European International Journal of Science and Technology
Vol. 2 No. 1
February 2013
Pyramidization
Javad Hamadani Zadeh
Department of Mathematics
Dalton State College
Dalton, Georgia 30720
[email protected]
Geometry and three-dimensional figures are rich areas for practicing visualization. Visualization can be taught
and practiced through illustrations in classroom settings. One way to teach and practice visualization of threedimensional figures is through dividing a polyhedron into pyramids, which we call “pyramidization.” A
polyhedron is a geometric solid in three dimensions with flat faces and straight edges [1].
This article will illustrate through figures and examples, two methods for dividing n-gon prisms (n ≥ 3
representing the number of sides of the base) into pyramids in three-dimensional geometry, analogous to
triangulation of polygons in the plane (two-dimensional geometry). As a refresher, a polygon consists of a
closed path of straight lines (edges or sides); the points where the lines meet are the polygon’s vertices. We
refer to an n-gone in this paper; where n refers to the number of sides of the polygon [2].
We will show that triangular pyramids serve the same role in three-dimensional geometry as do triangles
in two dimensions. The illustrated methods provide problem-solving and skill-building opportunities for
practicing three-dimensional visualization in geometry.
Triangulation of Polygons
To calculate the area of any polygon in two-dimensional geometry, customarily we divide the n-gon (n >
3) into triangles. Then we can more easily calculate the area of each of the triangles within the polygon and
sum them to obtain the total area of the polygon. For example, as illustrated in Figure 1 – a pentagon – there
are two general methods for obtaining these triangles within a polygon:
Method 1: Join any one vertex of a polygon to all other vertices. Figure 1(a).
Method 2: Join any interior point of a polygon to the vertices. Figure 1(b). (In this paper, we refer to
this interior point as the center of the figure.)
(a)
(b)
Figure 1
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In the first method, the number of triangles that we obtain will be n ─ 2 (or 5 – 2 = 3), where n > 3 is the
number of sides of the polygon. In the second method the number of triangles obtained will be n (or 5). Note
that if n = 3, the polygon is a triangle, and there is no need to triangulate it further. Also, method 1 is limited to
solids with parallel bases, like prisms; while method 2 can be extended to pyramidization of any polyhedron.
Pyramidization
Prisms generally have polygonal parallel bases and rectangular faces [3]. We show in three-dimensional
geometry that if we pyramidize an n-gon prism (n ≥ 3) by joining a vertex to all other vertices, there will be n –
1 polygonal pyramids. A pyramid’s base can have different number of sides; it can be trilateral, quadrilateral,
or any other polygon shape [4].
To further explain pyramidization of an n-gon prism, if n = 3, after joining a vertex to all other vertices,
one of the pyramids obtained will be triangular, and the other rectangular. If we select and then join an interior
point of an n-gon prism to the vertices, we will obtain n + 2 polygonal pyramids. If n = 3, the number of
pyramids we obtain is 3 + 2 = 5 (two triangular and three rectangular pyramids). These examples are further
illustrated in the “Pyramidization of Triangular Prisms” section of this paper.
Triangular Pyramidization of Pyramids (Tetrahedronization)
For a polygonal pyramid with n-sided base (n > 3) we can triangulate the base either using method 1,
Figure 1(a), or method 2, Figure 1(b). If we use method 1 to triangulate the base, the pyramid will be divided
into n − 2 triangular pyramids or tetrahedrons (where n is the number of sides of the base). See Figure 2(a).
Note that tetrahedrons are convex polyhedrons with four triangular faces, including the base [5]. If we use
method 2 to triangulate the base, and also join the interior point of the base, G, to the apex, F, the pyramid will
be divided into n triangular pyramids. See Figure 2(b).
F
F
D
D
E
E
G
C
C
A
B
(a)
Figure 2
A
B
(b)
In the following sections we use the analogous methods 1 and 2 to pyramidize n-gon prisms. We
illustrate these methods by first pyramidizing triangular prisms, then cubes, square prisms and finally
pentagonal prisms. We also indicate the number of polygonal and triangular pyramids that are obtained, in each
case. In the conclusion section we include concise formulas that give the number of polygonal and triangular
pyramids for each method in terms of n.
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February 2013
Pyramidization of Triangular Prisms
Analogous to the triangulation of polygons, we pyramidize an equilateral triangular prism with base side
3 2
s , or
s, and vertical height h, Figure 3(a). Its volume is, height h times the area of the base
4
3 2
V =
hs .
(1)
4
We divide the triangular prism into 2 and 5 pyramids, using methods 1 and 2, respectively.
Method 1
We can join the vertex P in Figure 3(a), to only two vertices, U and S.
T
T
S
S
S
U
U
U
h
h
h
h
P
P
s
R
R
P
s
s
s
s
s
Q
Q
(a)
Figure 3
(b)
(c)
We get the rectangular pyramid P - QRSU, Figure 3(b), and the triangular pyramid S - TUP, Figure 3(c). They
3
both have the vertical height equal to the altitude of the lower and upper triangular bases of the prism,
s.
2
Using Figure 2(a) or method 1, we can triangulate the base of the rectangular pyramid, Figure 3(b), and
divide it into two triangular pyramids. See Figure 4(a) which illustrates how the diagonal RU is drawn to obtain
the two triangular pyramids, 4(b) and 4(c). We have three triangular pyramids; P – SUR, Figure 4(b), P – QRU,
Figure 4(c), and S – TUP, Figure 4(d) with equal volumes [6, 12]. Note that Figure 4(d) is the same as Figure
3(c).
T
T
S
S
S
U
U
U
U
s
h
h
h
s
h
P
P
P R
P
s
R
s
R
s
s
Q
s
Q
(a)
(b)
(c)
(d)
Figure 4
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To further explain, in figure 4(a), UR divides the base of the rectangular pyramid P - QRSU into two congruent
triangles, ∆QRU and ∆SUR which will be the bases of two triangular pyramids with the same vertical height,
3
s , the altitude of triangle PQR. The third triangular pyramid 4(d) has the base ∆TUP, congruent to triangles
2
3
∆QRU and ∆SU R. These three triangular pyramids have the same vertical heights,
s , and hence equal
2
1
volumes. The volume of each is of the volume of the triangular prism (1), that is
3
1 3 2 1 3 1
3 2
hs = ⋅
s ⋅ hs =
hs ,
3 4
3 2
2
12
which is one third of the altitude,
1 3
1
⋅
s , times the area of the base,
hs , for each of the three triangular
3 2
2
pyramids.
If we use Figure 2(b) to triangulate the base of the rectangular pyramid in Figure 3, we get 5 triangular
pyramids; specifically, the rectangular pyramid 3(b) will be divided into 4 triangular pyramids and pyramid 3(c)
is triangular already.
Method 2
To illustrate method 2, we join an interior point of the triangular prism to all the vertices to obtain 5
pyramids; two triangular and three rectangular, one on each face. See Figure 5.
T
S
U
O
h
h
h/2
s
P
s
R
s
Q
Figure 5
Triangular Pyramidization
We can triangulate the bases of the three rectangular pyramids obtained in Figure 5. If we use Figure
2(a) or method 1 to triangulate the bases, we get 8 triangular pyramids. Each of the three rectangular pyramids
will be divided into two triangular pyramids, giving 6 triangular pyramids plus the 2 triangular pyramids on the
lower and upper bases. However, if we use Figure 2(b) or method 2, we get 14 triangular pyramids. Each of
the three rectangular pyramids will be divided into four triangular pyramids, giving 3 ⋅ 4 = 12 triangular
pyramids plus the two triangular pyramids on the lower and upper bases, a total of 14, altogether.
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Pyramidization of Cubes
Next, we can pyramidize the cube with base ABCD, Figure 6(a). In method 1, we join one vertex of the
upper base, say, C’ to all other vertices. The cube can be divided into three identical oblique square pyramids
of base length s and vertical height s. See Figure 6(a). The three pyramids are C’ – ABCD, C’ – ABB’A’ and
C’ – AA’D’D, where the square bases have equal areas, s 2 .
C'
D'
B'
A'
s
C
D
s
B
s
A
Figure 6(a)
The volume of the cube with side s is Vcube = s 3 . The three oblique square pyramids in Figure 6(a) are
1
congruent, hence have equal volumes. The volume of each square pyramid in Figure 6(a), is V pyramid = s 3 or
3
1
s ⋅ s 2 , i.e., one third of the altitude s times area of the base, s 2 . Photo 1 shows a paper model of Figure 6(a)
3
and photo 2 shows Figure 6(a) opened up.
Photo 1
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Photo 2
If we triangulate the bases of the above three pyramids, using Figure 2(a) or method 1, and we get 6
triangular pyramids, since the square bases will be divided into 2 triangles. If we use figure 2(b) or method 2,
there will be 12 triangular pyramids, since each base is divided into 4 triangles.
In method 2, Figure 1(b), we join an interior point of the cube; center O, to each vertex. Six congruent
1
1 s
pyramids are formed, one on each face, and the volume of each is: s 3 , or ⋅ ⋅ s 2 , one third of the altitude
6
3 2
times the area of the base.
1
of its vertical height times the
So far, the above examples illustrated that the volume of a pyramid is
3
area of its base.
C'
D'
A'
B'
O
s
s/2
C
D
s
s
A
Figure 6(b)
If we triangulate the square bases of each of the six pyramids using Figure 2(a) or method 1, we get
2 ⋅ 6 = 12 triangular pyramids, and if we use Figure 2(b) or method 2, we get 4 ⋅ 6 = 24 triangular pyramids.
Pyramidization of Square Prisms
The following describes the two analogous methods for pyramidizations of square prisms which are
similar to the methods described for pyramidization of the cube, but are presented to illustrate volume
computations.
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February 2013
Method 1
For the following square prism, Figure 7, we join a vertex, say, H, to all other vertices. Figure 7 uses
this method to illustrate how a square right prism with base side length s, vertical height h, and volume V = hs 2
is divided into three oblique pyramids.
F
G
E
H
h
B
C
A
D
s
Figure 7
The three resulting pyramids have bases ABCD, AEFB, and BFGC. The areas of the bases are s 2 , sh , and sh ,
respectively. The corresponding vertical heights of these pyramids are h, s and s , respectively.
Similarly to the cube, if we triangulate the bases of the resulting pyramids using Figure 2(a) or method
1, we get 6 triangular pyramids (each base is divided into two triangles). If we use Figure 2(b) or method 2, we
get 12 triangular pyramids (each base is divided into 4 triangles).
Method 2
We can join an interior point of the square right prism to all its vertices and divide it into six pyramids,
one on each face. See Figure 8. This is analogous to the polygonal division into triangles, Figure 1(b). In
Figure 8, for simplicity, we have joined the midpoint, O, of the vertical height h, to all vertices.
H
G
E
F
h
O
I
s/2
h/2
D
A
s
C
B
Figure 8
If we triangulate the bases of each of the six pyramids using Figure 2(a) or method 1, we get 12
triangular pyramids, since each base is divided into two triangles. If we use Figure 2(b) or method 2, we get 24
triangular pyramids, since each base is divided into 4 triangles.
Pyramidization of Pentagonal Prisms
Next, let us consider pyramidization of a regular pentagonal prism with base side length s, apothem a,
Figure 9(b) and vertical height h. Its volume is:
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V =
5
has .
2
ISSN: 2304-9693
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(2)
Regularity is not required, but we assume a regular pentagonal prism to simplify volume computations.
Method 1
We divide the pentagonal prism by joining a vertex, say, D’, to all other vertices. See Figure 9(a).
E'
A'
D'
E
B'
C'
A
E
A
D
O
r
D
a
s
B
F
B
C
C
(a)
(b)
Figure 9
There are four pyramids in Figure 9(a); a pentagonal pyramid with base ABCDE, the lower base of the prism,
apex D’, and vertical height h, (the height of the prism); and the other three are rectangular pyramids, with
bases, BCC’B’, ABB’A’ and AEE’A’ (all with base sides s and h and apex D’). Each of the rectangular
pyramids has a vertical face in the upper base of the prism, that is, one of the three triangles, B’C’D’, A’B’D’
and A’D’E’. If the base has n sides, the number of pyramids will be (n − 2) + 1 = n − 1.
Let us triangulate the bases of the above pyramids using Figure 2(a) or method 1. We get 9 triangular
pyramids, since the pyramid with pentagonal base is divided into 3 triangular pyramids, and each of the three
rectangular pyramids will be divided into 2 triangular pyramids ( 3 ⋅ 2 + 3 = 9 ). If we use Figure 2(b) or method
2, we get 17 triangular pyramids. The pentagonal pyramid is divided into 5 triangular pyramids, and each of the
three rectangular pyramids is divided into 4 triangular pyramids, a total of 17.
Method 2
We can divide a pentagonal prism into polygonal pyramids by joining any interior point, for example the
midpoint of the vertical height, O, to all other vertices, as illustrated in Figure 10.
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February 2013
C'
D'
B'
E'
A'
h
O
C
h/2
D
B
a
M
r
E
s
A
Figure 10
h
, and five
2
rectangular pyramids with base lengths s and h, and heights specified by the apothem a of the pentagon; which
give a total of seven pyramids. In a polygonal prism with n base sides, this method results in n + 2 pyramids.
The two pentagonal and the five rectangular pyramids are shown in Figure 10.
Finally, if we triangulate the bases of the above 7 polygonal pyramids, using Figure 2(a) or method 1,
we get 16 triangular pyramids. Each of the two pentagonal pyramids is divided into 3 triangular pyramids, and
each of the five rectangular pyramids is divided into 2 triangular pyramids, giving a total of 16. If we use
Figure 2(b) or method 2, we get 30 triangular pyramids. Each of the two pentagonal pyramids is divided into 5
triangular pyramids, and each of the five rectangular pyramids is divided into 4 triangular pyramids, resulting in
a total of 30 triangular pyramids.
In the following conclusion, for each method, Table 1 gives concise formulas, and the total numbers of
polygonal pyramids and triangular pyramids in terms of n, the number of sides of the base of n-gon prisms (n >
3).
Consequently, the pentagonal prism is divided into two pentagonal pyramids each with height
Conclusion
Pyramidization of n-gon prisms (n ≥ 3) presented in this paper can be used for teaching and learning problemsolving and skill-building exercises in three-dimensional geometric visualization and volume computation
similar to polygonal triangulation and area calculations in two dimensions. We can use two general analogous
methods for illustrating such visualization techniques: Method 1 joins any one vertex to all other vertices, and
Method 2 joins any interior point of the n-gon prism to all vertices. Method 2 can be extended easily to any
polyhedron. Table 1 summarizes the results for n-gon prisms, n > 3.
The presented methods also can be used in geometry courses to illustrate the analogy from triangulation
of polygons in two-dimensions to pyramidization of polyhedra in general, and n-gon prisms, specifically, in
three-dimensions.
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Table1. Pyramidization of n-gon
Prisms, (n > 3)
Method
Num. of
Num. of
Triangular
Triangular
Pyramids
Pyramids
Num. of
Pyramids Using Fig. 2(a) Using Fig. 2(b)
1
n−1
3(n − 2)
5n − 8
2
n+2
4(n − 1)
6n
Reference
1. Wikipedia: definition of polyhedron; accessed at http://en.wikipedia.org/wiki/polyhedron.
2. Wikipedia: definition of polygon; accessed at http://en.wikipedia.org/wiki/Polygon.
3. Wikipedia: definition of prism; accessed at http://en.wikipedia.org/wiki/Prism_(optics)
4. Wikipedia: definition of pyramid; accessed at http://en.wikipedia.org/wiki/Pyramid#cite_note-0.
5. Eric W. Weisstein. “Tetrahedron” from Math World.
6. G. F. Simmons, Precalculus Mathematics in a Nutshell: Geometry, Algebra,
Trigonometry, Janson Publications, 1987; 12.
*"The author thanks Mina Zadeh- Badakhsh for looking up some of the online references of this paper, and
Robert Ford for using this paper in his geometry course as enrichment."
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