Calculating the height of an irregular tetrahedron

Hi All,

I am attempting to calculate the volume of an irregular tetrahedron and have found the are of the base via Heron's formula but cannot find out how to calculate the height. I'm assuming law of cosines is involved somehow.

You have triangle ABC whose sides measure 11, 20 and 21. P, Q, and R are the midpoints of AB, BC and AC respectively. Fold along PR, RQ and PQ until the vertices A B and C touch and calculate the volume. It is easy to see that the four triangles inside triangle ABC are congruent which is how I calculated the area of the base (~27). Yet I am remain uncertain how to find the height so I can find the volume.