The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
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 $DE+EF=4+5=\boxed{\textbf{(C)}\ 9}$.