2013 Gauss 8 Problem 22

In right-angled, isosceles triangle FGH, FH = \sqrt{8}. Arc FH is part of the circumference of a circle with centre G and radius GH, as shown. The area of the shaded region is

\textbf{(A)}\ \pi - 2\quad \textbf{(B)}\ 4\pi - 2\quad \textbf{(C)}\ 4\pi - \frac{1}{2}\sqrt{8}\quad \textbf{(D)}\ 4\pi - 4\quad \textbf{(E)}\ \pi - \sqrt{8}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Gauss contests here.