Height of irregular tetrahedron, center of sphere I know the lengths of all edges of a tetrahedron. How can i connect that with the height of the irregular tetrahedron? Is there a relation between the height of tetrahedron and the radius o circumsphere?

If a tetrahedron is not regular, it does not make sense to talk about _the_ heigth, since there are four heights, one for each face. On the other hand, if you know the edge lengths you also know the volume $V$ through the Cayley-Menger determinant and the area of any face through Heron's formula, so you know the length of every height. The circumradius $R$ can be found through $$ R = \frac{\Xi}{6V} $$ where $\Xi$ is the area of a triangle with side lengths $a\cdot a', b\cdot b', c\cdot c'$, with $a\leftrightarrow a',b\leftrightarrow b',c\leftrightarrow c'$ being opposite edges in the tetrahedron.