An $8$ by $2\sqrt{2}$ rectangle has the same center as a circle of radius $2$. The area of the region common to both the rectangle and the circle is
$\textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2$
We draw the diagram above. Dropping an altitude from the center of the circle to the top side of the rectangle, yields a segment of length $\sqrt{2}$. Since the hypotenuse is the radius of the circle, it has length 2 and by Pythagoras the other leg of the triangle also has length $\sqrt{2}$.  We deduce that the two triangles formed by two radii of circle $O$ and the segment of the rectangle are 45-45-90 triangles. 
The overlapping area consists of two sectors with central angle 90 and four 45-45-90 triangles with base $\sqrt{2}$ and height $\sqrt{2}$.  The area of the circle is $4\pi$ so the area of the sectors is $2 \cdot 4\pi/4 = 2\pi$.  The area of the triangles is  $4\cdot (\sqrt{2})^2 /2 = 4$.  The combined area is \[\boxed{\textbf{(D) }2\pi+4.}\]
--Solution by [TheMaskedMagician](https://artofproblemsolving.comhttp://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685)