InterMath | Workshop Support | Write Up Template
Title
Apothem and Area
Problem Statement
Determine an equation of the area of a regular hexagon in terms of its apothem and
perimeter. The apothem is the shortest distance from the center of the polygon to one of
the sides.
Will this equation be true for other regular polygons? Explain.
Problem setup
This exercise reminds me of the “Folded Square” problem (a problem involving the finding of
the area of a square from the given total perimeter of that square folded into a rectangle), because
only select criteria can be used to determine the answer. For this particular problem, we are given
critical pieces of information:
1.
The terminology
a. Regular hexagon – a polygon whose six-sided perimeter lengths are
equilaterial – or all the sides individually are the exact same measure.
b. Apothem – a new term for me! Apothem is the shortest distance from the
center of the polygon to one of the sides. For a regular polygon, once that
length is determined for one…it becomes the determiner for all. In our
exercise, the midpoint of each perimeter line segment to the center of the
polygon is the apothem. The midpoint is the shortest distance from the center
in a regular polygon
2.
The apothem length. The problem allows me to know the apothem as one of the
givens.
3.
The total perimeter of the polygon. We aren’t given the size of a single side; we
are given the total perimeter. From the known lengths of the apothem and the total
perimeter of the hexagon, we are to construct an equation that will give us the
area of any regular hexagon! I look forward to that challenge!
Plans to Solve/Investigate the Problem
Using the Geometry Sketch Pad (GSP), we can construct a regular polygon, which will
give us a pictorial representation of the problem, as well as correlate the relationships of sides
and apothem. I have a feeling (gained from the representation in my head) that triangle areas are
involved. We can, of course, get the total area of the hexagon from GSP and then find a way to
figure an alternative (but equal) area by breaking the hexagon down to smaller equal areas and
manipulating those bits of information to reflect the total area.
Investigation/Exploration of the Problem
My favorite activity is the construction of a model in GSP! How can I construct a regular
hexagon?
1.
We can begin by constructing an equilateral triangle, which will represent one
equilateral “chunk” of area of a hexagon. We need to do this to keep the integrity of a
“regular” hexagon intact. By creating a line segment, then constructing a circle on
both ends with that same line segment as the radius, we can construct the next point
of the hexagon (maybe two). The construct looks like this:
A
m DB = 2.70 cm
m AD = 2.70 cm
C
D
m AB = 2.70 cm
m CA = 2.70 cm
B
m BC = 2.70 cm
As you can see, after I constructed line segment AB, I highlighted points A and B,
constructed circles by points and radius, and then constructed the line segments from points
on the intersection of the circles. This was an easy way to construct two equal portions of the
hexagon.
2.
Now, to quickly finish the hexagonal construct, all I have to do is create lines through
line segments AC, AB, and AD. This will ensure equilateral line segments which will
extend via the newly constructed lines and intersect on Circle A. Then, we can finish
the line segments for the remaining portions of the hexagon and label our points. GSP
makes this easy!
m AF = 2.70 cm
E
m FG = 2.70 cm
D
F
m EF = 2.70 cm
m AG = 2.70 cm
m GB = 2.70 cm
A
m BC = 2.70 cm
C
G
m DC = 2.70 cm
m AD = 2.70 cm
B
m CA = 2.70 cm
m AB = 2.70 cm
m DE = 2.70 cm
m AE = 2.70 cm
As you can see, every line segment in this regular polygon is congruent. Now, if we remove
our circles and lines, our regular hexagon should be wonderfully displayed!
m AF = 2.70 cm
E
m FG = 2.70 cm
D
F
m EF = 2.70 cm
m AG = 2.70 cm
m GB = 2.70 cm
A
m BC = 2.70 cm
C
G
m DC = 2.70 cm
m AD = 2.70 cm
m CA = 2.70 cm
B
m AB = 2.70 cm
m DE = 2.70 cm
m AE = 2.70 cm
Next we need to construct an apothem, which is the shortest distance from the center of
Circle A to any side of the hexagon. Since we know our “mini” triangles (like triangle ABC)
are all equilateral, either a bisected angle (BAC) or a bisected exterior line segment (BC) will
not only produce an apothem, but a height (or altitude) for the triangle. This revelation is
significant because we use the height in a formula for finding the area of a triangle. Since, the
problem stated that one of our known parameters is the apothem, the knowledge that the
apothem in a hexagon is the height of triangle gives me an overwhelming clue on how to
complete the problem!
m AF = 2.70 cm
E
m FG = 2.70 cm
D
F
m EF = 2.70 cm
m AG = 2.70 cm
m GB = 2.70 cm
A
m BC = 2.70 cm
C
G
m DC = 2.70 cm
m AD = 2.70 cm
H
B
m CA = 2.70 cm
m AB = 2.70 cm
m DE = 2.70 cm
m AE = 2.70 cm
Line segment AH is one of six possible apothems for hexagon BCDEFG, and is also the
height of triangle ABC. Since we know beyond a doubt that triangle ABC contains one-sixth
of the hexagon’s area (by SSS method! All the line segments are the same…and if one
triangle’s sides are congruent to another triangle’s sides, then the areas will also be
congruent), we can use that knowledge to help us derive a formula for the area of any
hexagon…just by knowing the total perimeter and the apothem. Let’s look to the math for
proof. Notice, I haven’t even found the area of our hexagon in GSP. I want to see if my
hypothesis works!
For the perimeter of a hexagon, I only need one of the six bases, so for the area
of triangle = 1/2bh, the base (by just the TOTAL perimeter of the hexagon
given) is P/6 where P = Perimeter. Our “h” in the triangle area formula can
simply be denoted as “Apoth” for the apothem. Also, we need to multiply the
area of one triangle area in the hexagon by 6, since the hexagon contains 6
equal areas inscribed by the triangles.
1
P
P(Apoth)
P(Apoth)
6(----(---- )(Apoth) = 6(--------------) = -------------2 6
12
2
Whew! What a loooooong way to a very short solution for regular hexagons! Now, let’s
measure our apothem in our GSP construction (measure of line segment HA):
TOTAL PERIMETER
m BC+m DC+m DE+m EF+m FG+m GB = 16.19 cm
m AF = 2.70 cm
E
m FG = 2.70 cm
D
F
m EF = 2.70 cm
m AG = 2.70 cm
m GB = 2.70 cm
A
m BC = 2.70 cm
C
G
m AD = 2.70 cm
H
m HA = 2.34 cm
m DC = 2.70 cm
B
m CA = 2.70 cm
m AB = 2.70 cm
m DE = 2.70 cm
m AE = 2.70 cm
…and use our very short formula to speculate what the total area of hexagon BCDEFG is:
P(Apoth)
16.19(2.34)
---------- =
-------------- = 18.92cm2
2
2
Let’s let GSP determine what the area is and see if we are close:
TOTAL PERIMETER
m BC+m DC+m DE+m EF+m FG+m GB = 16.19 cm
m AF = 2.70 cm
E
m FG = 2.70 cm
F
D
m EF = 2.70 cm
m AG = 2.70 cm
m GB = 2.70 cm
A
m BC = 2.70 cm
G
C
m DC = 2.70 cm
m AD = 2.70 cm
H
m CA = 2.70 cm
B
m HA = 2.34 cm
m AB = 2.70 cm
m DE = 2.70 cm
m AE = 2.70 cm
m BC+m DC+m DE+m EF+m FG+m GBm HA
2
= 18.92 cm2
Area BCDEFG = 18.92 cm2
Wonderful! The hypothesis proved to be correct! Therefore, I conclude, based on algebraic
proof, that we can calculate the area of a regular hexagon by knowing only the total perimeter
and the apothem. I love the way “apothem” rolls from the tongue!
Can we use this information to construct a formula for any regular polygon? I don’t know. Let’s
try by playing with our original formula for a moment. Instead of using a six to multiply and
divide at select places in the formula, let’s insert an “N” to represent the number of sides in the
regular polygon of choice:
1
P
P(Apoth)
P(Apoth)
N(----(---- )(Apoth)) = N(--------------) = -------------2
N
2N
2
Hmmmm…according to my calculations, the answer derived from the regular hexagon is
identical to the “any” regular polygon scenario! I constructed an octagonal polygon this
time…and the legs of the triangles were NOT congruent with the sides. I learned that the
triangles constructed within regular polygons only need be isosceles. The interior legs need to be
congruent to all the other interior legs…which by the SAS theorem makes all the triangles of
such a polygon congruent (making the perimeter sides congruent…since they do make up the
third side of all the triangles). Here is my GSP construction with the math:
m LM = 3.35 cm
J
m MI = 4.38 cm
m MJ = 3.35 cm
M
K
m JK = 3.35 cm
m IJ = 4.38 cm
m KI = 4.38 cm
m KN = 3.35 cm
m NI = 4.38 cm
L
P1
N
I
m RI = 4.05 cm
m NO = 3.35 cm
m MJ8 = 26.81 cm
m OM = 8.76 cm
m MJ8m RI
m PO = 3.35 cm
m PI = 4.38 cm
Q
m QP = 3.35 cm
m LQ = 3.35 cm
m IL = 4.38 cm
O
2
= 54.23 cm2
R
P
Area P1 = 54.23 cm2
So, we can conclude that…for any regular polygon, we can calculate the area of that polygon
with the formula (P/Apothem)/2.
Extensions of the Problem
What about 3-D polygons? If we know the surface area of 3-D regular polygon, the height, and
perimeter, can we determine the apothem? Suppose we know the surface area, height and
apothem? Can we determine the total perimeter of the 2-D polygon? Can we find a single side?
I am so used to GSP to help me solve this, so I am going to rely purely on an algebraic proof this
time to determine if we can do calculations. Let’s start with formulas in need of consideration:
Author & Contact
Lindell Dillon
lindelldillon@yahoo.com
Link(s) to resources, references, lesson plans, and/or other materials
Link 1
Link 2