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In the shaded region above, \(\angle{KOL}=120^{\circ}\), and the area of the entire circle is \(A=144\pi\)...

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by swerve » Tue Oct 13, 2020 8:05 am
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In the shaded region above, \(\angle{KOL}=120^{\circ}\), and the area of the entire circle is \(A=144\pi\). The perimeter of the shaded region is

A. \(12 + 8\pi\)
B. \(12 + 16\pi\)
C. \(24 + 8\pi\)
D. \(24 + 16\pi\)
E. \(24 + 24\pi\)

The OA is C

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Re: In the shaded region above, \(\angle{KOL}=120^{\circ}\), and the area of the entire circle is \(A=144\pi\)...

by Scott@TargetTestPrep » Thu Oct 22, 2020 9:23 am
swerve wrote: ↑
Tue Oct 13, 2020 8:05 am
 c2_img2.png

In the shaded region above, \(\angle{KOL}=120^{\circ}\), and the area of the entire circle is \(A=144\pi\). The perimeter of the shaded region is

A. \(12 + 8\pi\)
B. \(12 + 16\pi\)
C. \(24 + 8\pi\)
D. \(24 + 16\pi\)
E. \(24 + 24\pi\)

The OA is C

Solution:

The perimeter of the shaded region is KO + OL + arc KL.

KO and OL are both radii of the circle, and arc KL is 1/3 of the circumference of the circle since 120 degrees is 1/3 of 360 degrees.

Since the area of the circle is 144π, the radius of the circle is 12, and hence the circumference of the circle is 2 x π x 12 = 24π. Therefore, the perimeter of the shaded region is:

12 + 12 + 1/3 x 24π = 24 + 8π

Answer: C

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