[ Rhombuses. Properties and characteristics ] $[$ Cosine Theorem $]$

The sides of the rhombus $E F G H$ are the hypotenuses of isosceles right triangles $E A F, F D G, G C H$, $H B E$, and all these triangles have common interior points with the rhombus $E F G H$. The sum of the areas of the quadrilateral $A B C D$ and the rhombus $E F G H$ is 12. Find $G H$.
Using the symmetries of the obtained figure with respect to the diagonals of the rhombus, prove that the quadrilateral $A B C D$ is a rectangle.

## Solution

The sides of the rhombus, and thus the triangles based on them, are symmetric with respect to each diagonal of the rhombus. Therefore, the quadrilateral $A B C D$ is a rectangle. If the side of the rhombus is $a$, and its acute angle is $\alpha$, then

$$
\angle D F A=45^{\circ}+45^{\circ}-\alpha=90^{\circ}-\alpha, \angle D C G=\left(180^{\circ}-\alpha\right)-45^{\circ}-45^{\circ}=90^{\circ}-\alpha
$$

Let's find the sides of the rectangle $A B C D$:

$$
\begin{aligned}
A D^{2}=A F^{2}+F D^{2}-2 A F \cdot F D \cdot \cos \angle D F A= & \left(\frac{a}{\sqrt{2}}\right)^{2}+\left(\frac{a}{\sqrt{2}}\right)^{2}-2 \cdot \frac{a}{\sqrt{2}} \cdot \frac{a}{\sqrt{2}} \cdot \cos \left(90^{\circ}-\alpha\right)= \\
& =a^{2}(1-\sin \alpha) \\
C D^{2}=D G^{2}+G C^{2}-2 D G \cdot G C \cdot \cos \angle D G C= & \left(\frac{a}{\sqrt{2}}\right)^{2}+\left(\frac{a}{\sqrt{2}}\right)^{2}-2 \cdot \frac{a}{\sqrt{2}} \cdot \frac{a}{\sqrt{2}} \cdot \cos \left(90^{\circ}-\alpha\right)= \\
& =a^{2}(1-\sin \alpha)
\end{aligned}
$$

Therefore,

$$
12=S_{\mathrm{ABCD}}+S_{\mathrm{EFGH}}=A D \cdot C D+a^{2} \sin \boldsymbol{\alpha}=a^{2}(1-\sin \boldsymbol{\alpha})+a^{2} \sin \boldsymbol{\alpha}=a^{2}
$$

Thus, $a=\sqrt{12}=2 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-28.jpg?height=946&width=523&top_left_y=-1&top_left_x=765)

## Answer

$2 \sqrt{3}$.