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### Volume of the Frustum of a Pyramid

```
Date: 10/31/2001 at 15:21:37
From: Kerry
Subject: Formula of frustum of a pyramid

I am trying to figure out how to derive the formula for the volume of
a frustum of a pyramid.  As it states in your formulas:

V = (h(B1+B2+sqrt[B1B2])/3

I understand everything except where you get the sqrt[B1B2] - what
does that part represent in the frustum?
```

```
Date: 10/31/2001 at 22:51:27
From: Doctor Peterson
Subject: Re: Formula of frustum of a pyramid

Hi, Kerry.

There are two main ways to derive this formula: dissection and
subtraction. Let's try both.

I'll use the dissection method specifically for a square frustum, from
which you can apply the formula to other cross-sections using
Cavalieri's theorem. Here is a top view of a square frustum:

+---+----------+---+
| \ |         d| / |
+---+----------+---+
|   |          | d |
|   |          |   |
|   |        s1|   |s2
|   |          |   |
|   |          |   |
+---+----------+---+
| / |         d| \ |
+---+----------+---+

I have cut vertically through the sides of the top square, dividing
the pyramid into nine parts:

a central square prism with volume s1^2*h

four corners that together form a pyramid of volume (2d)^2*h/3

four triangular prisms (on their sides), each with volume s1*d*h/2

The total volume is then

V = [s1^2 + 1/3 (s2-s1)^2 + s1(s2-s1)]h

= [s1^2 + 1/3 s2^2 - 2/3 s1*s2 + 1/3 s1^2 + s1*s2 - s1^2]h

= [1/3 s1^2 + 1/3 s2^2 + 1/3 s1*s2]h

= [s1^2 + s1*s2 + s2^2]h/3

Since B1 = s1^2 and B2 = s2^2, this is

V = [B1 + sqrt(B1*B2) + B2]h/3

Now let's do it by finding the difference between the whole pyramid
and the part cut off. This doesn't depend on the shape of the bases at
all. Just look at the similar triangles formed in a side view:

+ -------------------
/  \                ^
/     \              |k
/        \            v
+-----------+ -----------
/      a       \        ^
/                 \      |h
/                    \    v
+-----------------------+ ---
b

It doesn't matter what a and b actually are; since horizontal cross-
sections are all similar, any linear measurement is proportional to
the square root of the area, so we know

b/a = sqrt(B2/B1)

First we have to find the height of the original pyramid, using
similar triangles:

b/a = (h+k)/k = h/k + 1
so
h/k = sqrt(B2/B1) - 1

1
k/h = ---------------
sqrt(B2/B1) - 1

sqrt(B1)
= -------------------
sqrt(B2) - sqrt(B1)

sqrt(B1) (sqrt(B2) + sqrt(B1))
= ------------------------------
B2 - B1

Now the volume of the frustum is the volume of the whole pyramid minus
the volume of the top part:

V = B2(h+k)/3 - B1 k/3

= B2 h/3 + (B2-B1)k/3

= B2 h/3 + (B2-B1)h/3 * k/h

= B2 h/3 + sqrt(B1) (sqrt(B2) + sqrt(B1)) h/3

= [B2 + B1 + sqrt(B1B2)]h/3

The latter method is equivalent to that given in the Dr. Math archives
for a frustum of a cone (since a pyramid is just a special cone):

Deriving the Volume of a Frustum
http://mathforum.org/dr.math/problems/taylor5.6.98.html

Derivation of the Formula for the Frustum
http://mathforum.org/dr.math/problems/rizza08.09.99.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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