Answer:
A = (16π -32) in²
P = (4π +8√2) in
Explanation:
The area is that of a quarter-circle of radius 8 inches less half the area of a square with side length 8 inches. Two formulas are useful:
area of a circle = πr² . . . . .r = radius
area of a square = s² . . . . s = side length
Then your area is ...
A = (1/4)π(8 in)² - (1/2)(8 in)² = (64 in²)(π/4 -1/2)
A = (16π -32) in²
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The applicable formulas for the side lengths of your figure are ...
arc BD = (1/4)(2πr) = π(r/2) = π(8 in)/2 = 4π in
segment BD = (8 in)√2
The perimeter is the sum of these lengths, so is ...
P = (4π +8√2) in
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Of course, you are very familiar with the fact that an isosceles right triangle with side lengths 1 has a hypotenuse of length √(1²+1²) = √2. Scaling the triangle by a factor of 8 inches means the segment AB will be 8√2 inches long.