11.003. The side of the base of a regular triangular pyramid is equal to $a$, and the dihedral angle at the base is $45^{\circ}$. Determine the volume and the total surface area of the pyramid.
Solution.

For a regular triangular pyramid $S A=S B=S C$ (Fig. 11.5); $S H$ is the height, and $H$ is the center of the circumscribed circle; $D$ is the midpoint of $A C$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-651.jpg?height=404&width=386&top_left_y=286&top_left_x=127)

Fig. 11.5

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-651.jpg?height=596&width=660&top_left_y=94&top_left_x=572)

Fig. 11.6

This means that $H D \perp A C$ (since $\triangle A B C$ is equilateral). $\angle S D H=45^{\circ}$. The area of the base $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. From $\triangle S D H$ we have $D H=S H \operatorname{tg} 45^{\circ}=S H$. On the other hand, $D H$ is the radius of the inscribed circle $-r=\frac{a \sqrt{3}}{6}$. This means that $S H=\frac{a \sqrt{3}}{6}$. Then the slant height $S D=\frac{S H}{\sin 45^{\circ}}=\frac{a}{\sqrt{6}}$. The semiperimeter of the base $p=\frac{3}{2} a$. The lateral surface area $S_{\text {lateral }}=S D \cdot p=\frac{a^{2} \sqrt{6}}{4}$. The total surface area $S=S_{\text {base }}+S_{\text {lateral }}=\frac{a^{2} \sqrt{3}(1+\sqrt{2})}{4}$, and the volume $V=\frac{1}{3} S H \cdot S_{\text {base }}=\frac{a^{3}}{24}$.

Answer: $\frac{a^{3}}{24} ; \frac{a^{2} \sqrt{3}(1+\sqrt{2})}{4}$.