4B. On the sides of a square with side length 2, isosceles trapezoids are constructed outward, such that the vertices of all trapezoids are simultaneously vertices of a regular dodecagon.

What is the perimeter of the dodecagon?
Solution. Let's consider a trapezoid from the dodecagon. The interior angle of the dodecagon is

![](https://cdn.mathpix.com/cropped/2024_06_05_9fe49ffb18e1d6225559g-03.jpg?height=326&width=331&top_left_y=1017&top_left_x=1149)

$$
\frac{12-2}{12} \cdot 180^{\circ}=150^{\circ}
$$

It follows that $\alpha=30^{\circ}$. From $|B C|=|C D|=|D A|=x$

and $|A B|=2$, it follows that $|A E|=1-\frac{x}{2}$. On the other hand, for the right triangle $A E D$,

![](https://cdn.mathpix.com/cropped/2024_06_05_9fe49ffb18e1d6225559g-03.jpg?height=211&width=567&top_left_y=1440&top_left_x=935)
with angles $30^{\circ}$ and $60^{\circ}$, we have $|A E|=x \frac{\sqrt{3}}{2}$. Therefore, $x \frac{\sqrt{3}}{2}=1-\frac{x}{2}$, which means $x=\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$. Thus, the perimeter of the dodecagon

$$
L=12 x=12(\sqrt{3}-1)
$$

## Second Year