11.064. Find the volume of a regular triangular pyramid, the height of which is $h$, and all plane angles at the vertex are right angles.
## Solution.

Draw $B D \perp A C$ (Fig. 11.59). Let the side of the base be $a$: Since the pyramid is regular, then $O D=\frac{a \sqrt{3}}{6}, A D=D C=\frac{a}{2}$, $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. Express $a$ in terms of $h$. In $\triangle S A C$ we have $S A=S C$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-689.jpg?height=578&width=536&top_left_y=95&top_left_x=122)

Fig. 11.59

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-689.jpg?height=578&width=536&top_left_y=95&top_left_x=686)

Fig. 11.60

$\angle C S A=90^{\circ}$, hence $\angle A S D=45^{\circ}$ and $S D=A D=\frac{a}{2}$. Then

$$
\begin{aligned}
& h=S O=\sqrt{S D^{2}-O D^{2}}=\sqrt{\frac{a^{2}}{4}-\frac{3 a^{2}}{36}}=\frac{a}{\sqrt{6}} \Rightarrow a=h \sqrt{6} \\
& \text { We have } V=\frac{1}{3} S_{\text {base }} h=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot h=\frac{1}{3} \cdot \frac{6 h^{2} \sqrt{3}}{4} \cdot h=\frac{h^{3} \sqrt{3}}{2} .
\end{aligned}
$$

Answer: $\quad h^{3} \sqrt{3} / 2$.