21 points and brainliest for correct answer!

the figures below are made out of circles, semicircles, quarter circles, and a square. find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations).

A = 5π/2 and P = 3π+2

The figure in the image has been constructed by piecing together two semicircles - one with a diameter of 2 in (AD), and one with a diameter of 4 in (AC). To get the perimeter of the big figure, we'll cut the perimeters of each circle in half and put them together, and we'll find the area in the same way.

π, the circle constant, is defined as the ratio between a circle's circumference (perimeter) and its diameter. In other words,

π = c / d

Multiplying both sides by the diameter d, we find that the circumference c is

c = dπ

Making the circumference of the smaller circle 2π in and the circumference of the larger circle 4π in. This isn't totally accurate, though. We're dealing with semicircles, not whole circles, so we need to cut each of those circumferences in half, getting π in for the small semicircle and 2π in for the large semicircle, giving the figure a perimeter of π + 2π = 3π in total.

Edit: I actually missed that last 2 inch segment connecting points D and C, so adding that in should give us a perimeter of 3π + 2 in, not 3π.

To find the area, we do the same thing. The area of any circle is πr², where r is the radius of the circle. We can find the radii of those semicircles by cutting their diameters in half, getting a radius of 1 in for the small one and 2 in for the larger one. Remember that we'll also have to cut whatever areas we get in half, since we're only dealing with half circles! Here are the calculations:

Small: (1/2)π(1)² = π/2 in²

Large: (1/2)π(2)² = (1/2)π(4) = 2π in²

Putting the two together:

A = 5π/2 and P = 3π+2

My answer is right, pick brainliest! The other one is wrong!! They are trying to trick you so be careful, kids! <33

Tysm! <3

its the fourth option

8

step-by-step explanation: