Question

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?

$\dfrac{3}{2}$

$\sqrt 3$

$2\sqrt 3 - 1$

$8\sqrt 3 - 12$

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

Solution:

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 Problem 14

Solution:

Problem 14 in Other Years