imc

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 $\boxed{\frac{28}{3}}$.