At the vertices of a unit square, perpendiculars are erected to its plane, and on them, on one side of the square's plane, points are taken at distances of 3, 4, 6, and 5 from this plane (in the order of traversal). Find the volume of the polyhedron whose vertices are the specified points and the vertices of the square.

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Let $A B C D$ be a unit square, and $A_{1}, B_{1}, C_{1}$, and $D_{1}$ be the specified points on the perpendiculars to its plane, such that $A A_{1}=3, B B_{1}=4, C C_{1}=6, D D_{1}=5$.

Extend the perpendiculars to segments $A A_{2}=B B_{2}=C C_{2}=D D_{2}=9$. Note that $A_{1} B_{1} C_{1} D_{1}$ is a parallelogram (for example, because the midpoints of segments $A_{1} C_{1}$ and $B_{1} D_{1}$ coincide). Therefore, the points $A_{1}, B_{1}$, $C_{1}, D_{1}$ lie in the same plane, which divides the parallelepiped $A B C D A_{2} B_{2} C_{2} D_{2}$ into two equal figures.

Since the volume of the specified parallelepiped is 9, the volume of each of the halves is $9 / 2$.

## Answer

4.5 .