10. In tetrahedron $ABCD$, $AB=CD=6$, $AC=$ $AD=BC=BD=5$. Then the radius $r$ of the inscribed sphere is $\qquad$.
10. $\frac{3 \sqrt{7}}{8}$.

Let the midpoint of $C D$ be $E$. In $\triangle A B E$,
$$
\begin{array}{l}
A E=B E=\sqrt{5^{2}-3^{2}}=4, \\
E H=\sqrt{4^{2}-3^{2}}=\sqrt{7} .
\end{array}
$$

Therefore, the height $h$ on $B E$ is $h=\frac{6 \sqrt{7}}{4}=\frac{3 \sqrt{7}}{2}$, and $h$ is also the height of the tetrahedron $A B C D$.
Since the four faces of the tetrahedron $A B C D$ are equal, we have $r=\frac{h}{4}=\frac{3 \sqrt{7}}{8}$.