11.138. The base of the pyramid is a parallelogram with sides equal to 10 and $18 \mathrm{~cm}$, and the area is $90 \mathrm{~cm}^{2}$. The height of the pyramid passes through the point of intersection of the diagonals of the base and is 6 cm. Determine the lateral surface area of the pyramid.
## Solution.

Draw $S O$ - the height of the pyramid $S A B C D$ (Fig. 11.35). Drop perpendiculars $S M, S L, S N, S K$ to $A B, B C, C D$, and $A D$ respectively. By the condition, $A B C D$ is a parallelogram, so point $O$, the intersection of the diagonals, is the midpoint of segments $M N$ and $K L$, which are the heights of the parallelogram. $L K=\frac{S_{A B C D}}{A D}=\frac{90}{18}=5$ (cm); $M N=\frac{S_{A B C D}}{A B}=\frac{90}{10}=9$ (cm).

Therefore, $O K=\frac{1}{2} L K=\frac{5}{2}$ cm, $M O=\frac{1}{2} M N=\frac{9}{2}$ cm.

From $\triangle S O K\left(\angle S O K=90^{\circ}\right): S K=\sqrt{S O^{2}+O K^{2}}=\frac{13}{2}$ cm.

From $\triangle S O M\left(\angle S O M=90^{\circ}\right): S M=\sqrt{S O^{2}+O M^{2}}=\frac{15}{2}$ cm. The lateral surface area of the pyramid $S_{6}=2 S_{\triangle A S B}+2 S_{\triangle A S D}=A B \cdot S M+A D \cdot S K=10 \cdot \frac{15}{2}+$ $+18 \cdot \frac{13}{2}=192\left(\mathrm{~cm}^{2}\right)$.

Answer: $192 \mathrm{~cm}^{2}$.