12.128. A regular hexagonal prism with equal edges is inscribed in a cone with a generatrix of length $l$. Find the lateral surface area of the prism if the angle between the generatrix and the height of the cone is $\boldsymbol{\alpha}$.
## Solution.

The lateral surface area of the prism $S_{\text {bok }}=P \cdot h$ (Fig. 12.128). Let the edge of the prism be $x$. Then $h=x$ and $P=6 x$. From $\triangle S O A \quad S O=l \cos \alpha$. From $\Delta S O_{1} B \quad S O_{1}=\frac{x}{\operatorname{tg} \alpha}$. We find $S O=O O_{1}+O_{1} S=x+\frac{x}{\operatorname{tg} \alpha}=x\left(1+\frac{1}{\operatorname{tg} \alpha}\right)$. Solving the equation: $x\left(1+\frac{1}{\operatorname{tg} \alpha}\right)=l \cos \alpha \Rightarrow x=\frac{l \cos \alpha \sin \alpha}{\cos \alpha+\sin \alpha}$. Then $S_{\text {bok }}=6 x^{2}=$ $=\frac{6 l^{2} \cos ^{2} \alpha \sin ^{2} \alpha}{(\cos \alpha+\sin \alpha)^{2}}=\frac{3 l^{2} \cos ^{2} \alpha \sin ^{2} \alpha}{\left(\frac{1}{\sqrt{2}}\right)^{2}(\cos \alpha+\sin \alpha)^{2}}=\frac{3 l^{2} 4 \cdot \cos ^{2} \alpha \sin ^{2} \alpha}{4\left(\sin \frac{\pi}{4} \cos \alpha+\cos \frac{\pi}{4} \sin \alpha\right)^{2}}=$
$=\frac{3 l^{2} \sin ^{2} 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right)}$. Answer: $\frac{3 l^{2} \sin ^{2} 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right)}$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-813.jpg?height=498&width=466&top_left_y=93&top_left_x=157)

Fig. 12.129

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-813.jpg?height=252&width=456&top_left_y=344&top_left_x=730)

Fig. 12.130