Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
7. Areas of Circles and Sectors
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Exercise 43 Page 665

To find the area of a segment for a minor arc, draw radii to form a sector. The area of the segment equals the area of the sector minus the area of the triangle formed.

4.4m^2

Practice makes perfect

A part of a circle bounded by an arc and the segment joining its endpoints is a segment of a circle.

To find the area of a segment for a minor arc, we need to draw radii to form a sector. The area of the segment equals the area of the sector minus the area of the triangle formed.

For the given diagram, we will find the area of the sector and then the area of the triangle. Finally, we will find their difference to find the area of the segment.

Area of the Sector

The area of a sector of a circle is the product of the measure of the arc divided by 360 and the area of the circle.

With this in mind, let's consider the given diagram.

We can see that the radius of the circle is 7m. Also, we can see that the measure of the major arc is 300. Recall the arc addition postulate, which tells us that the measure of a minor arc is the measure of the related major arc subtracted from 360. Let's find the measure of the minor arc. 360 - 300= 60 Let's consider how these pieces of information fit in the diagram.

We have all the information we need to calculate the area of the sector. Let's do it!
Measure of the arc/360* π r^2
60/360 * π ( 7^2)
Simplify
1/6 * π (7^2)
1/6 * π (49)
1/6 * 49π
49π/6
The area of the sector is 49π6 m^2.

Area of the Triangle

The area of a triangle can be found by taking half the product of two side lengths and the sine of the included angle. With this in mind, let's consider the triangle in our diagram.

The length of both of the known sides is 7m and the included angle measures 60^(∘). This is enough information to find the area of the triangle.
1/2( 7)( 7)sin 60^(∘)
Simplify
1/2(49)sin 60^(∘)
49/2sin 60^(∘)

Replace sin 60^(∘) with sqrt(3)/2

49/2 * sqrt(3)/2
49sqrt(3)/4
49/4*sqrt(3)
12.25sqrt(3)
The area of the triangle is 12.25sqrt(3)m^2.

Area of the Segment

Finally, to obtain the area of the segment we need to subtract the area of the triangle from the area of the sector.
49π/6-12.25sqrt(3)
4.43871...
4.4
The area of the segment, to the nearest tenth, is 4.4m^2.