The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\triangle ABG$ and subtract this from $200$, the total area of the two squares.

Since $G$ is the center of $ABCD$, $BG$ is half of the diagonal of the square. The diagonal of $ABCD$ is $10\sqrt{2}$ so $BG=5\sqrt{2}$. Since $EFGH$ is a square, $\angle G=90^\circ$. So $\triangle ABG$ is an isosceles right triangle. Its area is $\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25$. Therefore, the area of the region is $200-25=\boxed{\textbf{(E) }175.}$
--Solution by TheMaskedMagician