Answer: $P=44 \pi \sqrt{2-\sqrt{3}}$

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_25c5d6100f66704b4a85g-2.jpg?height=245&width=651&top_left_y=757&top_left_x=768)

The circles intersect since $r>1$. In the figure, three of the thirteen circles are depicted.

$$
A C=2, A B=B C=r=2 \sqrt{2-\sqrt{3}}, \square B A E=\alpha, \cos \alpha=\frac{A E}{A B}=\frac{1}{2 \sqrt{2-\sqrt{3}}}=\frac{\sqrt{2+\sqrt{3}}}{2}
$$

We will calculate $\cos 2 \alpha$:

$$
\sin ^{2} \alpha=1-\cos ^{2} \alpha=1-\frac{2+\sqrt{3}}{4}=\frac{2-\sqrt{3}}{4} \rightarrow \cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=\frac{\sqrt{3}}{2} \rightarrow 2 \alpha=30^{\circ}
$$

The length of the arc $B D: L_{B D}=\frac{2 \pi r}{12}=\frac{\pi}{6} \sqrt{2-\sqrt{3}}$

The perimeter of the figure $P=13 \cdot 2 \pi r-24 L_{B D}=26 \pi r-24 \cdot \frac{\pi r}{6}=22 \pi r=44 \pi \sqrt{2-\sqrt{3}}$.
2. (mathematics) Thirteen circles of the same radius $r=2 \sqrt{2-\sqrt{3}}$ are cut out of colored paper and laid out on a plane so that their centers lie on one straight line. The distance between the centers of adjacent circles is the same and equals 2. Find the perimeter of the figure on the plane formed by these circles.