This is a multi-variable prob.
We know that a tetrahedron is a solid which has four vertices and has four triangular faces. For every side of a tetrahedron, you can take a vector whose length then is equal to the area of the face and further where its direction is pointing outward perpendicular to the face. Mark these vectors as v_1, v_2, v_3, and v_4.
1.) Show that v_1 + v_2 + v_3 + v_4 = 0
2.) Suppose that we have that 3 of the faces of the tetrahedron are perpendicular to each other (now it will look as though it could fit in the very corner of a box. Label these faces A,B,C and the area of the 4th face as D. Prove that D^2 = A^2 + B^2 + C^2
HINTS: you will need to use vector products - also, you may want to use a coordinate system such that one vertex will lie at the origin (and then from there consider edges as vectors)