Punched!

We can find the area by finding the area of the semi-circle and subtracting the area of triangle $DEG$ .

The area of the semi-circle is just $\frac{1}{2}\pi r^2=\frac{1}{2}\pi(4)^2=8\pi$ .

$\overline{EG}$ is a diameter of the circle, so $EG=8$ . Since triangle $DEG$ is a $45^\circ-45^\circ-90^\circ$ triangle, $ED=DG=4\sqrt{2}$ . Thus, the area of triangle $DEG$ is $\frac{1}{2} (4\sqrt{2})^2=16$ .

Thus, our desired area is $\boxed{8\pi-16}$

EG is the diameter of the circle as ang'DEG is 90 degrees DF & EG being diameters intersect at the center. BY completing tri angle DEG we get an isoceles one with base angles =45 degrees using this find ED =DG { ED cos45= 4} find area of triangle DEG SUBTRACT IT FROM THE AREA OF THE SEMICIRCLE AND there is the ans.......

Let's call the area we are evaluating "area $A$ " which is composed of equal areas $A_1$ under the square, and $A_2$ to the left of the square. Let's also draw $\overline{EF},\ \overline{FG}$ forming areas $A_3$ above and $A_4$ to the right. Finally, let's define area $B = 2A = A_1 + A_2 +A_3 + A_4,$ area $C = \pi r^2,$ and area $D = s^2,$ where $s$ is the side length of square $DEFG.$ $A = \frac{1}{2}B \\ B = C - D = 16\pi - (\frac{8}{\sqrt{2}})^2 \\ B = 16\pi - 32 \\ \\ A = \boxed{8\pi - 16}$