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7. Area of a Sector
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Chapter 5
7. 

Area of a Sector

A sector in a circle is like a piece of pie, and understanding how much space it occupies is essential in various fields. The lesson elaborates on how to determine the area of these sectors. Knowing the area of a sector is particularly useful in designing objects like clocks, fans, and protractors. It also has practical applications in sectors like agriculture, where farmers might need to measure the area of a partial circular plot. By delving into the concept, one can grasp the nuances of spaces in circular forms, aiding in both academic pursuits and real-world problem-solving.
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8 Theory slides
11 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
Area of a Sector
Slide of 8
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

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.
Sector of a Circle
In the diagram, sector ACB is created by AC, BC, and AB.
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 360^(∘).


Area of Sector = θ/360^(∘) * π r^2

From the fact that 2πrad equals 360^(∘), an equivalent formula can be written if the central angle is given in radians.


Area of Sector &= θ/2π * π r^2 &⇓ Area of Sector &= θ/2* 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 = mAB/360^(∘) * π r^2
or
Area of Sector = mAB/2* r^2

Proof

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

sector of a circle

Since a circle measures 360^(∘), this sector represents θ360^(∘) of ⊙ C. Therefore, the ratio of the area of a sector to the area of the whole circle is proportional to θ360^(∘). Area of Sector/Area of Circle = θ/360^(∘)

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. Area of Sector/π r^2 = θ/360^(∘) Therefore, the area of a sector of a circle can be found by using the following formula.


Area of Sector = θ/360^(∘) * 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 ∠ 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.
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 = θ/360^(∘) * π r^2 Substituting 135^(∘) for θ and 4 for r into the formula will give the result.
Area of Sector = θ/360^(∘) * π r^2
Area of Sector = 135^(∘)/360^(∘) * π ( 4)^2
Evaluate right-hand side
Area of Sector = 0.375 * π ( 4)^2
Area of Sector = 0.375 * π 16
Area of Sector = 6π
Area of Sector = 18.849555 ...
Area of Sector ≈ 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.
Area of Sector
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 = 1/2h(b_1+b_2)
Area of Trapezoid = 1/2 4( 5+ 9)
Evaluate right-hand side
Area of Trapezoid = 1/2 4(14)
Area of Trapezoid = 1/256
Area of Trapezoid = 56/2
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.

Pac-Man
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 = θ/360^(∘)* π r^2 The area of the sector and the radius are given. Substitute these values into the formula.
Area of Sector = θ/360^(∘)* π r^2
66.5 = θ/360^(∘)* π ( 5)^2
Solve for θ
66.5 = θ/360^(∘)* π * 25
66.5 = θ/360^(∘)* 25π
66.5/25 π = θ/360^(∘)
0.846704 ... = θ/360^(∘)
304.813547 ... ^(∘) = θ
θ = 304.813547 ... ^(∘)
θ ≈ 305^(∘)
The corresponding central angle is about 305^(∘).
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.
Gears
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 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^(∘).
Furthermore, the length of the arc is equal to the circumference of ⊙ S. Recall that the circumference is given by the formula C= 2π r. Substituting r= 2 into the formula will give the circumference C_S of ⊙ S. C_S & = 2 π ( 2) & = 4 π 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.
Arc length/Circumference of ⊙ L= Arc measure/360^(∘)
4π/2π R= 144^(∘)/360^(∘)
Solve for R
2/R= 144^(∘)/360^(∘)
2/R= 0.4
2 = 0.4 R
5 = R
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 288^(∘).
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.
Area of Sector = θ/360^(∘)* π R^2
Area of Sector = 288^(∘)/360^(∘)* π ( 5)^2
Evaluate right-hand side
Area of Sector = 0.8 * π (5)^2
Area of Sector = 0.8 * π (25)
Area of Sector = 20 π
The area of the sector of ⊙ L is 20π 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.

Sectors

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.
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^(∘) for θ into the formula.
A = θ/360^(∘)* π r^2
18 = 108^(∘)/360^(∘)* π r^2
Solve for r
18 = 0.3 * π r^2
18/0.3 * π = r^2
sqrt(18/0.3 * π) = r
4.370193 ... = r
r= 4.370193 ...
r ≈ 4.4
Since 10 identical radii form the figure, the perimeter is 10r. 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.
Sectors
In Part A, the radii of the sectors were found to be about 4.4 millimeters. The measure of the arc is 108^(∘) because the corresponding central angle measures 108^(∘). Now, the Arc Length Formula can be used.
Arc Length = θ/360^(∘)* 2 π r
Arc Length = 108^(∘)/360^(∘)* 2 π ( 4.4)
Evaluate right-hand side
Arc Length = 0.3 * 2π (4.4)
Arc Length = 2.64 π
Arc Length = 8.293804 ...
Arc Length ≈ 8.3
There are 5 of these arc. 5 * 8.3 ≈ 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 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.

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 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) = 1/2* 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) = 1/2* MP * NP * sin(θ)
A_(triangle) = 1/2* 6 * 6 * sin( 100^(∘))
Evaluate right-hand side
A_(triangle) = 18 sin(100^(∘))
A_(triangle) = 17.726539 ...
A_(triangle) ≈ 17.73
Next, the area of the sector MNP will be calculated.
A_(sector) = θ/360^(∘)* π r^2
A_(sector) = 100^(∘)/360^(∘)* π ( 6)^2
Evaluate right-hand side
A_(sector)= 100^(∘)/360^(∘)* π (36)
A_(sector) = 100^(∘) * 36/360^(∘)* π
A_(sector)= 3600^(∘)/360^(∘) * π
A_(sector)= 10 * π
A_(sector) = 31.415926 ...
A_(sector) ≈ 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)
A_(segment) ≈ 31.42 - 17.73
A_(segment) ≈ 13.69



Area of a Sector
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