Area of a Sector

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 $4\text{-}$meter radius. as shown.

Tiffaniqua knows that $\angle ABC$ measures $135^{\circ }.$

Solution

a The sprinkler covers a $135\text{-}$degree sector of a circle with radius $4$ meters. The area of a sector of a circle can be calculated using the following formula.

$$\begin{array}{r}\text{Area of Sector}=\frac{\theta }{360^{\circ }}\cdot \pi r^2\end{array}$$

Substituting $135^{\circ }$ for $\theta$ and $4$ for $r$ into the formula will give the result.

$\text{Area of Sector}=\frac{\theta }{360^{\circ }}\cdot \pi r^2$

SubstituteII

$\theta =135^{\circ }$, $r=4$

$\text{Area of Sector}=\frac{135^{\circ }}{360^{\circ }}\cdot \pi (4{)}^2$

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

$\text{Area of Sector}=0.375\cdot \pi (4{)}^2$

CalcPow

Calculate power

$\text{Area of Sector}=0.375\cdot \pi 16$

Multiply

Multiply

$\text{Area of Sector}=6\pi$

UseCalc

Use a calculator

$\text{Area of Sector}=18.849555\dots$

RoundInt

Round to nearest integer

$\text{Area of Sector}\approx 19$

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.

$\text{Area of Trapezoid}=\frac{1}{2}h(b_1+b_2)$

SubstituteValues

Substitute values

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

▼

Evaluate right-hand side

AddTerms

Add terms

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

Multiply

Multiply

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

MoveRightFacToNumOne

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

$\text{Area of Trapezoid}=\frac{56}{2}$

CalcQuot

Calculate quotient

$\text{Area of Trapezoid}=28$

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.

$$\begin{array}{r}28-19=9\end{array}$$

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

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.

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.

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 $108^{\circ }$ for $\theta$ into the formula.

$A=\frac{\theta }{360^{\circ }}\cdot \pi r^2$

SubstituteII

$A=18$, $\theta =108^{\circ }$

$18=\frac{108^{\circ }}{360^{\circ }}\cdot \pi r^2$

▼

Solve for $r$

CalcQuot

Calculate quotient

$18=0.3\cdot \pi r^2$

DivEqn

$\text{LHS}/0.3\cdot \pi =\text{RHS}/0.3\cdot \pi$

$\frac{18}{0.3\cdot \pi }=r^2$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$\sqrt{\frac{18}{0.3\cdot \pi }}=r$

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 $10r.$

$$\begin{array}{r}r=4.4\iff 10r=44\end{array}$$

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 $108^{\circ }$ because the corresponding central angle measures $108^{\circ }.$ Now, the Arc Length Formula can be used.

$\text{Arc Length}=\frac{\theta }{360^{\circ }}\cdot 2\pi r$

SubstituteII

$\theta =108^{\circ }$, $r=4.4$

$\text{Arc Length}=\frac{108^{\circ }}{360^{\circ }}\cdot 2\pi (4.4)$

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

$\text{Arc Length}=0.3\cdot 2\pi (4.4)$

Multiply

Multiply

$\text{Arc Length}=2.64\pi$

UseCalc

Use a calculator

$\text{Arc Length}=8.293804\dots$

RoundDec

Round to $1$ decimal place(s)

$\text{Arc Length}\approx 8.3$

There are $5$ of these arc.

$$\begin{array}{r}5\times 8.3\approx 41.5\end{array}$$

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

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.

Solution

From the diagram, it can be seen that the region bounded by $\overline{MP},$ $\overline{NP},$ and $\overgroup{MN}$ is a sector of $\odot P.$

The region bounded by $\overline{MN}$ and $\overgroup{MN}$ 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 $MPN.$

$$\begin{array}{r}{A}_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}\end{array}$$

The area of $△MPN$ is half the product of the lengths of any two sides and the sine of the included angle.

$$\begin{array}{c}{A}_{\text{triangle}}=\frac{1}{2}\cdot MP\cdot NP\cdot \sin (\theta )\end{array}$$

Since $\overline{MP}$ and $\overline{NP}$ are radii of $\odot P,$ $MP$ and $NP$ are $6$ meters. Moreover, the included angle $MPN$ measures $100^{\circ }$ because it intercepts a $100^{\circ }$ arc. Substitute these values.

$A_{\text{triangle}}=\frac{1}{2}\cdot MP\cdot NP\cdot \sin (\theta )$

SubstituteValues

Substitute values

$A_{\text{triangle}}=\frac{1}{2}\cdot 6\cdot 6\cdot \sin (100^{\circ })$

▼

Evaluate right-hand side

Multiply

Multiply

$A_{\text{triangle}}=18\sin (100^{\circ })$

UseCalc

Use a calculator

$A_{\text{triangle}}=17.726539\dots$

RoundDec

Round to $2$ decimal place(s)

$A_{\text{triangle}}\approx 17.73$

Next, the area of the sector $MNP$ will be calculated.

$A_{\text{sector}}=\frac{\theta }{360^{\circ }}\cdot \pi r^2$

SubstituteII

$r=6$, $\theta =100^{\circ }$

$A_{\text{sector}}=\frac{100^{\circ }}{360^{\circ }}\cdot \pi (6{)}^2$

▼

Evaluate right-hand side

CalcPow

Calculate power

$A_{\text{sector}}=\frac{100^{\circ }}{360^{\circ }}\cdot \pi (36)$

MoveRightFacToNum

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

$A_{\text{sector}}=\frac{100^{\circ }\cdot 36}{360^{\circ }}\cdot \pi$

Multiply

Multiply

$A_{\text{sector}}=\frac{3600^{\circ }}{360^{\circ }}\cdot \pi$

CalcQuot

Calculate quotient

$A_{\text{sector}}=10\cdot \pi$

UseCalc

Use a calculator

$A_{\text{sector}}=31.415926\dots$

RoundDec

Round to $2$ decimal place(s)

$A_{\text{sector}}\approx 31.42$

Finally, the area of the region $A_R$ is the difference of $A_S$ and $A_T.$

$A_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}$

SubstituteValues

Substitute values

$A_{\text{segment}}\approx 31.42-17.73$

SubTerm

Subtract term

$A_{\text{segment}}\approx 13.69$