Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$? 
$\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad \textbf{(B)}\ \frac{12-5\sqrt3}{36} \qquad \textbf{(C)}\ \frac{\sqrt3}{18} \qquad \textbf{(D)}\ \frac{3-\sqrt3}{9} \qquad \textbf{(E)}\ \frac{\sqrt3}{12}$
The region is the shaded area:



We can find the area of the shaded region by subtracting the pentagon from the middle third of the square.  The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$.  The pentagon can be split into a rectangle and an equilateral triangle.
The base of the equilateral triangle is $\frac13$ and the height is $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$.  Thus, the area is $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$.
The base of the rectangle is $\frac13$ and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle.  This is:
$\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$
Therefore, the area of the shaded region is

$\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$