In the tetrahedron $A-BCD$, $AB=CD=5$, $AC=BD=\sqrt{34}$, $AD=BC=\sqrt{41}$, then the radius of the circumscribed sphere of the tetrahedron $A-BCD$ is
(8) $\frac{5 \sqrt{2}}{2}$ Hint: Construct a rectangular prism with edge lengths of $3$, $4$, and $5$, then the circumscribed sphere of the tetrahedron $A-B C D$ is the same as the circumscribed sphere of the rectangular prism, so
$$
2 R=\sqrt{3^{2}+4^{2}+5^{2}}=5 \sqrt{2},
$$

thus $R=\frac{5 \sqrt{2}}{2}$.