[Solution] The four parts where the circle overlaps with the rectangle are shown in the figure. Since $2^{2}+2^{2}=(2 \sqrt{2})^{2}$, we know that the two chords $A B \perp C D$, thus
$$
\begin{aligned}
& S_{I}+S_{\mathbb{I}}=\frac{1}{2} S_{\odot 0}=2 \pi . \\
\text { Also } & S_{\mathbb{I}}=S_{\mathbb{I}}=\frac{1}{2} \times 2 \times 2=2, \\
\therefore \quad & S_{\mathrm{I}}+S_{\llbracket}+S_{\mathbb{I}}+S_{\mathbb{I}}=2 \pi+4 .
\end{aligned}
$$

Therefore, the answer is $(D)$.
20.92 A rectangle of size $8 \times 2 \sqrt{2}$ and a circle of radius 2 have a common center. The area of the overlapping part of the two figures is
(A) $2 \pi$.
(B) $2 \pi+2$.
(C) $4 \pi-4$.
(D) $2 \pi+4$.
(E) $4 \pi-2$.
(45th American High School Mathematics Examination, 1994)