Vertex $E$ of isosceles right triangle $△ABE$ is in the interior of rectangle $ABCD$ with side lengths $AB=1$ and $AD=2$. $\angle AEB=90^{\circ }$ and $AE=BE$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $△ABE$ whose distance from $\overline{AD}$ is between $\frac{1}{4}$ and $\frac{1}{2}$. What is the area of $R$?