In right triangle $PQR$, we have $\angle Q = \angle R$ and $QR = 6\sqrt{2}$.  What is the area of $\triangle PQR$?
I recognize that this is an isosceles right triangle, since the two acute angles are congruent. That means the two legs are also congruent, so I can let $PQ = PR = x$. I can use the Pythagorean theorem to relate $x$ and $QR$: $x^2 + x^2 = (6\sqrt{2})^2$. Simplifying, I get $2x^2 = 72$, so $x^2 = 36$, and $x = 6$. Now I have the base and the height of the triangle, so I can find the area by multiplying them and dividing by 2: $A = \frac{1}{2} \cdot 6 \cdot 6 = 18$. 