11.023. In a regular quadrilateral pyramid, the side of the base is 6 dm, and the height is 4 dm. Find the lateral surface area of the truncated pyramid cut off from the given one by a plane parallel to its base and at a distance of 1 dm from it.
Solution.

Let $A D=D C=B C=A B=a_{1}$ (6 dm), $S H$-height, $S H=4$ (dm) (Fig. 11.18). By the condition $H H_{1}=1$ (dm). From the similarity of $\triangle H E S$ and $\Delta H_{1} E_{1} S$ (where $E$ and $E_{1}$ are the midpoints of $C D$ and $C_{1} D_{1}$ respectively): $\frac{H E}{H_{1} E_{1}}=\frac{S H}{S H_{1}}$, but $S H_{1}=S H-H H_{1}=4-1=3$ (dm). Therefore, $H_{1} E_{1}=H E: \frac{4}{3}=\frac{9}{4}$ (dm). Therefore, $A_{1} D_{1}=2 H_{1} E_{1}=\frac{9}{2}$ (dm). $S E_{1}=\sqrt{S H_{1}^{2}+H_{1} E_{1}^{2}}=\sqrt{3^{2}+\left(\frac{9}{4}\right)^{2}}=\frac{15}{4}$ (dm). $S E=\sqrt{S H^{2}+H E^{2}}=\sqrt{4^{2}+3^{2}}=5$. Then $S_{\text {side } 1}=p_{1} \cdot S E_{1}, S_{\text {side } 2}=p_{2} \cdot S E$, where $p_{1}, p_{2}$ are the semiperimeters of $A_{1} B_{1} C_{1} D_{1}$ and $A B C D$ respectively; $p_{2}=2 \cdot 6=12$ (dm), $p_{1}=2 \cdot \frac{9}{2}=9$ (dm). Then $S_{\text {side1 }}=9 \cdot \frac{15}{4}=\frac{135}{4}$ (dm ${ }^{2}$ ); $S_{\text {side } 2}=12 \cdot 5=60$ (dm²). The lateral surface of the truncated pyramid $S_{6 \text { side }}=S_{6 \text { side } 2}-S_{6 \text { side } 1}=60-\frac{135}{4}=\frac{105}{4}=26.25\left(\right.$ dm $\left.{ }^{2}\right)$.

Answer: 26.25 dm $^{2}$.