Volume of Frustum
Yue Kwok Choy
A frustum is a truncated cone or pyramid; the part that is left when a cone or pyramid is cut by
a plane parallel to the base with the apical part removed. We concentrate on the right frustum and not
slant frustum as in the right-most diagram on the top.
We begin with :
1
Volume of prism  Bh
3
, where B is the base area and h is the height of prism.
Since the apical part is removed, we have:
Volume of frustum, V 
The height of the frustum
1
B2 h 2  B1h1 
3
…. (1)
 h  h 2  h1
…. (2)
2
B2 h 2

B1 h12
We also have :
…. (3)
Our aim is to remove the variables
From (2),
h 2  h  h1
(4)  (1) ,
V 
h1, h2
using (1) – (3) .
…. (4)
1
B2 h  h1   B1h1   1 B2 h  B2 h1  B1h1 
3
3

1
B2 h  B2  B1 h1 
3
2
h1 B 2
B1

B2
 h1 
From (3),
h2 
(6)  (2) ,
h  h1
2
B1
h 2  h1
B 2  B1
B1
…. (5)
B2
…. (6)
B1
h1

h1 
B1
B 2  B1
h
…. (7)
1
(7)  (5) ,
B1
1
V  B2 h  B2  B1 
3 
B2  B1

h

1
 B 2 h 
3 



B2  B1

B2  B1

h
B2  B1 
B1



1
B2 h  B1
3

h
B1  B1B 2  B 2
3
B2  B1 h


…. (8)
From (8), we can derive the followings:
(a) Volume of circular cone frustum,
V

h 2
2
r1  r1r2  r2
3

, where
r1 , r2
are the radii of the top and bottom circular bases.
(b) Volume of square frustum,
V

h 2
2
a 1  a 1a 2  a 2
3

, where
a1 , a 2
are the sides of the top and bottom square bases.
(c) Since the area of equilateral triangle is given by
3 2
a , where a is the side of the triangle, we have:
4
Volume of equilateral triangular frustum,
A
V

h 2
2
a 1  a 1a 2  a 2
16
, where

a1 , a 2
are the sides of the top and bottom triangular bases.
2