3. As shown in the figure, 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 $\qquad$ .
Ronggui $\frac{205 \sqrt{3}}{2}$.
Analysis As shown in the analysis diagram, 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 $G K=5-x$. By
$$
\triangle K G H \sim \triangle B G A \Rightarrow \frac{K H}{A B}=\frac{G K}{B G} \Rightarrow \frac{15}{10}=\frac{5-x}{x} \Rightarrow x=2, G K=3
$$

Then
$$
\begin{array}{c}
S_{\text {regular hexagon }}=(10+20) \times \frac{\sqrt{3}}{2} \times 10=150 \sqrt{3}, \quad S_{\text {trapezoid CKHD }}=\frac{1}{2}(10+15) \times \frac{5 \sqrt{3}}{2}=\frac{125 \sqrt{3}}{4} \\
S_{\triangle K G H}=\frac{1}{2} \times 3 \times 15 \times \frac{\sqrt{3}}{2}=\frac{45 \sqrt{3}}{4}, \quad S_{\triangle B G A}=\frac{1}{2} \times 2 \times 10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3}
\end{array}
$$

Therefore
$$
S_{\text {pentagon } A F E H G}=150 \sqrt{3}-\frac{125 \sqrt{3}}{4}-\frac{45 \sqrt{3}}{4}-5 \sqrt{3}=\frac{205 \sqrt{3}}{2}
$$