7.1. Find the volume of a regular triangular pyramid, the midpoint of the height of which is at a distance of 2 and $\sqrt{12}$ from the lateral face and the lateral edge, respectively. Round your answer to two decimal places if necessary.
Answer: 374.12 (exact value: $216 \sqrt{3}$).

Solution. Consider the section of the pyramid $S A B C$ passing through the lateral edge $S A$ and the apothem of the opposite face $S D$. Then $S H$ is the height of the pyramid, the distance from $H$ to the line $S D$ is $H N=2 x$, where $x=2$, and the distance from $H$ to the line $S A$ is $H M=2 y$, where $y=2 \sqrt{3}$. Let $S H=h, A B=a$ and $H D=d=a \sqrt{3} / 6$, then $A H=2 d, V=\frac{a^{2} h}{4 \sqrt{3}}=\sqrt{3} d^{2} h$ and, using the relationship between the height of a right triangle and its legs, we have

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \frac { 1 } { d ^ { 2 } } + \frac { 1 } { h ^ { 2 } } = \frac { 1 } { ( 2 x ) ^ { 2 } } } \\
{ \frac { 1 } { ( 2 d ) ^ { 2 } } + \frac { 1 } { h ^ { 2 } } = \frac { 1 } { ( 2 y ) ^ { 2 } } }
\end{array} \Rightarrow \left\{\begin{array}{l}
d^{2}=\frac{3 x^{2} y^{2}}{y^{2}-x^{2}} \\
h^{2}=\frac{12 x^{2} y^{2}}{4 x^{2}-y^{2}}
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow V=\frac{18 x^{3} y^{3}}{\left(y^{2}-x^{2}\right) \sqrt{4 x^{2}-y^{2}}}=216 \sqrt{3} \approx 374.12
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c22e465a9d4d7e156a81g-4.jpg?height=411&width=459&top_left_y=2367&top_left_x=1478)