5. In the figure alongside, $A B C D$ is a square with side length $2 \, \text{cm}$ and the triangles $A B E$ and $C D F$ are equilateral. How many $\text{cm}^2$ is the area of the quadrilateral $F G E H$?
(A) $\frac{8 \sqrt{3}}{4}$
(B) $\frac{2 \sqrt{3}}{3}$
(C) $\frac{8 \sqrt{3}}{3}-4$
(D) $\frac{2 \sqrt{3}}{3}-1$
(E) $\frac{1}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_17_e21ecd5de9c42efa6d1eg-1.jpg?height=280&width=302&top_left_y=457&top_left_x=2415)
(5) The correct answer is $(C)$.

Let $M$ be the midpoint of $A B$, the measure (in $\mathrm{cm}$) of $E M$ is $\sqrt{3}$. Let $N$ be the midpoint of $G H$, the measure of $E N$ (which is the height of the equilateral triangle $E G H$) is $\sqrt{3}-1$. Therefore, the sides of the triangle $E G H$ measure $(\sqrt{3}-1) \cdot \frac{2}{\sqrt{3}}$. It follows that the area of $E G F H$ (which is twice that of $E G H$) is (in $\mathrm{cm}^{2}$) equal to $(\sqrt{3}-1)^{2} \cdot \frac{2}{\sqrt{3}}=\frac{8 \sqrt{3}}{3}-4$.

Problem proposed by Antonio Fanelli.