Right triangles $T_1$ and $T_2$, have areas of 1 and 2, respectively. A side of $T_1$ is congruent to a side of $T_2$, and a different side of $T_1$ is congruent to a different side of $T_2$. What is the square of the product of the lengths of the other (third) sides of $T_1$ and $T_2$?
$\textbf{(A) }\frac{28}{3}\qquad\textbf{(B) }10\qquad\textbf{(C) }\frac{32}{3}\qquad\textbf{(D) }\frac{34}{3}\qquad\textbf{(E) }12$
First of all, let the two sides which are congruent be $x$ and $y$, where $y > x$. The only way that the conditions of the problem can be satisfied is if $x$ is the shorter leg of $T_{2}$ and the longer leg of $T_{1}$, and $y$ is the longer leg of $T_{2}$ and the hypotenuse of $T_{1}$.
Notice that this means the value we are looking for is the square of $\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}$, which is just $y^{4}-x^{4}$.
The area conditions give us two equations: $\frac{xy}{2} = 2$ and $\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1$.
This means that $y = \frac{4}{x}$ and that $\frac{4}{x^{2}} = y^{2} - x^{2}$.
Taking the second equation, we get $x^{2}y^{2} - x^{4} = 4$, so since $xy = 4$, $x^{4} = 12$.
Since $y = \frac{4}{x}$, we get $y^{4} = \frac{256}{12} = \frac{64}{3}$.
The value we are looking for is just $y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}$ so the answer is $\boxed{\textbf{(A) }\frac{28}{3}}$.