We see that side $BE$, which we know is $1$, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$. Then $HB = HE + BE = HE + 1$, and $HE$ is one of the sides of the square whose area we want to find. So:
\[1^2 + (HE+1)^2=\sqrt{50}^2\]
\[1 + (HE+1)^2=50\]
\[(HE+1)^2=49\]
\[HE+1=7\]
\[HE=6\] 
So, the area of the square is $6^2=\boxed{\textbf{(C) }36}$
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 $EFGH$?
[AMC102005Aq.png](https://artofproblemsolving.com/wiki/index.php/File:AMC102005Aq.png)
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$