12.210. The base of a right prism is an isosceles triangle, the base of which is equal to $a$, and the angle at the base is equal to $\alpha$. Find the volume of the prism if its lateral surface area is equal to the sum of the areas of the bases.
## Solution.

Let $\triangle A B C$ be the base of the right prism $A B C A_{1} B_{1} C_{\mathrm{i}}$ (Fig. 12.77). $A B=B C, A C=a$, $\angle B A C=\alpha, B D$ - height.

$$
\text { In } \triangle A D B\left(\angle A D B=90^{\circ}\right): B D=\frac{a}{2} \operatorname{tg} \alpha, A B=\frac{a}{2 \cos \alpha}
$$

If $H$ is the height of the prism, then the lateral surface area is

$$
S_{6}=\left(\frac{a}{\cos \alpha}+a\right) H=a H \cdot \frac{1+\cos \alpha}{\cos \alpha}=\frac{2 a H \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}
$$

The area of the base of the prism $S_{\text {base }}=\frac{1}{2} A C \cdot B D=\frac{a^{2}}{4} \operatorname{tg} \alpha$.

By the condition $S_{6}=2 S_{\text {base }}$, therefore $\frac{2 a H \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}=\frac{a^{2}}{2} \operatorname{tg} \alpha \Rightarrow$ $\Rightarrow H=\frac{a \sin \alpha}{4 \cos ^{2} \frac{\alpha}{2}}=\frac{a \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}}=\frac{1}{2} a \operatorname{tg} \frac{\alpha}{2}$.

The volume of the prism $V=S_{\text {base }} \cdot H=\frac{1}{8} a^{3} \operatorname{tg} \alpha \operatorname{tg} \frac{\alpha}{2}$.

Answer: $\frac{1}{8} a^{3} \operatorname{tg} \alpha \operatorname{tg} \frac{\alpha}{2}$.