Vertex $E$ of equilateral triangle $\triangle ABE$ 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 $\overline{AD}$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$? $\textbf{(A)}\ \frac{12-5\sqrt{3}}{72} \qquad$ $ \textbf{(B)}\ \frac{12-5\sqrt{3}}{36} \qquad$ $ \textbf{(C)}\ \frac{\sqrt{3}}{18} \qquad$ $ \textbf{(D)}\ \frac{3-\sqrt{3}}{9} \qquad$ $ \textbf{(E)}\ \frac{\sqrt{3}}{12}$
Solution
The strip has area $\frac13$. The portion inside $\triangle ABE$ consists of trapezoid $HFGJ$, with area $\frac{\sqrt3}{9}$, and triangle $EFG$, with area $\frac{\sqrt3}{36}$.
Thus the shaded area is $\frac13-\frac{5\sqrt3}{36}=\boxed{\frac{12-5\sqrt3}{36}}$.
Vertex $E$ of equilateral triangle $\triangle ABE$ 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 $\overline{AD}$ is between $\frac14$ and $\frac34$. What is the area of $R$?
