Unlocking the Secrets of Sector Area in Circles

This lesson will explore the regions bounded by two radii of a circle and their intercepted arc. Additionally, the formula for the area of such regions will be derived considering the part-whole relationships.

Catch-Up and Review

Here is some recommended reading before getting started with this lesson.

Circle
Arc
Arc Measure
Arc Length
Circumference
Area of Circle
Ratio
Proportion
Similarity

Challenge

Investigating the Area of a Slice of Pizza

Three friends are sharing a $16 -$ inch pizza equally. Emily is becoming a nutritionist and is curious about how many calories are in a slice. To find out, she will calculate the area of one slice.

Help Emily to find the area of one slice. How can the measure of a central angle be used to find the area?

Discussion

Investigating the Area of Sectors of a Circle

Concept

Sector of a Circle

A sector of a circle is a portion of the circle enclosed by two radii and their intercepted arc.

In the diagram, sector

A C B

is created by

\overline{A C}, \overline{B C},

and

A B .

Rule

Area of a Sector of a Circle

The area of a sector of a circle is calculated by multiplying the circle's area by the ratio of the measure of the central angle to $36 0^{\circ} .$

$Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}$

From the fact that $2 π rad$ equals $36 0^{\circ},$ an equivalent formula can be written if the central angle is given in radians.

Area of Sector Area of Sector = \frac{θ}{2 π} \cdot π r^{2} ⇓ = \frac{θ}{2} \cdot r^{2}

Since the measure of an arc is equal to the measure of its central angle, the arc $A B$ measures $θ .$ Therefore, by substituting $m A B$ for $θ,$ another version of the formula is obtained which can also be written in degrees or radians.

$Area of Sector = \frac{m A B}{3 6 0 ^{\circ}} \cdot π r^{2}$
or
$Area of Sector = \frac{m A B}{2} \cdot r^{2}$

Proof

Consider sector $A C B$ bounded by $\overline{A C}, \overline{B C},$ and $A B .$

Since a circle measures

36 0^{\circ},

this sector represents

\frac{θ}{3 6 0 ^{\circ}}

⊙ C .

Therefore, the ratio of the area of a sector to the area of the whole circle is proportional to

\frac{θ}{3 6 0 ^{\circ}} .

\frac{Area of Sector}{Area of Circle} = \frac{θ}{3 6 0 ^{\circ}}

Recall that the area of a circle is

π r^{2} .

By substituting it into the equation and solving for the area of a sector, the desired formula can be obtained.

\frac{Area of Sector}{π r ^{2}} = \frac{θ}{3 6 0 ^{\circ}}

Therefore, the area of a sector of a circle can be found by using the following formula.

$Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}$

Example

Using Sectors of Circles to Solve Problems

Like the pizza problem, numerous real-life problems can be modeled by sectors of a circle.

Tiffaniqua has a trapezoid-shaped yard whose side lengths are shown on the diagram. To water the lawn, she sets up a water sprinkler that can water the grass within a $4 -$ meter radius. as shown.

Trapezoidal Yard: Vertices labeled A, B, C, D counterclockwise. Length of AB and AD is 4 meters each, DC is 9 meters. A water sprinkler at point B can water the grass within a radius of 4 meters.

Tiffaniqua knows that $∠ A B C$ measures $13 5^{\circ} .$

a Help Tiffaniqua calculate the area of the lawn covered by the sprinkler. If necessary, round the answer to the nearest square meter.

b What is the area of the lawn that is not watered? If necessary, round the answer to the nearest square meters.

Hint

a The sprinkler covers a

135 -

degree sector of a circle with a radius of

4

meters.

b The area of a trapezoid is half the product of the height and the sum of the two bases.

Solution

a The sprinkler covers a

135 -

degree sector of a circle with radius

4

meters. The area of a sector of a circle can be calculated using the following formula.

Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

Substituting

13 5^{\circ}

for

θ

and

4

for

r

into the formula will give the result.

Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

SubstituteII

$θ = 13 5^{\circ}$ , $r = 4$

Area of Sector = \frac{1 3 5 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (4)^{2}

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

Area of Sector = 0.375 \cdot π (4)^{2}

CalcPow

Calculate power

Area of Sector = 0.375 \cdot π 16

Multiply

Area of Sector = 6 π

UseCalc

Use a calculator

Area of Sector = 18.849555 \dots

RoundInt

Round to nearest integer

Area of Sector \approx 19

The area of the lawn covered is about

19

square meters.

b To find the area of the lawn that is not watered, the area found in Part A should be subtracted from the area of the trapezoid.

On the diagram, the trapezoid's bases are

5

and

9

meters, and its height is

4

meters. Substitute these values into the formula for the area of a trapezoid.

Area of Trapezoid = \frac{1}{2} h (b_{1} + b_{2})

SubstituteValues

Substitute values

Area of Trapezoid = \frac{1}{2} 4 (5 + 9)

▼

Evaluate right-hand side

AddTerms

Add terms

Area of Trapezoid = \frac{1}{2} 4 (14)

Multiply

Area of Trapezoid = \frac{1}{2} 56

MoveRightFacToNumOne

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

Area of Trapezoid = \frac{5 6}{2}

CalcQuot

Calculate quotient

Area of Trapezoid = 28

This means that the area that Tiffaniqua should water, initially, is

28

square meters. By subtracting the area of the sector from the total area, the lawn area that is not watered can be found.

28 - 19 = 9

Tiffaniqua needs to water an area of

9

square meters.

Example

Calculating the Central Angle of Pac-Man's Mouth

When the area of a sector is given, the measure of the corresponding central angle can be calculated

Consider a two-dimensional image of Pac-Man. The area covered by Pac-Man is about $66.5$ square millimeters.

If the radius of the circle used to draw Pac-Man is

5

millimeters, find the measure of the central angle formed in the colored region. If necessary, round the answer to the nearest degree.

Hint

Pac-Man is essentially a sector of a circle. Use the formula for the area of a sector of a circle to find the measure of the angle.

Solution

To find the measure of the corresponding central angle, the formula for the area of a sector will be used.

Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

The area of the sector and the radius are given. Substitute these values into the formula.

Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

SubstituteValues

Substitute values

66.5 = \frac{θ}{3 6 0 ^{\circ}} \cdot π (5)^{2}

▼

Solve for

θ

CalcPow

Calculate power

66.5 = \frac{θ}{3 6 0 ^{\circ}} \cdot π \cdot 25

Multiply

66.5 = \frac{θ}{3 6 0 ^{\circ}} \cdot 25 π

DivEqn

$LHS / 25 π = RHS / 25 π$

\frac{6 6 . 5}{2 5 π} = \frac{θ}{3 6 0 ^{\circ}}

UseCalc

Use a calculator

0.846704 \dots = \frac{θ}{3 6 0 ^{\circ}}

MultEqn

$LHS \cdot 36 0^{\circ} = RHS \cdot 36 0^{\circ}$

304.813547 \dots^{\circ} = θ

RearrangeEqn

Rearrange equation

θ = 304.813547 \dots^{\circ}

RoundInt

Round to nearest integer

θ \approx 30 5^{\circ}

The corresponding central angle is about

30 5^{\circ} .

Example

Comparing the Radius of Two Circles

The diagram below models the motion of two gears

S

and

L .

Gear

S

has a radius of

2

inches.

a Find the radius of the larger gear.

b Find the area of the sector of Gear

L

formed when Gear

S

completes two revolutions. Write the answer in terms of

π .

Hint

a How can the measure of the central angle formed in Gear

L

be used?

b Use the formula for the area of a sector of a circle.

Solution

a When the smaller gear completes a single revolution, the central angle measured in the larger gear becomes

14 4^{\circ} .

Since the measure of an arc is the same as its corresponding central angle, the measure of the colored arc of

⊙ L

is also

14 4^{\circ} .

Furthermore, the length of the arc is equal to the circumference of

⊙ S .

Recall that the circumference is given by the formula

C = 2 π r .

Substituting

r = 2

into the formula will give the circumference

C_{S}

⊙ S .

C_{S} = 2 π (2) = 4 π

The length of the arc of

⊙ L

is, therefore,

4 π

inches. Now that the measure and length of the arc is known, the radius

R

⊙ L

can be found. To do so, substitute the values into the formula for the arc length.

\frac{Arc length}{Circumference of ⊙ L} = \frac{Arc measure}{3 6 0 ^{\circ}}

SubstituteValues

Substitute values

\frac{4 π}{2 π R} = \frac{1 4 4 ^{\circ}}{3 6 0 ^{\circ}}

▼

Solve for

R

ReduceFrac

$\frac{a}{b} = \frac{a / 2 π}{b / 2 π}$

\frac{2}{R} = \frac{1 4 4 ^{\circ}}{3 6 0 ^{\circ}}

UseCalc

Use a calculator

\frac{2}{R} = 0.4

MultEqn

$LHS \cdot R = RHS \cdot R$

2 = 0.4 R

DivEqn

$LHS / 0.4 = RHS / 0.4$

5 = R

RearrangeEqn

Rearrange equation

R = 5

The radius of the larger gear is

5

inches.

b As can be seen on the diagram, when Gear

S

makes two complete revolutions, the central angle measures

28 8^{\circ} .

Note that the arc created due to revolutions also measures

28 8^{\circ} .

In the previous part, the radius of

⊙ L

is found as

5

inches. Using the formula for the area of a sector, the sector of

⊙ L

can be calculated.

Area of Sector = \frac{θ}{3 6 0 ^{\circ}} \cdot π R^{2}

SubstituteII

$θ = 28 8^{\circ}$ , $R = 5$

Area of Sector = \frac{2 8 8 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (5)^{2}

▼

Evaluate right-hand side

UseCalc

Use a calculator

Area of Sector = 0.8 \cdot π (5)^{2}

CalcPow

Calculate power

Area of Sector = 0.8 \cdot π (25)

Multiply

Area of Sector = 20 π

The area of the sector of

⊙ L

20 π

square inches.

Example

Using Areas Of Sectors to Solve Problems

In his free time, Dylan enjoys making decorative figures by hand. He has $5$ identical sectors and brings these sectors together as shown.

Dylan knows that the area of each sector is $18$ square millimeters.

a Find the perimeter of the star-shaped figure. If necessary, round the answer to one decimal place.

b Find the perimeter of the figure. If necessary, round the answer to one decimal place.

Hint

a Notice that each side of the star-shaped figure is a radius.

b Start by finding the corresponding arc length of a sector.

Solution

a Notice that the sides of the star-shaped figure are the radii of the sectors.

The value of the radius

r

can be found using the formula for the area of a sector. Substitute

18

for the area of the sector

A,

and

10 8^{\circ}

for

θ

into the formula.

A = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

SubstituteII

$A = 18$ , $θ = 10 8^{\circ}$

18 = \frac{1 0 8 ^{\circ}}{3 6 0 ^{\circ}} \cdot π r^{2}

▼

Solve for

r

CalcQuot

Calculate quotient

18 = 0.3 \cdot π r^{2}

DivEqn

$LHS / 0.3 \cdot π = RHS / 0.3 \cdot π$

\frac{1 8}{0 . 3 \cdot π} = r^{2}

SqrtEqn

$LHS = RHS$

\frac{1 8}{0 . 3 \cdot π} = r

UseCalc

Use a calculator

4.370193 \dots = r

RearrangeEqn

Rearrange equation

r = 4.370193 \dots

RoundDec

Round to $1$ decimal place(s)

r \approx 4.4

Since

10

identical radii form the figure, the perimeter is

10 r .

r = 4.4 \Leftrightarrow 10 r = 44

The perimeter of the star-shaped figure is

44

millimeters.

b The perimeter of the figure created by Dylan is the sum of

5

identical arc lengths.

In Part A, the radii of the sectors were found to be about

4.4

millimeters. The measure of the arc is

10 8^{\circ}

because the corresponding central angle measures

10 8^{\circ} .

Now, the Arc Length Formula can be used.

Arc Length = \frac{θ}{3 6 0 ^{\circ}} \cdot 2 π r

SubstituteII

$θ = 10 8^{\circ}$ , $r = 4.4$

Arc Length = \frac{1 0 8 ^{\circ}}{3 6 0 ^{\circ}} \cdot 2 π (4.4)

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

Arc Length = 0.3 \cdot 2 π (4.4)

Multiply

Arc Length = 2.64 π

UseCalc

Use a calculator

Arc Length = 8.293804 \dots

RoundDec

Round to $1$ decimal place(s)

Arc Length \approx 8.3

There are

5

of these arc.

5 \times 8.3 \approx 41.5

Therefore, the figure has a perimeter of approximately

41.5

millimeters.

Closure

Solving Real Life Problems Using Sectors of Circles

Mark set up a lamp in his courtyard. He uses a light bulb that illuminates a circular area with a radius of $6$ meters. The diagram shows a bird's eye view of Mark's house.

If the measure of arc

M N

10 0^{\circ},

what is the area of the region that is illuminated outside of the courtyard area? If necessary, round the answer to two decimal places.

Hint

The area of a triangle is half the product of the lengths of any two sides and the sine of the included angle.

Solution

From the diagram, it can be seen that the region bounded by $\overline{M P},$ $\overline{N P},$ and $M N$ is a sector of $⊙ P .$

The region bounded by

\overline{M N}

and

M N

is called segment of the circle

P .

To find the area of the segment, the area of the triangle

M P N

should be subtracted from the area of the sector

M P N .

A_{segment} = A_{sector} - A_{triangle}

The area of

△ M P N

is half the product of the lengths of any two sides and the sine of the included angle.

A_{triangle} = \frac{1}{2} \cdot M P \cdot N P \cdot sin (θ)

Since

\overline{M P}

and

\overline{N P}

are radii of

⊙ P,

M P

and

N P

are

6

meters. Moreover, the included angle

M P N

measures

10 0^{\circ}

because it intercepts a

10 0^{\circ}

arc. Substitute these values.

A_{triangle} = \frac{1}{2} \cdot M P \cdot N P \cdot sin (θ)

SubstituteValues

Substitute values

A_{triangle} = \frac{1}{2} \cdot 6 \cdot 6 \cdot sin (10 0^{\circ})

▼

Evaluate right-hand side

Multiply

A_{triangle} = 18 sin (10 0^{\circ})

UseCalc

Use a calculator

A_{triangle} = 17.726539 \dots

RoundDec

Round to $2$ decimal place(s)

A_{triangle} \approx 17.73

Next, the area of the sector

M N P

will be calculated.

A_{sector} = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

SubstituteII

$r = 6$ , $θ = 10 0^{\circ}$

A_{sector} = \frac{1 0 0 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (6)^{2}

▼

Evaluate right-hand side

CalcPow

Calculate power

A_{sector} = \frac{1 0 0 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (36)

MoveRightFacToNum

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

A_{sector} = \frac{1 0 0 ^{\circ} \cdot 3 6}{3 6 0 ^{\circ}} \cdot π

Multiply

A_{sector} = \frac{3 6 0 0 ^{\circ}}{3 6 0 ^{\circ}} \cdot π

CalcQuot

Calculate quotient

A_{sector} = 10 \cdot π

UseCalc

Use a calculator

A_{sector} = 31.415926 \dots

RoundDec

Round to $2$ decimal place(s)

A_{sector} \approx 31.42

Finally, the area of the region

A_{R}

is the difference of

A_{S}

and

A_{T} .

A_{segment} = A_{sector} - A_{triangle}

SubstituteValues

Substitute values

A_{segment} \approx 31.42 - 17.73

SubTerm

Subtract term

A_{segment} \approx 13.69

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Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution