10,11

In a regular truncated quadrilateral pyramid, the height is 2, and the sides of the bases are 3 and 5. Find the diagonal of the truncated pyramid.
Let's make a section through the opposite lateral edges $A A_{1}$ and $C C_{1}$ of the given truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ with bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}\left(A B=5, A_{1} B_{1}=3\right)$. Let $O$ and $O_{1}$ be the centers of the bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, respectively. The cutting plane passes through the height $O O_{1}$ of the truncated pyramid. In the section, we obtain an isosceles trapezoid $A A_{1} C_{1} C$ with bases $A C=5 \sqrt{2}$ and $A_{1} C_{1}=3 \sqrt{2}$. Let $A_{1} K$ be the height of the trapezoid. Then

$$
\begin{gathered}
A_{1} K=O O_{1}=2 \\
A K=\frac{1}{2}\left(A C-A_{1} C_{1}\right)=\frac{1}{2}(5 \sqrt{2}-3 \sqrt{2})=\sqrt{2} \\
C K=A C-A K=5 \sqrt{2}-\sqrt{2}=4 \sqrt{2}
\end{gathered}
$$

From the right triangle $A_{1}$ КС, we find that

$$
A_{1} C=\sqrt{A K^{2}+C K^{2}}=\sqrt{4+32}=6
$$

## Answer

6.

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