7. A. As shown in Figure 4, the side length of square $ABCD$ is $2 \sqrt{15}, E, F$ are the midpoints of sides $AB, BC$ respectively, $AF$ intersects $DE, DB$ at points $M, N$. Then the area of $\triangle DMN$ is $\qquad$.
7. A. 8 .

Connect $D F$. Let the side length of the square $A B C D$ be $2 a$. From the problem, we easily know
$$
\begin{array}{l}
\triangle B F N \backsim \triangle D A N \\
\Rightarrow \frac{A D}{B F}=\frac{A N}{N F}=\frac{D N}{B N}=\frac{2}{1} \\
\Rightarrow A N=2 N F \Rightarrow A N=\frac{2}{3} A F .
\end{array}
$$

In the right triangle $\triangle A B F$, given $A B=2 a, B F=a$, we have
$$
A F=\sqrt{A B^{2}+B F^{2}}=\sqrt{5} a \text {. }
$$

Thus, $\cos \angle B A F=\frac{A B}{A F}=\frac{2 \sqrt{5}}{5}$.
From the problem, we know
$$
\begin{array}{c}
\triangle A D E \cong \triangle B A F \Rightarrow \angle A E D=\angle A F B \\
\Rightarrow \angle A M E=180^{\circ}-\angle B A F-\angle A E D \\
=180^{\circ}-\angle B A F-\angle A F B=90^{\circ} .
\end{array}
$$

Therefore, $A M=A E \cos \angle B A F=\frac{2 \sqrt{5}}{5} a$,
$$
M N=A N-A M=\frac{2}{3} A F-A M=\frac{4 \sqrt{5}}{15} a \text {. }
$$

Thus, $\frac{S_{\triangle M N D}}{S_{\triangle A F D}}=\frac{M N}{A F}=\frac{4}{15}$.
Since $S_{\triangle A F D}=\frac{1}{2}(2 a)(2 a)=2 a^{2}$, then
$$
S_{\triangle M N D}=\frac{4}{15} S_{\triangle A F D}=\frac{8}{15} a^{2} .
$$

Given $a=\sqrt{15}$, we have $S_{\triangle M N D}=8$.