Question

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE?$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\sqrt{3}$

$2$

$1 + \sqrt{3}$

Answer: D

Concepts:equilateral triangle area ratio Pythagorean Theorem

Difficulty rating: 1790

Solution:

Let the square have side $1,$ and by symmetry let $ED = DF = x,$ so $AE = 1 - x.$

Since $\triangle BEF$ is equilateral, $EF^2 = EB^2,$ giving $2x^2 = 1 + (1 - x)^2,$ which simplifies to $x^2 = 2(1 - x).$

The right triangles have areas $[DEF] = \tfrac12 x^2$ and $[ABE] = \tfrac12(1 - x),$ so $\begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned}$

Thus, the correct answer is D.

2004 AMC 10A Problem 20

Answer: D

Difficulty rating: 1790

Solution:

Problem 20 in Other Years