4. The side length of square $A B C D$ is $2 \sqrt{15}, E, F$ are the midpoints of $A B, B C$ respectively, $A F$ intersects $D E, D B$ at $M, N$. Then the area of $\triangle D M N$ is
(A) 8
(B) 9
(C) 10
(D) 11
4. (A).

It is easy to know that $\triangle A M E \backsim \triangle A B F$, thus,
$$
S_{\triangle A M E}=\left(\frac{A E}{A F}\right)^{2} S_{\triangle A B F}=\frac{1}{5} S_{\triangle A B F}
$$
Since $N$ is the centroid of $\triangle A B C$,
$\therefore S_{\triangle B F N}=\frac{1}{3} S_{\triangle A B F}$.
Since $S_{\triangle D E B}=S_{\text {A.1PY }}$.
$$
\begin{aligned}
\therefore S_{\triangle A K N} & =S_{\triangle A M K}-S_{\triangle B F N}=\frac{8}{15} S_{\triangle A B F} \\
& =\frac{8}{15} \times \frac{1}{2} \times 2 \sqrt{15} \times \frac{2 \sqrt{15}}{2}=8 .
\end{aligned}
$$