12.214. The base of the pyramid is a rhombus with side $a$ and acute angle $\alpha$. Two lateral faces are perpendicular to the base, while the other two are inclined to it at an angle $\varphi$. Find the volume and lateral surface area of the pyramid.
Solution.

Let the rhombus $ABCD$ be the base of the pyramid $SABCD$ (Fig. 12.81), $AB=a$, and the acute angle of the rhombus $\alpha$.

Then the area of the rhombus $S=a^{2} \sin \alpha$, and its height $h=a \sin \alpha$.

The lateral faces $ABS$ and $CBS$ of the pyramid are perpendicular to the plane of the base, so their

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0962.jpg?height=616&width=586&top_left_y=540&top_left_x=690)

Fig. 12.81 common edge $SB$ is the height of the pyramid. Draw perpendiculars $BK$ and $BM$ from point $B$ to lines $AD$ and $CD$. Then $BK=BM=h=a \sin \alpha, SK \perp AD, SM \perp CD, \angle SKB=\angle SMB=\varphi$. From $\triangle SBK\left(\angle SBK=90^{\circ}\right): BS=BK \operatorname{tg} \angle SKB=a \sin \alpha \operatorname{tg} \beta$.

The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} \cdot BS=\frac{1}{3} a^{3} \sin ^{2} \alpha \operatorname{tg} \varphi$.

$\triangle ABD$ and $\triangle CBD$ are the projections of the faces $ASD$ and $CSD$ on the plane of the base, so $S_{\triangle ASD}=S_{\triangle CSD}=\frac{S_{\triangle ABD}}{\cos \varphi}=\frac{a^{2} \sin \alpha}{2 \cos \varphi}$.

$$
S_{\triangle ABS}=S_{\triangle CBS}=\frac{1}{2} BS \cdot AB=\frac{1}{2} a^{2} \sin \alpha \operatorname{tg} \varphi=\frac{a^{2} \sin \alpha \sin \varphi}{2 \cos \varphi}
$$

The area of the lateral surface of the pyramid

$$
\begin{aligned}
& S_{\text {lateral }}=2\left(S_{\triangle ASD}+S_{\triangle ABS}\right)=\frac{a^{2} \sin \alpha}{\cos \varphi}+\frac{a^{2} \sin \alpha \sin \varphi}{\cos \varphi}=\frac{a^{2} \sin \alpha(1+\sin \varphi)}{\cos \varphi}= \\
& =\frac{a^{2} \sin \alpha\left(1+\cos \left(90^{\circ}-\varphi\right)\right)}{\cos \varphi}=\frac{2 a^{2} \sin \alpha \cos ^{2}\left(45^{\circ}-\frac{\varphi}{2}\right)}{\cos \varphi} \\
& \text { Answer: } \frac{1}{3} a^{3} \sin ^{2} \alpha \operatorname{tg} \varphi ; \frac{2 a^{2} \sin \alpha \cos ^{2}\left(45^{\circ}-\frac{\varphi}{2}\right)}{\cos \varphi}
\end{aligned}
$$