InterMath | Circles
Title
Equal Areas
Problem Statement
Construct a circle that has five regions of equal area where none of the perimeters of the regions
can pass through the center of the circle.
Problem setup
Construct a circle that contains five regions that are equal in area to each other but where none of
these regions pass through or touch the center of the circle.
Plans to Solve/Investigate the Problem
I am going to first construct a circle. Then I will divide the circle into five equal pie pieces
where the perimeters of the regions can pass through the center of the circle. After having
constructed this figure I will explore the figure and construct equal areas out of the first five pie
pieces. Using the area calculation tool in GSP I will measure the areas of the five figures.
Investigation/Exploration of the Problem.
To construct the figure I will:
1. Use the circle tool to draw a circle in GSP.
2. I draw a segment from the center of the circle to the outside of the circle. (This is otherwise
known as a radius.
3. I will use the rotation tool to rotate the radius 360 degrees around the circle by integrals of 36
degrees.
The below figure on the left is what you should have so far:
4. Select every other of the ten radii and construct midpoints of each of these lines.
5. Connect the five midpoints so that they form a regular pentagon
The above figure on the right is an example of what the figure should look like.
6. Connect the points F, I, J, L, F1 to form a larger pentagon.
7. Going all around the circle connect the points so that you now have five larger pentagons.
The following is what the figure should look like if the pentagons were colored in.
F'''''
j
F''''''
G
F'''
F'''''''
k
K
F''''''''
F''
I
J
H
l
F'
F
8. Select each of the pentagons and measure by using the area tool to find the area of each
pentagon.
The following figure is what you should have:
As can clearly be seen the areas of the five pentagons are equal. None of the perimeters of the
pentagons though cross through the center of the circle. Therefore, the figure meets the
requirements of problem.
The reason that constructing the figure in this way creates five regions of equal area is because of
some properties of circles. First, the angles formed by the radii are consistent and equal
throughout the whole circle. For one thing all of the radii in the circle are equal in length. That
means that when the midpoints were constructed on the radii we have equal line segments.
Therefore, when the first pentagon is constructed all of its segments are equal in length. All the
leftover area across the circle will be equal then. When constructed the other pentagons all of
their segments are equal also. Therefore, since all the segments are equal in length, all of the
pentagons will be equal in area. The reason that this is true in any circle is because the definition
of a circle ensures that the distance from the center of the circle to any point on the outside of the
circle will be the same length. Add this to the fact that the angles formed by the radii are the
same, we are able to construct pentagons equal in area.
Also, the areas of the arc sectors are equal throughout the figure. This property exists because
for example in the figure below line segment AP bisects arc KPB. Therefore, arc KPB is split
into two smaller arc sectors equal in area. Since the area of the pentagons (for example pentagon
KPBEL in the figure below) is equal throughout the circle, the area of the ten small arc segments
is equal in area to each other.
B
Area ELKPB = 25.81 cm 2
Area EBCDF = 25.81 cm 2
P
C
E
K
D
F
L
Area JIOKL = 25.81 cm 2
A
Area FDMGH = 25.81 cm 2
O
M
J
H
I
G
Area HGNIJ = 25.81 cm 2
N
Extensions of the Problem
If the circle was inscribed within a square so that points on the exterior of the circle lay on the
sides of square, how do the areas of the pentagons inscribed within the circle relate to the area of
the square?
The length of the segments that construct the pentagons are ¼ the length of the sides of the
square. The area of the square is 109.77 cm2. The area of one of the pentagons is equal to 8.53
cm 2. The area of all of the pentagons added together is equal to 64.34 cm2. The relation
between the square and the circle inscribed within is ¼ * pi. This is because each radii is ½ the
length of the side. Since Asquare = sides2 and Acircle = pi*r2 . Therefore, the area of the square is
related to the area of the circle by ¼ * pi. However, I could not determine a relation between the
area of the pentagons inscribed within the circle and the square the circle is inscribed within.
The below figure shows the pentagons inscribed within the circle inscribed within the square.
F'''''
P
M
F''''''
j
F''''
G
F'''
F'''''''
K
k
m'''''
L
A
F''''''''
F''
I
J
H
l
O
F'
F
N
Author & Contact
Tabitha Davis
Tabbydavis240@yahoo.com
I am a junior in the middle grades cohort concentrating in mathematics and English/language
arts.