2.10. Determine the volume of a regular truncated square pyramid if its diagonal is $18 \mathrm{~cm}$, and the side lengths of the bases are 14 and 10 cm.
2.10. The desired volume is expressed by the formula $V=\frac{h}{3}\left(S_{1}+S_{2}+\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-460.jpg?height=294&width=496&top_left_y=1568&top_left_x=658)

Fig. 2.10
$\left.+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}=196 \mathrm{~cm}^{2}, S_{2}=100 \mathrm{~cm}^{2}$. Let's find $h=B_{1} K$ (Fig. 2.10). 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 $B K=\frac{1}{2}\left(B D-B_{1} D_{1}\right)=\frac{1}{2}(14 \sqrt{2}-10 \sqrt{2})=2 \sqrt{2}$ (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,

$$
V=\frac{6}{3}(196+100+14.0)=872\left(\mathrm{~cm}^{3}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-461.jpg?height=461&width=415&top_left_y=660&top_left_x=46)

Fig. 2.11

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-461.jpg?height=471&width=423&top_left_y=1241&top_left_x=42)

Fig. 2.12

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