9. As shown in Figure 3, the center of the equilateral triangle $\triangle A_{1} B_{1} C_{1}$ with side length 1 is $O$. Rotate $\triangle A_{1} B_{1} C_{1}$ around the center $O$ to $\triangle A_{2} B_{2} C_{2}$ such that $A_{2} B_{2} \perp B_{1} C_{1}$. Then the area of the common part (i.e., the hexagon $A B C D E F$) of the two triangles is
9. $\frac{3}{4}-\frac{\sqrt{3}}{4}$.

As shown in Figure 8, connect
$$
O B_{1}, O B_{2}, B_{1} B_{2} \text{. }
$$

Then
$$
\begin{array}{l}
O B_{1}=O B_{2}, \\
\angle O B_{1} B_{2} \\
=\angle O B_{2} B_{1} .
\end{array}
$$

Also, $\angle O B_{1} C=30^{\circ}=\angle O B_{2} C$, so,
$$
\angle C B_{1} B_{2}=\angle C B_{2} B_{1} \text{. }
$$

Therefore, $C B_{1}=C B_{2}$.
Similarly, $B_{2} D=D C_{1}$.
Let $C B_{1}=x$. Then
$$
C B_{2}=x, C D=\sqrt{3} x, D C_{1}=D B_{2}=2 x \text{. }
$$

Thus, $x+\sqrt{3} x+2 x=1 \Rightarrow x=\frac{1}{3+\sqrt{3}}$.
Hence, $S_{\text {hexagon } A B C D E F}=S_{\triangle \mathrm{H}_{2} B_{2} c_{2}}-3 S_{\triangle B_{2} C D}$
$$
=\frac{\sqrt{3}}{4}-3 \times \frac{1}{2} x \times \sqrt{3} x=\frac{\sqrt{3}}{4}-\frac{3 \sqrt{3}}{2} x^{2}=\frac{3}{4}-\frac{\sqrt{3}}{4} .
$$