Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is
Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed sphere. Let the three edges at the same vertex of the rectangular parallelepiped be $x$, $y$, and $z$, respectively. Then, from the conditions, we have $y^{2}+z^{2}=9$, $z^{2}+x^{2}=10$, and $x^{2}+y^{2}=13$. According to this, the radius of the required circumscribed sphere is
$$
\frac{1}{2} \sqrt{x^{2}+y^{2}+z^{2}}=2 \text {. }
$$