I am looking for a formula: $V=f(S_1,S_2,S_3,S_4)$, where $S_1$, $S_2$, $S_3$, and $S_4$ are the areas of the four faces.
We know $V=\dfrac{S_1.h_1}{3}=\dfrac{S_2.h_2}{3}=\dfrac{S_3.h_3}{3}=\dfrac{S_4.h_4}{3}$, where $h_1$, $h_2$, $h_3$, and $h_4$ are the corresponding altitudes.
So we need to find
$h_1=g(S_2,S_3,S_4)$
$h_2=g(S_1,S_3,S_4)$
$h_3=g(S_1,S_2,S_4)$
$h_4=g(S_1,S_2,S_3)$
Also, if all points of the tetrahedron are on a plane, the volume should be zero. Thus
Firstly, if the projection of point is out of $S1$ Area, $S_1+S_2=S_3+S_4$ then $V=0$
If the projection of point is in $S1$ Area, $S_1=S_2+S_3+S_4$ then $V=0$
So can we determine from the areas if volume is zero or not ?
Is it possible to find $V=f(S_1,S_2,S_3,S_4)$? Is the surface information enough to create a unique closed volume?
Thanks a lot for advice and answers.