3. As shown in Figure 1, in a regular hexagon $A B C D E F$ with side length 10, $H$ is the midpoint of side $D E$, and $G$ is a point on side $B C$ such that $\angle A G B = \angle C G H$. Then the area of pentagon $A F E H G$ is $\qquad$
3. $\frac{205 \sqrt{3}}{2}$.

As shown in Figure 3, draw a line through point $H$ parallel to $B E$, intersecting $G C$ at point $K$.
It is easy to see that $K H=15$,
$B K=K C=5$.
Let $B G=x$. Then
$$
G K=5-x \text {. }
$$

Since $\triangle K G H \backsim \triangle B G A$
$$
\begin{array}{l}
\Rightarrow \frac{K H}{A B}=\frac{G K}{B G} \Rightarrow \frac{15}{10}=\frac{5-x}{x} \\
\Rightarrow x=2, G K=3 .
\end{array}
$$

Then $S_{\text {hexagon }}=(10+20) \times \frac{\sqrt{3}}{2} \times 10=150 \sqrt{3}$,
$S_{\text {trapezoid } C K H D}=\frac{1}{2}(10+15) \times \frac{5 \sqrt{3}}{2}=\frac{125 \sqrt{3}}{4}$,
$S_{\triangle K G H}=\frac{1}{2} \times 3 \times 15 \times \frac{\sqrt{3}}{2}=\frac{45 \sqrt{3}}{4}$,
$S_{\triangle B G A}=\frac{1}{2} \times 2 \times 10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3}$.
Thus, $S_{\text {pentagon }}$. $A E H G$
$$
\begin{array}{l}
=150 \sqrt{3}-\frac{125 \sqrt{3}}{4}-\frac{45 \sqrt{3}}{4}-5 \sqrt{3} \\
=\frac{205 \sqrt{3}}{2} .
\end{array}
$$