11.117. The base of a right prism is an isosceles trapezoid $A B C D ; A B=C D=13$ cm, $B C=11$ cm, $A D=21$ cm. The area of its diagonal section is $180 \mathrm{~cm}^{2}$. Calculate the total surface area of the prism.
Solution.

$C K$ is the height of the base of the right prism $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.14). Given $A B=C D$, then $K D=\frac{A D-B C}{2}=5 \text{ cm}, A K=\frac{A D+B C}{2}=16 \text{ cm}$.

From $\triangle C K D\left(\angle C K D=90^{\circ}\right): C K=\sqrt{C D^{2}-K D^{2}}=12 \text{ cm}$.

From $\triangle A K C\left(A K C=90^{\circ}\right): A C=\sqrt{A K^{2}+C K^{2}}=20 \text{ cm}$. The diagonal section of the prism $A A_{1} C_{1} C$ is a rectangle with an area of $S_{1}=180 \text{ cm}^{2}$,
thus the height of the prism $H=C C_{1}=\frac{S_{1}}{A C}=9$ cm. The perimeter of the base of the prism $P=58$ cm, the area of the base $S_{\text {base }}=\frac{A D+B C}{2} \cdot C K=192 \text{ cm}^{2}$, the lateral surface area $S_{6}=P H=522 \text{ cm}^{2}$. The total surface area of the prism $S=S_{6}+2 S_{\text {base }}=906 \text{ cm}^{2}$.

Answer: $906 \text{ cm}^{2}$.