3.35. 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.
3.35. By condition, $S A B C D$ is a pyramid, $A B C D$ is a rhombus, $\angle B A D=$ $=\alpha\left(\alpha<90^{\circ}\right), A B=a$,

$((S A B) ;(A B C))=$

$=((S B C) ;(A B C))=$

$=((S C D) ;(A B C))=$

$=((S D \widehat{A}) ;(A B C))=\beta$,

$S O \perp(A B C)$ (Fig. 3.37). Let's 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 three perpendiculars) and, therefore, $\angle S E O=$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-520.jpg?height=728&width=654&top_left_y=1082&top_left_x=500)
$=\angle S F O=\angle S K O=$ $=\angle S L O=\beta$. Further, we have

$$
\begin{aligned}
& S_{\text {bot }}=\frac{S_{A B C D}}{\cos \beta}=\frac{a^{2} \sin \alpha}{\cos \beta} . \text { Finally, } \\
& S_{\text {full }}=\frac{a^{2} \sin \alpha}{\cos \beta}+a^{2} \sin \alpha=\frac{a^{2} \sin \alpha(1+\cos \beta)}{\cos \beta}=\frac{2 a^{2} \sin \alpha \cos ^{2} \frac{\beta}{2}}{\cos \beta} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-521.jpg?height=518&width=515&top_left_y=537&top_left_x=58)

Fig. 3.38