If we have a right tetrahedron, where one of the corners matches a cube's corner, there are three leg-faces and one hypote-face. It turns out that the square of the hypoteface area is equal to the sum of the squares of the three legface areas. (I don't know why, or if it works in more dimensions.) Any triangle can be a hypoteface, and the lengths of the X,Y,Z legs are easy to calculate: If the hypoteface edges are A,B,C, then X2+Y2 = A2, etc. and X2 = (A2+B2-C2)/2. The legface areas are XY/2 etc., the area squares are X2Y2/4, and it's not hard to check that the sum of squares of the three legface areas matches Hero^2.

it's already been pointed out that only non-obtuse triangles can occur as a "hypoteface". however, we can reduce to that case as follows. let K(a, b, c) = sqrt(s(s - a)(s - b)(s - c)) , which is heron's formula. then 16 K^2 = c^2 (2 a^2 + 2 b^2 - c^2) - (a^2 - b^2)^2 , which shows that K(a, b, c) = K(a, b, c') , where (c')^2 = 2 a^2 + 2 b^2 - c^2 . this corresponds to a simple geometric transformation: attach two (a, b, c) triangles along their c edge to make a parallelogram. then divide the parallelogram into two congruent (a, b, c') triangles using the other diagonal. the relationship c^2 + (c')^2 = 2 (a^2 + b^2) is the usual parallelogram law. it is easy to show that any obtuse triangle can be transformed into a non-obtuse triangle by a sequence of transformations of this ilk. (always attach along the edge opposite the obtuse angle; repeat as needed ... have patience ... you will get there!) mike