20. Figure 1 is a shape formed by 11 line segments each of length 2, and the area of pentagon $A B C D E$ is $\sqrt{m}+\sqrt{n}\left(m, n \in \mathbf{Z}_{+}\right)$. Then $m+n=$ ( ).
(A) 20
(B) 21
(C) 22
(D) 23
(E) 24
20. D.

In $\triangle A B C$,
$$
A B=B C=2, \angle A B C=120^{\circ} \text {. }
$$

Then $A C=2 \sqrt{3}, \angle B A C=30^{\circ}$.
Similarly, $A D=2 \sqrt{3}, \angle D A E=30^{\circ}$.
Thus, $\angle C A D=30^{\circ}$.
Therefore, $S_{\text {pentagon } A B C D E}=S_{\triangle A B C}+S_{\triangle A C D}+S_{\triangle A D E}$
$$
\begin{array}{l}
=2 \times \frac{\sqrt{3}}{4} \times 2^{2}+\sqrt{(2 \sqrt{3}+1) \times 1 \times 1 \times(2 \sqrt{3}-1)} \\
=2 \sqrt{3}+\sqrt{11} .
\end{array}
$$

Therefore, $m+n=12+11=23$.