2014 AMC 10B Problem 15

Let $AD=h.$ Since $DC=AB=2h,$ the area of rectangle $ABCD$ is $2h^2.$ The rays $DE$ and $DF$ make angles of $60^\circ$ and $30^\circ,$ respectively, with $DC.$ Therefore, $$AE=h\tan30^\circ=\dfrac{h}{\sqrt3}$$ and $$AF=h\tan60^\circ=h\sqrt3.$$ Hence $$\begin{aligned}EF&=AF-AE\\&=h\sqrt3-\dfrac{h}{\sqrt3}\\&=\dfrac{2h\sqrt3}{3}.\end{aligned}$$ The area of $\triangle DEF$ is $$\dfrac12\cdot EF\cdot h=\dfrac{h^2\sqrt3}{3}.$$ The desired ratio is $$\dfrac{\frac{h^2\sqrt3}{3}}{2h^2}=\dfrac{\sqrt3}{6}.$$ Thus, the correct answer is A.