Properties of Cones

Cone-shaped figures can be recognized in many places in everyday life, such as when eating ice cream. This lesson will focus on the formulas for the volume and the surface area of a cone.

Catch-Up and Review

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

Circle
Circumference of a Circle
Area of a Circle
Volume
Cylinder
Volume of a Cylinder
Pythagorean Theorem

Explore

Relating the Volumes of Cones and Cylinders

Heichi knows how to calculate the volume of a cylinder. He thinks that the volume of a cone can be found using cylinders. To do so, he makes a cone-shaped mold with a height and radius of $r .$ Then, he fills it with sand and pours it into a cylinder with the same radius and height.

Animation to fill up a cylinder using a cone

How can the relationship between the volumes of a cone and a cylinder be expressed? What can Heichi assume at the end of this experiment?

Discussion

Cones

Before proceeding to the volume of a cone, the definition of a cone and its characteristics will be examined.

A cone is a three-dimensional solid with a circular base and a point, called the vertex or apex, that is not in the same plane as the base. The altitude of a cone is the segment from the vertex perpendicular to the plane of the base.

The length of the altitude is called the height of the cone. If the altitude meets the base at its center, then the cone is a right cone. In a right cone, the distance from the vertex to a point on the edge of the base is called the slant height of a cone.

Interactive Cone where the height and radius can be change. The Slant height is highlighted.

If a cone is not a right cone, it is called oblique. Oblique cones do not have a uniform slant height.

Interactive Cone changing from oblique to right

Considering Heichi's experiment, the formula for the volume of a cone will be one third of the volume of a cylinder with the same radius and height.

Rule

Volume of a Cone

The volume of a cone is one third of the product of its base area and height.

The base area $B$ is the area of the circle, and the height $h$ is measured perpendicular to the base.

$V = \frac{1}{3} B h$

As the base is a circle, its area is dependent on its radius. Therefore, the base area can also be expressed in terms of the radius $r .$

$V = \frac{1}{3} π r^{2} h$

Example

Finding the Volume of the Cathedral of Maringá

The Cathedral of Maringá, one of the tallest churches in the world, was designed in the form of a cone by José Augusto Bellucci.

The cathedral reaches

114

meters in height, excluding the cross. Furthermore, its circular base has a radius of

50

meters. Calculate its volume. Round the answer to the nearest cubic meter.

Hint

The volume of a cone is one third of the product of its base area and height.

Solution

To find the volume of the cathedral, the formula for the volume of a cone will be used.

V = \frac{1}{3} π r^{2} h

In this formula,

r

is the radius of the base and

h

is the height of the cone. The height of the conical cathedral is

114

meters and its radius is

50

meters. By substituting these values into the formula, the volume can be found.

V = \frac{1}{3} π r^{2} h

SubstituteII

$r = 50$ , $h = 114$

V = \frac{1}{3} π (50)^{2} (114)

Simplify right-hand side

CalcPow

Calculate power

V = \frac{1}{3} π (2500) (114)

Multiply

V = \frac{1}{3} 285000 π

MoveRightFacToNumOne

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

V = \frac{2 8 5 0 0 0 π}{3}

UseCalc

Use a calculator

V = 298451.302091 \dots

RoundInt

Round to nearest integer

V \approx 298451

The volume of the cone is approximately

298451

cubic meters.

Example

Finding the Volume a Chinese Conical Hat

Tadeo is learning how to make a traditional Chinese conical hat. He notices that the craftsman uses $20 -$ centimeter bamboo sticks to make the framework.

If the radius of the base is $16$ centimeters, find the volume of the hat. Round the answer to the nearest cubic centimeter.

Hint

Use the Pythagorean Theorem to find the height of the cone.

Solution

To find the volume of the conical hat, its height will be calculated first. Then, the formula for the volume of a cone will be used.

Finding Height of the Hat

The distance from the vertex of the cone to its base is the height of the cone. Each stick if the frame represents the slant height of the cone. Therefore, the cone has a slant height of $20$ centimeters and a radius of $16$ centimeters.

These three segments form a right triangle,

△ A B C .

From here, the height

A B

can be determined by using the Pythagorean Theorem.

A C^{2} = A B^{2} + B C^{2}

SubstituteII

$A C = 20$ , $B C = 16$

20^{2} = A B^{2} + 16^{2}

Solve for

A B

CalcPow

Calculate power

400 = A B^{2} + 256

SubEqn

$LHS - 256 = RHS - 256$

144 = A B^{2}

RearrangeEqn

Rearrange equation

A B^{2} = 144

SqrtEqn

$LHS = RHS$

A B^{2} = 144

SqrtPowToNumber

$a^{2} = a$

A B = 12

The principal root was taken because negative values do not make sense for these measurements.

Finding the Volume of the Hat

Recall the formula for the volume of a cone.

V = \frac{1}{3} π r^{2} h

Here,

r

is the radius and

h

is the height of the cone. The height was found to be

12

centimeters, and the radius of the base is

16

centimeters. The volume of the hat can be found by substituting these values into the formula.

V = \frac{1}{3} π r^{2} h

SubstituteII

$r = 16$ , $h = 12$

V = \frac{1}{3} π (16^{2}) (12)

Simplify right-hand side

CalcPow

Calculate power

V = \frac{1}{3} π (256) (12)

Multiply

V = \frac{1}{3} 3072 π

MoveRightFacToNumOne

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

V = \frac{3 0 7 2 π}{3}

UseCalc

Use a calculator

V = 3216.990877 \dots

RoundInt

Round to nearest integer

V \approx 3217

The volume of the cone is about

3217

cubic centimeters.

Example

Finding the Radius of a Traffic Cone

The diagram shows a traffic cone, which has a volume of $3128$ cubic inches.

The height of the cone part is $28$ inches. The prism below it is a square prism with side lengths of $14$ inches and height of $1$ inch. Find the radius of the cone. Write the answer to the nearest inch.

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. Therefore, the volume occupied by the traffic cone is the sum of the volume of the prism $V_{p}$ and the volume of the cone $V_{c} .$ $V = V_{p} + V_{c}$ First the volume of the cone will be found. Then, the volume formula of a cone will be used to calculate the radius of the base of the cone.

Finding the Volume of the Cone

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

Given that the total volume of the traffic cone is

3128

cubic inches, the volume of the cone part can be found now.

V = V_{p} + V_{c}

SubstituteII

$V = 3128$ , $V_{p} = 196$

3128 = 196 + V_{c}

Solve for

V_{c}

SubEqn

$LHS - 196 = RHS - 196$

2932 = V_{c}

RearrangeEqn

Rearrange equation

V_{c} = 2932

Finding the Radius of the Cone

Finally, using the formula for the volume of a cone, the radius of the cone can be found.

V_{c} = \frac{1}{3} π r^{2} h

To do so, substitute

2932

for

V_{c}

and

28

for

h

into the formula and solve for

r .

V_{c} = \frac{1}{3} π r^{2} h

SubstituteII

$V_{c} = 2932$ , $h = 28$

2932 = \frac{1}{3} π r^{2} (28)

Solve for

r

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

8796 = π r^{2} (28)

CommutativePropMult

Commutative Property of Multiplication

8796 = 28 π r^{2}

DivEqn

$LHS / 28 π = RHS / 28 π$

\frac{8 7 9 6}{2 8 π} = r^{2}

RearrangeEqn

Rearrange equation

r^{2} = \frac{8 7 9 6}{2 8 π}

SqrtEqn

$LHS = RHS$

r^{2} = \frac{8 7 9 6}{2 8 π}

SqrtPowToNumber

$a^{2} = a$

r = \frac{8 7 9 6}{2 8 π}

UseCalc

Use a calculator

r = 9.999738 \dots

RoundInt

Round to nearest integer

r \approx 10

The radius of the cone is about

10

inches.

Example

Cylinder With a Cone Inside

For the Jefferson High Science Fair, Ali is thinking about a chemistry experiment in which he will need a cylinder with a radius of $20$ centimeters and a height $81$ centimeters, with a cone inside. The cylinder must be open on both ends, and the cone must have an open bottom.

To conduct the experiment, Ali needs to answer some questions first. Help him find the answers in order to win the first prize in the fair!

a Ali will fill the cone with water and the interior portion of the cylinder not occupied by the cone with foam. How many cubic centimeters of water and foam does Ali need? Write the answers to the nearest integer.

b What percent of the interior of the cylinder is not occupied by the cone? 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 obtained in Part A.

Solution

a The volumes will be calculated one at a time.

Volume of Water

Ali will fill the cone with water. Therefore, the volume of the cone is needed. The height and radius of the cone are the same as the height and radius of the cylinder. Therefore, the height of the cone is $81$ centimeters and its radius is $20$ centimeters.

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

In the this formula,

81

and

20

can be substituted for

h

and

r,

respectively.

V_{cone} = \frac{1}{3} π r^{2} h

SubstituteII

$h = 81$ , $r = 20$

V_{cone} = \frac{1}{3} π (20^{2}) (81)

Evaluate right-hand side

CalcPow

Calculate power

V_{cone} = \frac{1}{3} π (400) (81)

Multiply

V_{cone} = \frac{1}{3} π (32400)

CommutativePropMult

Commutative Property of Multiplication

V_{cone} = \frac{1}{3} (32400) π

MoveRightFacToNumOne

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

V_{cone} = \frac{3 2 0 0 0}{3} π

CalcQuot

Calculate quotient

V_{cone} = 10800 π

Therefore, the volume of water that Ali needs is

10800 π

cubic centimeters. Finally, this number will be approximated to the nearest integer.

10800 π

UseCalc

Use a calculator

33929.200658 \dots

RoundInt

Round to nearest integer

33929

Ali needs about

33929

cubic centimeters of water.

Volume of Foam

Ali will fill the part of the cylinder not occupied by the cone with foam. Therefore, the volume of this portion is needed. It is given that the height of the cylinder is $81$ centimeters and that its radius is $20$ 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

One more time,

81

and

20

can be substituted for

h

and

r,

respectively.

V_{cylinder} = π r^{2} h

SubstituteII

$h = 81$ , $r = 20$

V_{cylinder} = π (20^{2}) (81)

Evaluate right-hand side

CalcPow

Calculate power

V_{cylinder} = π (400) (81)

Multiply

V_{cylinder} = 32400 π

The volume of the cylinder is

32400 π

cubic centimeters. The volume of the space inside the cylinder that is not occupied by the cone is the difference between the volume of the cylinder and the volume of the cone, which is

10800 π

cubic centimeters.

32400 π - 10800 π = 21600 π

The volume of the cylinder that is not occupied by the cone is

26600 π

cubic centimeters and is the volume of foam that Ali will need. This number will be approximated to the nearest integer.

21600 π

UseCalc

Use a calculator

67858.401317 \dots

RoundInt

Round to nearest integer

67858

Ali needs about

67858

cubic centimeters of foam.

b In Part

A

it was found that the volume of the cylinder is

32400 π

cubic centimeters. It was also found that the volume of the portion of the cylinder not occupied by the cone is

21600 π

cubic centimeters. The ratio of the second value to the first value will result in the desired percentage.

\frac{2 1 6 0 0 π}{3 2 4 0 0 π}

Evaluate

CancelCommonFac

Cancel out common factors

\frac{2 1 6 0 0 π}{3 2 4 0 0 π}

SimpQuot

Simplify quotient

\frac{2 1 6 0 0}{3 2 4 0 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0 8 0 0}{b / 1 0 8 0 0}$

\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 ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$

66.7 %

It can be concluded that the volume of the cylinder not occupied by the cone represents

66.7 %

of the interior of the cylinder.

Now, suppose that Ali wants to make the figure by himself. How many square meters of material does he need to use? To answer that question, Ali needs to know how to calculate the area of the surfaces of a cone.

Discussion

Surface Area of a Cone

Consider a right cone with radius $r$ and slant height $ℓ .$

The surface area of a right cone is the sum of the base area and the lateral area. The area of the base is given by $π r^{2}$ and the lateral area has the area $π r ℓ .$

$S = π r^{2} + π r ℓ$

Example

Cylinder With a Cone Inside

For his experiment for the science fair, Ali plans to make the figure by himself.

The material to be used to create the figure costs

$ 125

per square meter. What is the cost of the material for this object? Write the answer to two decimal places.

Hint

The area of the lateral surface of a cone is the product of $π,$ the radius, and the slant height. A cylinder's lateral surface area is twice the product of $π,$ the radius, and the height.

Solution

Since the cost is given per square meter, all the measures will be converted from centimeters to meters. To do this, the measures need to be multiplied by a conversion factor, $\frac{1 m}{1 0 0 cm} .$ $Height : 81 cm \cdot \frac{1 m}{1 0 0 cm} Radius : 20 cm \cdot \frac{1 m}{1 0 0 cm} = 0.81 m = 0.2 m$ With this information, update the measures on the diagram.

The lateral areas of the cylinder and the cone will be calculated one at a time. Then, their sum will be multiplied by the cost per square meter.

Lateral Area of the Cylinder

Recall that the cylinder is open in both ends. The lateral area of a cylinder is twice the product of

π,

the radius, and the height.

L A_{cylinder} = 2 π r h

In the above formula,

0.81

and

0.2

can be substituted for

h

and

r,

respectively.

L A_{cylinder} = 2 π r h

SubstituteII

$h = 0.81$ , $r = 0.2$

L A_{cylinder} = 2 π (0.2) (0.81)

Evaluate right-hand side

CommutativePropMult

Commutative Property of Multiplication

L A_{cylinder} = 2 (0.2) (0.81) π

Multiply

L A_{cylinder} = 0.324 π

The lateral area of the cylinder is about

0.324 π

square meters.

Lateral Area of the Cone

The lateral area of a cone is the product of $π,$ the radius, and the slant height. $L A_{cone} = π r ℓ$ The slant height $ℓ$ is the hypotenuse of the right triangle formed by the radius, the height, and the segment that connects 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

0.81^{2} + 0.2^{2} = ℓ^{2}

Solve for

ℓ

CalcPow

Calculate power

0.6561 + 0.04 = ℓ^{2}

AddTerms

Add terms

0.6961 = ℓ^{2}

SqrtEqn

$LHS = RHS$

0.6961 = ℓ

RearrangeEqn

Rearrange equation

ℓ = 0.6961

The slant height of the cone is about

0.6961

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

0.83

and

0.2

for

ℓ

and

r,

respectively.

L A_{cone} = π r ℓ

SubstituteII

$ℓ = 0.6961$ , $r = 0.2$

L A_{cone} = π (0.2) (0.6961)

CommutativePropMult

Commutative Property of Multiplication

L A_{cone} = 0.2 0.6961 π

Total Cost

Finally, the sum of the lateral areas will be multiplied by the cost per square meter.

(0.324 π + 0.2 0.6961 π) 125

Evaluate

UseCalc

Use a calculator

(1.017876 \dots + 0.524222 \dots) 125

AddTerms

Add terms

(1.542098 \dots) 125

Multiply

192.76

To make the shape that Ali needs, he has to spend about

$ 192.76 .

Example

Finding the Surface Area of a Cone

Tiffaniqua's teacher gives her a piece of paper on which a circle and a sector are drawn. The paper is a square of side length $10$ centimeters.

A square piece of paper with a circle and a sector drawn on it.

The teacher also gives the following set of information.

The circle is tangent to the top and right sides of the paper as well as to the sector.
The centers of the circle and sector are on the diagonal of the paper.
The circle and the sector can form a cone.

Help Tiffaniqua answer the following questions.

a Find the radii of the circle and the sector. Write the answers to one decimal place.

b Find the surface area of the cone. Round the answer to the nearest square centimeter.

Hint

a The circumference of the circle is equal to the arc length of the sector.

b The surface area of a cone is the sum of the base area and the lateral area.

Solution

a Let

ℓ

be the radius of the sector and

r

be the radius of the circle. To find these values, a system of two equations should be written first. The equations can be written using the trigonometric ratios and the properties of cones. Then, the system of equations will be solved using the Substitution Method.

Writing the System of Equations

Start by drawing a right triangle whose hypotenuse is

ℓ + r

centimeters.

A square paper and a right triangle with hypotenuse r+l

Since the diagonal of the square bisects the angle, the right triangle is also an isosceles triangle. Using the sine ratio of

4 5^{\circ},

an equation describing the relationship between

r

and

ℓ

can be written.

sin 4 5^{\circ} = \frac{1 0 - r}{ℓ + r}

Since

sin 4 5^{\circ} = \frac{1}{2}

, the above equation can be solved for

ℓ .

sin 4 5^{\circ} = \frac{1 0 - r}{ℓ + r}

Substitute

$sin 4 5^{\circ} = \frac{1}{2}$

\frac{1}{2} = \frac{1 0 - r}{ℓ + r}

Solve for

ℓ

MultEqn

$LHS \cdot (ℓ + r) = RHS \cdot (ℓ + r)$

\frac{1}{2} \cdot (ℓ + r) = 10 - r

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

ℓ + r = 10 - r (2)

Distr

Distribute $2$

ℓ + r = 102 - r 2

SubEqn

$LHS - r = RHS - r$

ℓ = 102 - r 2 - r

The first equation is obtained. It is also known that the sector of circle with radius

ℓ

and the circle with radius

r

form a cone.

As shown above,

ℓ

will be the slant height of the cone and

r

will be the radius of its circular base. Recall that in a cone, the circumference of the circular base is equal to the arc length of the sector. Since the measure of the sector is

9 0^{\circ},

its arc length can be found using the relationship between arc length and arc measure.

Arc Length : Circumference of Base : \frac{9 0}{3 6 0} \cdot 2 π ℓ 2 π r

Because these two lengths are equal, another equation can be obtained.

\frac{9 0}{3 6 0} \cdot 2 π ℓ = 2 π r

Solve for

ℓ

DivEqn

$LHS / 2 π = RHS / 2 π$

\frac{9 0}{3 6 0} \cdot ℓ = r

ReduceFrac

$\frac{a}{b} = \frac{a / 9 0}{b / 9 0}$

\frac{1}{4} \cdot ℓ = r

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

ℓ = 4 r

Now, a system of equations can be written using the two values of

ℓ .

{ℓ = 102 - r 2 - r ℓ = 4 r (I) (II)

Solving the System of Equations

The Substitution Method will be used to find the unknowns. First use the value of

ℓ

to find the value of

r .

{ℓ = 102 - r 2 - r ℓ = 4 r (I) (II)

Substitute

$(I):$ $ℓ = 4 r$

{4 r = 102 - r 2 - r ℓ = 4 r

Solve for

r

AddEqn

$(I):$ $LHS + r 2 + r = RHS + r 2 + r$

{5 r + r 2 = 102 ℓ = 4 r

FactorOut

$(I):$ Factor out $r$

{r (5 + 2) = 102 ℓ = 4 r

DivEqn

$(I):$ $LHS / (5 + 2) = RHS / (5 + 2)$

⎩ ⎪ ⎨ ⎪ ⎧ r = \frac{1 0 2}{5 + 2} ℓ = 4 r

UseCalc

$(I):$ Use a calculator

{r = 2.204812 \dots ℓ = 4 r

RoundDec

$(I):$ Round to ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$

{r \approx 2.2 ℓ = 4 r

Next, the value of

r

can be substituted into the other equation to find the value of

ℓ .

{r \approx 2.2 ℓ = 4 r

$(II):$ $r \approx 2.2$

{r \approx 2.2 ℓ \approx 4 (2.2)

Multiply

$(II):$ Multiply

{r \approx 2.2 ℓ \approx 8.8

The radii of the circle and the sector are about

2.2

and

8.8

centimeters, respectively.

b The surface area of a cone is the sum of the base area and the lateral area.

S = π r^{2} + π r ℓ

By substituting

ℓ = 8.8

and

r = 2.2,

the surface area of the cone will be determined.

S = π r^{2} + π r ℓ

SubstituteII

$ℓ = 8.8$ , $r = 2.2$

S = π (2.2)^{2} + π (2.2) (8.8)

Evaluate right-hand side

CalcPow

Calculate power

S = π (4.84) + π (2.2) (8.8)

Multiply

S = π (4.84) + π (19.36)

AddTerms

Add terms

S = π (24.2)

UseCalc

Use a calculator

S = 76.026542 \dots

RoundInt

Round to nearest integer

S \approx 76

The surface area of the cone is about

76

square centimeters.

Pop Quiz

Practice Finding the Volume and Surface Area of a Cone

The applet shows right cones. Use the given information to answer the question. If necessary, round the answer to one decimal place.

Closure

Comparing the Volumes of a Pyramid and a Cube

In this lesson, the characteristics of cones have been studied, including their relationship with cylinders. It has been shown that the volume of a cone is one third the volume of a cylinder with the same radius and perpendicular height.

Comparing the volumes of a cylinder and cone

How can the volume of a cuboid with side lengths

h

be related to the volume of a pyramid with the same base and height? The answer will be found in the next lesson.