[ Quadrilateral pyramid ] [ Volume of a tetrahedron and pyramid $]$

The base of the pyramid is a parallelogram, with adjacent sides equal to 9 and 10, and one of the diagonals equal to 11. The opposite lateral edges are equal, and each of the larger edges is equal to $10^{\frac{1}{2}}$. Find the volume of the pyramid.
Let $P A B C D$ be the given pyramid with vertex $P, P O$ - the height of the pyramid. Since $A P = C P$, point $O$ is equidistant from points $A$ and $C$. Therefore, point $O$ lies on the perpendicular bisector of segment $A C$. Similarly, point $O$ lies on the perpendicular bisector of segment $B D$, and since the perpendicular bisectors of the diagonals of a parallelogram intersect at its center, $O$ is the point of intersection of the diagonals of parallelogram $A B C D$. Let $A B = 10, A D = 9, B D = 11$. By the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
B D^2 + A C^2 = 2 \cdot A B^2 + 2 \cdot A D^2
$$

from which we find that

$$
A C^2 = 2 \cdot A B^2 + 2 \cdot A D^2 - B D^2 = 200 + 162 - 121 = 241.
$$

Therefore, $A C = \sqrt{241}, A O = C O = \frac{\sqrt{241}}{2}$. Since $\sqrt{241} > 11$, $A C$ is the larger diagonal of parallelogram $A B C D$, so $A P$ and $C P$ are the larger lateral edges of pyramid $P A B C D$. Therefore, $A P = C P = \frac{21}{2}$. From the right triangle $A O P$ we find that

$$
P O = \sqrt{A P^2 - A O^2} = \sqrt{\left(\frac{21}{2}\right)^2 - \left(\frac{\sqrt{241}}{2}\right)^2} = \sqrt{\frac{441 - 241}{4}} = 5 \sqrt{2}
$$

Using Heron's formula, we find that

$$
S_{\triangle A B D} = \sqrt{15 \cdot 6 \cdot 5 \cdot 4} = 30 \sqrt{2}
$$

Therefore,

$$
V_{P A B C D} = \frac{1}{3} S_{A B C D} \cdot P O = \frac{1}{3} \cdot 2 \cdot 30 \sqrt{2} \cdot 5 \sqrt{2} = 200
$$

## Answer

200.00