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 $m + n ?$ $\textbf{(A)} ~20 \qquad$ $\textbf{(B)} ~21 \qquad$ $\textbf{(C)} ~22 \qquad$ $\textbf{(D)} ~23 \qquad$ $\textbf{(E)} ~24$
Solution
Connect $AC, AD$. Triangles $ABC$ and $AED$ have two sides of length $2$ with included angle $120^\circ$ (the figure is built from equilateral triangles), so $[AED]=[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3$, and $AC=AD=2\sqrt3$.
$\triangle ACD$ has sides $2, 2\sqrt3, 2\sqrt3$; its height to $CD$ is $\sqrt{12-1}=\sqrt{11}$, so its area is $\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}$.
So the area is $\sqrt3+\sqrt3+\sqrt{11}=2\sqrt3+\sqrt{11}=\sqrt{12}+\sqrt{11}$.
The figure is constructed from $11$ line segments, each of which has length $2$ except for $CD=4$, as shown in the figure. The area of pentagon $ABCDE$ can be written as $\sqrt m+\sqrt n$, where $m,n$ are positive integers. What is $m+n$?
$\textbf{(A)} ~20 \qquad$ $\textbf{(B)} ~21 \qquad$ $\textbf{(C)} ~22 \qquad$ $\textbf{(D)} ~23 \qquad$ $\textbf{(E)} ~24$ 
