Tetrahedron volume relation to parallelepiped and pyramid

In the unit cube, consider the vertices with $x \le y \le z$; these form a tetrahedron. Indeed, for any of the six possible orderings of the variables, you get a tetrahedron, and the interiors of these tets are disjoint, and every point of the unit cube lies in one of the tets. In fact, all the tets have the same shape (the long edge is the main diagonal from $(0,0,0)$ to $(1,1,1)$, etc.) So the volume of each is 1/6 the volume of the cube.

That's only for a cube, but other cases follow by applying linear transformations.