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379D Chapter 4 Three-Dimensional Figures
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4.5 Volume of a Sphere 379
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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
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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
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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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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
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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.
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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
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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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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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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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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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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
Sphere4pr2
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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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. 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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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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Problem 1
A standard sheet of paper is used to create two different sized cylinders.
Students !rst predict which cylinder contains the greater volume and then use
mathematics to verify their answer. Next, students write a strategy to determine
the approximate volume of an odd shaped vase and then use mathematics to
support their estimate.
?
PROBLEM 1 On a Roll
1. A standard sized sheet of paper measures 8.5 inches by 11 inches. Use two standard
sized sheets of paper to create two cylinders. One cylinder should have a height of
11 inches and the other cylinder should have a height of 8.5 inches.
2. Carol predicts that the cylinder with a height of 11 inches has a greater volume.
Lois predicts that the cylinder with a height of 8.5 inches has a greater volume.
Stu predicts that the two cylinders have the same volume.
Predict which cylinder has the greatest volume.
Answers will vary.
Student response may include:
I predict both cylinders will have the same volume because the same sized paper
was used to create both cylinders.
3. Determine the radius and the height of each cylinder without using a measuring tool.
Radius of cylinder with height of 8.5 inches:
C 5 2pr
11 5 2pr
11 ___ 2p
5 r
r < 1.75
The cylinder with a height of 8.5 inches has a radius of approximately 1.75 inches.
Radius of cylinder with height of 11 inches:
C 5 2pr
8.5 5 2pr
8.5 ___ 2p
5 r
r < 1.35
The cylinder with a height of 11 inches has a radius of approximately 1.35 inches.
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Grouping
Have students complete
Questions 1 through 6 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 through 6
If the two cylinders are made
from the same sized sheet
of paper, how can they be
different sizes?
What formula is used to
determine the length of
the radius of the base of
the cylinder?
What formula is used to
determine the height of
the cylinder?
What formula is used to
determine the volume of
the cylinder?
Is the volume exact or
approximate? Why?
Which variable was squared
and used as a multiplier in
the volume formula?
What equation was used to
determine the height of a
cylinder with equal volume?
Is the height exact or
approximate? Why?
Why does the radius have a
more signi!cant impact on
the volume of the cylinder
than the height? Explain.
4. Calculate the volume of each cylinder to prove or disprove your prediction and
determine who was correct.
Volume of the cylinder with height of 8.5 inches:
V 5 pr2h
5 p(1.75)2(8.5)
< 81.74
Volume of the cylinder with height of 11 inches:
V 5 pr2h
5 p(1.35)2(11)
< 62.95
Lois is correct. The cylinder with a height of 8.5 inches has a greater volume.
5. Does the radius or the height have a greater impact on the magnitude of the volume?
Explain your reasoning.
The radius has a greater impact on the volume than the height because the radius is
squared, whereas the height is not squared. The height would have to have been
substantially longer to create a larger volume.
6. Consider the volume of the cylinder with a height of 8.5 inches. What radius would be
required to create a cylinder with a height of 11 inches that has the same volume?
V 5 pr2h
81.74 5 p(1.35)2(h)
h < 14.3
The height would need to be approximately 14.3 inches to create a cylinder of
equal volume.
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Problem 2
Students deconstruct a
composite !gure into simpler
solid !gures in order to
determine the volume and
total and lateral surface area
of the !gure.
Grouping
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 there more than one
correct strategy to
approximate the volume of
the vase?
How do you decide which
strategy will produce a more
accurate result?
Which volume formula(s)
are used to approximate the
volume of the vase?
PROBLEM 2 Let’s Vase It
1. Describe a strategy for approximating the volume of the vase shown.
4 in.
3 in.
2 in. 12 in.
The volume of the upper and lower portions of the vase could be approximated as
cylinders. I could average the length of the radii on both bases of each cylinder to
get an approximate volume.
2. Determine the approximate volume of the vase.
One possible solution is using the volume formula for a cylinder as shown.
12"
4"
3"
2"
I used 2.5 as the length of the bases of the upper cylinder and 3 as the length of the
bases of the lower cylinder.
V 5 pr2h
V (upper cylinder) 5 (3.14)(2.5)2(6) 5 117.75
V (lower cylinder) 5 (3.14)(3)2(6) 5 169.56
117.75 1 169.56 5 287.31
The approximate volume of the vase is 287.31 cubic units.
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4.7 Applying Surface Area and Volume Formulas 405
4
Grouping
Discuss the de!nition of
composite figure. Have
students complete Question
3 with a partner. Then have
students share their responses
as a class.
Guiding Questions for Share Phase, Question 3
What three-dimensional
!gures can you eliminate as
possibilities for making up
the vase?
What faces do you not need
to consider when formulating
your answers?
How can you check the
reasonableness of
your answers?
You can think of the vase in this problem as a composite figure. A composite figure is
a combination of two or more solid !gures.
3. Use your work in Questions 1 and 2 to determine the total and lateral surface area
of the composite !gure you used to model the vase.
a. Describe the three-dimensional !gures that make up your model of the vase. Include
the dimensions.
I modeled the vase using two cylinders, each with a height of 6 inches. I estimated
that the upper cylinder has a radius of 2.5 inches and the lower cylinder has a
radius of 3 inches.
b. List the formulas you will need.
Total surface area of a cylinder: 2pr2 1 2prh
Lateral surface area of a cylinder: 2prh
c. Describe what steps you will take to determine the total and lateral surface areas.
Note any faces you will not need to include in your calculations.
To determine the total surface area of the composite figure, I do not include the
upper base of the lower cylinder or any of the bases of the upper cylinder. To
determine the lateral surface area of the composite figure, I can add the lateral
surface areas of each cylinder.
Total surface area(upper cylinder)
5 2p(2.5)(6)
¯ 94.25 in.2
Lateral surface area(upper cylinder)
5 2p(2.5)(6)
¯ 94.25 in.2
Total surface area(lower cylinder)
5 p(32) 1 2p(3)(6)
¯ 141.37 in.2
Lateral surface area(lower cylinder)
5 2p(3)(6)
¯ 113.1 in.2
The total surface area of the composite figure is approximately 235.62 square inches.
94.25 1 141.37 5 235.62
The lateral surface area of composite figure is approximately 207.35 square inches.
94.25 1 113.1 ¯ 207.35
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What is the volume formula for a sphere?
Problem 3
Students apply formulas for
volume, total surface area, and
lateral surface area to solve
problems involving solid !gures
and composite solid !gures.
Grouping
Have students complete
Questions 1 through 7 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 through 7
Is a hot-air balloon spherical?
Why or why not?
When approximating the
volume, would it be better to
think of the balloon as one
solid or divide the hot-air
balloon into two solids?
A hot-air balloon can
be divided into which
familiar solids?
What is the radius of the
widest part of the balloon?
What formulas were used to
determine the volume of
the balloon?
Is the volume exact or
approximate? Why?
Is a unit conversion
necessary to determine the
time it takes to empty
the lake?
How do you convert cubic
miles into cubic inches?
What operation is used to
determine the time it takes to
empty the lake?
PROBLEM 3 Balloons, Lakes, and Graphs
1. A typical hot-air balloon is about 75 feet tall and about 55 feet in diameter at its widest
point. About how many cubic feet of hot air does a typical hot-air balloon hold? Explain
how you determined your answer.
The top of the balloon can be approximated by a hemisphere with a diameter of
55 feet and a radius of 27.5 feet. The formula for the volume of a hemisphere is 2 __ 3 pr3.
So, the top would have a volume of 2 __ 3 p(27.5)3, or approximately 43,556.87 cubic feet.
The remaining part of the balloon can be approximated by a cone with the same
diameter but with a height of 75 2 27.5, or 47.5, feet. The formula for the volume of a
cone is 1 __ 3 pr2h. So, the bottom of the balloon would have a volume of 1 __
3 p(27.5)2(47.5),
or approximately 37,617.3 cubic feet.
The total volume of hot air the balloon could contain would be 81,174.17 cubic feet.
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2. Determine the approximate total and lateral surface area of the balloon.
Total and lateral surface area of sphere: 4pr2
Total and lateral surface area of hemisphere: 2pr2
Total surface area of cone: pr2 1 prs
Lateral surface area of cone: p rs
I can use the Pythagorean Theorem to approximate the slant
height, s, of the cone:
27.52 1 47.52 5 s2
3012.5 5 s2
54.89 ¯ s
To determine the total surface area of the balloon, I need to add the lateral surface
area of the cone to the surface area of the hemisphere.
The total surface area of the balloon is approximately 9493.81 square feet.
prs 1 2pr2 5 p(27.5)(54.89) 1 2p(27.52)
¯ 9493.81
The lateral surface area of the balloon is the same, approximately 9493.81 square feet.
27.5 ft
47.5
ft
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3. Lake Erie, the smallest of the Great Lakes by volume, still holds an impressive 116 cubic
miles of water. Suppose you start today dumping out the entire volume of Lake Erie
using a cone cup. A typical cone cup has a diameter of 2 3 __ 4 inches and a height of
4 inches. About how long would it take you to empty the lake if you could dump out one
cup per second?
It would take me 15,467 minutes, or 257 hours, or 10.7 days working around the
clock to empty the lake.
Volume of cone cup: 1 __ 3 p ( 11 ___
8 )
2
(4) 5 7.92 cubic inches
Volume of Lake Erie: 116 cubic miles is equal to 7,349,760 cubic inches
7,349,760 4 7.92 5 928,000 cups, or 928,000 seconds
4. Lake Erie has an average depth of 62 feet. Suppose the volume of Lake Erie were
contained in a cylinder. What would be the radius of the cylinder?
The radius of the cylinder would be approximately 55.47 miles.
The volume of Lake Erie is 116 cubic miles. I need to determine the radius of
the cylinder.
62 feet 5 62 _____
5280 mi
¯ 0.012 mi
V 5 pr2(0.012)
116 5 0.012pr2
9666.67 ¯ pr2
3077 ¯ r2
r ¯ 55.47 mi
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4.7 Applying Surface Area and Volume Formulas 409
4
5. Suppose that the entire volume of Lake Erie were contained in a giant cone with a
diameter of 16 miles and a height of 6 miles. Determine the slant height of this cone.
The slant height of the cone is 10 miles.
I can use the Pythagorean Theorem to determine the slant height, s, of the cone.
62 1 82 5 s2
100 5 s2
s 5 10
6. A cone and a sphere each have a radius of r units. The cone and the sphere also have
equal volumes. Describe the height of the cone in terms of the radius. Show your work.
The height of any cone that has the same radius and volume of a sphere would be
4 times the radius of the sphere.
4 __ 3 pr3 5 1 __
3 pr2h
4pr3 5 pr2h
4r3 5 r2h
4r 5 h
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7. The diagram shows a cylinder and a square pyramid with the same height. The width
of the pyramid is equal to the radius of the cylinder. Suppose that the radius of the
cylinder is gradually increased and the side lengths of the square pyramid are also
gradually increased. Which solid’s volume would increase more rapidly? Explain
your reasoning.
r
r
hh
The formula for the volume of a cylinder is p r 2 h. The formula for the volume of a
pyramid is 1 __ 3 s 2 h. In this problem, s 5 r and the heights are equal. So, I want to
compare y 5 p r 2 and y 5 1 __ 3 r 2 . If I graph these two functions, I can see that the
volume of the cylinder is growing more rapidly as the radius increases.
1.5 2.0
10
210
220
230
240
20
30
40
20.5 1.00.521.021.522.0x
0
y I’m going to use a
graphing calculator to help me.
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Problem 4
Students investigate
how proportional and
non-proportional changes
in linear dimensions of solid
!gures affects their surface area
and volume.
Grouping
Discuss the paragraph above
Question 1. 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
What patterns do you see in
the table?
Why are doubling and
tripling considered
“proportional” changes?
PROBLEM 4 Proportional and Non-Proportional Change
You have learned previously that when the linear dimensions of a two-dimensional !gure
change proportionally by a factor of k, its perimeter changes proportionally by a factor of k,
and its area changes proportionally by a factor of k2.
When the dimensions of a three-dimensional !gure change, its surface area and volume
change as well. Let’s investigate how both proportional and non-proportional changes in a
solid !gure’s dimensions affect its surface area and volume.
1. Consider the following right rectangular prisms with the dimensions shown.
Prism 2 Prism 3Prism 1
3 in.
3 in.
3 in.
4 in.
9 in.
5 in.
2 in.
6 in.
3 in.
Complete the table to determine how doubling or tripling each original prism’s length,
width, and height affects its total surface area and volume. The proportional changes to
Prism 1 have been done for you.
Original PrismPrism Formed by
Doubling Dimensions
Prism Formed by
Tripling Dimensions
Prism 1
Linear
Dimensions
(in.)
ℓ 5 9
w 5 4
h 5 5
ℓ 5 18
w 5 8
h 5 10
ℓ 5 27
w 5 12
h 5 15
Surface Area
(in.2)
2(9 3 4) 1 2(9 3 5)
1 2(4 3 5) 5 202
2(18 3 8) 1 2(18 3 10)
1 2(8 3 10) 5 808
2(27 3 12) 1 2(27 3
15) 1 2(12 3 15) 5
1818
Volume
(in.3)(9)(4)(5) 5 180 (18)(8)(10) 5 1440 (27)(12)(15) 5 4860
Prism 2
Linear
Dimensions
(in.)
ℓ 5 2
w 5 3
h 5 6
ℓ 5 4
w 5 6
h 5 12
ℓ 5 6
w 5 9
h 5 18
Surface Area
(in.2)
2(2 3 3) 1 2(2 3 6) 1
2(3 3 6) 5 72
2(4 3 6) 1 2(4 3 12) 1
2(6 3 12) 5 288
2(6 3 9) 1 2(6 3 18) 1
2(9 3 18) 5 648
Volume
(in.3)(2)(3)(6) 5 36 (4)(6)(12) 5 288 (6)(9)(18) 5 972
Prism 3
Linear
Dimensions
(in.)
ℓ 5 3
w 5 3
h 5 3
ℓ 5 6
w 5 6
h 5 6
ℓ 5 9
w 5 9
h 5 9
Surface Area
(in.2)
2(3 3 3) 1 2(3 3 3) 1
2(3 3 3) 5 54
2(6 3 6) 1 2(6 3 6) 1
2(6 3 6) 5 216
2(9 3 9) 1 2(9 3 9) 1
2(9 3 9) 5 486
Volume
(in.3)(3)(3)(3) 5 27 (6)(6)(6) 5 216 (9)(9)(9) 5 729
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2. Describe how a proportional change in the linear dimensions
of a right rectangular prism affects its surface area.
a. Describe the change to the surface area when the
dimensions of a prism change proportionally by a factor
of 2.
The surface area of the new prism increases by a
factor of 4.
b. Describe the change to the surface area when the
dimensions of a prism change proportionally by a factor
of 3?
The surface area of the new prism increases by a
factor of 9.
c. Describe how you think the surface area of the resulting prism would compare to the
surface area of Prism 1 if the dimensions of Prism 1 were each quadrupled. Then,
determine the surface area of the resulting prism.
The surface area of the resulting prism, 3232 square inches, would be 16 times
the surface area of the original prism, 202 square inches.
Surface Area of Resulting Prism (in.2):
2(36 3 16) 1 2(36 3 20) 1 2(16 3 20) 5 3232
d. Describe how you think the surface area of the resulting prism would compare to
the surface area of Prism 1 if the dimensions of Prism 1 were each decreased by a
factor of 1 __
2 . Then, determine the surface area of the resulting prism.
The surface area of the resulting prism, 50.5 square inches, would be 1 __
4 times the
surface area of the original prism, 202 square inches.
Surface Area of Resulting Prism (in.2):
2(4.5 3 2) 1 2(4.5 3 2.5) 1 2(2 3 2.5) 5 50.5
e. What happens to the surface area when the dimensions of a prism change
proportionally by a factor of k?
The surface area of the prism changes by a factor of k2.
Doubling and tripling are both examples of proportional
changes.
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Grouping
Have students complete
Question 3 with a partner.
Then have students share their
responses as a class.
Guiding Questions for Share Phase, Question 3
How does the change
in volume differ from the
change in surface area?
What units are used to
describe volume?
surface area?
3. Describe how a proportional change in the linear dimensions of a right rectangular
prism affects its volume.
a. Describe the change in the volume when the dimensions of a prism change
proportionally by a factor of 2.
The volume of the new prism increases by a factor of 8.
b. What happens to the volume when the dimensions of a prism change proportionally
by a factor of 3?
The volume of the new prism increases by a factor of 27.
c. Describe how you think the volume of the resulting prism would compare to the
volume of Prism 1 if the dimensions of Prism 1 were each quadrupled. Then,
determine the volume of the resulting prism.
The volume of the resulting prism, 11,520 cubic inches, would be 64 times the
volume of the original prism, 180 cubic inches.
Volume of Resulting Prism (in.3):
(36)(16)(20) 5 11,520
d. Describe how you think the volume of the resulting prism would compare to the
volume of Prism 1 if the dimensions of Prism 1 were each decreased by a factor
of 1 __
2 . Then, determine the volume of the resulting prism.
The volume of the resulting prism, 22.5 cubic inches, would be 1 __
8 times the
volume of the original prism, 180 cubic inches.
Volume of Resulting Prism (in.3):
(4.5)(2)(2.5) 5 22.5
e. What happens to the volume when the dimensions of a prism change proportionally
by a factor of k?
The volume of the prism changes by a factor of k3.
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Grouping
Have students complete
Questions 4 and 5 with a
partner. Then have students
share their responses
as a class.
Guiding Questions for Share Phase, Questions 4 and 5
Why are addition and
subtraction to
dimensions considered
“non-proportional” changes?
Can you predict whether the
surface area will increase
or decrease?
When you doubled and tripled the dimensions of the prism, you changed the dimensions
proportionally. You can also change the dimensions non-proportionally by adding or
subtracting from each linear dimension. Let’s see what effect a non-proportional change in
the dimensions has on the surface area and volume.
4. The right rectangular prisms from Question 1 are shown.
Prism 2 Prism 3Prism 1
3 in.
3 in.
3 in.
4 in.
9 in.
5 in.
2 in.
6 in.
3 in.
Complete the table to determine how adding 2 or adding 3 inches to each original
prism’s length, width, and height affects its total surface area and volume.
Original Prism
Prism Formed by
Adding 2 Inches to
Each Dimension
Prism Formed by
Adding 3 Inches to
Each Dimension
Prism 1
Linear
Dimensions
(in.)
ℓ 5 9
w 5 4
h 5 5
ℓ 5 11
w 5 6
h 5 7
ℓ 5 12
w 5 7
h 5 8
Surface Area
(in.2)
2(9 3 4) 1 2(9 3 5)
1 2(4 3 5) 5 202
2(11 3 6) 1 2(11 3 7)
1 2(6 3 7) 5 370
2(12 3 7) 1 2(12 3 8)
1 2(7 3 8) 5 472
Volume
(in.3)(9)(4)(5) 5 180 (11)(6)(7) 5 462 (12)(7)(8) 5 672
Prism 2
Linear
Dimensions
(in.)
ℓ 5 2
w 5 3
h 5 6
ℓ 5 4
w 5 5
h 5 8
ℓ 5 5
w 5 6
h 5 9
Surface Area
(in.2)
2(2 3 3) 1 2(2 3 6) 1
2(3 3 6) 5 72
2(4 3 5) 1 2(4 3 8) 1
2(5 3 8) 5 184
2(5 3 6) 1 2(5 3 9) 1
2(6 3 9) 5 258
Volume
(in.3)(2)(3)(6) 5 36 (4)(5)(8) 5 160 (5)(6)(9) 5 270
Prism 3
Linear
Dimensions
(in.)
ℓ 5 3
w 5 3
h 5 3
ℓ 5 5
w 5 5
h 5 5
ℓ 5 6
w 5 6
h 5 6
Surface Area
(in.2)
2(3 3 3) 1 2(3 3 3) 1
2(3 3 3) 5 54
2(5 3 5) 1 2(5 3 5) 1
2(5 × 5) 5 150
2(6 3 6) 1 2(6 3 6) 1
2(6 3 6) 5 216
Volume
(in.3)(3)(3)(3) 5 27 (5)(5)(5) 5 125 (6)(6)(6) 5 216
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5. Describe how a non-proportional change in the linear dimensions of a right rectangular
prism affects its surface area.
a. Describe the change to the surface area when the dimensions of a prism are each
increased by 2 inches.
It depends on the prism.
The surface area of Prism 1 increased by 168 cubic inches.
The surface area of Prism 2 increased by 112 cubic inches.
The surface area of Prism 3 increased by 96 cubic inches.
b. Describe the change to the surface area when the dimensions of a prism are each
increased by 3 inches.
It depends on the prism.
The surface area of Prism 1 increased by 270 cubic inches.
The surface area of Prism 2 increased by 186 cubic inches.
The surface area of Prism 3 increased by 162 cubic inches.
c. Determine the surface area of the resulting prism, given that the dimensions of Prism
1 are all increased by 4 inches.
Surface Area of Resulting Prism (in.2):
2(13 3 8) 1 2(13 3 9) 1 2(8 3 9) 5 586
d. Determine the surface area of the resulting prism, given that the dimensions of
Prism 1 are all decreased by 1 inch.
Surface Area of Resulting Prism (in.2):
2(8 3 3) 1 2(8 3 4) 1 2(3 3 4) 5 136
e. What happens to the surface area when the dimensions of a prism
change non-proportionally?
The surface area of the prism increases or decreases as the dimensions increase
or decrease, but it does not increase or decrease by a factor k.
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Grouping
Have students complete
Questions 6 and 7 with a
partner. Then have students
share their responses
as a class.
Guiding Questions for Share Phase, Questions 6 and 7
Can you predict whether
the volume will increase
or decrease?
How does changing
dimensions proportionally
differ from changing
dimensions
non-proportionally?
6. Describe how a non-proportional change in the linear dimensions of a right rectangular
prism affects its volume.
a. Describe the change to the volume when the dimensions of a prism are each
increased by 2 inches.
It depends on the prism.
The volume of Prism 1 increased by 282 cubic inches.
The volume of Prism 2 increased by 124 cubic inches.
The volume of Prism 3 increased by 48 cubic inches.
b. Describe the change to the volume when the dimensions of a prism are each
increased by 3 inches.
It depends on the prism.
The volume of Prism 1 increased by 156 cubic inches.
The volume of Prism 2 increased by 234 cubic inches.
The volume of Prism 3 increased by 189 cubic inches.
c. Determine the volume of the resulting prism, given that the dimensions of Prism 1
are all increased by 4 inches.
Volume of Resulting Prism (in.3):
(13)(8)(9) 5 936
d. Determine the volume of the resulting prism, given that the dimensions of Prism 1
are all decreased by 1 inch.
Volume of Resulting Prism (in.3):
(8)(3)(4) 5 96
e. What happens to the volume when the dimensions of a prism
change non-proportionally?
The volume of the prism increases or decreases as the dimensions increase or
decrease, but it does not increase or decrease by a factor k.
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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
Do the surface area and
volume of a sphere change
proportionally when its radius
is changed proportionally?
Do the surface area and
volume of a sphere change
non-proportionally when
its radius is changed
non-proportionally?
7. What conjecture can you make about the change in surface area and volume of
any solid !gure, given a proportional or non-proportional change in its dimensions?
Answers will vary.
When I change the dimensions of a solid figure proportionally (multiplying or dividing
one or more dimensions by a number), its surface area and volume change
proportionally.
When I change the dimensions non-proportionally (adding or subtracting a number
to each dimension), the surface area and volume change non-proportionally.
You can test your conjecture using various solid !gures.
8. Test your conjecture from Question 6 using a sphere.
a. Label appropriate dimensions for the sphere shown.
Answers will vary.
3 ft
b. Determine and describe the change in total surface area and volume of the !gure,
given that the radius decreases proportionally by a factor of 1 __ 3 . Write your measures
in terms of p.
The surface area of the new sphere is 1 __ 9 of the surface area of the original sphere.
The volume of the new sphere is 1 ___ 27
of the surface area of the original sphere.
Surface area of original sphere (f t 2 ): 4p(32) 5 36p
Volume of original sphere (f t 3 ): 4 __
3 p(33) 5 36p
Surface area of new sphere (f t 2 ): 4p(12) 5 4p
Volume of new sphere (f t 3 ): 4 __
3 p(13) 5
4 __
3 p
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4 Grouping
Have students complete
Questions 9 and 10 with a
partner. Then have students
share their responses as a class.
Guiding Questions for Share Phase, Question 9
How can you represent the
new radius algebraically?
How can you represent the
new height algebraically?
c. Determine and describe the change in total surface area and volume of the !gure,
given that the radius decreases by 1 __ 3 foot. Write your measures in terms of p.
The new surface area decreased by 28 8 __
9 square feet.
The new volume decreased by approximately 10.72 cubic feet.
Surface area of original sphere (f t 2 ): 4p(32) 5 36p
Volume of original sphere (f t 3 ): 4 __
3 p(33) 5 36p
Surface area of new sphere (f t 2 ): 4p( 8 __
3 )2 5 256 ____
9 p
Volume of new sphere (f t 3 ): 4 __
3 p(
8 __
3 )3 5
2048 _____
81 p
The surface area of the new sphere is 7 5 __ 9 square feet less than the surface area of
the original sphere.
256 4 9 5 28 4 __
9
36 2 28 4 __ 9
5 7 5 __ 9
The volume of the new sphere is approximately 10.72 cubic feet less than the
volume of the original sphere.
2048 4 81 5 25 23
___ 81
36 – 25 23
___ 81
5 10 58
___ 81
¯ 10.72
9. Using the formula for the total surface area of a cylinder, show algebraically that
changing the radius and height of a cylinder by a factor of k changes the total surface
area by a factor of k2.
a. Write the formula for the total surface area of a cylinder.
Total surface area 5 2pr2 1 2prh
b. The radius and height both change by a factor of k. Write expressions to represent
the new radius and height.
New radius 5 rk
New height 5 hk
c. Substitute these expressions into the formula and simplify to show that the original
surface area is multiplied by k2.
New total surface area 5 2p(rk)2 1 2prkhk
5 2pk2r2 1 2pk2rh
5 k2(2pr2 1 2prh)
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Guiding Questions for Share Phase, Question 10
How can you represent the
new radius algebraically?
How can you represent the
new height algebraically?
Can you test your conjecture
with other !gures
using calculations?
Can you test your
conjecture with other
!gures using algebra?
10. Using the formula for the volume of a cylinder, show
algebraically that changing the radius and height of a
cylinder by a factor of k changes the volume by a factor
of k3.
a. Write the formula for the volume of a cylinder.
Volume 5 pr2h
b. The radius and height both change by a factor of k. Write
expressions to represent the new radius and height.
New radius 5 rk
New height 5 hk
c. Substitute these expressions into the formula and
simplify to show that the original volume is multiplied
by k3.
New volume 5 p(kr)2kh
5 pk2r2kh
5 k3(pr2h)
Be prepared to share your solutions and methods.
Can you test your
conjecture with other solid figures?
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Check for Students’ Understanding
Analyze the diagram of a piece of cylindrical pipe shown.
1. Determine the outer lateral surface area.
2p(6)(42) ¯ 1582.56 c m 2
2. Determine the inner lateral surface area.
2p(4)(42) ¯ 1055.04 c m 2
8 cm
42 cm
12 cm
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421A
ESSENTIAL IDEAS
A cross section of a three-dimensional solid is a two-dimensional !gure that is formed by the intersection of the solid and a plane.
The cross section of a sphere results in a great circle, a circle smaller than a great circle, or a single point.
The cross section of a cube results in a square, a rectangle that is not a square, a triangle, a pentagon, or a hexagon.
The cross section of a pyramid results in a square, isosceles trapezoid, quadrilateral, or single point.
The cross section of a cone results in a circle, ellipse, parabola, triangle, or a single point.
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
In this lesson, you will:
Determine the shapes of cross sections.
Determine the shapes of intersections of solids and planes.
LEARNING GOALS
Tree RingsCross Sections
4.8
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Overview
Geometric solids are sliced with planes drawn parallel to the base, perpendicular to the base, and on an
angle to the base which create a variety of cross sections. Students practice identifying solids given their
cross sections, and cross sections given their solids.
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Warm Up
1. Describe the shapes of the banana pieces after using the banana slicer shown.
The banana slicer slices the banana into circular shaped flat pieces exposing the interior of the
banana. Most pieces are the same size but taper off as either end slice is cut.
2. Draw several slices of the banana.
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3. Describe the shapes of the pieces egg after using the egg slicer shown.
The egg slicer slices the egg into oval shaped flat pieces exposing a round yellow yolk with a
white exterior. Each piece gradually gets smaller but is similar to all other pieces until the yolk
disappears.
4. Draw several slices of the egg.
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421
LEARNING GOALS
In this lesson, you will:
Determine the shapes of cross sections.
Determine the shapes of intersections of solids and planes.
Tree RingsCross Sections
Each year, a tree grows in diameter. The amount of growth of the diameter
depends on the weather conditions and the amount of water that is available to
the tree. If a tree is cut perpendicular to its trunk, you can see rings, one for each year
of growth. The wider the ring, the greater the amount of growth in one year.
Dendrochronology, or tree-ring dating, is the method of dating trees based on an
analysis of their rings. Often times, it is possible to date a tree to an exact year.
The oldest known tree in the world is a Great Basin bristlecone pine located in White
Mountains, California. It is estimated to be over 5000 years old!
4.8
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When the plane cuts through the cylinder perpendicular to the base, what is
the relationship between the plane and the altitude of the cylinder?
Problem 1
Students analyze the various
cross sections of a cylinder,
sphere, rectangular prism,
pyramid, and cone.
PROBLEM 1 Cutting a Tree Trunk
A section of a tree trunk is roughly in the shape of a cylinder as shown.
When a tree trunk is cut in order to see the tree rings, a cross section of the trunk is
being studied.
Suppose a plane intersects a cylinder parallel to its bases. The shape of the
cross section formed is a circle.
1. Suppose a plane intersects a cylinder perpendicular to its bases so that the plane
passes through the centers of the bases. What is the shape of this cross section?
Sketch an example of this cross section.
The cross section is a rectangle.
Have students work
in small groups to
answer the questions in
Problem 1. Have a mix
of expert and novice
students. Guide students
to articulate how they
are able to determine the
resulting planar �gures
from the intersections.
Suggest that students
can create models in
order to �nd intersections.
Students should talk
among themselves to
name planar �gures,
being as speci�c
as possible.
Grouping
Have students complete
Questions 1 through 7 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Question 1
When the plane cuts through
the cylinder perpendicular
to the altitude, what is the
relationship between the plane
and the base of the cylinder?
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Guiding Questions for Share Phase, Questions 2 through 7
What if the plane cuts
through the cylinder at an
angle along its height less
than 90 degrees?
What if the plane cuts
through the cylinder at an
angle along its height equal
to 90 degrees?
What if the plane cuts
through the cylinder at an
angle along its height greater
than to 90 degrees?
Can additional cross sections
be formed in Questions 1
through 6? If so, identify
them and explain how they
are formed.
Is a circular cross section
possible? If so, how?
Is an elliptical cross section
possible? If so, how?
Is a rectangular cross section
possible? If so, how?
Is a square cross section
possible? If so, how?
Is a trapezoidal cross section
possible? If so, how?
Is a triangular cross section
possible? If so, how?
Is a single point cross section
possible? If so, how?
Is a pentagonal cross section
possible? If so, how?
Is a hexagonal cross section
possible? If so, how?
Is a parabolic cross section
possible? If so, how?
Can additional cross sections
be formed in Questions 6
and 7? If so, identify them
and explain how they
are formed.
2. Suppose a plane intersects a cylinder so that it is not parallel to its bases. What is the
shape of this cross section? Sketch an example of this cross section.
The cross section formed is an ellipse.
You can also identify two-dimensional cross sections of spheres.
One cross section formed when a plane intersects a sphere is a great circle.
3. Sketch and identify two more cross sections formed when a plane intersects a sphere.
The cross section formed is a circle smaller than a great circle.
The cross section formed is a single point.
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Now let’s identify two-dimensional cross sections of a rectangular prism—in this case, a cube.
One cross section formed when a plane intersects a cube is a square.
4. Sketch and identify 4 more two-dimensional cross sections of the cube.
The cross section formed is a triangle. The cross section formed is a pentagon.
The cross section formed is a rectangle. The cross section formed is a hexagon.
Use models if you have them. They can help you “see” the cross
sections.
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4.8 Cross Sections 425
4
? 5. Sally says that it’s not possible to form a circle or an octagon as a cross section of a
cube. Is Sally correct? Explain your reasoning.
Sally is correct. When the cube is cut by a plane, each edge of the cross section
relates to an intersection on one of the cube’s faces with the plane. A cube has no
curved faces, so it is not possible to intersect a plane and a cube to create a cross
section with a curved segment on its perimeter.
A cube has six faces, so it is not possible to cut a cube with a plane and create an
octagonal cross section.
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426 Chapter 4 Three-Dimensional Figures
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Two dimensional cross sections are formed when a plane intersects a pyramid.
One cross section formed when a plane intersects a pyramid is a square.
6. Sketch and identify 3 more cross sections of the pyramid.
The cross section formed is a quadrilateral. The cross section formed is a
single point.
The cross section formed is an isosceles trapezoid.
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4.8 Cross Sections 427
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You can identify two-dimensional cross sections of a cone.
One cross section formed when a plane intersects a cone is a circle.
7. Sketch and identify 3 more cross sections of the cone.
The cross section formed is a parabola. The cross section formed is a
single point.
The cross section formed is an ellipse.
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Grouping
Have students complete
Questions 8 and 9 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 8 and 9
Identify the name of the given
solids in Question 8.
What other cross sections
can be formed with the solids
in Question 8?
Which solids have a circle
as a cross section but not
a triangle?
Which solids have a triangle
as a cross section but not
a circle?
Does more than one
solid have a triangle as a
cross section?
Which solids have a
rectangle as a cross section?
8. Use each solid to create two cross sections. Create one cross section parallel to a
base and a second cross section perpendicular to a base. Then, identify each of the
cross sections.
a.
The cross section parallel to the base is a hexagon congruent to the
hexagonal bases.
The cross section perpendicular to the base is a rectangle.
b.
The cross section parallel to the base is a pentagon similar to the pentagonal base.
The cross section perpendicular to the base is a triangle.
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4.8 Cross Sections 429
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9. Draw a solid that could have the given cross sections.
a. A cross section parallel to a base and a cross section perpendicular to a base
The solid is a cone.
b. A cross section parallel to a base and a cross section perpendicular to a base
The solid is a cylinder.
c. Consider a prism placed on a table such that the base is horizontal. If a plane
passes through the prism horizontally at any height, describe the cross sections.
All cross sections will be congruent to the bases of the prism.
d. De!ne a cylinder in terms of its cross sections.
A cylinder is a solid that has uniform circular cross sections.
Be prepared to share your solutions and methods.
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Check for Students’ Understanding
Which of the following are not possible cross sections of a cube? If it is not possible, explain why it is
not possible.
1. a hexagon
possible
2. an octagon
Not possible. When the cube is cut by a plane, each edge of the cross section relates to an
intersection on one of the cube’s faces with the plane. A cube has six faces, therefore it is not
possible to cut a cube with one plane and create an octagonal cross section.
3. a rectangle
possible
4. a equilateral triangle
possible
5. a circle
Not possible. A cube has no curved faces, so it is not possible to intersect a plane and a cube to
create a cross section with a curved segment on its perimeter.
6. a triangle that is not equilateral
possible
7. a pentagon
possible
8. a parallelogram that is not a rectangle
possible
9. a rectangle that is not a square
possible
10. a square
possible
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431A
4.9
In this lesson, you will:
Use the Pythagorean Theorem to determine the length of a diagonal of a solid.
Use a formula to determine the length of a diagonal of a rectangular solid given the lengths of three
perpendicular edges.
Use a formula to determine the length of a diagonal of a rectangular solid given the diagonal
measurements of three perpendicular sides.
LEARNING GOALS
Two Dimensions Meet Three DimensionsDiagonals in Three Dimensions
ESSENTIAL IDEAS
The Pythagorean Theorem is used to
determine the length of a diagonal of
a solid.
The formula to determine the length
of a diagonal of a rectangular solid is
d2 5 a2 1 b2 1 c2, where d represents the
length of a diagonal, and a, b, and c
represent the length, width and height of the
rectangular solid.
The formula to determine the length of a
diagonal of a rectangular solid is
d2 5 1 __ 2 (d
12 1 d
22 1 d
32), where d represents
the length of a three dimensional diagonal,
and d1, d
2, and d
3 represent the lengths of
the diagonals on three perpendicular faces
of the solid.
TEXAS ESSENTIAL KNOWLEDGE
AND SKILLS FOR MATHEMATICS
(2) Coordinate and transformational geometry.
The student uses the process skills to understand
the connections between algebra and geometry
and uses the one- and two-dimensional
coordinate systems to verify geometric
conjectures. The student is expected to:
(B) derive and use the distance, slope, and
midpoint formulas to verify geometric
relationships, including congruence
of segments and parallelism or
perpendicularity of pairs of lines
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Overview
The length of a diagonal of a rectangular prism is calculated using different methods. Student use the
Pythagorean Theorem and two other formulas to determine the length of a diagonal. One formula uses
the length, width, and height of the rectangular prism and the other formula uses the diagonal lengths
of three perpendicular faces.
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Warm Up
Rectangle DEHJ is drawn perpendicular to rectangle HEFG.
D
EF
GH
J
Suppose the length of lines segments DE, EF, and EH are known. Explain a strategy for calculating the
length of line segments DG and JF.
First connect points E and G to form line segment EG, and points H and F to form line segment HF.
Since opposite sides of a rectangle are congruent, we know DE 5 JH, DJ 5 EH 5 FG, and HG 5 EF.
Using the known values, the Pythagorean Theorem can be used to solve for the lengths of line
segments EG and HF. Since ____
DG is the hypotenuse of right triangle DEG and ___
JF is the hypotenuse of
right triangle JHF, we can use the Pythagorean Theorem again to solve for the length of line
segments DG and JF.
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431D Chapter 4 Three-Dimensional Figures
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4
431
In this lesson, you will:
Use the Pythagorean Theorem to determine the length of a diagonal of a solid.
Use a formula to determine the length of a diagonal of a rectangular solid given the lengths of three
perpendicular edges.
Use a formula to determine the length of a diagonal of a rectangular solid given the diagonal
measurements of three perpendicular sides.
LEARNING GOALS
Two Dimensions Meet Three DimensionsDiagonals in Three Dimensions
There are entire industries dedicated to helping people move from one location to
another. Some people prefer to hire a moving company that will pack, move, and
unpack all of their belongings. Other people choose to rent a van and do all the
packing and unpacking themselves.
One of the most common questions heard during a move is, “Will it fit?” Moving
companies pride themselves on being able to pack items as efficiently as possible.
Sometimes it almost seems impossible to fit so much into so little a space.
What strategies would you use if you had to move and pack yourself?
4.9
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Problem 1
A rectangular solid is used to
model a box of roses. Students
sketch three-dimensional
diagonals and determine their
lengths using the Pythagorean
Theorem. They also practice
using the formula d2 5 a2 1
b2 1 c2 where d represents
the length of a diagonal, and
a, b, and c represent the
length, width and height of the
rectangular solid.
Grouping
Ask students to read the
introduction to Problem 1.
Discuss as a class.
Have students complete
Questions 1 through 9 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 through 9
Do the roses have to be
positioned in the box parallel
to the sides of the box?
How would you describe
the position of a rose of
maximum length contained in
the box?
What is the de�nition of
a diagonal?
Does this de�nition apply to
both two-dimensional and
three-dimensional diagonals?
How many three-dimensional
diagonals can be drawn in
the box?
How many two-dimensional diagonals can be drawn on the bottom side of
the box?
What equation is used to determine the length of the second leg?
What equation is used to determine the length of the
three-dimensional diagonal?
What equation is used to determine if Norton is correct?
How do you suppose Norton made this discovery?
PROBLEM 1 A Box of Roses
The dimensions of a rectangular box for long-stem
roses are 18 inches in length, 6 inches in width, and
4 inches in height.
You need to determine the maximum length
of a long-stem rose that will !t in the box
without bended the rose’s stem. You can
use the Pythagorean Theorem to solve
this problem.
1. What makes this problem different
from all of the previous applications
of the Pythagorean Theorem?
I have always applied the Pythagorean
Theorem in two dimensions. This
problem will require me to use the
Pythagorean Theorem in a three-dimensional situation.
2. Compare a two-dimensional diagonal to a three-dimensional diagonal. Describe the
similarities and the differences.
2-D Diagonal
3-D Diagonal
A two-dimensional diagonal and a three-dimensional diagonal are both line segments
that connect any vertex to another vertex that does not share a face or an edge.
A two-dimensional diagonal shares the same plane as the sides on which the figure
is drawn. A three-dimensional diagonal does not share a plane with any side on
which the figure is drawn.
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3. Draw all of the sides in the rectangular solid you cannot see using dotted lines.
18 in.
6 in.
4 in.
4. Draw a three-dimensional diagonal in the rectangular solid shown.
See image.
5. If the three-dimensional diagonal is the hypotenuse and an edge of the rectangular solid
is a leg of the right triangle, where is the second leg?
The second leg is a diagonal drawn on the bottom of the box.
6. Draw the second leg using a dotted line.
See image.
7. Determine the length of the second leg.
d 2 5 182 1 62
d 2 5 324 1 36
d 2 5 360
d 5 √____
360 < 18.974
The length of the second leg is approximately 18.974 inches.
8. Determine the length of the three-dimensional diagonal.
d 2 5 18.9742 1 42
d 2 5 360.013 1 16
d 2 5 376.013
d 5 √________
376.013 < 19.39
The length of the three-dimensional diagonal is approximately 19.39 inches.
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?
9. Describe how you used the Pythagorean Theorem to calculate the length of the
three-dimensional diagonal.
I formed a right triangle on the plane in which the three-dimensional diagonal was
drawn such that it was the hypotenuse. However, I did not know the length of a leg
of this triangle. I first applied the Pythagorean Theorem to determine the length of a
leg using the plane on the bottom of the rectangular solid, and then I used the
Pythagorean Theorem a second time to determine the length of the three-
dimensional diagonal.
10. Norton tells his teacher he knows a shortcut for determining the length of a three-
dimensional diagonal. He says, “All you have to do is calculate the sum of the squares
of the rectangular solids’ three perpendicular edges [the length, the width, and the
height] and that sum would be equivalent to the square of the three-dimensional
diagonal.” Does this work? Use the rectangular solid in Question 3 and your answer in
Question 8 to determine if Norton is correct. Explain your reasoning.
SD 2 5 42 1 62 1 182
SD 2 5 16 1 36 1 324
SD 2 5 376
SD 5 √____
376 < 19.390
Norton is correct.
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4.9 Diagonals in Three Dimensions 435
4
Did you perform a substitution? Where?
What variable did you solve for next?
Was another substitution performed? Where?
What equation is used to determine if Norton is correct?
How do you suppose Norton made this discovery?
Problem 2
Rectangular solids are used
in different contexts. Students
determine the lengths of three-
dimensional diagonals. They
also practice using the formula
d2 5 1 __ 2
(d1
2 1 d2
2 1 d32), where d
represents the length of a three
dimensional diagonal, and d1,
d2, and d
3 represent the lengths
of the diagonals on three
perpendicular faces of the solid.
Grouping
Have students complete
Questions 1 through 4 with a
partner. Then have students
share their responses as
a class.
Guiding Questions for Share Phase, Questions 1 through 4
Which panel of the
rectangular solid is widest?
How do you know?
Why is it helpful to
sketch each panel
(front, side, top) separately?
What variables were used
to represent the unknown
lengths on each panel?
How many panels were
associated with the height of
the rectangular solid?
How many panels were
associated with the length of
the rectangular solid?
How many panels were
associated with the width of
the rectangular solid?
Which variable did you solve
for �rst?
PROBLEM 2 More Solids
1. A rectangular prism is shown. The diagonal across the front panel is 6 inches, the
diagonal across the side panel is 7 inches, and a diagonal across the top panel is
5 inches. Determine the length of a three-dimensional diagonal.
Draw a view of each side of the solid.
5 in.
7 in.
6 in.
W
H6"
L
H 7"
W
L5"
Write equations representing each side of the solid.
H 2 1 W 2 5 62 H 2 1 L2 5 72 L2 1 W 2 5 52
Solve for H 2 and L 2 and substitute into the third equation to calculate W.
H 2 1 W 2 5 36 36 2 W 2 1 L2 5 49 W 2 1 13 1 W 2 5 25
H 2 5 36 2 W2 L2 2 W 2 5 13 2W 2 5 12
L2 5 W 2 1 13 W 2 5 6
W 5 √__
6
Substitute the value of W into the equations to calculate L and H.
L2 5 W 2 1 13 H 2 5 36 2 ( √__
6 ) 2
L2 5 ( √__
6 ) 2 1 13 H 2 5 36 2 6
L2 5 6 1 13 H 2 5 30
L2 5 19 H 5 √___
30
L 5 √___
19
The length is √___
19 inches, the width is √__
6 inches, and the height is √___
30 inches.
Calculate the three-dimensional diagonal.
SD 2 5 52 1 ( √___
30 ) 2
SD 2 5 25 1 30
SD 2 5 55
SD 5 √___
55 < 7.42
The length of a three-dimensional diagonal is √___
55 inches, or approximately
7.4 inches.
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? 2. Norton tells his teacher that there is a much faster way to do Question 1. This time, he
says, all you have to do is take one-half the sum of the squares of the diagonals on
each dimension of the rectangular solid, and that would be equivalent to the square of
the three-dimensional diagonal. Is Norton correct this time?
SD 2 5 1 __ 2 (52 1 62 1 72)
SD 2 5 1 __ 2 (25 1 36 1 49)
SD 2 5 1 __ 2 (110)
SD 2 5 55
SD 5 √___
55 < 7.4
Norton is once again correct!
3. Write a formula to determine the three-dimensional diagonal in the rectangular prism in
terms of the dimensions of the rectangular prism.
f 2 5 b 2 1 c 2
d 2 5 a 2 1 f 2
d 2 5 a2 1 b2 1 c2
4. Write a formula to determine the three-dimensional diagonal in the rectangular prism in
terms of the lengths of the diagonals.
d12 5 a 2 1 b 2
d22 5 a 2 1 c 2
d32 5 b 2 1 c 2
2( a 2 1 b 2 + c 2 ) 5 d12 1 d
22 1 d
32
2 d 2 5 d12 1 d
22 1 d
32
d 2 5 1 __ 2
(d12 1 d
22 1 d
32)
b
c
ad
f
d3
d1a
b
c
d2
d
Can you see how Norton
figured this out based on your work in
Question 1?
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4.9 Diagonals in Three Dimensions 437
4
How many feet of striping are needed to design one pencil box?
How many feet of striping are needed to design two hundred pencil boxes?
How was 20% pro�t used to determine the amount of the bid?
Problem 3
Students apply what they have
learned in this lesson about
three-dimensional diagonals to
solve real-world problems.
Grouping
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
How did you determine
the formula for the three-
dimensional diagonal?
How will a three-dimensional
diagonal help to solve
this problem?
How many inches are in
eight feet?
Are you able to use a formula
to solve this problem?
Which formula?
Is the three-dimensional
diagonal longer or shorter
than the tree? How do
you know?
Under what constraints will
the tree �t into the car?
Are you able to use a formula
to solve this problem?
Which formula?
Which dimensions
are unknown?
Is the striping considered
two or three-dimensional
diagonals?
How did you determine the
length of each diagonal?
PROBLEM 3 Box It Up
1. Your new part-time job with a landscaping business requires you to transport small
trees in your own car. One particular tree measures 8 feet from the root ball to the top.
The interior of your car is 62 inches in length, 40 inches in width, and 45 inches
in height. Determine if the tree will !t inside your car. Explain your reasoning.
d 2 5 a2 1 b2 1 c2
d 2 5 622 1 402 1 452
d 2 5 3844 1 1600 1 2025
d 2 5 7469
d 2 5 √_____
7469 < 86.42
The height of the tree is 96 inches, and the three-dimensional diagonal is
approximately 86.42 inches, so the tree will not fit inside my car.
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2. Andy’s company is bidding on a project to create a decal design for pencil boxes. If the
client likes Andy’s design then they will order 200 boxes. The design plan with the
decal stripes is shown. The length of the three-dimensional diagonal is approximately
8.5 inches.
If the striping will cost Andy $0.59 per linear foot, how much should Andy bid to make
sure he gets at least a 20% pro!t?
2 in.
2.06 in.
8 in.
d1 d2
82 1 22 1 h2 5 8.52
64 1 4 1 h 2 5 72.25
68 1 h 2 5 72.25
h 2 5 4.25
h 5 2.06 inches
2.062 1 22 5 d12
8.25 5 d12
d1 5 √
_____ 8.25 < 2.87 inches
82 1 2.062 5 d22
68.2436 5 d22
d2 5 √
________ 68.2436 < 8.26 inches
8.26 inches 3 2 5 16.52 inches
2.87 inches 3 2 5 5.74 inches
16.52 inches 1 5.74 inches 5 22.26 inches or approximately 1.855 feet for each box.
1.855 ? 200 5 371 feet
371 3 0.59 5 $218.89
It will cost Andy $218.89 for decal striping on all 200 boxes.
218.89 3 1.2 5 $262.67
Andy will need to bid $262.67 in order to make a 20% profit.
Be prepared to share your solutions and methods.
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Check for Students’ Understanding
4"
2"
3"
The hole for a straw in a juice box is situated in the upper corner of the top of the box as shown. The
diagonal measurements of three perpendicular sides of the box are given. If the straw is 3.80, is it
possible for the straw slip inside the box? Use a formula to answer the question.
d2 5 1 __ 2
( d12 1 d
22 1 d
32 )
d2 5 1 __ 2
( 22 1 32 1 42 )
d2 5 1 __ 2
( 4 1 9 1 16 )
d2 5 1 __ 2
( 29 )
d2 5 14.5
d 5 √_____
14.5 < 3.8080
The 3-D diagonal of the juice box is approximately 3.8080 and the straw is 3.80 so it could possibly
slip inside the juice box.
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Chapter 4 Summary
4
KEY TERMS
disc (4.1)
isometric paper (4.2)
right triangular prism (4.2)
oblique triangular prism (4.2)
right rectangular prism (4.2)
oblique rectangular prism (4.2)
right cylinder (4.2)
oblique cylinder (4.2)
Cavalieri’s principle (4.3)
sphere (4.5)
radius of a sphere (4.5)
diameter of a sphere (4.5)
great circle of a sphere (4.5)
hemisphere (4.5)
annulus (4.5)
lateral face (4.6)
lateral surface area (4.6)
total surface area (4.6)
apothem (4.6)
composite !gure (4.7)
Creating and Describing Three-Dimensional Solids Formed by
Rotations of Plane Figures through Space
A three-dimensional solid is formed by rotating a two-dimensional !gure around an axis. As
the shape travels around in a somewhat circular motion through space, the image of a
three-dimensional solid is formed. A rotated rectangle or square forms a cylinder. A rotated
triangle forms a cone. A rotated disc forms a sphere. The radius of the resulting solid relates
to the distance of the edge of the plane !gure to the axis point.
Example
A cylinder is formed by rotating a rectangle around the axis. The width of the rectangle is
equal to the radius of the cylinder’s base.
4.1
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Creating and Describing Three-Dimensional Solids Formed by
Translations of Plane Figures through Space
A two-dimensional drawing of a solid can be obtained by translating a plane figure through
two dimensions and connecting the corresponding vertices. The solid formed by translating
a rectangle is a rectangular prism. The solid formed by translating a square is a cube. The
solid formed by translating a triangle is a triangular prism. The solid formed by translating a
circle is a cylinder.
Example
A triangular prism is formed by translating a triangle through space.
Building Three-Dimensional Objects by Stacking Congruent
Plane Figures
Congruent shapes or !gures are the same shape and the same size. A stack of congruent
!gures can form a solid shape. A stack of congruent circles forms a cylinder. A stack of
congruent squares forms a rectangular prism or cube. A stack of congruent triangles forms a
triangular prism. The dimensions of the base of the prism or cylinder are the same as the
original plane !gure.
Example
A rectangular prism can be formed by stacking rectangles.
4.2
4.2
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Chapter 4 Summary 441
4
Building Three-Dimensional Objects by Stacking Similar
Plane Figures
Similar shapes or !gures have the same shape, but they do not have to be the same size.
A stack of similar !gures that is incrementally smaller with each layer form a solid shape.
A stack of similar circles forms a cone. A stack of similar squares, rectangles, triangles,
pentagons, hexagons, or other polygons forms a pyramid. The dimensions of the base of the
cone or pyramid are the same as the original plane !gure.
Example
A square pyramid is formed by stacking similar squares.
4.2
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Applying Cavalieri’s Principles
Cavalieri’s principle for two-dimensional !gures states that if the lengths of one-dimensional
slices—just a line segment—of two !gures are the same, then the !gures have the same area.
Cavalieri’s principle for three-dimensional !gures states that if, in two solids of equal altitude,
the sections made by planes parallel to and at the same distance from their respective bases
are always equal, then the volumes of the two solids are equal.
Examples
Both !gures have the same area:
Both solids have the same volume:
h
< w
Right rectangular prism Oblique rectangular prism
h
< w
4.3
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Chapter 4 Summary 443
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Building the Cylinder Volume Formula
To build a volume formula for a cylinder, you can think of it as an in!nite stack of discs, each
with an area of p r 2 . These discs are stacked to a height of h (the height of the cylinder).
A cylinder can also be created by rotating a rectangle about one of its sides. To determine
the volume formula for the cylinder, determine the average, or typical, point of the rectangle,
which is located at ( 1 __ 2
r, 1 __ 2
h ) . Then, use this point as the radius of a circle and calculate the circumference of that circle:
2p ( 1 __ 2
r ) 5 pr. This is the average distance that all the points of the rectangle are rotated to
create the cylinder.
Finally, multiply this average distance by the area of the rectangle, rh, to determine the
volume formula: pr(rh) 5 p r 2 h.
Examples
The volume of any cylinder is p r 2 3 h, or p r 2 h.
h
h
r
4.4
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Building the Cone Volume Formula
You can think of the volume of a cone as a stacking of an in!nite number of similar discs
at a height of h. You can also think of a cone as being created by rotating a right triangle.
To determine the volume formula for the cone, determine the average, or typical, point
of the triangle, which is located at ( 1 __ 3
r, 1 __ 3
h ) . Then, use this point as the radius of a circle and
calculate the circumference of that circle: 2p ( 1 __ 3
r ) 5 2 __ 3
pr. This is the average distance
that all the points of the triangle are rotated to create the cone.
Finally, multiply this average distance by the area of the triangle, ( 1 __ 2
rh ) , to determine the
volume formula: 2 __ 3 pr 3 1 __
2 (rh) 5 1 __
3 p r 2 h.
Example
The volume of any cone is 1 __ 3
p r 2 h.
h
r
Determining the Pyramid Volume Formula
A pyramid is to a prism what a cone is to a cylinder. A prism is created by stacking
congruent polygons, which is similar to creating a cylinder by stacking congruent circles.
A pyramid is created by stacking similar polygons that are not congruent, which is similar to
creating a cone by stacking similar circles. The volume of any pyramid is 1 __ 3
the volume of a
prism with an equal base area and height.
Example
Volume of prism: V 5 Bh
Volume of pyramid: V 5 1 __ 3
Bh
4.4
4.4
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Chapter 4 Summary 445
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Calculating Volume of Spheres
A sphere is the set of all points in three dimensions that are equidistant from a given point
called the center.
The volume of a sphere is the amount of space contained inside the sphere. To calculate the
volume of a sphere, use the formula V 5 4 __ 3 p r 3 , where V is the volume of the sphere and r is
the radius of the sphere.
The volume of a sphere is equal to twice the volume of a cylinder minus the volume of a
cone with an equal base area and height.
Volume formulas can be used to solve problems.
Examples
V 5 4 __ 3 p r 3
5 4 __ 3 p (18 3 )
< 4 __ 3 (3.14)(5832)
< 24,416.6
The volume of the sphere is approximately 24,416.6 cubic yards.
Volume of sphere 5 2 3 Volume of cylinder 2 Volume of cone
2 3 p r 2 h 2 1 __ 3 p r 2 h 5 2 3 2 __
3 p r 2 h 5 4 __
3 p r 2 h
In a sphere, h and r are equal, so the volume formula for a sphere can be written as 4 __ 3 p r 3 .
4.5
18 yd
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Determining Total and Lateral Surface Area
You can apply formulas for the total and lateral surface area of a solid !gure or composite
!gure to solve problems.
Example
You can determine the total and lateral surface area of this
composite !gure by using the formulas for the surface area of a
rectangular pyramid and prism.
Surface Area of Exposed Faces of Bottom Rectangular Prism (i n. 2 ):
Total surface area – area of top base 5 (2ab 1 2ac 1 2bc) 2 (ac)
5 2ab 1 ac 1 2bc
5 2(21)(7) 1 (21)(7) 1 2(7)(7)
5 539
Surface Area of Exposed Faces of Cube (i n. 2 ):
4(72) 5 196
Surface Area of Exposed Faces of Triangular Prism (i n. 2 ):
First, I calculated the length of the hypotenuse of the right triangle.
7 2 1 1 4 2 5 c 2
15.65 ø c
Lateral surface area 2 area of cube face 5 [2 ( 1 __ 2
) (21 2 7) 1 7(15.65)] 2 [49]
5 158.55
Total surface area (i n. 2 ): 539 1 196 1 158.55 ¯ 893.55
Lateral surface area (i n. 2 ): 893.55 2 (21 3 7 1 7 3 7 1 7 3 15.65) ¯ 588
4.6
21 in.7 in.
7 in.
7 in.
7 in.7 in.
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Solving Problems Using Volume and Surface Area Formulas
Formulas for the surface areas and volumes of pyramids, cylinders, and cones can be used
to solve problems.
Examples
115 feet
90.59 feet70 feet
115 feet
V 5 1 __ 3 ( 115 2 )(70) < 308,583
The volume of this pyramid is about 308,583 cubic feet.
Total surface area (i n. 2 ): 1 __ 2 (460)(90.59) 1 1152 5 34,060.7
Lateral surface area (i n. 2 ): 1 __ 2 (460)(90.59) 5 20,835.7
2 feet
6.32 feet
12 feet
r 5 6, h 5 2
V 5 1 __ 3 p( 6 2 )(2) 5 1 __
3 (3.14)(36)(2) 5 75.36
The volume of the cone is about 75.36 cubic feet.
Total surface area (i n. 2 ): p(62) 1 p(6)(6.32) 5 232.23
Lateral surface area (i n. 2 ): p(6)(6.32) 5 119.13
4 inches
3.6 inches
4 inches
1 inch
This is a composite !gure formed by a prism and a pyramid.
Volumeprism
5 (4)(4)(1) Volumepyramid
5 1 __ 3
(4 ) 2 (3.6)
5 16 5 57.6
The volume of the composite !gure is 16 1 57.6, or 73.6 cubic inches.
4.7
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Determining Shapes of Cross Sections
A cross section of a solid is the two-dimensional !gure formed by the intersection of a plane
and a solid when a plane passes through the solid.
Examples
The cross sections of the triangular prism are a triangle and a rectangle.
The cross sections of the square pyramid are a triangle and a square.
The cross sections of the cylinder are a rectangle and a circle.
The cross sections of the cone are a triangle and a circle.
The cross sections of the sphere are a circle and a point.
4.8
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Determining Lengths of Spatial Diagonals
To determine the length of a spatial diagonal of a rectangular solid given the lengths of the
diagonals of the faces of the solid, use the following formula, where d represents the length
of a spatial diagonal, and d1, d
2, and d
3 represent the lengths of the diagonals of the faces of
the rectangular solid.
d 2 5 1 __ 2 ( d
1 2 1 d
2 2 1 d
3 2 )
Examples
10 ft
16.2 ft
17 ft
A
B
(AB)2 5 1 __ 2 ( d
1 2 1 d
2 2 1 d
3 2 )
(AB)2 5 1 __ 2 ( 102 1 172 1 16.22 )
(AB)2 5 1 __ 2 ( 100 1 289 1 262.44 )
(AB)2 5 1 __ 2 ( 651.44 )
AB < 18
The length of the spatial diagonal AB is approximately 18 feet.
4.9
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