12.114. In a regular hexagonal prism, a plane passing through a side of the base and the midpoint of the segment connecting the centers of the bases forms an acute angle $\alpha$ with the base plane. Find the area of the section formed by this plane if the side of the base of the prism is equal to $a$.
Solution.

The area of the section $S_{\mathrm{sec}}=2 S_{B A F E}=2\left(\frac{B E+A F}{2} \cdot P K\right)$ (Fig. 12.114). The radius of the inscribed circle in the base of the prism: $O K=r=\frac{a \sqrt{3}}{2}$. From $\triangle P O K\left(\angle P O K=90^{\circ}\right): P K=\frac{O K}{\cos \alpha}=\frac{a \sqrt{3}}{2 \cos \alpha}$. The radius of the circumscribed circle $R=B P=a$. Hence, $B E=2 a$. Finally, we get:

$$
S_{\mathrm{sec}}=(B E+A F) \cdot P K=(2 a+a) \frac{a \sqrt{3}}{2 \cos \alpha}=\frac{3 a^{2} \sqrt{3}}{2 \cos \alpha}
$$

Answer: $\frac{3 a^{2} \sqrt{3}}{2 \cos \alpha}$.