12.323. The base of a right prism is an equilateral triangle. A plane is drawn through one of its sides, cutting off a pyramid from the prism, the volume of which is equal to $V$. Find the area of the section, if the angle between the cutting plane and the base plane is $\alpha$.
Solution.

Let $\triangle B E C$ be the section of the regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.184),

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1079.jpg?height=696&width=701&top_left_y=76&top_left_x=352)

Fig. 12.185

$D$ is the midpoint of $B C$. Then $\angle E D A=\alpha, E A$ is the height of the pyramid $E A B C$. $V=\frac{1}{3} S_{\triangle A B C} \cdot E A=\frac{1}{3} S_{\triangle B E C} \cdot \cos \alpha \cdot E A$.

If $E A=h, A B=a$, then $S_{\triangle B E C}=\frac{3 V}{h \cos \alpha}, S_{\triangle A B C}=\frac{3 V}{h}=\frac{a^{2} \sqrt{3}}{4}, A D=\frac{a \sqrt{3}}{2}$.

In $\triangle E A D\left(\angle E A D=90^{\circ}\right): h=\frac{a \sqrt{3}}{2} \operatorname{tg} \alpha, a=\frac{2 h \operatorname{ctg} \alpha}{\sqrt{3}} \Rightarrow$ $\Rightarrow \frac{3 V}{h}=\frac{4 h^{2} \operatorname{ctg}^{2} \alpha \cdot \sqrt{3}}{12} \Leftrightarrow$

$\Leftrightarrow h^{3} \sqrt{3} \operatorname{ctg}^{2} \alpha=9 V \Leftrightarrow h=\sqrt[3]{\frac{3 \sqrt{3} V}{\operatorname{ctg}^{2} \alpha}}=\sqrt[3]{3 \sqrt{3} V \operatorname{tg}^{2} \alpha} \Rightarrow$

$\Rightarrow S_{\triangle B E C}=\frac{3 V}{h \cos \alpha}=\frac{3 V}{\sqrt[3]{3 \sqrt{3} V \operatorname{tg}^{2} \alpha} \cdot \cos \alpha}=\sqrt[3]{\frac{27 V^{3}}{3 \sqrt{3} V \operatorname{tg}^{2} \alpha \cos ^{3} \alpha}}=$

$=\sqrt[3]{\frac{3 \sqrt{3} V^{2}}{\sin ^{2} \alpha \cos \alpha}}$

Answer: $\sqrt[3]{\frac{3 \sqrt{3} V^{2}}{\sin ^{2} \alpha \cos \alpha}}$