Unizor - Geometry3D - Prisms - Volume of Parallelepiped
Channel: Zor Shekhtman on UNIZOR Education
Date: 09/07/2015

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The volume of a parallelepiped (as a volume of any solid body) is an additive measure that quantifies the space occupied by it.
The unit of volume is a cube with the length of each edge equaled to a linear unit. So, if linear unit is 1 meter, the unit of volume is 1 cubic meter (1m³).

Consider a rectangular parallelepiped with three edges sharing a vertex A having linear lengths of a, b (along the sides of a base) and c (height).
For now, consider these three lengths as integer numbers. Obviously, the number of unit cubes that fill the area of the base of this parallelogram in one single layer is S=a·b.
There are c such layers along the height of the parallelepiped, so the total number of unit cubes that fill this rectangular parallelepiped equals to V=S·c=a·b·c.

Next step is to consider the lengths of edges to be rational numbers. We can always assume them to have the same common denominator, so
a=p/n; b=q/n; c=r/n
A unit cube can be filled with smaller cubes with edges of 1/n. The number of these small cubes is, obviously, n³. So, it's natural to assume that the volume of a cube with the edge of 1/n is V=1/n³.
Since volume is an additive measure, we can count the number N of small cubes with the edge of 1/n that fit into our parallelepiped with above mentioned lengths of edges. Then the volume of a parallelepiped would be this number N multiplied by a volume of a small cube 1/n³. Obviously, N=pqr.Therefore, the volume of a parallelepiped is
V = (pqr)/n³ =
= (p/n)·(q/n)·(r/n) = abc
Again, as we see, the volume of a rectangular parallelepiped equals to a product of lengths of three edges sharing one vertex.

Extending without a rigorous proof this result to irrational lengths, we can state that, in general, the volume of a rectangular parallelepiped with lengths of three edges sharing the same vertex equaled to a, b and c is equal to
V=abc

Let's consider a more general case of a right parallelepiped. The difference between right and rectangular parallelepipeds is that the base of the right one is a general parallelogram, while the base of the rectangular one is a rectangle. In both cases the side edges are perpendicular to bases.

Assume the bottom base of our right parallelepiped is a parallelogram ABCD and the top base is, correspondingly, A'B'C'D'.
Within the bottom base draw perpendiculars BM and CN from vertices B and C onto side AD (or its continuation). Assume for definitiveness that point M lies in between points A and D, while point N is outside this segment.

Obviously, triangles ΔABM and ΔDCN are congruent - a trivial statement from plane geometry.

Now draw a plane through edge BB' and line BM on the bottom base. It cuts from our parallelepiped a right prism with a triangles ΔABM and ΔA'B'M' as bases and side edges AA', BB' and MM'.
Draw another plane through edge CC' and line CN. Consider a right prism with bases ΔDCN and ΔD'C'N' and side edges DD', CC' and NN'.

Two right prisms with triangles as bases are congruent since their bases and edges are congruent - an easy to prove statement. Therefore, their volumes are equal. What's interesting is that if we cut off the first prism from our original right parallelepiped and add the second, we convert our right parallelepiped into a rectangular one. This new rectangular parallelepiped has rectangles BCNM and B'C'N'M' as bases and segments BB', CC', NN' and MM' as side edges.

Since we formed this new rectangular parallelepiped by cutting one triangular prism from the original right parallelepiped and adding another, equal in volume, the volumes of the original right parallelepiped and a new rectangular one are the same.

The volume of the latter equals to a product of an area of the base S = BM · MN by the height BB'. But the area of the base (rectangle) BCNM equals to the area of the original base (parallelogram) ABCD. Therefore, the volume of the right parallelepiped with base ABCD also equals to a product of the area of the base (parallelogram) ABCD, that is S = AD · BM, and the height of the parallelepiped AA'.