11.159. The base of the pyramid is a parallelogram, the adjacent sides of which are 9 and $10 \mathrm{~cm}$, and one of the diagonals is $11 \mathrm{~cm}$. The opposite lateral edges are equal, and the length of each of the larger edges is 10.5 cm. Calculate the volume of the pyramid.
Solution.

Let the parallelogram $ABCD$ be the base of the pyramid $EABCD$ (Fig. 11.54), $AB=9 \text{~cm}, AD=10 \text{~cm}, BD=11 \text{~cm}, AE=CE, BE=DE$. $O$ is the point of intersection of the diagonals $AC$ and $BD$ of the parallelogram. Then $EO$ is the median of the isosceles triangles $AEC$ and $BED$.

Therefore, $EO \perp AC, EO \perp BD$, that is, the segment $EO$ is perpendicular to the plane of the base and is the height of the pyramid. $AC^2 + BD^2 = 2(AD^2 + AB^2); \quad AC = \sqrt{241}$ cm. Thus, $AC > BD$ and $AO > BO \cdot AO$ is the projection of $AE$, $BO$ is the projection of $BE$ on the plane of the base. Therefore, $AE > BE, AE = 10.5 \text{~cm} = \frac{21}{2} \text{~cm}$. From $\triangle AOE (\angle AOE = 90^\circ): EO = \sqrt{AE^2 - AO^2} = \sqrt{\frac{441}{4} - \frac{241}{2}} = 5\sqrt{2}$ (cm). By Heron's formula $S_{\triangle ABD} = 30\sqrt{2} \text{~cm}^2$, the area of the parallelogram $ABCD$ $A = 2S_{\triangle ABD} = 60\sqrt{2} \text{~cm}^2$. The volume of the pyramid $V = \frac{1}{3} S \cdot EO = 200 \text{~cm}^3$.

Answer: $200 \text{~cm}^3$.