2014 AMC 10A Problem 16

We can find the area of the shaded region by finding the area of $\triangle DHC$ and subtracting out the two unshaded triangles. Extend $\overline{DH}$ so that it hits $B.$ Let the intersection of $\overline{DB}$ and $\overline{AF}$ be $X.$ We have that $\triangle DXF\sim\triangle BXA.$ Since $AB=2\cdot DF$, corresponding sides give $BX=2\cdot DX$. This means that $DX = \dfrac{1}{3} \cdot DB,$ which means that the altitude of $\triangle DXF$ is $\dfrac{1}{3}$ the height of the rectangle. The area of $\triangle DXF$ is then $$\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{6}.$$ The area of both unshaded triangles is then $2 \cdot \dfrac{1}{6} = \dfrac{1}{3}.$ The area of $\triangle DHC$ is $$\dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2}.$$ The area of the shaded region is then $\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}.$ Thus, E is the correct answer.