volume of a sphere - ihs mathihsmathorms.weebly.com/uploads/8/7/0/1/87018438/... · 4.5 volume of a...

© C

arn

eg

ie L

earn

ing

4.5 Volume of a Sphere 379

4

379

LEARNING GOAL KEY TERMS

sphere

radius of a sphere

diameter of a sphere

great circle of a sphere

hemisphere

annulus

In this lesson, you will:

Derive the formula for the volume of

a sphere.

Archimedes of Syracuse, Sicily, who lived from 287 BC to 212 BC, was an ancient

Greek mathematician, physicist, and engineer. Archimedes discovered formulas

for computing volumes of spheres, cylinders, and cones.

Archimedes has been honored in many ways for his contributions. He has appeared

on postage stamps in East Germany, Greece, Italy, Nicaragua, San Marino, and Spain.

His portrait appears on the Fields Medal for outstanding achievement in mathematics.

You can even say that his honors are out of this world. There is a crater on the moon

named Archimedes, a mountain range on the moon named the Montes Archimedes,

and an asteroid named 3600 Archimedes!

Spheres à la ArchimedesVolume of a Sphere

4.5

© C

arn

eg

ie L

earn

ing

380 Chapter 4 Three-Dimensional Figures

4

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.

© C

arn

eg

ie L

earn

ing


4

The cone shown on the left has a height and a radius equal to b. The height and the

radius form two legs of a right triangle inside the cone. The hypotenuse lies along the side

of the cone.

The cone shown on the right is an enlargement of the !rst cone. It also has a height that is

equal to its radius, r. The smaller cone is shown inside the larger cone.

b

b

r

r

1. Write an expression to describe the area of:

a. the base of the smaller cone.

pb2

b. the base of the larger cone.

pr2

2. Place both cones inside a cylinder with the same radius and height as the larger cone.

r

r

Then, make a horizontal cross section through the cylinder just at the base of the

smaller cone.

r

r

What is the area of this cross section? Explain your reasoning.

The cross section is a circle with a radius of r. So, the area of the cross section is pr2.

© C

arn

eg

ie L

earn

ing


4

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!

© C

arn

eg

ie L

earn

ing


4


How does the shape of the cross section in the hemisphere compare to the

shape of the cross section in the cylinder?

Where is the center point of the hemisphere located?

What do you know about the distance from the center point of the hemisphere

to any point on the hemisphere?

What do the variables b and r represent?

Problem 2

Using a hemisphere and

cylinder with a cone with the

same radius and equal heights,

students write an algebraic

expression representing the

area of the circular cross

section of the hemisphere. They

conclude that this area is equal

to the area of the cross section

in the previous problem.

?

PROBLEM 2 And Now Hemispheres

The diagram shows a hemisphere with the same radius and height of the cylinder from

Problem 1.

The diagram also shows that there is a cross section in the hemisphere at the same height,

b, as that in the cylinder.

r

r

r

b

r

r

r2 2 b2

1. Describe the shape of the cross section shown in the hemisphere.

The cross section is a circle.

2. Analyze the hemisphere. Write expressions for the side lengths of the right triangle in

the diagram. Label the diagram with the measurements.

The length of the shortest leg is b, and the length of the hypotenuse is r, because

a sphere or hemisphere is defined by all the points that are equidistant from a

center point. Using the Pythagorean Theorem, I can determine that the horizontal

side length is √_______

r 2 2 b 2 .

3. Lacy says that the length of the horizontal side has a measure of r. Is Lacy correct?

Explain your reasoning.

Lacy is not correct. The hypotenuse of a right triangle is the longest side of that

triangle. The horizontal length could be close to the hypotenuse, but it can’t be as

long as the hypotenuse, which has a length of r.

4. Write an expression to describe the area of the cross section in the hemisphere.


The cross section has a radius of √_______

r2 2 b2 . So, the area of the cross section is

p 3 radius2, or p(r2 2 b2).

To better follow and

read Problem 2,

students should make

their own copy of the

cylinder and hemisphere

diagram. They should

then add the cross

section diagram from

Problem 1 underneath

the cylinder, and draw

a corresponding cross

section diagram for the

sphere. They should

carefully label the

radiuses. Check their

labels for understanding.

Grouping

Ask students to read the

introduction. Discuss as

a class.





a class.

© C

arn

eg

ie L

earn

ing


4


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 .

© C

arn

eg

ie L

earn

ing


4

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

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

a. Substitute the known values into the formula.

4 __

3 pr3 ¯ 268.08

b. Solve the equation for the radius.

4 __

3 pr3 ¯ 268.08

pr3 ¯ 201.06

r3 ¯ 64

r ¯ 4

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

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

could !t in the Sun? Explain your reasoning.

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

Volume(Earth)

5 4 __

3 p(39603)

¯ 260,120,252,602 cubic miles

Volume(Sun)

5 4 __

3 p(432,4503)

¯ 338,763,267,878,015,100 cubic miles

Volume

(Sun) ___________

Volume(Earth)

¯ 1,302,333

Be prepared to share your solutions and methods.

Use a calculator to

help you solve this problem.

r

© C

arn

eg

ie L

earn

ing


4

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.

© C

arn

eg

ie L

earn

ing

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

© C

arn

eg

ie L

earn

ing

387B Chapter 4 Three-Dimensional Figures

4

Overview

Students apply the formulas for the total and lateral surface areas of three-dimensional !gures

to solve problems.

© C

arn

eg

ie L

earn

ing

4.6 Total and Lateral Surface Area 387C

4

Warm Up

1. The right rectangular prism shown has a length of 8 feet, a width of 2 feet, and a height of 3 feet.

8 ft

2 ft

3 ft

Determine the total surface area of the prism.

The surface area of the prism is 92 square feet.

SA 5 2(8)(2) 1 2(8)(3) 1 2(2)(3)

5 32 1 48 1 12

5 92

© C

arn

eg

ie L

earn

ing

387D Chapter 4 Three-Dimensional Figures

4

© C

arn

eg

ie L

earn

ing

4.6 Total and Lateral Surface Area 387

4

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.

© C

arn

eg

ie L

earn

ing


4

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.”

© C

arn

eg

ie L

earn

ing


4

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

© C

arn

eg

ie L

earn

ing


4

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.

© C

arn

eg

ie L

earn

ing


4

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.

© C

arn

eg

ie L

earn

ing


4

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.

© C

arn

eg

ie L

earn

ing


4

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.

© C

arn

eg

ie L

earn

ing


4

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

© C

arn

eg

ie L

earn

ing


4


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.

© C

arn

eg

ie L

earn

ing


4

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


Sphere

4pr2

r 5 radius of sphere

4pr2

r 5 radius of sphere

base area 5 0 in.2

base area 5 0 in.2

r 5 5.2 in.

© C

arn

eg

ie L

earn

ing


4

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

© C

arn

eg

ie L

earn

ing


4

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

© C

arn

eg

ie L

earn

ing


4

4. A cylinder tube with a diameter of 18 inches will be painted on the outside and the

ends. The length of the tube is 110 inches. What is the surface area that will be

painted?

Approximately 6729.29 square inches will be painted.

I need to determine the total surface area of the cylinder.

2p(92) 1 2p(9)(110) ¯ 6729.29

5. A cone has a radius of 6 meters and a height of 8 meters as shown. Determine the

total and lateral surface area of the cone. Show your work and explain your

reasoning.

6 m

8 m

The surface area of the cone is approximately 301.6 square meters.

The lateral surface area of the cone is p(6)(10), or approximately

188.5 square meters.

First, I calculate the slant height, l, by applying the Pythagorean Theorem.

<2 5 62 1 82

<2 5 36 1 64

<2 5 100

< 5 10

The slant height of the cone is 10 meters.

Next, I can calculate the total surface area of the cone.

S 5 pr2 1 pr<

5 p(62) 1 p(6)(10)

5 36p 1 60p

5 96p

< 301.6

Be prepared to share your solutions and methods.

© C

arn

eg

ie L

earn

ing


4

Check for Students’ Understanding

The Great Pyramid of Giza also known as the Khufu’s Pyramid, Pyramid of Khufu, and Pyramid of

Cheops is located near Cairo, Egypt, and is one of the Seven Wonders of the Ancient World. The Great

Pyramid was the tallest man-made structure in the world for over 3,800 years.

A side of the square base originally measured approximately 230.4 meters. The original height was

approximately 146.7 meters and the slant height was approximately 186.5 meters.

Calculate the lateral surface area of the Great Pyramid.

SA 5 1 __ 2

(P)(s)

5 1

__ 2

(4 ? 230.4)(186.5)

5 85,939.2 square meters

© C

arn

eg

ie L

earn

ing

401A

ESSENTIAL IDEAS

The volume formulas for a pyramid, cylinder, cone, and a sphere are used to solve application problems.

Formulas for the total and lateral surface area of three-dimensional !gures are used to solve problems.

When a solid !gure’s dimensions are changed proportionally, its surface area and volume are changed proportionally.

When a solid !gure’s dimensions are changed non-proportionally, its surface area and volume are changed non-proportionally.

TEXAS ESSENTIAL KNOWLEDGE

AND SKILLS FOR MATHEMATICS

(10) Two-dimensional and three-dimensional

!gures. The student uses the process skills

to recognize characteristics and dimensional

changes of two- and three-dimensional !gures.

The student is expected to:

(A) identify the shapes of two-dimensional

cross-sections of prisms, pyramids,

cylinders, cones, and spheres and identify

three-dimensional objects generated by

rotations of two-dimensional shapes

(B) determine and describe how changes in

the linear dimensions of a shape affect

its perimeter, area, surface area, or

volume, including proportional and

non-proportional dimensional change

Turn Up the . . .Applying Surface Area and Volume Formulas

4.7

composite !gure

KEY TERM


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

© C

arn

eg

ie L

earn

ing

401B Chapter 4 Three-Dimensional Figures

4

(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.

© C

arn

eg

ie L

earn

ing

401D Chapter 4 Three-Dimensional Figures

4

Warm Up

Ice Cream Cone Piñata

Carly asked her parents to make a piñata for her birthday party. A piñata is a brightly-colored papier-

mâché, cardboard, or clay container, originating in Mexico, and !lled with any combination of candy or

small toys suspended from a height for blindfolded children to break with sticks. Her parents decided to

make the piñata in the shape of her favorite dessert, an ice cream cone. They stuffed only the cone

portion of the piñata.

The height of the cone is 340.

The length of the diameter of the base is 240.

Calculate the amount of space (cubic feet) in the cone that will be !lled with goodies.

(144 square inches equal 1 square foot)

(1728 cubic inches equals 1 cubic foot)

Volume of the cone:

V 5 1 __ 3

Bh

V 5 1 __ 3

(144p)(34) 5 1632p < 5124.5 in3 < 3 ft3

© C

arn

eg

ie L

earn

ing

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.

volume of a sphere - ihs mathihsmathorms.weebly.com/uploads/8/7/0/1/87018438/... · 4.5 volume of a...

Documents