3. As shown in Figure 1, the side length of square $A B C D$ is $10, M$ and $N$ are the midpoints of sides $A B$ and $A D$ respectively. Find the area of the shaded region in the figure.
3. The area of the shaded part in the figure is 35.

Let $C N$ and $D M$ intersect at point $G$.
It is easy to prove $\triangle C D N \cong \triangle D A M$
$\Rightarrow \triangle D N G \backsim \triangle D M A$.
Using the Pythagorean theorem, we know
$$
\begin{array}{l}
\frac{S_{\triangle D N G}}{S_{\triangle D M A}}=\frac{D N^{2}}{D M^{2}}=\frac{1}{1+4}=\frac{1}{5} \\
\Rightarrow S_{\triangle D N G}=\frac{1}{5} S_{\triangle D M A}=\frac{1}{5} \times \frac{1}{2} \times 5 \times 10=5 .
\end{array}
$$

Given $S_{\triangle C D N}=S_{\triangle D A M}=25$,
$$
S_{\triangle C D M}=\frac{1}{2} \times 10 \times 10=50 \text {, }
$$

then $S_{\triangle C G M}=S_{\triangle C D M}+S_{\triangle D N G}-S_{\triangle C D N}$
$$
\begin{array}{l}
=50+5-25=30 \\
\Rightarrow S_{\text {shaded }}=5+30=35 .
\end{array}
$$