Question

find the area of the smaller segment whose chord is 8" long in a circle with and 8" radius

Answer 1

find the area of the smaller segment whose chord is $8''$ long in a circle with and $8''$ radius

Solution. Let the chord is AB =8" and AC = CB= 8" are radiuses. As AB=AC=CB=8" so ABC is equilateral triangle with all angles equal 60°.ACB=60°=π/3

$S = \frac{1}{2} AC^2 (ACB - \sin ACB) = 32\left(\frac{\pi}{3} -\sin \frac{\pi}{3}\right) = 5.79$ square inch

Answer: area of segment is 5.79 square inch