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 $ABCD?$ \t\t

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

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

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

$\ \dfrac{1}{3}$

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

Solution:

The area of $EFB$ is equal to $\dfrac{EF\cdot AD}2 .$

Similarly, The area of $ABCD$ is equal to $AB\cdot AD .$

Thus, their ratio is $\dfrac{EF}{2\cdot AD} = \dfrac{EF}{4\cdot AB}.$

Then, $\begin{align*} EF &= AF - AE \\&= AB \tan(\angle ADE) \\&\quad - AB \tan(\angle ADF) \\&= AB( \tan(60^\circ) - \tan(30^ \circ)) \\&= AB\left(\sqrt 3 - \dfrac {\sqrt{3}}3\right) \\&= AB \left(\dfrac{2\sqrt 3}3\right)\end{align*}$ This makes our result $\dfrac{EF}{4\cdot AB} = \dfrac{AB(\frac{2 \sqrt 3}3)}{4\cdot AB} = \dfrac {\sqrt 3}6.$

Thus, the correct answer is A .

2014 AMC 10B Problem 15

Solution:

Problem 15 in Other Years