9. (12 points) In the figure, quadrilaterals $A B C D$ and $E F G H$ are both squares, $\triangle A E H$, $\triangle B E F$, $\triangle C F G$, and $\triangle D H G$ are all equilateral triangles. The area of square $A B C D$ is 360. What is the area of trapezoid $B E H D$?
90
【Solution】Solution: Extend $B E$ to intersect $A H$ at $M$, let the side length of the square $E F G H$ be $a$. It is easy to know that $S_{\triangle A B E}=S_{\triangle A H D}=\frac{1}{2} \cdot a \cdot \frac{1}{2} a=\frac{1}{4} a^{2}$,
$\therefore S_{\triangle A B E}+S_{\triangle A D H}=\frac{1}{2} a^{2}=S_{\text {quadrilateral } E N K H}$,
$\because S_{\triangle E N B}+S_{\triangle D J K}=S_{\triangle A E H}$,
$\therefore S_{\text {quadrilateral } E B D H}=\frac{1}{2} S_{\triangle A B D}=\frac{1}{4} S_{\text {square } A B C D}=\frac{1}{4} \times 360=90$.
Therefore, the answer is 90.