# How to find the volume of a irregular tetrahedron, when I know the length of all sides?

What I have been told: There's a irregular tetrahedron(pyramid with a base of a triangle), I know that three edges that form the tip are of length 2(a),3(b) and 4(c) and all edges at the one tip, where I have been given all the lengths are crosswise, so ∠ab = 90 ; ∠bc = 90 and ∠ac = 90.

What I know how to do: From here I can easily calculate all six edges of the tetrahedron and also the surface are.

What I am asked to do:find the volume.

Any tips and ideas would be awesome

Apologies for the poor english and if I wasn't clear, please tell me.

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That would be out of my league, as I am what would in the USA be junior high - i think, in advanced class, but still. When I apply the equation I get the given answer, but I do not fully understand how. –  user1506630 Feb 5 '14 at 21:25
The fact that you have three right angles at a vertex makes taking the Cayley-Menger determinant way overkill. Just take $V=abc/6$ –  Nate Feb 5 '14 at 21:26

Hint: If I read you correctly $c$ is already perpendicular on the plane generated by $a$ and $b$. No need to hunt for height. The base opposing this height is a solvable right triangle.

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Yes, the apothem of a and b , c, and a median form a right triangle... I just got it, embarrassingly slowly... thanks. –  user1506630 Feb 5 '14 at 21:42

If there are 3 edges length a,b and c perpendicular to each other, then

Volume = $1/3* c * a b/2 = a b c/ 6$

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