12.121. The base of the pyramid is an equilateral triangle with side $a$. Two lateral faces of the pyramid are perpendicular to the base plane, and the equal lateral edges form an angle $\alpha$. Find the height of a right triangular prism equal in volume to the given pyramid and having a common base.
## Solution.

Let $DABC$ be a pyramid, $ABC$ be an equilateral triangle, $(ABD) \perp (ABC), (BCD) \perp (ABC), \angle ADC = \alpha, AB = a$ (Fig. 12.121). We need to find the height of the prism with the base equal to $\triangle ABC$, such that $V_{\text{pr}} = V_{\text{pir}}$. The height of the pyramid coincides with the edge $DB$, since the perpendicular dropped from $D$ to $(ABC)$ must belong to both $(ABC)$ and $(CBD)$. Draw $DK \perp AC$; since $\triangle ADC$ is isosceles, $KA = KC$. From $\triangle DKC$ we find $DK = \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$, and from $\triangle DBK$ we get $DB = \sqrt{DK^2 - BK^2} = \sqrt{\frac{a^2 \operatorname{ctg} \frac{\alpha}{2}}{4} - \frac{3a^2}{4}} = \frac{a}{2} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - 3}$. Therefore, $V_{\text{pir}} = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot \frac{a}{2} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - 3} = \frac{a^3}{24} \sqrt{3\left(\operatorname{ctg}^2 \frac{\alpha}{2} - 3\right)}$. Since

$$
\begin{gathered}
V_{\text{pir}} = V_{\text{pr}} = \frac{a^2 \sqrt{3}}{4} H, \text{ then } \frac{a^2 \sqrt{3}}{4} H = \frac{a^3}{24} \sqrt{3\left(\operatorname{ctg}^2 \frac{\alpha}{2} - 3\right)}, \\
H = \frac{a}{6} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - \operatorname{ctg}^2 \frac{\pi}{6}} = \frac{a \sqrt{\sin \left(\frac{\pi}{6} + \frac{\alpha}{2}\right) \sin \left(\frac{\pi}{6} - \frac{\alpha}{2}\right)}}{3 \sin \frac{\alpha}{2}} \\
\text{Answer: } \frac{a}{3 \sin \frac{\alpha}{2}} \cdot \sqrt{\sin \left(\frac{\pi}{6} + \frac{\alpha}{2}\right) \sin \left(\frac{\pi}{6} - \frac{\alpha}{2}\right)}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-807.jpg?height=642&width=402&top_left_y=93&top_left_x=173)

Fig. 12.122

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-807.jpg?height=466&width=470&top_left_y=239&top_left_x=733)

Fig. 12.123