The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$

$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$
Let $M$ be the midpoint of $CD$. Noting that $AED$ and $ABC$ are $120$ - $30$ - $30$ triangles because of the equilateral triangles, \[AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}.\] Also, $[AED]=2\cdot2\cdot\frac{1}{2}\cdot\sin{120^\circ}=\sqrt{3}$ (or split $\triangle AED$ into two $30-60-90$ triangles) and so \[[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}.\]