A sphere can be inscribed in a regular quadrilateral truncated pyramid. The base edges are $18 \mathrm{~cm}$, and the edges of the top face are $8 \mathrm{~cm}$ long. What is the surface area of the truncated pyramid?
Connect the center of the sphere with the vertices of the pyramid. The combined volume of the 6 small pyramids thus obtained gives the volume of the frustum.

![](https://cdn.mathpix.com/cropped/2024_05_02_437ce2f85a2c4a7e2e31g-1.jpg?height=371&width=540&top_left_y=262&top_left_x=779)

As we know,

$$
V_{\mathrm{csg}}=\frac{m(T+t+\sqrt{T t})}{3}
$$

where $T$ is the area of the base, $t$ is the area of the top, and $m$ in our case is equal to the diameter $2R$ of the sphere inscribed in the pyramid.

The height of each of the small pyramids is $R$, and the sum of the areas of their bases gives the surface area of the frustum, $A$.

Thus, we can write the following equation:

$$
\frac{A R}{3}=\frac{2 R(T+t+\sqrt{T t})}{3}
$$

Substituting the known values and simplifying by $R / 3$, we find that $A=1064 \mathrm{~cm}^{2}$.