Question

两个等边三角形位于一个边长为 $2\sqrt 3$ 的正方形内。这两个三角形的底边分别是正方形的一对对边，它们的交集是一个菱形。这个菱形的面积是多少？

Two equilateral triangles are contained in a square whose side length is $2\sqrt 3.$ The bases of these triangles are opposite sides of the square, and their intersection is a rhombus. What is the area of the rhombus?

$\dfrac{3}{2}$

$\sqrt 3$

$2\sqrt 3 - 1$

$8\sqrt 3 - 12$

$\dfrac{4\sqrt 3}{3}$

答案：D

知识点：等边三角形菱形特殊直角三角形

难度评级：1670

解答：

这个菱形可分成两个全等的等边三角形。设每个小等边三角形的边长为 $s$ ，则一个面积为 $\dfrac{s^2 \sqrt 3}4$ ，两个总面积为 $\dfrac{s^2 \sqrt 3}2$ 。

大等边三角形边长为 $2 \sqrt 3$ ，其高为 $2 \sqrt 3 \sin(60^\circ)$ ，也就是 $3$ 。

正方形半边长为 $\sqrt 3$ ，所以角上被截出的小等边三角形高为 $3 - \sqrt 3$ 。因此 $s$ 与 $2 \sqrt 3$ 的比为 $\dfrac {3-\sqrt 3}3$ 。

因此 $s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2 .$

所以菱形面积为 $\begin{align*} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{align*}$

所以正确答案是 D。

This rhombus is created by placing two congruent equilateral triangles. Let the side length of it be $s.$ Then, the area of one of them is $\dfrac{s^2 \sqrt 3}4,$ making the total area $\dfrac{s^2 \sqrt 3}2.$

The side length of the larger equilateral triangle is $2 \sqrt 3.$ The height of it is $3$ since the height is equal to $2 \sqrt 3 \sin(60^\circ) .$

Half of the square is $\sqrt 3,$ so the height of the smaller triangle is $3 - \sqrt 3 .$ Thus, the ratio between $s$ and $2 \sqrt 3$ is $\dfrac {3-\sqrt 3}3 .$

As such, $s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2 .$

Therefore, the combined area is $\begin{align*} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{align*}$

Thus, the correct answer is D .

2012 AMC 10B 第 14 题

答案：D

难度评级：1670

解答：

其他年份的第 14 题