Question

Given a circle with an 8" radius, find the area of the smaller segment whose chord is 8" long.

Answer 1

Find the area of the smaller segment whose chord is 8" long in a circle with an 8" radius.
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Since all sides are 8", they form an equilateral triangle.
The area of the sector is pi*r^2*60/360

Subtract the area of the triangle, that leaves the sector.
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Sector = pi*64*/6 = 32pi/3 sq inches
Area of triangle = 4*8*cos(30) = 16sqrt(3)
Area of segment = (32pi/3) - 16sqrt(3)