4. Fold a triangle with side lengths of $10$, $12$, and $14$ along its three midlines to form a tetrahedron. Then the diameter of the circumscribed sphere of the tetrahedron is $\qquad$ .
4. $\sqrt{55}$.

The tetrahedron has equal opposite edges, and the lengths of the three pairs of opposite edges are 5, 6, and 7.
As shown in Figure 4, let $A B=C D=5, A C=B D=7$,
$$
A D=B C=6 \text {. }
$$

Take the midpoints $M$ and $N$ of $B C$ and $A D$ respectively.
It is easy to see that $M N$ is the common perpendicular segment of $A D$ and $B C$, and noting that $A D=B C$, the midpoint $O$ of $M N$ is equidistant from points $A, B, C, D$, i.e., $O$ is the center of the circumscribed sphere of the tetrahedron $A B C D$, and the radius of the sphere is
$$
R=O A=\sqrt{\left(\frac{d}{2}\right)^{2}+\left(\frac{A D}{2}\right)^{2}}(d=M N) .
$$

Draw $N E \perp M C$. It is easy to see that $C E \perp$ plane $A E D$.
Since $N A=N D=N E=3$, we know $\angle A E D=90^{\circ}$.
Then $A E^{2}=A C^{2}-C E^{2}=7^{2}-d^{2}$,
$D E^{2}=D C^{2}-C E^{2}=5^{2}-d^{2}$,
$A E^{2}+D E^{2}=A D^{2}=6^{2}$.
Thus $7^{2}-d^{2}+5^{2}-d^{2}=6^{2} \Rightarrow d^{2}=19$
$$
\Rightarrow 2 R=\sqrt{d^{2}+A D^{2}}=\sqrt{55} \text {. }
$$