[Geometry]
Problem: Let $T=2 \sqrt{2}$. In a right triangle, one leg has length $T^{2}$ and the other leg is 2 less than the hypotenuse. Compute the triangle's perimeter.

Solution: Let $c$ be the length of the hypotenuse. Then, by the Pythagorean Theorem, $\left(T^{2}\right)^{2}+(c-2)^{2}=$ $c^{2} \Rightarrow c=\frac{T^{4}}{4}+1$. With $T=2 \sqrt{2}, T^{4}=64$, and $c=17$. So the triangle is a $8-15-17$ triangle with perimeter 40 .

Final answer: 40