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 $R$?

1. The vertical strip has area $\frac12$. The part of $\triangle ABE$ in the strip is the whole triangle less two corner triangles of area $\frac{\sqrt3}{32}$ each, so it has area $\frac{3\sqrt3}{16}$.
2. Therefore, $[R]=\frac12-\frac{3\sqrt3}{16}=\boxed{\frac{8-3\sqrt3}{16}}$.