11.133. The base of the pyramid is an equilateral triangle with a side equal to $a$. One of the lateral faces is also an equilateral triangle and is perpendicular to the plane of the base. Determine the total surface area of the pyramid.
## Solution.

Let the face $A M C$ of the pyramid $M A B C$ be an equilateral triangle and perpendicular to the plane of the base (Fig. 11.30). Then its height and median $M O$ is the height of the pyramid. Drop a perpendicular $O E$ to $A B$. Then $A E=\frac{A B}{4}=\frac{a}{4}, M E \perp A B$.

From $\triangle A E M\left(\angle A E M=90^{\circ}\right): M E=\sqrt{A M^{2}-A E^{2}}=\sqrt{a^{2}-\frac{a^{2}}{16}}=\frac{a \sqrt{15}}{4}$, $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}, S_{\triangle A M B}=\frac{1}{2} A B \cdot M E=\frac{a^{2} \sqrt{15}}{8}$.

Since $\triangle A B C=\triangle A M C, \triangle A M B=\triangle C M B$, the total surface area of the pyramid $S=2\left(S_{\triangle A B C}+S_{\triangle A M B}\right)=\frac{a^{2} \sqrt{3}}{2}+\frac{a^{2} \sqrt{15}}{4}=\frac{a^{2} \sqrt{3}(2+\sqrt{5})}{4}$.

Answer: $\quad \frac{a^{2} \sqrt{3}(2+\sqrt{5})}{4}$.