11.005. The plane angle at the vertex of a regular triangular pyramid is $90^{\circ}$. Find the ratio of the lateral surface area of the pyramid to the area of its base.
Solution.

For a regular pyramid $AB=BC=AC=a \quad$ (Fig. 11.7). $\angle ASC=90^{\circ}, SA=SC=SB$. This means that $\angle SAC=45^{\circ}, AD=DC$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-652.jpg?height=434&width=486&top_left_y=727&top_left_x=733)

Fig. 11.7 Then from $\triangle SAC: SA=\frac{AD}{\cos 45^{\circ}}=\frac{a \sqrt{2}}{2}$ (since $\triangle ASC$ is isosceles). $SD$ is the apothem. Then $SD=AS \sin 45^{\circ}=\frac{a}{2}$. The lateral surface area $S_{\text {lat }}=p \cdot SD$, where $p$ is the semiperimeter of the base $-p=\frac{3}{2} a$; $S_{\text {lat }}=\frac{3}{2} a \cdot \frac{a}{2}=\frac{3 a^{2}}{4}$. The base area $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. The ratio $\frac{S_{\text {lat }}}{S_{\text {base}}}=\frac{3 a^{2} \cdot 4}{4 a^{2} \cdot \sqrt{3}}=\sqrt{3}$.

Answer: $\sqrt{3}$.