[Geometry]
Problem: In rectangle $P A U L$, point $D$ is the midpoint of $\overline{U L}$ and points $E$ and $F$ lie on $\overline{P L}$ and $\overline{P A}$, respectively such that $\frac{P E}{E L}=\frac{3}{2}$ and $\frac{P F}{F A}=2$. Given that $P A=36$ and $P L=25$, compute the area of pentagon $A U D E F$.

Solution: For convenience, let $P A=3 x$ and let $P L=5 y$. Then the given equations involving ratios of segment lengths imply that $P E=3 y, E L=2 y, P F=2 x$, and $F A=x$. Then $[P A U L]=(3 x)(5 y)=15 x y$ and

$$
\begin{aligned}
{[A U D E F] } & =[P A U L]-[P E F]-[E L D] \\
& =15 x y-\frac{1}{2}(3 y)(2 x)-\frac{1}{2}(2 y)\left(\frac{3 x}{2}\right) \\
& =15 x y-3 x y-\frac{3 x y}{2} \\
& =\frac{21 x y}{2} .
\end{aligned}
$$

Because $15 x y=36 \cdot 25$, it follows that $3 x y=36 \cdot 5=180$ and that $\frac{21 x y}{2}=\frac{7}{2}(3 x y)=\frac{7}{2} \cdot 180=\mathbf{6 3 0}$.

Final answer: 630