Problem Solving

In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC = 8, what is the area of square AFGE (area of the square of side s = s^2)?

$$A.\ 32\left(1-\sqrt{2}\right)$$
$$B.\ 32\left(3-2\sqrt{2}\right)$$
$$C.\ 64\left(\sqrt{2}-1\right)^2$$
$$D.\ 64-16\pi$$
$$E.\ 32-4\pi$$

The OA is B.

Please, can any expert explain this PS question for me? I tried to solve it in the following way,

I can get AC,
$$AC=\sqrt{8^2+8^2}=\sqrt{128}=8\sqrt{2}$$
$$Then,\ AG=AC-8=8\sqrt{2}-8=8\left(\sqrt{2}-1\right)$$
Now, how can I find the area of the square AFGE? I need your help. Thanks.

One formula for the area of the square is diagonal 1 * diagonal 2 / 2 because a square is a form of a rhombus.

I would not expect anyone to know that for the test. I can't help but wonder at the source of this question.