11.065. Find the lateral surface area of a regular triangular pyramid if the plane angle at its vertex is $90^{\circ}$, and the area of the base is $S$.
## Solution.

Let $a$ be the side of the base; then $S=\frac{a^{2} \sqrt{3}}{4}$, from which $a=\frac{2 \sqrt{S}}{\sqrt[4]{3}}$.

The lateral surface area is expressed as: $S_{\text {lateral }}=3 \cdot \frac{1}{2} a \cdot B D$ (Fig. 11.60). Since $B D=D C=\frac{a}{2}$, then $S_{6 \text { lateral }}=\frac{3 a^{2}}{4}=\frac{3}{4} \cdot \frac{4 S}{\sqrt{3}}=S \sqrt{3}$.

Answer: $\quad S \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-690.jpg?height=596&width=422&top_left_y=92&top_left_x=185)

Fig. 11.61

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-690.jpg?height=466&width=364&top_left_y=223&top_left_x=778)

Fig. 11.62