© Carnegie Learning - IHS Math- Satresatreihs.weebly.com/uploads/5/8/3/6/...4_part_2.pdfA sphere is the set of all points in three dimensions that are equidistant from a given point

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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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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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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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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.


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


introduction. Discuss as

a class.





a class.

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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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8. The sphere shown has an approximate volume of 268.08 cubic inches.

Apply the formula for the volume of a sphere to determine the radius.

a. Substitute the known values into the formula.

4 __

3 pr3 ¯ 268.08

b. Solve the equation for the radius.

4 __

3 pr3 ¯ 268.08

pr3 ¯ 201.06

r3 ¯ 64

r ¯ 4

9. The radius of the Earth is approximately 3960 miles. The Sun’s radius is approximately

432,450 miles. Assuming that both the Earth and the Sun are spheres, how many Earths

could !t in the Sun? Explain your reasoning.

Approximately 1,302,333 Earths would fit in the Sun.

Volume(Earth)

5 4 __

3 p(39603)

¯ 260,120,252,602 cubic miles

Volume(Sun)

5 4 __

3 p(432,4503)

¯ 338,763,267,878,015,100 cubic miles

Volume

(Sun) ___________

Volume(Earth)

¯ 1,302,333

Be prepared to share your solutions and methods.

Use a calculator to

help you solve this problem.

r

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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


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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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


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


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


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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Grouping

Discuss Question 3 as a class.





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


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


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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Grouping





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






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






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


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.


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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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





a class.


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




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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Describe how the total and

lateral surface area of a cone

is similar to the surface areas

of other solid !gures.


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



a class.


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


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


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


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





as a class.


How can you tell whether a

problem is asking for total or

lateral surface area?


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


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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.




!gures. The student uses the process skills

to recognize characteristics and dimensional

changes of two- and three-dimensional !gures.


(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


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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(11) Two-dimensional and three-dimensional !gures. The student uses the process skills in the

application of formulas to determine measures of two- and three-dimensional !gures. The student is

expected to:

(C) apply the formulas for the total and lateral surface area of three-dimensional !gures, including

prisms, pyramids, cones, cylinders, spheres, and composite !gures, to solve problems using

appropriate units of measure

(D) apply the formulas for the volume of three-dimensional !gures, including prisms, pyramids, cones,

cylinders, spheres, and composite !gures, to solve problems using appropriate units of measure

Overview

Students apply the surface area and volume formulas of a cylinder, cone, pyramid, and sphere in

different problem situations. Students investigate the effects of proportional and non-proportional

changes to the linear dimensions of solid !gures and their effect on surface area and volume.

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4.7 Applying Surface Area and Volume Formulas 401C

4

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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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4.7 Applying Surface Area and Volume Formulas 401

4

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


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





a class.


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?


determine the height of

the cylinder?


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?


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





a class.


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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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.


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?


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





a class.


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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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.


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.


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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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.


(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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as a class.


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.



b. Describe the change to the surface area when the dimensions of a prism are each






c. Determine the surface area of the resulting prism, given that the dimensions of Prism

1 are all increased by 4 inches.


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.


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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as a class.


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



The volume of Prism 1 increased by 282 cubic inches.



b. Describe the change to the volume when the dimensions of a prism are each






c. Determine the volume of the resulting prism, given that the dimensions of Prism 1

are all increased by 4 inches.


(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.


(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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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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share their responses as a class.


How can you represent the

new radius algebraically?


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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new radius algebraically?


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)


Can you test your

conjecture with other solid figures?

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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.




�gures. The student uses the process skills

to recognize characteristics and dimensional

changes of two- and three-dimensional �gures.


(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


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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4.8 Cross Sections 421C

4

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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4.8 Cross Sections 421

4

421

LEARNING GOALS


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





a class.


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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What if the plane cuts

through the cylinder at an

angle along its height less

than 90 degrees?



angle along its height equal

to 90 degrees?



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


Is a rectangular cross section


Is a square cross section


Is a trapezoidal cross section


Is a triangular cross section


Is a single point cross section


Is a pentagonal cross section


Is a hexagonal cross section


Is a parabolic cross section


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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? 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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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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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





a class.


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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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.


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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


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.



(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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4.9 Diagonals in Three Dimensions 431C

4

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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4.9 Diagonals in Three Dimensions 431

4

431


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


introduction to Problem 1.

Discuss as a class.





a class.


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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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





a class.


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?


associated with the length of



associated with the width of


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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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





a class.


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.


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4.9 Diagonals in Three Dimensions 438A

4


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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439

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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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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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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Documents

© Carnegie Learning - IHS Math- Satresatreihs.weebly.com/uploads/5/8/3/6/...4_part_2.pdfA sphere is the set of all points in three dimensions that are equidistant from a given point