Solution:

The equal length segments show that the side pieces near $B$ and $E$ are made from halves of equilateral triangles of side length $2$. An equilateral triangle of side length $2$ has altitude $\sqrt3$ and area $\sqrt3$, so the two side pieces together contribute area $2\sqrt3=\sqrt{12}$.

The remaining central triangle is $\triangle ACD$. From the same equilateral-triangle altitudes, $AC=AD=2\sqrt3=\sqrt{12}$, and $CD=2$. Its altitude to $CD$ is

$$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$$

Therefore the area of $\triangle ACD$ is $\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}$. The pentagon's total area is

$$\sqrt{12}+\sqrt{11},$$

so $m+n=12+11=23$.

Thus, the answer is D .