DELVING
deeper
Javad Hamadani Zadeh
Convex Polyhedra Are
Assembled of Pyramids
W
hen working with preservice teachers in
secondary school mathematics education, I
have found that they sometimes have difficulty recalling formulas for calculating the volume
of various polygonal prisms. To build their understanding at a conceptual level and make connections to the mathematics they knew, I introduced
them to the concept of pyramidization. It is similar
to triangulation of regular polygons in the plane,
where we could find the area of the polygon as the
sum of the areas of the triangles. Using the same
strategy, we can also decompose a convex polyhedron into pyramids and find the total volume by
adding the volumes of each of the individual pyramids that make up the convex polyhedron.
A polyhedron (plural polyhedra or polyhedrons)
is a three-dimensional figure that consists of flat
faces, straight edges, and vertices (see “Math Is Fun:
Polyhedrons”). If the faces are all the same regular
polygons, then the polyhedron is called regular. In
this article, we will explore how to decompose convex regular polyhedra into polygonal pyramids to
find their total volumes. Polygonal prisms and frustums (truncated pyramids) are pyramidized in Zadeh
(2013) and motivate us to investigate the relationships between convex polyhedra and pyramids.
Before we consider decomposition of volumes,
we review the analogous process for areas—triangulation. For example, see the regular pentagon in
figure 1. The center of the pentagon, T, has been
joined to each vertex to form five isosceles triangles. The area of each triangle is half the altitude,
r, times the base, s—that is, (1/2)rs. The area of the
pentagon, then, is (5/2)rs. (Note that the segment
TD, or alternatively its length r, is called the apothem and is the same as the radius of the incircle of
the regular polygon.)
Returning to three dimensions, we consider the
regular-polygon faces of the polyhedra and assume
that they have side length s. We also assume that
we can inscribe a sphere of center C and radius R
in all the regular polyhedra (Zazkis, Zinitsky, and
Leikin 2013). This inscribed sphere will be tangent to each face at its center, T. Note that s and R
as constants of the polyhedra are easy to describe
and use in the volume formulas. To find the volumes, we have to introduce two other constants,
such as and , as is done in Zazkis, Zinitsky,
and Leikin (2013), where is the central angle of
the regular polygonal faces and is the angle that
R makes with the edge of the pyramid built on
the faces.
Edited by Brian Dean
Delving Deeper offers a forum for new insights into mathematics engaging secondary school teachers to extend their own content knowledge; it appears in
every issue of Mathematics Teacher. Manuscripts for the department should be
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Department editors
Brian M. Dean, [email protected], District School Board of Pasco County, FL;
Daniel Ness, [email protected], St. John’s University, Jamaica, New York; and
Nick Wasserman, [email protected], Teachers College, Columbia
University, New York
Fig. 1 A polygon is decomposed into triangles to formulate
its area.
552 MATHEMATICS TEACHER | Vol. 109, No. 7 • March 2016
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This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
(a)
(b)
(c)
(d)
(e)
Source: https://www.wikipedia.org
Fig. 2 The Platonic solids are regular polyhedra: tetrahedron (a); hexahedron (b); octahedron (c); dodecahedron (d); and icosahedron (e).
REGULAR POLYHEDRA
Hexahedrons (Cubes)
Geometers have studied the mathematical beauty
and symmetry of the Platonic solids for thousands
of years. These regular polyhedra—tetrahedron,
hexahedron (cube), octahedron, dodecahedron,
and icosahedron—are named after the numbers of
faces (see fig. 2). In a Platonic solid, each face is
the same regular polygon and the same numbers of
polygons meet at each vertex. The center, C, of the
polyhedron or its inscribed sphere joins to each
vertex to form polygonal pyramids on the faces.
For example, if the polyhedron is a cube, the pyramidization is shown in figure 3. Six pyramids are
used, one on each face—upper, lower, left, right,
front, and back.
Let’s investigate the volumes for the Platonic
solids as the number of faces increases.
A hexahedron, the familiar cube, has six square
faces of side length s and area A = s2. On each
square face, we build a square pyramid of vertical
height R = (1/2)s, the radius of the inscribed sphere
Tetrahedrons
A tetrahedron has four equilateral triangular
faces of side length s and area A = ( 3 / 4 ) s 2 . The
inscribed sphere in the tetrahedron is tangent to
each face at its center, T. On each triangular face,
1 ⎛ 3 ⎞
we build a triangular pyramid byv joining
⎟ = 3 Rs 2
= R ⎜ thes 2center
4 radius
⎠ 12
of the sphere to each vertex (see fig.3 4).⎝ The
of the inscribed sphere, R, serves as the vertical
Fig. 3 A cube is decomposed into six square pyramids.
height of the triangular pyramid. The volume
⎛ 3 of the
⎞
3 2
2
2 height,
⎟=
pyramid is 1/3 the
area
V =R,
4 vtimes
Rs .
= 4 ⎜ theRs
( 3vertical
/ 4 ) s(1)
.
⎝ 12
⎠ 3
of the base, A:
⎛2
⎞
(v =3 /1 4R) ⎜s .3 s ⎟ =
3
⎝4
2
⎠
3 12
1⎛ 1 ⎞
1
v =Rs RA = ⎜ s⎟ • s 2 = s 3 .
12 3
3⎝2 ⎠
6
⎛ 3 up
⎞ of four of these pyramids,
A tetrahedron 1is made
⎜ s⎛2⎟ =3 3 Rs
⎞ 2 3 ⎛1 ⎞
= RHence,
2
2 tetraone on(1)
eachv face.
the
volume
3
4
12
⎝
⎠
⎜
⎟ == 6vof=Rsthe
V = 4 v = 4 (2) Rs V
6 ⎜. s 3⎟ = s 3
12
3
hedron is
⎝
⎠
⎝6 ⎠
(1)
⎛ 3
⎞
2
⎛ 12⎟ =⎞ 32( Rs
V = 4v =
1 4 ⎜ 1Rs
31 /.12) Rs 2 .
v = RA
= ⎜ ⎠s⎟ • s3 = s 3 .
⎝ 12
3
3⎝2 ⎠
6
⎛ 3
⎞ 2 3
2
1
1⎛ 1 ⎞
1
Rs 2 .
v = RA = ⎛⎜1(3)s3⎟⎞ • s 2V3= = 8sv3 .= 8 ⎜ Rs ⎟ = center.
12
3
⎝
⎠
V3= 6v = 63⎜⎝ 2s ⎠⎟ = s 6
(2)
⎝6 ⎠
Fig. 4 The inscribed sphere is tangent to each face at its
(2)
⎛1 ⎞ 3
V = 6v( =36 /⎜ 12s 3)⎟Rs
= 2s.
Vol. 109, No. 7 • March 2016 | MATHEMATICS TEACHER 553
3
(a)
(b)
Fig. 5 The radius of a cube’s inscribed sphere is the height of a constituent pyramid.
⎝4
⎠
1
1 5
5
v = RA = R ⎜ rs⎟ =
3
3 ⎝2 ⎠ 6
12
1 ⎛ 3 ⎞
3
v = R ⎜ ⎛ s 23⎟ = 2⎞ Rs 23 2
⎛5
⎞
⎟=
V =34 v =⎝ 44⎜ ⎠(4)
Rs12
Rs .
(1)
⎝ 12
⎠ 3 V = 12v = 12 ⎜ 6 Rrs⎟ = 1
⎝
⎠
(see fig. 5). Its volume,2v, is
( 3 / 4⎛) s . ⎞
3 2⎞
3 2
V = 41v = 4 ⎜ 1 ⎛Rs
(1)
1 ⎟ = 1Rs3 . s
v = RA ⎝=12⎜ s⎟⎠• s 2 =
3 sr .= • 1 = 1 s • cot
3 ⎛ 33 ⎝⎞2 ⎠ 3 6
1
2 tan 36° 2
v = R⎜
s 2⎟ =
Rs 2
3 ⎝ 4 ⎛ ⎠ ⎞ 12
A hexahedron is
pyramids,
1 made⎛ 1up1⎞of six2 of 1these
⎛1
⎞
v = RA
s⎟ • 3sside
s3 .
= 1 ⎜3 with
= length
so the(2)
volume
is ⎜the
⎜ 3 s⎝ 2⎟(5)
6vthe
V =of
= 6cube
=⎠ s
V = 10s R
s • cot 36°⎟ s =
3
6
3
⎝⎛6 3 ⎠ ⎞
⎝2
⎠
familiar s :
3 2
V = 4v = 4 ⎜
Rs 2⎟ =
Rs .
(1)
⎛⎝1123⎞ 3⎠ 3
⎜ ) Rs
V = (6v3=/612
s ⎟2=. s
(2)
1 ⎛ 3 ⎞
3
v = R ⎜ s2⎟ =
Rs
⎝6 ⎠
3
4
12
⎝
⎠
1
1⎛ 1 ⎞ 2 1 3
v = RA =⎛ ⎜ s⎟⎞• s = s .
Octahedrons 3
33⎝22 ⎠2
6
Rs
. ⎟ = 2 3 Rs 2 .
8 ⎜)triangular
V
Rs
=( 83v/=12
(3)
With eight equilateral
each of side⎛ 3 2⎞
3
⎝ 12 (6) ⎠ faces,
V into
= 20veight
= 20 ⎜ Rs ⎟ =
length s, the octahedron
is
decomposed
⎝ 12
⎠
⎛1 3⎞ 3
⎛⎜ vertical
triangular
V = 6v = 6of
(2) pyramids
3s ⎟ =2⎞s height
2 3 R,2 the radius
⎟ = 6). The
V = 8sphere
v = 8 ⎜⎝6(see
Rs
Rs volume
.
(3)inscribed
⎠ fig.
of the
of
3
⎝ 12
⎠
each triangular pyramid is, again, v = ( 3 / 12 ) Rs 2 .
⎞ pyramids,
An octahedron,( made
) Rsof⎛2 . eight such
3 / 12up
⎜
⎟
has volume
⎝ (7) ⎠ V = 16 ⎛⎜ 1 Rs2⎞⎟ + 8 ⎛⎜ 3 R
⎛ 3⎛ ⎞ 2⎞ 3
⎝3
⎠ ⎝ 12
V = 8v = 8 ⎜ Rs 2⎟ =
Rs 2 .
(3)
⎝ 12⎜ ⎠ ⎟3
( )
( )
⎝
Dodecahedrons
Fig. 6 The octagon’s volume will be calculated on the basis of eight triangular
pyramids.
Fig. 7 The twelve constituent pyramids of a dodecahedron
are pentagonal.
554 MATHEMATICS TEACHER | Vol. 109, No. 7 • March 2016
⎠
A dodecahedron, which has twelve regular pentagonal faces, each of side length s, and apothem (the
⎛ r (see⎞ fig. 7). Earlier,
radius of the inscribed circle)
⎜ pentagon
⎟
we found the area of a regular
to be
⎠ of s, so that
A = (5/2)rs. Later, we find⎝ r in terms
all the volumes will be functions of R and s.
Joining the dodecahedron’s center to each vertex
builds twelve pentagonal pyramids with vertical
height R, the radius of the inscribed sphere. The
volume of each pyramid is
( )
1
1 ⎛5 ⎞ 5
v = RA = R ⎜ rs⎟ = Rrs,
3
3 ⎝ 2 ⎛⎠ 6⎞
1
1 5
5
v = RA = R ⎜ rs⎟ = Rrs,
3
3 ⎝2 ⎠ 6
so the volume of the dodecahedron
is
⎛5
⎞
(4)
V = 12v = 12 ⎜ Rrs⎟ = 10 Rrs.
⎝ 6 ⎛ 5⎠ ⎞
(4)
V = 12v = 12 ⎜ Rrs⎟ = 10 Rrs.
⎝6
⎠
s
1
1
⎞ 365°.
r = •1
1= ⎛s5• cot
We can findv r2=in
terms
follows.
tan
36
⎟ = RrsThe
RA
, cen=°1ofR2s⎜ asrs
s
1
2each
6 36°. = 72°.
⎝=
⎠s • cot
tral angles of ther pentagon
360°/5
=3 • 3 are
2⎛ tan 36
2
° five
The pentagon decomposes
into
⎞ isosceles trian1
°
°
2
(5) as shown
V =in10figure
R ⎜ s •1,cot
36
s = 5Rsof
cot
3654°,
.
gles,
with
72°,
⎛ 5 ⎟⎞angles
2
⎝
⎠
⎛
⎞
. 2
V = 12v r,
Rrs54-36-90°
= 12
⎟ =of°10
1⎜oneRrs
and(4)
54°.
The
altitude,
is
leg
the
(5)
V = 10 R ⎜ ⎝s6• cot ⎠36 ⎟ s = 5Rs cot 36°.
right triangle ATD. Therefore,
tan36°
⎝2
⎠ = (s/2)/r, or
⎛
⎞
1
3
3
v = Rs ⎜ 1s 2 ⎟ = 1 Rs 2 .
3
4
12
⎝
⎠
⎛
⎞
•
r= 1
3 = s • cot
3 362 °.
v =2 tan
R ⎜ 36°s 2 ⎟2=
Rs .
3 ⎝ 4 ⎠ 12
⎛
⎞
2⎟ ⎞ r5in3formula
If(6)
we substitute
expression
⎛ 1⎜ 3 Rsfor
20v =
20
V = this
Rs2 2 . (4),
°=
°
• cot 36
V = 10 R ⎜ ⎝s12
⎛ ⎠ ⎟ s⎞=35Rs cot 36 .
we(5)
get
3
5
3
2
⎝
⎠
2
2
(6)
V = 20v = 20 ⎜ Rs ⎟ =
Rs .
⎝ 12
⎠
3
( 3 / 12
) Rs2 .
1 ⎛ 3 2⎞
3 2
v = R ⎜ s ⎟2 =
Rs .
( 33 / 12
⎝ 4) Rs⎠. 12
⎛
⎞
⎛
⎞
⎛
⎞ 5
1 1 15
s1
= 36Rrs
rv== •RA = R=⎜⎛ srs• ⎟cot
°. ,
31 tan 3631° ⎝225 ⎠⎞ 65
v =2 RA
= R ⎜ rs⎟ = Rrs,
3
3 ⎝2 ⎠ 6
the volume of a solid does not mean that students
have conceptual understanding. All students should
develop the strategies and understanding to decompose a solid into small pieces to find its volume,
thereby strengthening their visualizations of threedimensional figures.
⎛1 ⎛5
⎞⎞
V
s •⎜cotRrs
36⎟°⎟=s10
Rs.2 cot 36°.
= 5Rrs
V ==10
v ⎜= 12
12R
⎝ 2 ⎝⎛65 ⎠⎞⎠
V = 12v = 12 ⎜ Rrs⎟ = 10 Rrs.
⎝6
⎠
Icosahedrons
(5)
(4)
(4)
1s • ⎛ 1
3 2 ⎞ 1 3• faces,
With twenty equilateral
each of
⎜ triangular
cot236
vr == R
s ⎟ == s Rs
. °.
side length s, the icosahedron
2112 be decomposed
32s tan
4136°⎠ can
⎝
r= •
= s • cot 36°.
into twenty triangular
pyramids
2 tan
36° 2 of vertical height
R, the radius of the inscribed
⎛ 1 ⎛ sphere.
⎞ The volume,
°⎞
°
3
3 2 cot
• cot
(5)each triangular
⎜
⎟ sthe
V
R
s
Rs
10
36
.
=
=55section
2
2 36
v, of
pyramid
(as
in
(6)
V = 20v =⎝⎛20
Rs
. on°
21 ⎜⎝ Rs °⎠⎟⎠⎞ =
2
(5)
cot 36 ⎟ s = 53Rs cot 36 .
tetrahedrons)
isV = 10 R ⎜ s •12
⎝2
⎠
REFERENCES
HAIGWOOD STUDIOS PHOTOGRAPHY, INC.
“Math Is Fun: Polyhedrons.” http://www.mathsisfun
.com/geometry/polyhedron.html
Zadeh, Javad Hamadani. 2013. “Pyramidization.”
⎛ 3 ⎞
European International Journal of Science and
3 2
(v =3 1/ 12
R ⎜)⎛Rs 2s.2 ⎟⎞ =
Rs .
Technology 2 (1): 133–42. 31 ⎝ 43 ⎠ 123 2
v = R ⎜ s2⎟ =
Rs .
———.
2014. “Pyramidization of Polygonal Prisms
3 ⎝ 4 ⎠ 12
⎛
⎞
⎛
⎞
and
Frustums.”
European International Journal
Therefore, the volume1 of the icosahedron
3 2 is
⎟ =3 16 +2 2 3 Rs 2 of Science and Technology 3 (3): 88–102. (7)
V = 16 ⎜ Rs 2⎟⎛⎜ + 38 ⎜ 2⎞⎟Rs5
(6)
V = 20⎝v3= 20⎠⎛ ⎝Rs
12 = ⎠ Rs 3.
⎝ 123 2⎠⎞ 53 3 2
Zazkis, Rina, Ilya Zinitsky, and Roza Leikin.
⎜
(6)
V = 20v = 20
Rs ⎟ =
Rs .
2013. “Derivative of Area Equals Perimeter—
⎝ 12
⎠
3
Coincidence or Rule?” Mathematics Teacher 106
( 3 / 12) Rs2 .
(9): 686–92.
( 3 / 12) Rs .
STRENGTHENING
STUDENTS’ VISUALIZATION
⎛1
⎞ ⎛ 3
⎞ 16 + 2 3
2
⎟ + 8 ⎜ Rsvarious
⎟=
V =students
16 ⎜⎛ Rs 2memorize
Rs 2
Instead(7)
of having
⎞
⎛
⎝ 31 2⎠ ⎝ 123 2⎠⎞ 16 +32 3 2
⎜ Rs ⎟of
⎜ Rs ⎟convex
(7)to find
formulas
V the
Rs
= 16volume
+ 8different
=
3
3
⎠ ⎝ 12
⎠ find the
polyhedra, this article⎝ emphasizes
how to
2
volume at a conceptual level by decomposing the
polyhedra into pyramids. Memorizing a formula for
JAVAD HAMADANI ZADEH, jhzadeh@
daltonstate.edu, teaches in the Department of Mathematics at Dalton State
College in Dalton, Georgia. His research
interests are the history of mathematics and mathematics education.
i ♥ oblate spheroids.
MatheMatics
is all around us.
Vol. 109, No. 7 • March 2016 | MATHEMATICS TEACHER 555