12.276. One of the sides of the base of a right triangular prism is equal to $a$, and the angles adjacent to it are $\alpha$ and $\beta$. Find the lateral surface area of the prism if its volume is $V$.
## Solution.

If $\mathrm{b}$ and $c$ are the sides of the base of the prism, opposite to its angles $\beta$ and $\alpha$, and $\gamma$ is the angle of the base opposite to side $a$, then

$$
\begin{aligned}
& b=\frac{a \sin \beta}{\sin \gamma}=\frac{a \sin \beta}{\sin \left(180^{\circ}-(\alpha+\beta)\right)}=\frac{a \sin \beta}{\sin (\alpha+\beta)} \\
& c=\frac{a \sin \alpha}{\sin (\alpha+\beta)}
\end{aligned}
$$

The area of the base of the prism $S=\frac{1}{2} a b \sin \alpha \frac{a^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)}$; the perimeter of the base:

$$
\begin{aligned}
& P=a+b+c=a+\frac{a \sin \beta}{\sin (\alpha+\beta)}+\frac{a \sin \alpha}{\sin (\alpha+\beta)}=\frac{a}{\sin (\alpha+\beta)}(\sin (\alpha+\beta)+\sin \alpha+\sin \beta)= \\
& =\frac{a}{\sin (\alpha+\beta)}\left(2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}+\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\right)= \\
& =\frac{2 a \sin \frac{\alpha+\beta}{2}}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}}\left(\cos \frac{\alpha+\beta}{2}+\cos \frac{\alpha-\beta}{2}\right)=\frac{2 a \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{\cos \frac{\alpha+\beta}{2}}
\end{aligned}
$$

The height of the prism $H=\frac{V}{S}=\frac{2 V \sin (\alpha+\beta)}{a^{2} \sin \alpha \sin \beta}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1031.jpg?height=623&width=587&top_left_y=64&top_left_x=94)

Fig. 12.140

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1031.jpg?height=621&width=606&top_left_y=75&top_left_x=672)

Fig. 12.141

The lateral surface of the prism:

$$
\begin{aligned}
& S_{6}=P \cdot H=\frac{2 V \sin (\alpha+\beta)}{a^{2} \sin \alpha \sin \beta} \cdot \frac{2 a \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{\cos \frac{\alpha+\beta}{2}}= \\
& =\frac{8 V \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{a \cdot 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cdot 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2} \cos \frac{\alpha+\beta}{2}}=\frac{2 V \sin \frac{\alpha+\beta}{2}}{a \sin \frac{\alpha}{2} \sin \frac{\beta}{2}} \\
& \text { Answer: } \frac{2 V \sin \frac{\alpha+\beta}{2}}{a \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}
\end{aligned}
$$