Three sides $OAB, OAC$ and $OBC$ of a tetrahedron $OABC$ are right-angled triangles, $\angle AOB = \angle AOC = \angle BOC = 90^\circ$. Given that $OA = 7, OB = 2$ and $OC = 6$, find :-
$\sqrt{([\Delta OAB]^2 + [\Delta OAC]^2 + [\Delta OBC]^2 + [\Delta ABC]^2 + 37)}$
What I Tried: Here is a picture of the top view of the tetrahedron.
Because of the $3$rd dimension, this problem got more complicated and I cannot seem to figure out how to do it. For example is it fine to use Pythagoras Theorem on each of the triangles $OAB,OBC,OAC$ $?$ If yes, then I can find the sides of $\Delta ABC$ as in the $2$nd dimension, but that is not really going to help me finding the areas of $\Delta OAB$ , $\Delta OAC$ , $\Delta OBC$ , on the other hand I think this is getting complicated for me to understand what to be done here.
Can anyone help me? Thank You.