In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB=32$, $BC=15$, $CD=18$, and $DA=13$. What is the area of $ABCD$?

1. Drop perpendiculars from $D$ and $C$ to $\overline{AB}$, with horizontal legs $x$ (under $DA=13$) and $y$ (under $BC=15$). Then $x+y=32-18=14$ and $y^2-x^2=15^2-13^2=56$, so $y-x=4$, $x=5$, and the height is $\sqrt{13^2-5^2}=12$.
2. The trapezoid's area is $\dfrac{AB+CD}{2}\cdot12=\dfrac{32+18}{2}\cdot12=\boxed{300}$.