8. Given that the three sides of $\triangle A B C$ are $10$, $12$, and $14$, respectively, and that $\triangle A B C$ is folded into a tetrahedron along its three midlines. Then the diameter of the circumscribed sphere of the tetrahedron is $\qquad$
8. $\sqrt{55}$.

The folded quadrilateral forms an isosceles tetrahedron, with three pairs of opposite edges measuring $5, 6, 7$. The diameter of the circumscribed sphere is
$$
2 R=\sqrt{\frac{5^{2}+6^{2}+7^{2}}{2}}=\sqrt{55} .
$$