Answer:
((160/3)π + 16√3) in²
Step-by-step explanation:
Together with the radii of 8 inches, the 8-inch chord creates an equilateral triangle with a central angle of 60°, or π/3 radians.
In problem 1, you found the area of the smaller segment to be ...
A = (1/2)r²·(θ -sin(θ))
For θ = π/3 and r = 8 in, this is (1/2)(8 in)²(π/3 -(√3)/2) = ((32/3)π -16√3) in².
The remaining segment of the circle is the area of the circle less this amount, so is ...
π·(8 in)² -(32/3π -16√3) in² = (160/3π +16√3) in² . . . . larger segment area
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