InterMath | Geometry | Polygons
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
The apothem of a polygon is the shortest distance from the center of the polygon to one of its
sides. The perimeter of a polygon is the measurement of all the sides added together. What is
the area of a regular hexagon using the apothem and the perimeter of a hexagon? Does the
equation found work for other regular polygons?
Plans to Solve/Investigate the Problem
First I will, construct a hexagon inscribed within a circle. I will draw a circle then use the rotate
function to rotate points along the circle. Then I will connect the points to form a hexagon. To
determine the area of the hexagon in terms of the apothem and the perimeter I will first
determine the perimeter of the hexagon. I will then replace and substitute into the equation for
the area of a hexagon to establish the area in terms of the apothem and perimeter.
Investigation/Exploration of the Problem
Construct the figure:
1. Draw a circle in GSP
2. Connect the center point of the circle to a point on the outside of the circle.
3. Use the rotate function to rotate the line segment every 60 degrees around the circle.
4. Connect the six points on the circle to form a hexagon.
Below is what the figure should look like:
5. Hide all the extra objects to leave the hexagon alone
6. Create an apothem (line segment from the center to a side of the hexagon). To do this
you need to create a midpoint on line segment DE. Connect the midpoint to the center
point of the hexagon.
Below is what the figure should look like:
A
B
C
F
apothem
E
D
By definition a regular hexagon is a hexagon whose six sides are equal. Therefore, in the
above figure AB  BC  CD  DE  EF  FA so the perimeter of the figure can be
represented by 6 (because there are six sides) times AB .
A hexagon is composed of six triangles. The triangles are all equal because
AB  BC  CD  DE  EF  FA and the angles of the triangles are 60 degrees.
The below figure shows the six triangles inscribed within the hexagon:
A
B
C
F
apothem
E
D
1
The area of a triangle is b * h . Since there are six triangles in the hexagon, the area of the
2
1
hexagon is 6* b * h . This can be rewritten by 3* b * h .
2
The base of the figure would be one of its sides since all the sides are equal. There are six
sides to the hexagon. Therefore, in terms of the perimeter the base of the triangle would
perimeter
be
. The height of the triangle would be the apothem because it is the distance
6
from the center of the hexagon to one of its sides.
By replacing what we now know into the formula for the area of a hexagon we get:
area  3* b * h
 perimeter 
area  3* 
 * apothem
6


perimeter * apothem
area 
2
This is the equation in terms of the perimeter and apothem for hexagons.
To ensure that this is correct for the hexagon, I check it using measurements from the above
hexagon.
The measurement of line segment AB = 5.50. Since the line segments that compose the hexagon
are all equal, the perimeter of the hexagon is 5.50 * 6. This equals 33.02.
By using the measurement function, the apothem equals 4.77.
By plugging it into the equation I derived I get:
perimeter * apothem
2
33.02* 4.77
area 
2
area  78.69
area 
In GSP I use the measurement function to determine the area of the hexagon. GSP calculates the
area to be 78.69.
Therefore, the formula I derived is correct.
The formula will also work for other regular polygons. This occurs because all sides of the
polygon will be equal. The equations for the other polygons will simplify to be the same
equation as the one above for the hexagon. For example, an octagon has 8 sides. The hexagon
had six sides and six triangles inscribed within it. The octagon then has 8 triangles inscribed
within it. Below is a picture of the 8 triangles:
A
B
H
P1
G
C
B
F
apothem
D
E
C
1
Therefore, the area would be 8* * b * h . This can be rewritten by 4* b * h . Likewise the base
2
would equal the perimeter divided by 8 (the number of sides on the octagon). The height would
once again be the apothem.
area  4* b * h
 perimeter 
The area then equals: area  4* 
 * apothem
8


perimeter * apothem
area 
2
As you can see the formula for the area of the two polygons is equal. This happens because the
1
basic formula you are working with is the formula of the area of a triangle: b * h . The number
2
of triangles inscribed within the polygon is equal to the number of sides the perimeter is divided
by. Therefore, those two numbers cancel each other out. This leaves the perimeter * the
apothem divided by 2. The formula then is true for all regular polygons.
Extensions of the Problem
Given a 3-d version of a regular hexagon, find the surface area in terms of the perimeter and the
apothem of the figure. I’m using a hexagonal prism as the 3-d representation.
First I draw the figure in GSP. I constructed two hexagons just as I constructed them earlier.
Then I connected the corresponding points to form a 3-dimensional figure.
Below is what it could look like:
A
B
A'''
J
K
K'
B
F
C
A''''
F
apothem
O
L
K''
B
E
G
D
M
N
E
The surface area of a 3-dimensional figure is found by adding together the areas of each face of
the figure. Since it is a regular figure it would be found by 2 * the area of the hexagon + 6 * the
perimeter * apothem
area of the sides. The area of the hexagon is found with the formula
. Since
2
perimeter * apothem
= perimeter *apothem. The
2
other 6 faces of the 3-dimensional figure are rectangles. Therefore, they are found by the
formula l*w. The width of the rectangle is equal to one side of the hexagon. The width is then
perimeter
equal to
because there are six sides on the hexagon. So the area of one of the
6
perimeter
* l . There are 6 of the rectangle faces so you multiply
rectangles can be rewritten as
6
perimeter
* l by 6 to find the area of rectangular faces of the 3-dimensional shape. This can be
6
rewritten as perimeter*l.
there are two hexagons the area is found by 2 *
Perimeter  2l  2 w
Perimeter  2 w  2l
Perimeter
wl
2
perimeter
w
6
Perimeter Perimeter

l
2
6
1 1
Perimeter (  )  l
2 6
Perimeter
l
3
Surface area is the area of the hexagon faces + the area of the rectangular faces.
SA  perimeter * apothem  perimeter * length
SA  perimeter * apothem  perimeter *
SA  perimeter * apothem 
perimeter
3
perimeter 2
3
Author & Contact
Tabitha Davis-junior in the middle grades cohort.
tabbydavis240@yahoo.com