11.122. The sides of the bases of a regular truncated quadrilateral pyramid are 2 and 1 cm, and the height is $3 \mathrm{~cm}$. A plane is drawn through the point of intersection of the diagonals of the pyramid, parallel to the bases of the pyramid, dividing the pyramid into two parts. Find the volume of each of the resulting parts.
Solution.

$O$ and $O_{1}$ are the centers of the lower and upper bases of the right truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}, O_{2}$ is the point of intersection of its diagonals $A C_{1}$ and $A_{1} C$ (Fig. 11.19), $O O_{1}=3 \text{ cm}, A D=2 \text{ cm}$, $A_{1} D_{1}=1$ cm. The square $A_{2} B_{2} C_{2} D_{2}$ is the section of the pyramid by the plane mentioned in the problem condition.

The diagonal section of the truncated pyramid is an isosceles trapezoid $A A_{1} C_{1} C, \Delta A_{1} O_{2} C_{1} \sim \Delta A O_{2} C$. Then $\frac{O_{1} O_{2}}{O_{2} O}=\frac{A_{1} C_{1}}{A C}=\frac{A_{1} D_{1}}{A D}=\frac{1}{2}$.

Since $O O_{1}=3 \text{ cm}$, then $O_{1} O_{2}=1 \text{ cm}, O O_{2}=2 \text{ cm} . \Delta A_{1} A C_{1} \sim \triangle A_{2} A O_{2}$ and $\frac{A_{2} O_{2}}{A_{1} C_{1}}=\frac{O O_{2}}{O O_{1}}=\frac{2}{3}$. Then $A_{2} O_{2}=\frac{2}{3} A_{1} C_{1}=\frac{2 \sqrt{2}}{3}$ cm and $A_{2} C_{2}=\frac{4 \sqrt{2}}{3} \text{ cm}$, $A_{2} D_{2}=\frac{4}{3}$ cm. The areas of the squares $A B C D, A_{1} B_{1} C_{1} D_{1}, A_{2} B_{2} C_{2} D_{2}$ are respectively $S=4 \text{ cm}^{2}, S_{1}=1 \text{ cm}^{2}, S_{2}=\frac{16}{9} \text{ cm}^{2}$. The volume of the truncated pyramid with height $O_{1} O_{2}: V_{1}=\frac{O_{1} O_{2}}{3}\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3}\left(1+\frac{16}{9}+\frac{4}{3}\right)=\frac{37}{27}$ cm $^{3}$.

The volume of the truncated pyramid with height $\mathrm{O}_{2} \mathrm{O}$ :

$V_{2}=\frac{O_{2} O}{3}\left(S+S_{2}+\sqrt{S S_{2}}\right)=\frac{2}{3}\left(4+\frac{16}{9}+\frac{8}{3}\right)=\frac{152}{27} \text{ cm}^{3}$.

Answer: $\frac{37}{27} \text{ cm}^{3} ; \frac{152}{27} \text{ cm}^{3}$.