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Dominik [7]
3 years ago
5

Express answer in exact form. Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A

chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.)

Mathematics
1 answer:
Advocard [28]3 years ago
4 0

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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