Determine the area of the gray region in the image, given that the circles touch each other and have a radius of $1 \mathrm{~cm}$, and the segments that bound the region are their common tangents.

![](https://cdn.mathpix.com/cropped/2024_04_17_e907aea6a48a86a4c76eg-1.jpg?height=331&width=351&top_left_y=1305&top_left_x=887)

(P. Tlustý)
The area of the gray part is calculated as the difference between the area of the entire figure and the area of the white region (i.e., three circles). To determine the area of the figure, we will divide it into an equilateral triangle, whose vertices are the centers of the white circles (the side of this triangle is $2 \mathrm{~cm}$), three identical rectangles with sides measuring $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$, and finally, three identical circular sectors. The radius of these sectors is $1 \mathrm{~cm}$, and the central angle has a size of $360^{\circ}-2 \cdot 90^{\circ}-60^{\circ}=120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e907aea6a48a86a4c76eg-1.jpg?height=463&width=494&top_left_y=2073&top_left_x=812)

Area of the equilateral triangle: First, we calculate the height of this triangle using the Pythagorean theorem:

$$
v^{2}+1^{2}=2^{2},
$$

from which $v=\sqrt{3} \mathrm{~cm}$,

$$
S_{\triangle}=\frac{a \cdot v}{2}=\frac{2 \cdot \sqrt{3}}{2}=\sqrt{3} \mathrm{~cm}^{2} .
$$

Area of the rectangles:

$$
S_{\text {rect }}=3 \cdot(1 \cdot 2)=6 \mathrm{~cm}^{2} \text {. }
$$

Area of the sectors: Since the central angle of all three sectors is $120^{\circ}$, the area of these sectors together is equal to the area of one white circle.

Area of one white circle:

$$
S_{\circ}=\pi r^{2}=\pi \cdot 1^{2}=\pi \mathrm{cm}^{2} .
$$

Area of the gray region:

$$
S=\left(S_{\triangle}+S_{\text {rect }}+S_{\circ}\right)-3 \cdot S_{\circ}=(\sqrt{3}+6+\pi)-3 \cdot \pi=(6+\sqrt{3}-2 \pi) \mathrm{cm}^{2} .
$$