11.029. The volume of a regular triangular pyramid, whose lateral face is inclined to the base plane at an angle of $45^{\circ}$, is $9 \mathrm{~cm}^{3}$. Find the total surface area of the pyramid.
## Solution.

Let $PABC$ be the given pyramid (Fig. 11.24), $PH$ the height, where $H$ is the center of the circle inscribed in $\triangle ABC$; $HK \perp AC$, hence $PK \perp AC$, $HK = r = \frac{a \sqrt{3}}{6}, \angle PKH = 45^{\circ}$. Further, $PH = HK \tan 45^{\circ} = \frac{a \sqrt{3}}{6}, PK = \frac{HK}{\cos 45^{\circ}} = \frac{a \sqrt{3} \sqrt{2}}{6} = \frac{a \sqrt{6}}{6}; S_{\text{base}} = \frac{a^2 \sqrt{3}}{4}$. The lateral surface area $S_{\text{lat}} = \frac{3a}{2} \cdot PK = \frac{3a}{2} \cdot \frac{a \sqrt{6}}{6} = \frac{a^2 \sqrt{6}}{4}$. Then $S = S_{\text{lat}} + S_{\text{base}} = \frac{a^2 \sqrt{3}(1 + \sqrt{2})}{4}$. The volume $V = \frac{1}{3} S_{\text{base}} \cdot PH = \frac{a^3}{24} = 9$, from which $a = 6$ (cm). Thus, $S = \frac{36 \sqrt{3}(1 + \sqrt{2})}{4} = 9 \sqrt{3}(1 + \sqrt{2}) \text{cm}^2$.

Answer: $9 \sqrt{3}(1 + \sqrt{2}) \text{cm}^2$.