11.118. The base of the parallelepiped is a rhombus with side $a$ and an acute angle of $30^{\circ}$. The diagonal of one lateral face is perpendicular to the base plane, and the lateral edge forms an angle of $60^{\circ}$ with the base plane. Find the total surface area and volume of the parallelepiped.
Solution.

In the rhombus $ABCD$, which is the base of the parallelepiped $ABCD A_1 B_1 C_1 D_1$ (Fig. 11.15), $CD = a$, $\angle BCD = 30^\circ$. Then its area $S_1 = a^2 \sin 30^\circ = \frac{a^2}{2}$. The diagonal $C_1 B$ of the lateral face $BB_1 C_1 C$, perpendicular to the plane of the base, is the height of the parallelepiped, $BC$ is the projection of the lateral edge $CC_1$ on the plane of the base, $\angle BCC_1 = 60^\circ$.

From $\triangle C_1 BC \left(\angle C_1 BC = 90^\circ\right): BC_1 = BC \operatorname{tg} \angle BCC_1 = a \sqrt{3}$;

$CC_1 = \frac{BC}{\cos \angle BCC_1} = 2a$. The volume of the parallelepiped $V = S_1 \cdot BC_1 = \frac{a^3 \sqrt{3}}{2}$, $S_{BB_1 C_1 C} = a^2 \sqrt{3} \cdot CC_1$ - is the oblique to the plane of the base, $BC$ is its projection on this plane, $CD$ is a line lying in this plane. Therefore, $\cos \angle C_1 CD = \cos \angle C_1 CB \cdot \cos \angle BCD = \cos 60^\circ \cos 30^\circ = \frac{\sqrt{3}}{4}$.

From this,

$\sin \angle C_1 CD = \sqrt{1 - \cos^2 \angle C_1 CD} = \frac{\sqrt{13}}{4}$; $S_{CC_1 D_1 D} = CC_1 \cdot CD \sin \angle C_1 CD = \frac{a^2 \sqrt{13}}{2}$. The total surface area of the parallelepiped: $S = 2 \left(S_1 + S_{BB_1 C_1 C} + S_{CC_1 D_1 D}\right) = 2 \left(\frac{a^2}{2} + a^2 \sqrt{3} + \frac{a^2 \sqrt{13}}{2}\right) = a^2 (1 + 2 \sqrt{3} + \sqrt{13})$. Answer: $a^2 (1 + 2 \sqrt{3} + \sqrt{13}) ; \frac{a^3 \sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0832.jpg?height=522&width=733&top_left_y=74&top_left_x=336)

Fig. 11.15