Question

In rectangle $ABCD,$ $DC = 2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD?$

$\ \ \dfrac{\sqrt{3}}{6}$

$\ \dfrac{\sqrt{6}}{8}$

$\ \dfrac{3\sqrt{3}}{16}$

$\ \dfrac{1}{3}$

$\ \dfrac{\sqrt{2}}{4}$

Answer: A

Concepts:trigonometry area ratio rectangle

Difficulty rating: 1660

Solution:

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 $\dfrac{h^2\sqrt3/3}{2h^2}=\dfrac{\sqrt3}{6}.$ Thus, the correct answer is A.

2014 AMC 10B Problem 15

Answer: A

Difficulty rating: 1660

Solution:

Problem 15 in Other Years