From three identical gray triangles and three identical squares, we have composed (without overlapping) the figure shown in the image. In its center, there is a triangular hole with an area of $2 \mathrm{~cm}^{2}$.

a) Determine the size of the interior angles of the gray triangle.

b) What is the area of the gray region in the image?

![](https://cdn.mathpix.com/cropped/2024_04_17_5c5418c5b56e6e441411g-1.jpg?height=442&width=511&top_left_y=767&top_left_x=801)
It is evident that the gray triangles are isosceles. The triangle in the middle is equilateral (each of its internal angles is $60^{\circ}$).

a) All internal angles in the square are $90^{\circ}$. We will add these details to the diagram.

![](https://cdn.mathpix.com/cropped/2024_04_17_5c5418c5b56e6e441411g-1.jpg?height=306&width=614&top_left_y=1452&top_left_x=755)

From this, it is easy to calculate that $\alpha=120^{\circ}$ and $\beta=30^{\circ}$.

b) If we "cut" the gray triangle into two equal parts and move one of them, we get a gray equilateral triangle (see the diagram).
![](https://cdn.mathpix.com/cropped/2024_04_17_5c5418c5b56e6e441411g-1.jpg?height=274&width=1076&top_left_y=1988&top_left_x=523)

The area of the gray triangle is the same as the area of the triangular hole in the middle. Therefore, the area of the gray region is $6 \mathrm{~cm}^{2}$.

[2 points for determining the angles, 4 points for determining the gray area]