Volume and Surface Area of Composite Solids - Geometry in Three Dimensions (Pre-Algebra)

Sometimes, a solid may not have a recognizable shape, which makes determining its volume or surface area difficult. However, certain solids can be broken down into familiar shapes like prisms, pyramids, cones, spheres, and more, allowing the application of familiar formulas. This lesson delves into the computations of volumes and surface areas for this type of solids.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Constructing New Solids

Consider a hemisphere, a cone, and a cylinder, all of which have the same radius. Each solid can be dragged and rotated. Create new solids by combining the given ones.

Moveable hemisphere, cylinder, and cone.

Discussion

Composite Solid

A solid that is made up of more than one solid is called a composite solid. The individual solids can be combined either by adding or subtracting them from one another. For instance, a hemisphere can be combined with a cone to make something that resembles a snow cone, or it could be used to dig a bowl shape out of a cylinder.

Snow cone and cylinder with a hollow in the shape of a hemisphere.

The volume of a composite solid is either the sum or the difference between the volumes of the individual solids, whichever is applicable. The surface area of a composite solid is the sum of the faces that enclose the solid.

Example

Finding the Volume of a Traffic Cone

Ramsha has recently learned how to find the volume of composite solids. She is curious about finding the volumes of composite solids that she encounters in her daily life. Consider the diagram of a traffic cone she passed during her walk to school.

A traffic cone with a conical upper part having a height of 30 inches and a base radius of 5 inches, and a prism lower part with base side lengths of 14 inches and a height of 1 inch.

The height of the cone part is

30

inches and its radius is

5

inches. The prism below the cone is a square prism with side lengths of

14

inches and a height of

1

inch. Help Ramsha find the volume of the traffic cone. Use a calculator for calculations and round the result to the nearest whole number.

Hint

The volume occupied by the traffic cone is the sum of the volume of the prism base and the volume of the cone part.

Solution

The traffic cone is basically composed of two solids — a cone and a square prism. This means that the volume of the traffic cone is the sum of the volume of the prism

V_{p}

and the volume of the cone

V_{c} .

V = V_{p} + V_{c}

This solution will begin by calculating the volume of the prism. After that, it will find the volume of the cone.

Finding the Volume of the Prism

The base of the prism is a square with side lengths of

14

inches, so its area is the square of

14 .

B = 1 4^{2} \Leftrightarrow B = 196

Since the volume of a prism is its base area times its height, the volume of the square prism can be found as follows.

V_{p} = B h

SubstituteII

$B = 196$ , $h = 1$

V_{p} = 196 \cdot 1

MultByOne

$a \cdot 1 = a$

V_{p} = 196

The volume of the prism part of the traffic cone is

196

cubic inches.

Finding the Volume of the Cone

Use the formula for the volume of a cone to find the volume of the part that has a conic shape.

V_{c} = \frac{1}{3} π r^{2} h

Substitute

5

for

r

and

30

for

h

into the formula and solve for

V_{c} .

V_{c} = \frac{1}{3} π r^{2} h

SubstituteII

$r = 5$ , $h = 30$

V_{c} = \frac{1}{3} π (5)^{2} (30)

▼

Simplify right-hand side

CalcPow

Calculate power

V_{c} = \frac{1}{3} π (25) (30)

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

V_{c} = \frac{π ( 2 5 ) ( 3 0 )}{3}

UseCalc

Use a calculator

V_{c} = 785.398163 \dots

RoundInt

Round to nearest integer

V_{c} \approx 785

The volume of the cone is about

785

cubic inches. Now the volume of the traffic cone can be found.

Volume of the Traffic Cone

The volume of the prism part —

196

cubic inches — can be added to the volume of the cone part —

785

cubic inches — to determine the volume of the traffic cone.

+ 196785981

The volume of the traffic cone is about

981

cubic inches.

Example

A Double-Walled Glass Cup

A double-walled glass cup is a special cup with two layers of glass that help keep the drink at the right temperature, whether hot or cold. Ramsha has one of these cups. Her cup is cylindrical with a radius of $4$ centimeters and a height of $12$ centimeters. The second wall of the cup creates a cone.

A cylinderical shaped double-walled glass cup with a 12 cm height and a radius of 4 cm. The second layer is a cone with a 12 cm height and a radius of 4 cm.

Ramsha fills the cup with water.

a Ramsha wants to find the volume of the water filling her cup and the volume of the air between the cup walls. Help her in calculate these volumes. Round the answers to the nearest whole number.

b What percent of the volume of the entire cup is the volume of the air between the walls of the double-walled glass cup? Round the answer to one decimal place.

Hint

a The volume of a cone is one third the product of

π,

the square of the radius, and the height. The volume of a cylinder is the product of

π,

the square of the radius, and the height.

b Use the exact values for the volumes from Part A.

Solution

a The volumes of each solid will be calculated one at a time.

Volume of Water

Ramsha will fill the cone with water, so the volume of the cone is needed. The cone has the same height and radius as the cylinder, measuring $12$ centimeters and $4$ centimeters, respectively.

The volume of a cone is one third the product of

π,

the square of the radius, and the height.

V_{cone} = \frac{1}{3} π r^{2} h

To find the volume of this cone, substitute

12

and

4

into the formula for

h

and

r,

respectively, and evaluate.

V_{cone} = \frac{1}{3} π r^{2} h

SubstituteII

$h = 12$ , $r = 4$

V_{cone} = \frac{1}{3} π (4^{2}) (12)

▼

Evaluate right-hand side

CalcPow

Calculate power

V_{cone} = \frac{1}{3} π (16) (12)

Multiply

V_{cone} = \frac{1}{3} π (192)

CommutativePropMult

Commutative Property of Multiplication

V_{cone} = \frac{1}{3} (192) π

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

V_{cone} = \frac{1 9 2}{3} π

CalcQuot

Calculate quotient

V_{cone} = 64 π

The volume of the water filling the cup is

64 π

cubic centimeters. Using a calculator, find the result of

64 π

and then round it to the nearest whole number.

64 π

UseCalc

Use a calculator

201.061929 \dots

RoundInt

Round to nearest integer

201

Ramsha filled the cup with approximately

201

cubic centimeters of water.

Volume of Air

Next, Ramsha needs to find the volume of the air between the two walls of the cup. The first step is to find the volume of the shell of the cup. The cup has a cylindrical shape with a height of $12$ centimeters and a radius of $4$ centimeters.

The volume of a cylinder is the product of

π,

the square of the radius, and the height.

V_{cylinder} = π r^{2} h

Substitute

12

and

4

into the formula for

h

and

r,

respectively, again and evaluate.

V_{cylinder} = π r^{2} h

SubstituteII

$h = 12$ , $r = 4$

V_{cylinder} = π (4^{2}) (12)

▼

Evaluate right-hand side

CalcPow

Calculate power

V_{cylinder} = π (16) (12)

Multiply

V_{cylinder} = 192 π

The volume of the cylindrical shell of the cup is

192 π

cubic centimeters. The volume of the air between the walls of the double-walled glass cup is the difference between the volume of the cylinder and the volume of the cone. Remember, the volume of the cone was previously determined to be

64 π

cubic centimeters.

192 π - 64 π = 128 π

The volume of the region between the cone and the cylinder is

128 π

cubic centimeters. This implies that the volume of the air between the walls of the double-walled glass cup is also

128 π

cubic centimeters. Use a calculator to find the nearest integer value of the volume.

128 π

UseCalc

Use a calculator

402.123859 \dots

RoundInt

Round to nearest integer

402

The volume of the air between the walls of the double-walled glass cup is

402

cubic centimeters.

b In Part A it was found that the volume of the cylindrical-shaped cup is

192 π

cubic centimeters and the volume of the portion of the cylinder not occupied by the cone is

128 π

cubic centimeters. Calculating the ratio of the second value to the first value will provide the desired percentage.

\frac{1 2 8 π}{1 9 2 π}

▼

Evaluate

CancelCommonFac

Cancel out common factors

\frac{1 2 8 π}{1 9 2 π}

SimpQuot

Simplify quotient

\frac{1 2 8}{1 9 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 6 4}{b / 6 4}$

\frac{2}{3}

▼

Convert to percent

FracToDiv

$\frac{a}{b} = a \div b$

0.666666 \dots

WritePercent

Convert to percent

66.666666 \dots %

RoundDec

Round to $1$ decimal place(s)

66.7 %

The volume of the air between the walls of the double-walled glass cup makes up about

66.7 %

of the volume of the entire cup.

Example

Surface Area of the Double-Walled Glass Cup

Ramsha also wants to calculate the surface area of her double-walled glass cup.

A cylinderical shaped double-wall glass cup with a 12 cm height and a radius of 4 cm. The second layer is a cone with a 12 cm height and a radius of 4 cm.

Calculate the surface area of the cup with her. Round the answer to two decimal places.

Hint

The surface area of the cup consists of the lateral area of the cylinder, one of the bases of the cylinder, and the lateral area of the cone.

Solution

The double-walled glass cup is made up of a cylinder with a cone inside. To calculate its surface area, the lateral areas of the cylinder and the cone, along with one base area of the cylinder, need to be calculated.

Lateral and Base Areas of the Cylinder

Notice that only one end of the cylinder is closed, so only the sum of the lateral area and the base area of the cylinder will be calculated.

The lateral area of a cylinder is twice the product of

π,

the radius, and the height.

L A_{cylinder} = 2 π r h

To find the lateral area of this cylinder, substitute

12

and

4

for

h

and

r,

respectively.

L A_{cylinder} = 2 π r h

SubstituteII

$h = 12$ , $r = 4$

L A_{cylinder} = 2 π (4) (12)

▼

Evaluate right-hand side

CommutativePropMult

Commutative Property of Multiplication

L A_{cylinder} = 2 (4) (12) π

Multiply

L A_{cylinder} = 96 π

The lateral area of the cylinder is about

96 π

square centimeters. The base area of the cylinder is calculated by finding the area of a circle with a radius of

4

centimeters.

B A_{cylinder} = π r^{2}

Substitute

$r = 4$

B A_{cylinder} = π (4)^{2}

CalcPow

Calculate power

B A_{cylinder} = π (16)

CommutativePropMult

Commutative Property of Multiplication

B A_{cylinder} = 16 π

Next, calculate the total by adding the lateral area of the cylinder to one of its base areas.

Surface Area of the \underline{Cylindrical Part of the Cup} 96 π + 16 π = 112 π

Lateral Area of the Cone

The lateral area of a cone is the product of

π,

the radius, and the slant height of the cone.

L A_{cone} = π r ℓ

The slant height

ℓ

is the hypotenuse of the right triangle formed by the radius, the height, and the segment connecting the center of the base of the cylinder with a point on the circumference of the opposite base.

The missing value can be found by using the Pythagorean Theorem.

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

4^{2} + 12^{2} = ℓ^{2}

▼

Solve for

ℓ

CalcPow

Calculate power

16 + 144 = ℓ^{2}

AddTerms

Add terms

160 = ℓ^{2}

SqrtEqn

$LHS = RHS$

160 = ℓ

RearrangeEqn

Rearrange equation

ℓ = 160

The slant height of the cone is

160

centimeters. Now the formula for the lateral area of a cone can be used. Substitute

160

for

ℓ

and

4

for

r

and simplify.

L A_{cone} = π r ℓ

SubstituteII

$ℓ = 160$ , $r = 4$

L A_{cone} = π (4) (160)

CommutativePropMult

Commutative Property of Multiplication

L A_{cone} = 4160 π

Total Surface Area

Finally, the combined areas of the cylinder and the cone will provide the total surface area of the cup. Use a calculator to make the calculations and round the result to two decimal places.

112 π + 4160 π

▼

Evaluate

UseCalc

Use a calculator

351.858377 \dots + 158.953412 \dots

AddTerms

Add terms

510.811789 \dots

RoundDec

Round to $2$ decimal place(s)

\approx 510.81

The surface area of the double-walled glass cup is about

510.81

square centimeters.

Example

Volume of a Pencil

Ramsha bought a pencil with a radius of $3$ millimeters. The total length of the pencil, excluding the eraser, is $160$ millimeters. Moreover, the tip of the pencil is a $10 -$ millimeter high cone.

A pencil that has a radius of 3 millimeters, and the length until the eraser is $160$ millimeters. The length of the tip is 10 millimeters.

Assuming the eraser is half of a sphere, what is the volume of the pencil? Round the answer to one decimal place.

Hint

The pencil is made of a cone, a cylinder, and a hemisphere, all with the same radius. The volume of the pencil equals the sum of the volumes of each solid.

Solution

The given pencil can be seen to consist of a cone, a cylinder, and half of a sphere — all with the same radius.

As such, the volume of the pencil equals the sum of the volumes of each of these solids.

Volume of a Cone	Volume of a Cylinder	Volume of a Hemisphere
$V_{1} = \frac{1}{3} π r^{2} h$	$V_{2} = π r^{2} h$	$V_{3} = \frac{2}{3} π r^{3}$

Use a calculator to make the calculations easier.

Volume of the Pencil's Tip

The tip of the pencil is a cone with a radius of

3

millimeters and a height of

10

millimeters. Substituting these values into the first formula will give the volume of the tip.

V_{1} = \frac{1}{3} π r^{2} h

SubstituteII

$r = 3$ , $h = 10$

V_{1} = \frac{1}{3} π (3)^{2} (10)

▼

Simplify right-hand side

CalcPow

Calculate power

V_{1} = \frac{1}{3} π (9) (10)

Multiply

V_{1} = \frac{1}{3} π (90)

CommutativePropMult

Commutative Property of Multiplication

V_{1} = \frac{1}{3} (90) π

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

V_{1} = \frac{9 0}{3} π

CalcQuot

Calculate quotient

V_{1} = 30 π

The tip of the pencil has a volume of

30 π

cubic millimeters.

Volume of the Pencil's Body

The body of the pencil is a cylinder with a radius of

3

millimeters. To find the height of the cylinder, subtract the height of the tip of the pencil from the original length of the pencil.

160 mm - 10 mm = 150 mm

Next, substitute

r = 3

and

h = 150

into the formula for the volume a cylinder.

V_{2} = π r^{2} h

SubstituteII

$r = 3$ , $h = 150$

V_{2} = π (3)^{2} (150)

▼

Simplify right-hand side

CalcPow

Calculate power

V_{2} = π \cdot 9 (150)

Multiply

V_{2} = 1350 π

The body of the pencil has a volume of

1350 π

cubic millimeters.

Volume of the Pencil's Eraser

The eraser is a hemisphere with a radius of

3

millimeters. To find its volume, substitute

r = 3

into the hemisphere volume formula.

V_{3} = \frac{2}{3} π r^{3}

Substitute

$r = 3$

V_{3} = \frac{2}{3} π (3)^{3}

▼

Simplify right-hand side

CalcPow

Calculate power

V_{3} = \frac{2}{3} π (27)

CommutativePropMult

Commutative Property of Multiplication

V_{3} = \frac{2}{3} (27) π

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{3} = \frac{2 ( 2 7 )}{3} π

Multiply

V_{3} = \frac{5 4}{3} π

CalcQuot

Calculate quotient

V_{3} = 18 π

The eraser of the pencil has a volume of

18 π

cubic millimeters.

Pencil's Volume

Finally, the total volume of the pencil is equal to the sum of the volumes of its parts.

V_{P} = V_{1} + V_{2} + V_{3}

SubstituteValues

Substitute values

V_{P} = 30 π + 1350 π + 18 π

▼

Evaluate right-hand side

FactorOut

Factor out $π$

V_{P} = π (30 + 1350 + 18)

AddTerms

Add terms

V_{P} = π (1398)

CommutativePropMult

Commutative Property of Multiplication

V_{P} = 1398 π

UseCalc

Use a calculator

V_{P} = 4391.946529 \dots

RoundDec

Round to $1$ decimal place(s)

V_{P} \approx 4391.9

In conclusion, the volume of Ramsha's pencil is approximately

4391.9

cubic millimeters.

Example

Finding the Lateral Area of a Roof

Ramsha's house is a rough composite solid consisting of a square pyramid with a height of $8$ feet and a base side length of $30$ feet on top of a square prism.

A house with a roof in the shape of a square pyramid with a height of $8$ feet and a base side length of $30$ feet and a square prism flat.

Ramsha's father decides to cover the roof of their house with waterproof insulation material. Help Ramsha and her father calculate how many square feet of insulation material are needed.

Hint

Ramsha only needs to know the lateral area of the pyramid. Use the Pythagorean Theorem to find the slant height.

Solution

The amount of insulation material to cover the roof corresponds to the lateral area of the pyramid. Recall the formula for the lateral area of a pyramid.

L A = \frac{1}{2} p ℓ

In this formula,

p

is the perimeter of the base and

ℓ

is the slant height. The slant height of this pyramid can be found by using the Pythagorean Theorem.

The height

h

of the pyramid is the distance between the vertex and the base, so

h = 8

for this pyramid. The value of

b

is half the base side length, so

b = \frac{3 0}{2} = 15

feet.

ℓ^{2} = b^{2} + h^{2}

SubstituteII

$b = 15$ , $h = 8$

ℓ^{2} = 15^{2} + 8^{2}

▼

Solve for

ℓ

CalcPow

Calculate power

ℓ^{2} = 225 + 64

AddTerms

Add terms

ℓ^{2} = 289

SqrtEqn

$LHS = RHS$

ℓ = 289

CalcRoot

Calculate root

ℓ = 17

Since a negative value does not make sense in this context, only the principal root is considered. This means that the slant height is

17

feet. The next step is to find the perimeter of the base. Since the base is a square, its perimeter

p

4

times the base side length.

p = 4 \cdot 30 = 120

Finally, the lateral area of the pyramid can be found by substituting

p = 120

and

ℓ = 17

into the formula.

L A = \frac{1}{2} p ℓ

SubstituteII

$p = 120$ , $ℓ = 17$

L A = \frac{1}{2} (120) (17)

▼

Evaluate right-hand side

Multiply

L A = \frac{1}{2} \cdot 2040

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

L A = \frac{2 0 4 0}{2}

CalcQuot

Calculate quotient

L A = 1020

Ramsha's father will need

1020

square feet of insulation material to cover the entire roof.

Example

Finding the Volume of a Deck Prism

While her father is busy installing the insulation material, Ramsha decides to explore the attic. She discovers her grandfather's old deck prism, a captivating object designed to illuminate cabins below the deck of a ship before electric lighting. The deck prism is a composite solid made up of a base prism and a pyramid, both with regular hexagonal bases.

A deck prism with a regular hexagonal prism base with sides of 4 centimeters and a height of 2 centimeters. The hexagonal pyramid has side lengths of 3 centimeters and a height of 4 centimeters.

Feeling a connection with her grandfather, Ramsha takes a closer look at the deck prism. Find the volume of the deck prism. Round the answer to the two decimal places.

Hint

The formula for the area of a regular hexagon with side lengths $a$ is $B = \frac{3 a ^{2} 3}{2} .$

Solution

The deck prism is composed of two solids:

a regular hexagonal prism with an edge length of $4$ centimeters and a height of $2$ centimeters, and
a regular hexagonal pyramid with an edge length of $3$ centimeters and a height of $4$ centimeters.

This means that the volume of the deck prism is the sum of the volumes of the two solids. The volume of each solid will be found one at a time.

Volume of the Base Prism

The base of the prism is a regular hexagon with a side length of

4

centimeters. Recall the formula for the area of a regular hexagon with side lengths

a .

B = \frac{3 a ^{2} 3}{2}

Substitute

4

for

a

into the formula and evaluate its value.

B = \frac{3 a ^{2} 3}{2}

Substitute

$a = 4$

B = \frac{3 ( 4 ) ^{2} 3}{2}

▼

Evaluate right-hand side

CalcPow

Calculate power

B = \frac{3 ( 1 6 ) 3}{2}

Multiply

B = \frac{4 8 3}{2}

CalcQuot

Calculate quotient

B = 243

The area of the hexagonal base is

243

square centimeters. Now the volume of the prism can be found by multiplying the base area by the height of the hexagonal prism.

V_{1} = 243 \cdot 2 \Rightarrow V_{1} = 483

Volume of the Pyramid

The base of the pyramid is a regular hexagon with side lengths of

3

centimeters. Use the formula for the area of the hexagonal base again, this time substituting

3

for

a .

B_{2} = \frac{3 a ^{2} 3}{2}

Substitute

$a = 3$

B_{2} = \frac{3 ( 3 ) ^{2} 3}{2}

▼

Evaluate right-hand side

CalcPow

Calculate power

B_{2} = \frac{3 ( 9 ) 3}{2}

Multiply

B_{2} = \frac{2 7 3}{2}

Now the volume can be found. Recall that the volume of a pyramid is one third of the product of its base area and height. The height of the pyramid is

4

centimeters.

V_{2} = \frac{1}{3} B h

SubstituteII

$B = \frac{2 7 3}{2}$ , $h = 4$

V_{2} = \frac{1}{3} \cdot \frac{2 7 3}{2} \cdot 4

▼

Evaluate right-hand side

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{2} = \frac{1}{3} \cdot \frac{1 0 8 3}{2}

MultFrac

Multiply fractions

V_{2} = \frac{1 0 8 3}{6}

SimpQuot

Simplify quotient

V_{2} = 183

Volume of the Deck Prism

The sum of the volumes of the solids will give the total volume of the deck prism.

V = V_{1} + V_{2}

SubstituteII

$V_{1} = 483$ , $V_{2} = 183$

V = 483 + 183

▼

Evaluate right-hand side

AddTerms

Add terms

V = 663

UseCalc

Use a calculator

V = 114.315353 \dots

RoundDec

Round to $2$ decimal place(s)

V \approx 114.32

The deck prism has a volume of about

114.32

cubic centimeters. A sense of wonder washes over Ramsha as she holds the relic of maritime history in her hands and thinks about the stories her grandfather told her about his life at sea.

Closure

Composite Solids in Real Life

This lesson explored a few real-life examples of composite solids. The calculation of volumes and surface areas for these combined shapes were examined. However, more composite solids can be found everywhere in daily life.

External credits: freepic.diller, xb100, wirestock

As the lesson comes to a close, take a look around and see what composite solids are nearby. Just for fun, try to find their volumes and surface areas!

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Catch-Up and Review

Hint

Solution

Finding the Volume of the Prism

Finding the Volume of the Cone

Volume of the Traffic Cone

Hint

Solution

Volume of Water

Volume of Air

Hint

Solution

Lateral and Base Areas of the Cylinder

Lateral Area of the Cone

Total Surface Area

Hint

Solution

Volume of the Pencil's Tip

Volume of the Pencil's Body

Volume of the Pencil's Eraser

Pencil's Volume

Hint

Solution

Hint

Solution

Volume of the Base Prism

Volume of the Pyramid

Volume of the Deck Prism