I see that this is an isosceles right triangle, since $\angle Q = \angle R = 45^\circ$. That means that $PQ = PR = 6\sqrt{2}$. To find the area of a triangle, I need to multiply the base and the height and divide by 2. In this case, the base and the height are both $6\sqrt{2}$, so the area is $(6\sqrt{2})(6\sqrt{2})/2$. I can simplify this expression by using the property that $\sqrt{a}\sqrt{b} = \sqrt{ab}$. So, the area is $(\sqrt{2}\sqrt{2})(6\cdot 6)/2 = 2\cdot 36/2 = 36$. 
In right triangle $PQR$, we have $\angle Q = \angle R$ and $PR = 6\sqrt{2}$.  What is the area of $\triangle PQR$?