3.31. The height of a regular triangular prism is $H$. A plane passing through the midline of the lower base and the parallel side of the upper base forms an acute dihedral angle $\alpha$ with the plane of the lower base. Find the area of the section formed by this plane.
3.31. Given that $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $A A_{1}=H, A N=N B, B M=M C$; $\left(\left(A_{1} C_{1} M\right) ;(A B C)\right)=\alpha$ (Fig. 3.33). Draw $B E \perp A C$ and $E F \| A A_{1} ; \angle F O E=\alpha$ as the linear angle of the dihedral angle. Let $B E \cap M N=O$; then $F O \perp M N$ (by the theorem of three perpendiculars). The section $A_{1} N M C_{1}$ is a trapezoid, since $M N \mid A C$. From $\triangle F E O$ we find

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-518.jpg?height=464&width=433&top_left_y=568&top_left_x=718)

Fig. 3.33 $F O=\frac{F E}{\sin \alpha}=\frac{H}{\sin \alpha}, E O=H \operatorname{ctg} \alpha$, hence $B E=2 H \operatorname{ctg} \alpha$. From $\triangle B E C$ we get

$$
E C=M N=B E \operatorname{ctg} 60^{\circ}=2 H \operatorname{ctg} \alpha \cdot \frac{\sqrt{3}}{3}=\frac{2 H \sqrt{3} \operatorname{ctg} \alpha}{3}
$$

thus, $A C=A_{1} C_{1}=\frac{4 H \sqrt{3} \operatorname{ctg} \alpha}{3}$. Therefore,

$$
S_{\mathrm{A} N M C_{1}}=\frac{A C+M N}{2} \cdot F O=H \sqrt{3} \operatorname{ctg} \alpha \cdot \frac{H}{\sin \alpha}=\frac{H^{2} \sqrt{3} \operatorname{ctg} \alpha}{\sin \alpha} .
$$