Find the perimeter and area of this figure
it is made up of semicircles and quarter circles
use the actual symbol pi, do not simplify

A = (16π -32) in²

P = (4π +8√2) in

Step-by-step 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

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

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.

anser is a

step-by-step explanation:

negitive slope

answer:

your answer is: c. 196 7/8 in^3.