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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.**

Challenge

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

Concept

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

In the diagram, sector $ACB$ is created by $AC,BC,$ and $AB.$

Rule

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 $360_{∘}.$

$Area of Sector=360_{∘}θ ⋅πr_{2}$

From the fact that $2πrad$ equals $360_{∘},$ an equivalent formula can be written if the central angle is given in radians.

$Area of SectorArea of Sector =2πθ ⋅πr_{2}⇓=2θ ⋅r_{2} $

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

$Area of Sector=360_{∘}mAB ⋅πr_{2}$

or

$Area of Sector=2mAB ⋅r_{2}$

Consider sector $ACB$ bounded by $AC,BC,$ and $AB.$

Since a circle measures $360_{∘},$ this sector represents $360_{∘}θ $ of $⊙C.$ Therefore, the ratio of the area of a sector to the area of the whole circle is proportional to $360_{∘}θ .$$Area of CircleArea of Sector =360_{∘}θ $

Recall that the area of a circle is $πr_{2}.$ By substituting it into the equation and solving for the area of a sector, the desired formula can be obtained.
$πr_{2}Area of Sector =360_{∘}θ $

Therefore, the area of a sector of a circle can be found by using the following formula. $Area of Sector=360_{∘}θ ⋅πr_{2}$

Example

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.

Tiffaniqua knows that $∠ABC$ measures $135_{∘}.$

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

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

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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.

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=360_{∘}θ ⋅πr_{2} $

Substituting $135_{∘}$ for $θ$ and $4$ for $r$ into the formula will give the result.
$Area of Sector=360_{∘}θ ⋅πr_{2}$

SubstituteII

$θ=135_{∘}$, $r=4$

$Area of Sector=360_{∘}135_{∘} ⋅π(4)_{2}$

$Area of Sector≈19$

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.

$Area of Trapezoid=21 h(b_{1}+b_{2})$

SubstituteValues

Substitute values

$Area of Trapezoid=21 4(5+9)$

Evaluate right-hand side

AddTerms

Add terms

$Area of Trapezoid=21 4(14)$

Multiply

Multiply

$Area of Trapezoid=21 56$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$Area of Trapezoid=256 $

CalcQuot

Calculate quotient

$Area of Trapezoid=28$

$28−19=9 $

Tiffaniqua needs to water an area of $9$ square meters.
Example

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["305"]}}

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.

To find the measure of the corresponding central angle, the formula for the area of a sector will be used.
The corresponding central angle is about $305_{∘}.$

$Area of Sector=360_{∘}θ ⋅πr_{2} $

The area of the sector and the radius are given. Substitute these values into the formula.
$Area of Sector=360_{∘}θ ⋅πr_{2}$

SubstituteValues

Substitute values

$66.5=360_{∘}θ ⋅π(5)_{2}$

Solve for $θ$

CalcPow

Calculate power

$66.5=360_{∘}θ ⋅π⋅25$

Multiply

Multiply

$66.5=360_{∘}θ ⋅25π$

DivEqn

$LHS/25π=RHS/25π$

$25π66.5 =360_{∘}θ $

UseCalc

Use a calculator

$0.846704…=360_{∘}θ $

MultEqn

$LHS⋅360_{∘}=RHS⋅360_{∘}$

$304.813547…_{∘}=θ$

RearrangeEqn

Rearrange equation

$θ=304.813547…_{∘}$

RoundInt

Round to nearest integer

$θ≈305_{∘}$

Example

The diagram below models the motion of two gears $S$ and $L.$ Gear $S$ has a radius of $2$ inches.
### Hint

### Solution

Furthermore, the length of the arc is equal to the circumference of $⊙S.$ Recall that the circumference is given by the formula $C=2πr.$ Substituting $r=2$ into the formula will give the circumference $C_{S}$ of $⊙S.$
The radius of the larger gear is $5$ inches.
Note that the arc created due to revolutions also measures $288_{∘}.$ 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.
The area of the sector of $⊙L$ is $20π$ square inches.

a Find the radius of the larger gear.

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b Find the area of the sector of Gear $L$ formed when Gear $S$ completes two revolutions. Write the answer in terms of $π.$

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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.

a When the smaller gear completes a single revolution, the central angle measured in the larger gear becomes $144_{∘}.$ Since the measure of an arc is the same as its corresponding central angle, the measure of the colored arc of $⊙L$ is also $144_{∘}.$

$C_{S} =2π(2)=4π $

The length of the arc of $⊙L$ is, therefore, $4π$ inches. Now that the measure and length of the arc is known, the radius $R$ of $⊙L$ can be found. To do so, substitute the values into the formula for the arc length.
$Circumference of⊙LArc length =360_{∘}Arc measure $

SubstituteValues

Substitute values

$2πR4π =360_{∘}144_{∘} $

Solve for $R$

ReduceFrac

$ba =b/2πa/2π $

$R2 =360_{∘}144_{∘} $

UseCalc

Use a calculator

$R2 =0.4$

MultEqn

$LHS⋅R=RHS⋅R$

$2=0.4R$

DivEqn

$LHS/0.4=RHS/0.4$

$5=R$

RearrangeEqn

Rearrange equation

$R=5$

b As can be seen on the diagram, when Gear $S$ makes two complete revolutions, the central angle measures $288_{∘}.$

$Area of Sector=360_{∘}θ ⋅πR_{2}$

SubstituteII

$θ=288_{∘}$, $R=5$

$Area of Sector=360_{∘}288_{∘} ⋅π(5)_{2}$

$Area of Sector=20π$

Example

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.

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

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

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

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

$r=4.4⇔10r=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.

$Arc Length=360_{∘}θ ⋅2πr$

SubstituteII

$θ=108_{∘}$, $r=4.4$

$Arc Length=360_{∘}108_{∘} ⋅2π(4.4)$

$Arc Length≈8.3$

$5×8.3≈41.5 $

Therefore, the figure has a perimeter of approximately $41.5$ millimeters.
Closure

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 $MN$ is $100_{∘},$ what is the area of the region that is illuminated outside of the courtyard area? If necessary, round the answer to two decimal places.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["13.69"]}}

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

From the diagram, it can be seen that the region bounded by $MP,$ $NP,$ and $MN$ is a sector of $⊙P.$

The region bounded by $MN$ and $MN$ is called segment of the circle $P.$ To find the area of the segment, the area of the triangle $MPN$ should be subtracted from the area of the sector $MPN.$

$A_{segment}=A_{sector}−A_{triangle} $

The area of $△MPN$ is half the product of the lengths of any two sides and the sine of the included angle. $A_{triangle}=21 ⋅MP⋅NP⋅sin(θ) $

Since $MP$ and $NP$ are radii of $⊙P,$ $MP$ and $NP$ are $6$ meters. Moreover, the included angle $MPN$ measures $100_{∘}$ because it intercepts a $100_{∘}$ arc. Substitute these values.
$A_{triangle}=21 ⋅MP⋅NP⋅sin(θ)$

SubstituteValues

Substitute values

$A_{triangle}=21 ⋅6⋅6⋅sin(100_{∘})$

$A_{triangle}≈17.73$

$A_{sector}=360_{∘}θ ⋅πr_{2}$

SubstituteII

$r=6$, $θ=100_{∘}$

$A_{sector}=360_{∘}100_{∘} ⋅π(6)_{2}$

Evaluate right-hand side

$A_{sector}≈31.42$

$A_{segment}=A_{sector}−A_{triangle}$

SubstituteValues

Substitute values

$A_{segment}≈31.42−17.73$

SubTerm

Subtract term

$A_{segment}≈13.69$