jmc

Notice that $AF+DE=BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\text{Area}=52=8\cdot 9-EF\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\boxed{9}$.