11.056. The diagonal of a rectangular parallelepiped is 10 cm and forms an angle of $60^{\circ}$ with the base plane. The area of the base is $12 \mathrm{~cm}^{2}$. Find the lateral surface area of the parallelepiped.
Solution.

From $\triangle D_{1} D B$ (Fig. 11.51) we find $B D=5 \text{ cm}, D D_{1}=5 \sqrt{3} \text{ cm}$. The lateral surface area $S_{\text {lateral }}=2\left(D D_{1} \cdot D C+D D_{1} \cdot A D\right)=2 D D_{1}(D C+A D)$. But $D C^{2}+A D^{2}=$ $=B D^{2}=25$, and by the condition $D C \cdot A D=12$. Therefore, $(D C+A D)^{2}=$ $=D C^{2}+A D^{2}+2 D C \cdot A D=25+24=49$, from which $D C+A D=7$ (cm). Thus, $S_{\text {lateral }}=2 \cdot 5 \sqrt{3} \cdot 7=70 \sqrt{3}\left(\text{cm}^{2}\right)$.

Answer: $70 \sqrt{3} \text{ cm}^{2}$.