I discovered this way of finding the formula for a truncated pyramid or cone, known as a frustum, by using geometric series.  It is certainly not the standard way of deriving the equation, but I liked the way it works out.  I hope you will also.


  The two dimensional version of this is to find the area of a trapezoid, so let's start with finding the area of a trapezoid using geometric series and then extend the result to 3 dimensions.  The procedure used to find the area of a trapezoid will be a bit roundabout, but it will be very easy to extend the results to 3 dimensions.

The strategy for finding the area of the trapezoid is to build a triangle by layering copies of the trapezoid as show in the illustration.  We will use the geometric series to express the area T of the triangle in terms of the base trapezoid .  We will then find another geometric series to find the area of the triangle.  We will then equate the two expressions for the area of the triangle to allow us to solve for the area of the trapezoid.

Trapezoid Geometric Series
The trapezoid whose area A we want to find is ABCD given the lengths of the two bases b1 and b2 height h.  The trapezoid BEFC above ABCD has b2 for its bottom base instead of b1.  We want trapezoid BEFC to be similar to ABCD.  Since b2 is the bottom base for BEFC, the scaling factor r = b2/b1. By taking the height of BEFC to be rh, it will be similar to ABCD.  The scaling factor for the next trapezoid applies the scaling factor r to BEFC and so has a scaling factor of r2. b3 = r2 b1 and for the nth trapezoid the scaling factor will be rn.

The next thing we want to do is to find the area of the nth trapezoid in terms of A, r and n.  Since the scaling factor is rn and area is proportional to the square of the scaling factor, the area of the nth trapeazoid is (rn)2A =  (r2)nA .  The area of the triangle is therefore:
T = A(1 + r2 + (r2)2 + (r2)3 + ...).  This is just an infinite geometric series in r2, so we get area of of triangle =
T = (1 - r2)    ( equation 1)

Next, let's find the area of the triangle in terms of b1 and h, using a geometric series to find the altitude H of the triangle in terms of h.  The scaling factor of r again applies, and so we get H = h(1 + r + r2 + r3 + ...) = h/(1 - r).  

The area of the triangle is therefore T = 1/2 b1 H = 1/2 b1 h /(1 - r). (equation 2)

We could at this point substitute  b2/b1 for r in the last expression for the area of the triangle, but let's keep it in terms of r and equate the two expressions for the area of the triangle.  Equating equation 1 with equation 2 gives

A/(1 - r2) =  1/2 b1 h /(1 - r)
A = 1/2 b1 h (1 - r2)/ (1 - r)

(1 - r2)/ (1 - r) simplifies to (1 + r).  Notice that (1 - r2)/ (1 - r)  is a speical case of the formula for geometric series.  (1 - r(n+1))/ (1 - r). In this case, n = 1 giving the very short geometric series 1 + r.  This idea will be more useful when we get to the three dimensional case.

Using the simplification, we get
A = 1/2 b1 h (1 + r)
Substituting b2/b1 for r,  A = 1/2 b1 h (1 + b2/b1) = 1/2 h (b1 + b2), which is the standard formula for the area of a trapezoid.

It is now relatively easy to find the volume of a square truncated pyramid using the technique we used for trapezoids.

Truncated pyramid series

 We have b1 for the lower base sides and b2 for the upper base sides.  h is again the height and V is the volume that we want to find.  When we stack up the truncated bases in an infinte series we get a pyramid. The scaling factor is again  r = b2/b1 and once again the scaling factor for the nth figure is rn.  The volume of the nth figure is proportional to the cube of the scaling factor, and so will be V(rn)3 = V(r3)n.   The volume P of the pyramid is V(1 + r3 + (r3)2 + (r3)3 + (r3)4 + ...), which is a geometric series in r3, so P = V/ (1 - r3).

The height of the pyramid is h/(1 - r) ,  Using this in the formula for the volume of a pyramid, P = 1/3 b12 h /(1 - r)

Equating the two equations for P:
V/ (1 - r3) = 1/3 b12 h/(1 - r)
V = 1/3 b12 h(1 - r3)/(1 - r)
Recognizing (1 - r3)/(1 - r) as the formula for a geometric series, substituting for r and simplifying gives us:
V = 1/3 b12 h (1 + r + r2) =  1/3 b12 h (1 + b2/b1  +b22/b12)  =  1/3 h (b12+ b1b2 + b22)

Do you see a pattern in going from two dimensions to three dimensions?  What would you expect the formula to be in four dimensions?

Solution Without Infinite Series

Here is an alternative way to get the  two equations for the volume of the pyramid.  It avoids using the two infinte geometric serices to derive the two equations.  We still make use of finite geometric series when combining thte two equations.  

The basic idea is actually pretty simple.  The small pyramid above the frustum is similar to the whole pyramid, with scaling factor r =
b2/b1. If, for example, r = 1/3 then h' is 1/3 of the height of the whole pyramid and therefore h is 1 - 1/3 = 2/3 of the total height, which to say that the total height is 3/2 h.  A similar argument applies to volumes, except that the scaling factor is r3

Let V' = volume of small pyramid lying on top of frustum.
P = V + V'
V'/(V+V')  = r3,
1 - V'(V + V')  = 1 -  
(V+V')/(V+V') - V'/(V+V') = 1 -  r3   
V/(V+V') =   1 -  r3  
V/P = 1 -  r3,  P = V/(1 -  r3), as we got before.

Let h' = height of small pyramid lying on top of the frustum. The height of the whole pyramid is h + h'.

 h'/(h + h')  = r.
 1 - h'/(h + h') = 1 - r
(h + h')/(h + h')
- h'/(h+h') =  1 - r
h/(h + h') = 1 - r
h + h' = h/(1 - r)
We can now plug this value into the second  equation for the volume P of the pyramid.

More General Case
The formula given for the frustum only covers the case of square pyramids.  Suppose instead we are given a base with any general shape having area B1 and the second base has area B2.   We know that B2/ B1  = r2, where r is the scaling factor.   Solve for r in terms of  B1 and B2 .  Then use this formula to find the volume of the frustum in terms of h, B1,  and  B2. Note that the two formulas for volume that were used are both in terms of r and are still valid, so the substitution need only be done at the end, when we substutue for r.    Use the formual V = 1/3 Bh for the volume of a pyramid with base area B and  height h.