Question 901283
Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius.
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Find 1/2 the angle at the center of the right triangle with hypotenuse = 8 and side = 4 --> 30 degrees.
Area of the sector = Area of the circle/6 = CA/6 = pi*r^2/6 = 32pi/3
Area of the 2 30-60-90 triangles = 2*bh/2 = 4*8*cos(30) = 16sqrt(3)
Subtract the area of the 2 right triangles for CA/6
--> 32pi/3 - 16sqrt(3)  [Area of small segment]
Large segment area = 64pi - small area
= 160pi/3 + 16sqrt(3) sq units