11.129. The height of a regular truncated quadrilateral pyramid is 3 cm, its volume is $38 \mathrm{~cm}^{3}$, and the areas of the bases are in the ratio $4: 9$. Determine the lateral surface area of the truncated pyramid.
Solution.

Let the area of the upper base be $S_{1}=4 x$ cm $^{2}$. Then the area of the lower base $S_{2}=9 x \mathrm{~cm}^{2}$, and since the volume of the truncated pyramid

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0840.jpg?height=434&width=557&top_left_y=954&top_left_x=717)

Fig. 11.25 $V=\frac{H}{3}\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, then $4 x+9 x+\sqrt{36 x^{2}}=38, x=2 ; S_{1}=8 \mathrm{~cm}^{2} ;$ $S_{2}=18 \mathrm{~cm}^{2}$. Then the side of the lower base $a=\sqrt{S_{2}}=3 \sqrt{2}$ cm, the side of the upper base $b=2 \sqrt{2}$ cm, the perimeters of the bases are respectively $p_{2}=12 \sqrt{2}$ cm, $p_{1}=8 \sqrt{2}$ cm. $N$ is the midpoint of the edge $C D, K$ is the edge $C_{1} D_{1}$ respectively of the lower and upper bases of the given truncated

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0841.jpg?height=668&width=650&top_left_y=78&top_left_x=79)

Fig. 11.26

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0841.jpg?height=615&width=572&top_left_y=76&top_left_x=703)

Fig. 11.27

pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.25). Then $K N-$ is its apothem, $O N=\frac{a}{2}, O_{1} K=\frac{b}{2}$.

Draw the height $K P=O O_{1}=3 \mathrm{cm} . P N=O N-O_{1} K=\frac{a-b}{2}=\frac{\sqrt{2}}{2}$ cm. From $\triangle K P N\left(\angle K P N=90^{\circ}\right): K N=\sqrt{K P^{2}+P N^{2}}=\sqrt{\frac{19}{2}}$ cm. The lateral surface $S_{6}=\frac{p_{1}+p_{2}}{2} \cdot K N=10 \sqrt{2} \cdot \sqrt{\frac{19}{2}}=10 \sqrt{19} \mathrm{~cm}^{2}$.

Answer: $10 \sqrt{19} \mathrm{~cm}^{2}$.