11.042. Determine the volume of a regular truncated square pyramid if its diagonal is 18 cm, and the lengths of the sides of the bases are 14 and 10 cm.
## Solution.

The desired volume is expressed by the formula $V=\frac{h}{3}\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}=196\left(\mathrm{~cm}^{2}\right), S_{2}=100\left(\mathrm{~cm}^{2}\right)$. Let's find $h=B_{1} K$ (Fig. 11.37). We have $B_{1} K=$ $=\sqrt{B_{1} D^{2}-K D^{2}}$. Since $B B_{1} D_{1} D$ is an isosceles trapezoid, then $B K=0.5\left(B D-B_{1} D_{1}\right)=0.5(14 \sqrt{2}-10 \sqrt{2})=2 \sqrt{2}$ (cm) and $K D=B D-B K=$ $=12 \sqrt{2}(\mathrm{~cm})$, i.e., $h=\sqrt{18^{2}-(12 \sqrt{2})^{2}}=6(\mathrm{~cm})$. Therefore,

$$
V=\frac{6}{3}(196+100+140)=872\left(\mathrm{~cm}^{3}\right)
$$

Answer: $872 \mathrm{~cm}^{3}$.