The explanation on the screen is pretty straightforward.

What do you not understand?

My answer 36pi

I got 36pi as well. I calculated the area of the semicircle, which is 32. After that, we have to add the area of the shaded cemicircle below the line, and subtract the area of the two cemicircles above the line.

32pi + 8pi - 4pi = 36pi

I think the confusion lies in the question of is this thing a circle or not?

resilient wrote:I think the confusion lies in the question of is this thing a circle or not?

That's the real problem with this question.

On the actual GMAT, the question would clearly state that the big arc is a semi-circle. Without that explicit information, there's no way to solve the question.

Where is this question from?

This is from REA.

If they said it was a circle, I see to be getting an area of 144pi for the whole circle and 64 Pi for semi circle before factoring in the cut outs. I find the r to equal 12. Is that were I am off? Thanks.

If they said the thing was a circle, then the area of the largest semi circle is just 1/2 the area of the circle (if it is so). The area is calculated with pi*r^2.

So, the largest semi circle is 1/2 * pi * 8^2 = 32pi (r = 8 because the entire length, diameter, is 16)

Subtract two identical semi circles of radius 2 = 2 * 1/2 * pi * 2^2 = 4pi

Add the single semicircle extending below the line, area = 1/2 * pi * 4^2 = 8pi.

As other posters had mentioned the shaded region = 32pi - 4pi + 8pi = 36pi if it is a true circle.

How do you know that the diameter is 16?

BC = 8, and the other portions equal each other. I guess I am missing this point. Thanks.

"How do you know that the diameter is 16? "

AB = 4
BC = 8
CD = 4

4 + 8 + 4 = 16

Basically (if the question mentioned circles) ... this diagram is just a collection of semi-circles.
Start with the big semi-circle with the diameter of 16, and work your way from there.