In rectangle $A B C D$, $A B=C D=2, B C=A D=$ 8, point $O$ is 1 unit away from $A B$, $B C$, and $A D$, and $O$ is inside the rectangle. If the rectangle is rotated $45^{\circ}$ about $O$, then the area of the overlapping part of the two rectangles is $\qquad$
$3.6 \sqrt{2}-4$.
As shown in Figure 5, the common part of the two figures after rotation is the hexagon $E F G H L S$, whose area can be obtained by $S_{\text {rhombus } A_{1} B_{1} S}-2 S_{\triangle M_{1} F G}$.

Connecting $O A_{1}$ intersects $A B$ at $K$, then
$$
\begin{array}{l}
O A_{1} \perp A B \text { and } F K=K A_{1}=K G=\sqrt{2}-1, \\
A_{1} F=\sqrt{2} A_{1} K=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}=A_{1} G .
\end{array}
$$

Note that it can be proven that
$\triangle A F E \cong \triangle A_{1} F G \cong \triangle B G H \cong \triangle B_{1} H L$, then $A_{1} E=A_{1} F+F E=2-\sqrt{2}+2(\sqrt{2}-1)=\sqrt{2}$,
$$
\begin{aligned}
B_{1} S & =B_{1} L+L S=B_{1} L+\sqrt{2} S M \\
& =2-\sqrt{2}+2 \sqrt{2}=2+\sqrt{2} .
\end{aligned}
$$
$$
=\frac{1}{2}(\sqrt{2}+2+\sqrt{2}) \times 2=2+2 \sqrt{2} \text {. }
$$

And $S_{\triangle A_{1} F G}=S_{\triangle H B_{1} L}=\frac{1}{2} A_{1} F^{2}=\frac{1}{2}(2-\sqrt{2})^{2}$
$$
=3-2 \sqrt{2} \text {, }
$$

Therefore, $S=(2+2 \sqrt{2})-2(3-2 \sqrt{2})=6 \sqrt{2}-4$.