In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$? $\textbf{(A)}\ 25\qquad$ $\textbf{(B)}\ 32\qquad$ $\textbf{(C)}\ 36\qquad$ $\textbf{(D)}\ 40\qquad$ $\textbf{(E)}\ 42$
Solution
In the right triangle $BEC$ (right angle at $E$), $BE=1$ and $BC=\sqrt{50}$, so $CE=\sqrt{50-1}=7$ and $[BEC]=\frac12\cdot1\cdot7=\frac72$. The four right triangles around $EFGH$ are congruent, with total area $14$.
$\textbf{(A)}\ 25\qquad$ $\textbf{(B)}\ 32\qquad$ $\textbf{(C)}\ 36\qquad$ $\textbf{(D)}\ 40\qquad$ $\textbf{(E)}\ 42$ 
