11.053. The height of a regular tetrahedron is $h$. Calculate its total surface area.
Solution.

$$
S=S_{\mathrm{och}}+3 S_{\triangle B S A}=4 S_{\triangle A B C}=4 \frac{A B^{2} \sqrt{3}}{4}=A B^{2} \sqrt{3} \text { (Fig. 11.48); }
$$

$A O=A B \sqrt{3} / 3$, since $S A=A B=S B=B C=S C=A C$ (by condition); $\triangle A O S, \angle A O S=90^{\circ} ; A S^{2}=O A^{2}+S O^{2} A S^{2}=\frac{A B^{2}}{3}+h^{2}$;

$$
\frac{2 A B^{2}}{3}=h^{2} ; A B^{2}=\frac{3 h^{2}}{2} ; S=\frac{3 h^{2} \sqrt{3}}{2}
$$

Answer: $\frac{3 h^{2} \sqrt{3}}{2}$.