12.099. The base of a quadrilateral pyramid is a rhombus with side $a$ and acute angle $\alpha$. All lateral faces are inclined to the base plane at the same angle $\beta$. Find the total surface area of the pyramid.
Solution.

Let $S A B C D$ be a pyramid, $A B C D$ be a rhombus, $\angle B A D=\alpha\left(\alpha<90^{\circ}\right)$, $A B=a,((S A B)(A B C))=((S B C) ;(A B C))=((S C D) ;(A B C))=((S D A) ;(A B C))=\beta$, $S O \perp(A B C)$ (Fig. 12.99). Draw the apothems of the pyramid $S E, S F, S K$, $S L$; then $O E \perp A B, O F \perp B C, O K \perp C D, O L \perp A D$ (by the theorem of the $3$ perpendiculars) and, therefore, $\angle S E O=\angle S F O=\angle S K O=\angle S L O=\beta$. Further, we have $S_{\text {lateral }}=\frac{S_{A B C D}}{\cos \beta}=\frac{a^{2} \sin \alpha}{\cos \beta}$. Hence,

$$
S_{\text {total }}=\frac{a^{2} \sin \alpha}{\cos \beta}+a^{2} \sin \alpha=\frac{a^{2} \sin \alpha(1+\cos \beta)}{\cos \beta}=\frac{2 a^{2} \sin \alpha \cos ^{2} \frac{\beta}{2}}{\cos \beta}
$$

Answer: $\frac{2 a^{2} \sin \alpha \cos ^{2} \frac{\beta}{2}}{\cos \beta}$.