B is 16( √___
97 ), or approximately
157.58 square centimeters.
S 5 B 1 1 __ 2 P< S 5 B 1 1 __
2 P<
5 62 1 1 __ 2 (24)( √
____
265 ) 5 82 1 1 __ 2 (32)( √
___
97 )
5 36 1 12( √____
265 ) 5 64 1 16( √___
97 )
< 231.35 < 221.58
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4.6 Total and Lateral Surface Area 399
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4. A cylinder tube with a diameter of 18 inches will be painted on the outside and the
ends. The length of the tube is 110 inches. What is the surface area that will be
painted?
Approximately 6729.29 square inches will be painted.
I need to determine the total surface area of the cylinder.
2p(92) 1 2p(9)(110) ¯ 6729.29
5. A cone has a radius of 6 meters and a height of 8 meters as shown. Determine the
total and lateral surface area of the cone. Show your work and explain your
reasoning.
6 m
8 m
The surface area of the cone is approximately 301.6 square meters.
The lateral surface area of the cone is p(6)(10), or approximately
188.5 square meters.
First, I calculate the slant height, l, by applying the Pythagorean Theorem.
<2 5 62 1 82
<2 5 36 1 64
<2 5 100
< 5 10
The slant height of the cone is 10 meters.
Next, I can calculate the total surface area of the cone.
S 5 pr2 1 pr<
5 p(62) 1 p(6)(10)
5 36p 1 60p
5 96p
< 301.6
Be prepared to share your solutions and methods.
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Check for Students’ Understanding
The Great Pyramid of Giza also known as the Khufu’s Pyramid, Pyramid of Khufu, and Pyramid of
Cheops is located near Cairo, Egypt, and is one of the Seven Wonders of the Ancient World. The Great
Pyramid was the tallest man-made structure in the world for over 3,800 years.
A side of the square base originally measured approximately 230.4 meters. The original height was
approximately 146.7 meters and the slant height was approximately 186.5 meters.
Calculate the lateral surface area of the Great Pyramid.
SA 5 1 __ 2
(P)(s)
5 1
__ 2
(4 ? 230.4)(186.5)
5 85,939.2 square meters
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401A
ESSENTIAL IDEAS
The volume formulas for a pyramid, cylinder, cone, and a sphere are used to solve application problems.
Formulas for the total and lateral surface area of three-dimensional !gures are used to solve problems.
When a solid !gure’s dimensions are changed proportionally, its surface area and volume are changed proportionally.
When a solid !gure’s dimensions are changed non-proportionally, its surface area and volume are changed non-proportionally.
TEXAS ESSENTIAL KNOWLEDGE
AND SKILLS FOR MATHEMATICS
(10) Two-dimensional and three-dimensional
!gures. The student uses the process skills
to recognize characteristics and dimensional
changes of two- and three-dimensional !gures.
The student is expected to:
(A) identify the shapes of two-dimensional
cross-sections of prisms, pyramids,
cylinders, cones, and spheres and identify
three-dimensional objects generated by
rotations of two-dimensional shapes
(B) determine and describe how changes in
the linear dimensions of a shape affect
its perimeter, area, surface area, or
volume, including proportional and
non-proportional dimensional change
Turn Up the . . .Applying Surface Area and Volume Formulas
4.7
composite !gure
KEY TERM
In this lesson, you will:
Apply the volume formulas for a pyramid, a cylinder, a cone, and a sphere to solve problems.
Apply surface area formulas to solve problems involving composite !gures.
Determine and describe how proportional and non-proportional changes in the linear dimensions of a shape affect its surface area and volume.
LEARNING GOALS
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401B Chapter 4 Three-Dimensional Figures
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(11) Two-dimensional and three-dimensional !gures. The student uses the process skills in the
application of formulas to determine measures of two- and three-dimensional !gures. The student is
expected to:
(C) apply the formulas for the total and lateral surface area of three-dimensional !gures, including
prisms, pyramids, cones, cylinders, spheres, and composite !gures, to solve problems using
appropriate units of measure
(D) apply the formulas for the volume of three-dimensional !gures, including prisms, pyramids, cones,
cylinders, spheres, and composite !gures, to solve problems using appropriate units of measure
Overview
Students apply the surface area and volume formulas of a cylinder, cone, pyramid, and sphere in
different problem situations. Students investigate the effects of proportional and non-proportional
changes to the linear dimensions of solid !gures and their effect on surface area and volume.
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4.7 Applying Surface Area and Volume Formulas 401C
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401D Chapter 4 Three-Dimensional Figures
4
Warm Up
Ice Cream Cone Piñata
Carly asked her parents to make a piñata for her birthday party. A piñata is a brightly-colored papier-
mâché, cardboard, or clay container, originating in Mexico, and !lled with any combination of candy or
small toys suspended from a height for blindfolded children to break with sticks. Her parents decided to
make the piñata in the shape of her favorite dessert, an ice cream cone. They stuffed only the cone
portion of the piñata.
The height of the cone is 340.
The length of the diameter of the base is 240.
Calculate the amount of space (cubic feet) in the cone that will be !lled with goodies.
(144 square inches equal 1 square foot)
(1728 cubic inches equals 1 cubic foot)
Volume of the cone:
V 5 1 __ 3
Bh
V 5 1 __ 3
(144p)(34) 5 1632p < 5124.5 in3 < 3 ft3
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401
A mnemonic is a device used to help you remember something. For example, the
mnemomic “My Very Energetic Mother Just Served Us Noodles” can be used to
remember the order of the planets in the Solar System.
What about volume formulas? Can you come up with mnemonics to remember these
so you don’t have to look them up?
Maybe you can use this one to remember the formula for the volume of a cylinder:
Cylinders are: “Perfectly Ready 2 Hold”
↓ ↓ ↓
p ? r2 ? h
Try to come up with mnemonics for the other volume formulas that you have learned!
Turn Up the . . .Applying Surface Area and Volume Formulas
4.7
In this lesson, you will:
Apply the volume formulas for a pyramid, a
cylinder, a cone, and a sphere to solve
problems.
Apply surface area formulas to solve problems
involving composite !gures.
Determine and describe how proportional
and non-proportional changes in the linear
dimensions of a shape affect its surface area
and volume.
LEARNING GOALS KEY TERM
composite !gure
After students work
through the problems,
have them go to stores
and look for items with
unusual geometric
packaging that may be a
prism, pyramid, cylinder,
cone, sphere, or even a
composite of those types
of !gures. In particular,
they should look for
interesting or unusual
packaging. They should
read the label and see if
the name of the product
or the manufacturer
can be related to the
shape of the packaging.
Students should then
take measurements and
!nd the volume.
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