$21 \cdot 7$ In rectangle $A B C D$, $A B=C D=2, B C=A D=8$. There is a point $O$ inside the rectangle, which is 1 unit away from $A B, B C, A D$. If the rectangle $A B C D$ is rotated $45^{\circ}$ around point $O$, then the area of the overlapping part after rotation is
(A) $\frac{6 \sqrt{2}+1}{2}$.
(B) $2(3 \sqrt{2}-2)$.
(C) $4 \frac{3}{4}$.
(D) $4 \frac{1}{18}$.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational Competition, 1985)
[Solution] Let $A M=x$, according to the problem, $M N=\sqrt{2} x$, and $B N=x$. Therefore, $x+\sqrt{2} x+x=2$, then
$$
x=\frac{2}{2+\sqrt{2}} \text {. }
$$

In the right triangle $\triangle K F H$, $K F=\sqrt{2} F H=2 \sqrt{2}$.
$$
\begin{array}{l}
\text { Also, } S_{M N L K F E}=S_{A^{\prime} E F B^{\prime}}-S_{\triangle A^{\prime} M N}- \\
S_{\triangle L B^{\prime} K} \\
=\frac{1}{2}\left(A^{\prime} E+B^{\prime} F\right) \cdot A^{\prime} B^{\prime}- \\
\frac{1}{2} x^{2}-\frac{1}{2} x^{2} \\
=\frac{1}{2}\left[\frac{2(\sqrt{2}+1)}{2+\sqrt{2}}+\left(\frac{2}{2+\sqrt{2}}+2 \sqrt{2}\right)\right] \\
\text { - } 2-\left(\frac{2}{2+\sqrt{2}}\right)^{2} \\
=\frac{1}{2} \cdot(2+2 \sqrt{2}) \cdot 2-2(3-2 \sqrt{2}) \\
=2(3 \sqrt{2}-2) \text {. } \\
\end{array}
$$

Therefore, the answer is $(B)$.