Track / Stage 4 / 121 of 340 #381 of 1964

Problem 381

AMC 12 late, AIME early

Geometry Difficulty 4.7 Multiple choice

6. As shown in Figure 2, in the right trapezoid $A B C D$ , $\angle B=$ $\angle C=90^{\circ}, A B=B C$ , point $E$ is on side $B C$ , and makes $\triangle A D E$ an equilateral triangle. Then the ratio of the area of $\triangle A D E$ to the area of trapezoid $A B C D$ is:

Pick one

Official solution

6. D.

As shown in Figure 6, extend trapezoid $ABCD$ to form square $ABCF$ , with side length 1, and let $CD = CE = x$ . Then,
$BE = FD = 1 - x.$

By the Pythagorean theorem, we have $2x^2 = 1 + (1 - x)^2$ .
Solving for $x$ gives $x = \sqrt{3} - 1$ .
$\text{Therefore, } \frac{S_{\triangle ADE}}{S_{\text{trapezoid } ABCD}} = \frac{\frac{\sqrt{3}}{4} \times 2x^2}{\frac{1}{2}(1 + x) \times 1} = (\sqrt{3} - 1)^2.$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.