11.119. The side of the base of a regular triangular pyramid is equal to $a$, and the height dropped from the vertex of the base to the opposite lateral face is equal to $b$. Determine the volume of the pyramid.
## Solution.

Let $M O$ be the height of the regular pyramid $M A B C$ (Fig. 11.16). $P$ is the midpoint of $B C$. Then the plane $M P A$ is perpendicular to the plane $B M C$, and the perpendicular $A K$, dropped onto the plane $B M C$, lies in the plane $M P A$, and the point $K$ belongs to $M P$. Let $M O=h$.

Then $M P=\sqrt{P O^{2}+M O^{2}}=\sqrt{\left(\frac{a \sqrt{3}}{6}\right)^{2}+h^{2}}=\sqrt{\frac{a^{2}}{12}+h^{2}}$. $S_{\triangle M P A}=\frac{1}{2} A P \cdot M O=\frac{1}{2} A K \cdot M P$.

Then $A P \cdot M O=A K \cdot M P$

$$
\begin{aligned}
& \frac{a \sqrt{3}}{2} \cdot h=b \sqrt{\frac{a^{2}}{12}+h^{2}} \\
& \frac{3 a^{2} h^{2}}{4}=\frac{b^{2}\left(a^{2}+12 h^{2}\right)}{12} \\
& 9 a^{2} h^{2}=a^{2} b^{2}+12 b^{2} h^{2} \\
& h=\frac{a b}{\sqrt{3\left(3 a^{2}-4 b^{2}\right)}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0832.jpg?height=567&width=525&top_left_y=1255&top_left_x=686)

Fig. 11.17

Volume of the pyramid

$$
V=\frac{1}{3} S_{\triangle A B C} \cdot h=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4} \cdot \frac{a b}{\sqrt{3\left(3 a^{2}-4 b^{2}\right)}}=\frac{a^{3} b}{12 \sqrt{3 a^{2}-4 b^{2}}}
$$

Answer: $\frac{a^{3} b}{12 \sqrt{3 a^{2}-4 b^{2}}}$