THE POSSIBILITY OF ESTIMATING THE VOLUME OF A SQUARE FRUSTRUM
USING THE KNOWN VOLUME OF A CONICAL FRUSTRUM
SAMUEL OLU OLAGUNJU
Adeyemi College of Education
NIGERIA
Email: [email protected]
ABSTRACT
This paper considers the calculation of the volumes of a Conical Frustrum and a Square Frustrum, noting
how related they are and how the knowledge of one can help in estimating the other. An earlier paper
had noted from records that the Egyptians used a process of dividing the pyramid into two portions first,
calculating the areas as A1 and A2, and then obtaining the volume of the pyramid as one-third the height
multiplied by the sum of the two different areas A1 and A2 added to the square-root of the product of
1
the two areas {i.e. V  h A1  A2  A1 A2 }. This paper reduces the cumbersome nature of such
3
calculations. It was noted that Macrae et al (2001) gave the volume of a Conical Frustrum, here
1
designated as VCF as VCF  h( R 2  Rr  r 2 ) ; VCF being the Volume of Square-based Frustrum; R,
3
the Radius of the large Circular base; r, the Radius of the small Circular top; and h = the height of the
Conical Frustrum. In the process of developing a less-cumbersome model for the volume of a Squarebased frustrum, Olagunju (2011) considered formulas for complete Pyramids, including that of Circular
1
Frustrum, to arrive at a proven formula for a Square Frustrum as VSF  h( D 2  Dd  d 2 ) Where
6
VSF  Volume of Square Frustrum, D = Diagonal of the large Square base, d = diagonal of the small
Square top, and h = the height of the Square Frustrum. The consideration process confirmed the
possibility of estimating the volume of one given the other, if the Top and Base Diagonals of the Square
Frustrum respectively have equal lengths with the Top and Base Diameters of the Conical Frustrum,


Keywords: Volume, Pyramids, Frustrum, Conical, Square, Diameters, Diagonals
1.0 Introduction
The basis of progress in any endeavor in life, especially scientific progress, lies in improvement. This
is why everybody yarns for improvement. Similarly, the essence of education is to find a way of
improving on an earlier situation. Thus, learning and mastering the use of existing models will not be
sufficient, as it is more useful to see how such models could be improved upon for the benefit of
mankind, and at times it could be useful to establish the relationship between two models. This forms
the basis on which the established formulae for the Volume of a Conical (or Circular-based) Frustum
and that of a Square (or Square-based) Frustum were considered for the purpose of establishing a
relationship between the two, approaching the issue from their respective Diameters and Diagonals.
2.0 Purpose of the Study
The essence of this work is to establish the possibility of estimating the volume of Square frustums from
that of a known Conical frustum (and vise-versa), provided that certain given condition is fulfilled. This
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will help teachers and students of Mathematics as well as constructing establishments to safe some time
in their effort in estimating such.
3.0 Some Necessary Clarifications
The following clarifications should please be noted
3.1 Pyramids and Frustums
According to Hart (2005), as noted in Olagunju (2011), a Pyramid is a Polyhedron having one polygonal
face (called ‘base’) and all other faces as Triangles, meeting at the Vertex (called ‘Apex’). A special
kind of Pyramid whose base is circular and all slant-edge lines meet at the vertex is referred to as a Cone
or a Conical Pyramid. When a part of a Pyramid is chopped off through the apex, it becomes a Truncated
Pyramid, usually referred to as a Frustum.
3.2 Classification of Pyramids
Pyramids are classified by their dimensions. While a Regular pyramid is one with a base with regular
polygon (e.g. Square-Based, Rectangular-Based), a Right pyramid is one whose apex is joined to the
center of the base by a perpendicular line. Another type with one single cross-sectional shape having
lengths scaling linearly with its height is referred to as an Arbitrary pyramid.
3.3 Pyramidal Frustums
A Truncated Pyramid is one whose part has been chopped-off to a given height. Frustums are named
after the shape of their base. While a Square Frustum is one whose Base and Top are both in the form
of Squares (usually, the length of one end-face is smaller than the other), a Conical Frustum (Truncated
Cone) is one whose Base and Top are circular (usually, the radius of one end-face is smaller than the
other).
4.0 Related Existing Models
4.1 Volume of a Conical Frustum
Considering an existing model of the Conical Frustum where the volume of the Frustum is obtained by
subtracting the chopped-off top volume from the big cone volume,
1
VBC  R 2 H ;
we have the Volume of a big cone as
3
1
and the Volume of the chopped-off cone as VSC  r 2 x ,
3
Going by Weisstein’s (2006) Concise Encyclopedia of Mathematics, irrespective of the base shape or
position of the apex relative to the base,
1
Pyramidal volume is V  Ab h
3
where Ab = Area of the base ,
and h = the height (perpendicular distance of the apex from the base) remembering that the capacity
of the pyramid equals one-third of a cylinder of same height and same base-radius.
Then, according to Kalejaiye et’al (2001), the volume of the Conical Frustum ‘ VCF ’ so formed is given
1
1
by VCF  R 2 H  r 2 x
3
3
where R = radius of Big Cone,
H = Height of Big Cone,
r = radius of Small Cone, and x = height of Small Cone.
1
This leads to Hero’s formula, VCF  h( R 2  Rr  r 2 ) .
3
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4.2 Volume of Pyramids
As earlier discussed, and as observed by Harris and Stocker (1998), the Volume of a pyramid is given
as one-third of the product of base-area and perpendicular height.
1
i.e. Volume = (base-area x  height)
3
Thus, considering a Square Pyramidal Frustum, if l = b (square base.) and base area = l x b,
Then, VP  13 l x b x h  13 l 2h  13 b2h
From the above, we obtain the volume of a Truncated Square Pyramid thus:
If the volume of a Big Square-based Pyramid is VBP  13 L2 H
And if the volume of the chopped-off small Pyramid is VSP  13 l 2 x
Then, the volume of the Truncated Square-based Pyramid (i.e. Pyramidal Frustum) so formed is given
by
VPF = VBP - VSP
1
VPF  13 L2 H  13 l 2 x  ( L2 H  l 2 h)
3
Where: l = length of Small-Square-Top (Base of chopped-off top pyramid),
x = height of Small Pyramid
L = Length of Big-Square-Base Pyramid,
H = Height of Big Pyramid [(H = x + h), h = height of Frustum],.
4.3 Some Guiding Principles for the Square-Based Model
Axiom I: Since the original Pyramid has a Square base, then, the top chopped-off Small Pyramid also
has a Square base.
Axiom II: The ratio of the height of the top chopped-off pyramid to the height of the original big
pyramid equals the ratio of the diagonal of the top chopped-off pyramid to the Diagonal of the original
big pyramid.
x:H =d:D
Lemma I: Since the Big Base-Square Length and the Small Base-Square length are in the ratio L : l ,
and this affects their diagonals, then, D2 : d2 = L : l
4.4 Volume of a Square Frustum
Let the base and top diagonals of the pyramidal frustum be D and d respectively. If its height is h, and
the Volume is designated as VSF ,
 Volume of Square Frustum will be the difference between the
Volume of large Square Pyramid and Volume of chopped Square Pyramid.
Considering Diagram 3.41 below, where
L = Length of Big-Square-Base,
l = length of Small-Square-Base,
x = height of Small Pyramid,
h = height of Frustum,
H = Height of Large Pyramid, (height of small pyramid + height of Frustum)
Which implies that the
Height of Pyramid PEFGI = h + x = H
Area of Square base EFGI = L. L = L2
Height of Pyramid PJKMN = x
Area of Square base JKMN = l. l = l2
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4.41: Diagram I: Pyramid
P
x
d
N
H
J
M
K
h
F
D
E
G
I
If the volume of a Large Square Pyramid is VBP  13 L2 H
And if the volume of the chopped smaller Pyramid is VSP  13 l 2 h
1
Then, volume of the Square Frustum formed is VSF  VBP  VSP  ( L2 H  l 2 h)
3
Now,
Volume of Frustum JKMNEFGI is given by
Volume of Pyramid PEFGI – Volume of Pyramid PJKMN
= Volume of PEFGI – Volume of PJKMN
i.e. VJKMNEFGI = VPEFGI - VPJKMN
Thus,
1
1
VJKMNEFGI = L2 ( x  h)  l 2 x
(4.1)
3
3
If D is the diagonal of the larger square EFGI above,
And d is the diagonal of the smaller square JKMN,
Then, by Pythagoras,
1
From PJM , d  2l 2  l 2  d 2
(4.2)
2
And
1
From PEG , D  2L2  L2  D 2
(4.3)
2
Thus, substituting (4.2) and (4.3) in (4.1), we have
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1  D2
d2 
x
VJKMNEFGI =  x  h  
3 2
2 
11 2
=
D x  h   d 2 x
32
1
= D 2h  D 2 x  d 2 x
6
1
= D 2 h  x( D 2  d 2 )
6






(4.4)
But considering similar triangles PJM and PEG,
 xD  xd  hd or
x( D  d )  hd
x xh

d
D
hd
(4.5)
Dd
Substituting (4.5) in (4.4), we have
1 2
hd

D h
( D  d )(D  d )
VJKMNEFGI
=

6
Dd

1 2
D h  hd ( D  d )
=
6
1 2
D h  hDd  hd 2
=
6
h 2
= D  Dd  d 2
(4.6)
6
Equation (4.6) is the lagsamolu formula obtained for the Square Frustum.
This was well illustrated and found useful and less-cumbersome.
Hence, x 






5.0 Establishing a Relationship between Square Frustum and Conical Frustum
5.1 Additional Guiding Principles for the Relationship
Axiom III: The respective Top and Base Diameters of the Conic Frustum must be equal to the respective
Top and Base Diagonals of the Square Frustum.
Axiom IV: From axiom III, it follows that the ratio of the respective Top Diameter to Base Diameter of
the Conic Frustum must be same as the ratio of the Top Diagonal to the Base Diagonal of the Square
Frustum.
Lemma II: Since the Big Radius Length and the Small Radius length are in the ratio R : r, this affects
their diameters. i.e. , then, R2 : r2 =D2 : d2
5.2 Relationship between the Volumes of the Two Frustums:
Considering diagrams (5.31) and (5.32) where the Top and Base Diagonals of the Square Frustum in
(5.31) are equal to the Top and Base Diameters of the Conical Frustum in (5.32), we consider their
Volumes as follows:
From diagram (5.31), we note that
h
Volume of Square Frustum = D 2  Dd  d 2
(5.1)
6
From diagram (5.32), we also note that
1
Volume of Conical Frustum =  h( R 2  Rr  r 2 )
(5.2)
3

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5.3: Diagrams II
Consider the diagrams below:
5.31 Square Frustum:
d
h
D
5.32 Conical Frustum:
d
h
D
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1
D
2
Hence, equation (5.2) becomes
But R 
Conical Frustum Volume
2
2
1  D  D d  d  
=  h  
  
3   2 
2 2  2  
2
2
1 D
Dd d 
= h

 
3 4
4
4 
11
=
 h  D 2  Dd  d 2 
43
11
=
 h  D 2  Dd  d 2 
26
(5.3)
By Axiom III,
Top Diagonal of Square Frustum = Top Diameter of Conical Frustum
Base Diagonal of Square Frustum = Base Diameter of Conical Frustum
11
Thus, Conical Frustum Volume =
 h  D 2  Dd  d 2 
26
1 1

=   h  D 2  Dd  d 2  
2 6

1
=  Volume of Square Frustrum 
(5.4)
2
Hence, it is noted from equation (5.4) that the Volume of a Conical Frustum equals half the product of
Pi (  ) and the Volume of a Square Frustum.
6.0 Illustrations
Here, we attempt to illustrate this finding by considering certain situations below:
6.1 Illustration I
Consider the situation where r:R = d:D = 1:2
If h = 5, r 2  8 , R 2  (2r ) 2  32 , Rr  R2 r 2  16 ,
Then, h = 5, d 2  (2r ) 2  32 , D 2  (2 R)2  128 , Dd  D2 d 2  64
Then, we have it that
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
5
280

=  (32  16  8) =
3
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
5
1120
= (128  64  32) =
6
6
1120
1
Multiplying
by  , we have
2
6
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(6.11)
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280
1  1120 
(6.12)
VS   

 
2  6 
3
Hence, from (6.11) and (6.12), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
6.2 Illustration II
Consider the situation where r:R = d:D = 1:3
If h = 5, r 2  8 , R 2  (3r ) 2  72 , Rr  R 2 r 2  24 ,
Then, h = 5, d 2  (2r ) 2  32 , D 2  (2 R)2  288 , Dd  D2 d 2  96
Then, we have it that
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
5
520
=  (72  24  8) =
(6.21)

3
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
5
2080
= (288  96  32) =
6
6
2080
1
Multiplying
by  , we have
6
2
1  2080 
520
VS   

(6.22)
 
2  6 
3
Hence, from (6.21) and (6.22), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
6.3 Illustration III
Consider the situation where r:R = d:D = 1:4
If h = 5, r 2  8 , R 2  (4r )2  128 , Rr  R2 r 2  32 ,
Then, h = 5, d 2  (2r ) 2  32 , D 2  (2 R) 2  512 , Dd  D2 d 2  128
Then, we have it that
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
5
840

=  (128  32  8) =
3
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
5
3360
= (512  128  32) =
6
6
3360
1
Multiplying
by  , we have
6
2
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(6.31)
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1  3360 
840
(6.32)
VS   

 
2  6 
3
Hence, from (6.31) and (6.32), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
6.4 Illustration IV
Consider the situation when h = 0.8, d 2  0.18 , D 2  0.72 , Dd  0.36
Then, we have it that h = 0.85, r 2  0.045 , R 2  0.18 , Rr  0.09
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
0.8
=
(6.41)
 (0.18  0.09  0.045) = 0.084
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
0.8
=
(0.72  0.36  0.18) = 0.168
6
1
Multiplying 0.168 by  , we have
2
1
VS    0.168   0.084
(6.42)
2
Hence, from (6.41) and (6.42), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
6.5 Illustration V
Consider the situation involving measurements less than 1.
If h = 2.7, d 2  0.72 , D 2  11.52 , Dd  2.88
Then, we have it that h = 2.7, r 2  0.18 , R 2  2.88 , Rr  0.72
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
2.7
 (2.88  0.72  0.18)
=
= 3.402
(6.51)
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
2.7
(11.52  2.88  0.72) = 6.804
=
6
1
Multiplying 6.804 by  , we have
2
1
VS    6.804   3.402
(6.52)
2
Hence, from (6.51) and (6.52), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
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5.6 Illustration VI
Consider the situation with decimals greater than 1.
when h = 10.2, d 2  18.0 , D 2  112.5 , Dd  45.0
Then, we have it that h = 10.2, r 2  4.5 , R 2  28.125 , Rr  11.25
For the Conical Frustum, we have
1
VC   h( R 2  Rr  r 2 )
3
10.2
=
= 149.175
(6.61)
 (28.125  11.25  4.5)
3
For the Square Frustum, we have
h
VS   D 2  Dd  d 2 
6
10.2
=
(112.5  45.0  18.0) = 298.35
6
1
Multiplying 298.35 by  , we have
2
1
VS    298.35   149.175
(6.62)
2
Hence, from (6.61) and (6.62), it is clear that the Volume of a Conical Frustum equals half the product
of Pi (  ) and the Volume of a Square Frustum.
Conclusion:
Given the above analysis and illustrations, this now confirms the existence of a strong relationship
between the Volume of a Conical Frustum and that of a Square Frustum, indicating that half the product
of Pi (  ) and Volume of a square Frustum equals the Volume of a Conical Frustum, and that the known
volume of one can therefore be used to estimate the unknown volume of the other, provided that the
Top and Base radii of the Conical Frustum are of the same ratio as the respective Top and Base diagonals
of the Square Frustum.
Precaution:
Since every model has its own precaution(s), it should be noted that this finding can only be used
successfully if it is ascertained that the Top and Base radii/diameters of the Conical Frustum are
respectively of equal ratio as the Top and Base diagonals of the Square Frustum. Otherwise, it may fail.
Recommendation:
It is recommended that the students and establishments willing to estimate the volume Square frustum
may now do so conveniently through the volume of a Conical frustum which they already know, and
vise-versa provided that the said ratios are as given above.
References
Harris, J. W. and Stocker, H. (1998): "Pyramid." Handbook of Mathematics and Computational
Science. (P. 98 – 99). Springer-Verlag, New York.
Hart, G. (2005): Pyramids, Dipyramids, and Trapezohedra. http://www.georgehart. com/virtualpolyhedra/pyramids-info.html.
Kern, W. F. & Bland, J. R. (1948): Pyramid and Regular Pyramid: Solid Mensuration with Proofs (2nd
ed.). New York, NY: Wiley & Sons.
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Macrae, M. F.; Kalejaye, A. O.; Chima, Z. I.; Garba,G. U.; Ademosu, M.; Chanon, J. B.; Smith, A. M.;
Head, H. C. (2001): New General Mathematics for Senior Secondary Schools Bk I (3rd Edition).
England: Pearson Education Limited.
Olagunju, S. O. (2011) Volume of a Square-Based Frustum: Alternative Formula (lagsamolu Equation).
In Nwakpa, Izuagie and Akinbile
(Eds) Meeting the Challenges in Science Education. (P.81 – 93).
Babson Press, Ondo.
Weisstein (2006): CRC Concise Encyclopedia of Mathematics. (P. 525, 1115, 2404).
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