There are two natural ways to inscribe a square in a given isosceles right triangle. 
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$. 
What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?

$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$
We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$. 

We now have that $3S=42\sqrt{2}$, so $S=14\sqrt{2}$. But we want the area of the square which is $S^2=(14\sqrt{2})^2= \boxed{\mathrm{(B)}\ 392}$