3. The length of the side of a regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ is $3 \mathrm{~cm}$. Its (8) sides $\overline{A_{1} A_{2}}, \overline{A_{2} A_{3}}, \overline{A_{3} A_{4}}, \overline{A_{4} A_{5}}, \overline{A_{5} A_{6}}, \overline{A_{6} A_{1}}$ are extended through the vertices $A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$, $A_{1}$ by $5 \mathrm{~cm}$ to the vertices $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ of a new regular hexagon. What is the length of its side?
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_fac652c8a845a57c6fa7g-13.jpg?height=700&width=669&top_left_y=975&top_left_x=702)

By the Law of Cosines, we have:

$$
\begin{aligned}
b^{2} & =\left|B_{1} B_{6}\right|^{2}=\left|A_{1} B_{6}\right|^{2}+\left|A_{1} B_{1}\right|^{2}-2\left|A_{1} B_{6}\right| \cdot\left|A_{1} B_{1}\right| \cdot \cos \angle B_{1} A_{1} B_{6} \\
& =5^{2}+8^{2}-2 \cdot 5 \cdot 8 \cdot \frac{1}{2}=49
\end{aligned}
$$

from which \( b = 7 \, \text{cm} \).

(full solution: 8 points)

(correct application of the Law of Cosines, but with a calculation error: 4 points)