Question

点 $E$ 和 $F$ 位于正方形 $ABCD$ 上，使得 $\triangle BEF$ 是等边三角形。 $\triangle DEF$ 的面积与 $\triangle ABE$ 的面积之比是多少？

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 $\triangle ABE?$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\sqrt{3}$

$2$

$1 + \sqrt{3}$

答案：D

知识点：等边三角形面积比勾股定理

难度评级：1790

解答：

设正方形边长为 $1$ ， $ED = DF = x$ ， $AE = 1 - x$ 。

由于 $\triangle BEF$ 是等边三角形， $EF^2 = EB^2$ ，所以化简得 $x^2 = 2(1 - x)$ 。 $2x^2 = 1 + (1 - x)^2,$

此外 $[DEF] = \tfrac12 x^2$ ， $[ABE] = \tfrac12(1 - x)$ ，所以 $\begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned}$

所以正确答案是 D。

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 第 20 题

答案：D

难度评级：1790

解答：

其他年份的第 20 题