InterMath
Title
Folding Squares
Problem Statement
A square is folded in half to form a rectangle. If the resulting rectangle has a perimeter
of x inches, what is the area of the original square?
Problem setup
I am trying to determine the relationship between the perimeter of a rectangle which is
half the size of a square and the area of a square. I am trying to come up with a
formula that will allow me to find the area of a square given the perimeter x of a
rectangle. To do this I will first construct a square and then divide the square into two
equal rectangles.
This problem is sort of similar to finding the basic areas and perimeters of squares and
rectangles. To find the perimeter for a rectangle you simple add (width + width + length
+ length) or this could be stated (2width x 2length). To find the perimeter for a square
add the four sides together (side + side + side + side) or this could also be stated (4 x
side). To find the area of both a rectangle and a square the formula is length x width.
The illustrations below are simple examples how of to find the perimeter and area of a
rectangle and square.
m DC+m CB+m AB+m DA = 13.76 cm
m DAm AB = 11.83 cm 2
E
F m EDm DC = 5.92 cm 2
m FC+m EF+m DC+m ED = 10.32 cm
D
C
Plans to Solve/Investigate the Problem
Prediction: I believe that the area is going to be (1/ 2 x) 2 where x represents the
perimeter of any given rectangle. The reason I believe this is because I know that area
also results in a measurement that is squared.
First I plan to construct a square in GSP. Then I will construct a midpoint to divide the
square into two equal rectangles. I will use GSP’s measurement feature and the area
formulas of each shape. Below I will note some formulas that might come in handy
when solving the problem.
Area of a square equals—A=length x width; perimeter= side + side + side + side
Area of a rectangle equals---A=length x width; perimeter= length + length + width +
width
I will begin to investigate this problem to see if my prediction is true which I have a
feeling will be incorrect.
Investigation/Exploration of the Problem
In order to discuss how I went about solving this problem I will do it in a step by step
process.
1. Create a square: draw a line segment, then I rotated the line segment 90
degrees to form my first right angle, I drew a perpendicular line to line
segment AB and a perpendicular line to line segment BC. I then put a point
where the two perpendicular lines intersected and created a line segment
from point D to point A and point D to point C. (See figure below)
m DC = 4.50 cm
mABC = 90.00
m CB = 4.50 cm
A
B
mBCD = 90.00
m AB = 4.50 cm
mCDA = 90.00
m DA = 4.50 cm
mDAB = 90.00
D
C
2. Next I will create a midpoint on line segment AD to demonstrate the square
folded in half to form a rectangle. The figure on the right shows another way
that the square could be folded in half to create a rectangle
A
B
F
m FC = 2.25 cm
E
F
m DC = 4.50 cm
represents recta ngle thats
been folde d
D
m EF = 4.50 cm
H
B
I
G
represents
rectang le
thats been
folded
m ED = 2.25 cm
C
m FC+m EF+m DC+m ED = 13.49 cm
D
3. The next step is to see how the perimeter of the rectangle relates back to the
area of the original square. To do this, I could simply use my original square
and use the area formula of length x width. Below shows the illustration and
area of the original square.
B
A
m DAm AB = 20.23 cm 2
D
C
4. I will now try to relate the perimeter of the rectangle and area of the square to
each other to come up with a formula where I can find the area of the original
square in terms of the perimeter, x, of the folded rectangle to use with other
problems. The illustrations below give the area for the square and the
perimeter for the folded rectangle.
A
B
E
F
A
B
m DC = 4.50 cm
represents recta ngle thats
been folde d
m CB = 4.50 cm
m AB = 4.50 cm
D
C
m FC = 2.25 cm
D
C
m DA = 4.50 cm
m DAm AB = 20.23 cm 2
m EF = 4.50 cm
m DC = 4.50 cm
m ED = 2.25 cm
m FC+m EF+m DC+m ED = 13.49 cm
5. As I can see, when I divided the square in half the width of the square was
halved to form the width of one side of the rectangle. Therefore, I know because I these
widths are divided in half I am missing one whole side of the square. The widths of the
original square have been halved each resulting in half the measurement of the original
square. What a am saying is that the width of the original square was 4.50 but when I
have the width it becomes 2.25 for the width of the rectangle. So both widths of the
rectangle are 2.25 whereas the both of the squares widths were 4.50. Therefore, I have
lost a total of 4.50cm when I folded the square to become a rectangle. So since I had 4
sides that measured 4.50cm in the original square, I now have lost a total of one of
those sides of 4.50cm when I half the width of the square to form my rectangle.
Therefore I only have 3 sides of the original square. To get the area of the original
square from the perimeter of the rectangle, I will simply use the formula below. I used
my data from above and then plugged it into the formula below to see if I would get the
same area. The formula for finding the Area= (1/ 3x) 2 , where x is the perimeter of the
rectangle. Since I only have 3 sides of the original square present in my rectangle, I
must multiply the perimeter by 1/3 and the square this product to get the area of the
original square. The reason I only have 3 sides is because when I halved the widths of
the square, I lost half of both widths which would equal one whole side of a square.
formula for fi ndi ng area of ori gi nal square
where x= perimeter of folded rectangl e
1/3x2
(1/ 3*13.49cm)2  20.22cm2
The formula shows that the formula holds true when trying to find the perimeter, x, of a
rectangle to find the area of a given square.
6. I think this relationship can be seen better with a simpler problem. I will use
illustrations and the formula I came up with below to show this relationship holds true for
another situation below.
m JD = 2.00 cm
J
m DC = 4.00 cm
represents
rectang le tha ts
been folde d
K
A
B
m CK = 2.00 cm
m KJ = 4.00 cm
D
C
Perimeter of rectangle = 12 cm
D
C
m DC = 4.00 cm
2
Here we can take the formula, A= (1/ 3 x) , and
plug in the perimeter for the rectangle and see if
we get the area of the original square.
m CB = 4.00 cm
m AB = 4.00 cm
m DA = 4.00 cm
Area of origi nal square = 16 cm 2
(1/ 3*12cm) 2  16cm2
The results above indicate that the formula does hold true. For this problem, I conclude
that when given a perimeter of a rectangle that has been formed from a square, a
person can find the area of a square by using the formula that follows below.
formula for finding area of original square
1/3x2, where A=Area of the original
square, and x=the perimeter of the
rectangle folded from the square
Extension
I wonder what would happen if you took the square and folded it in half to form a
triangle with a perimeter x. How would you find the area of the original square?
B
A
m AB = 4.50 cm
m CB = 4.50 cm
m DC = 4.50 cm
D
C
m ABm CB = 20.23 cm 2
m DA = 4.50 cm
A
m DA = 4.50 cm
m AC = 6.36 cm
D
C
Peri meter
ADC = 15.36 cm
m DC = 4.50 cm
m DAm DC = 20.23 cm 2
After looking at this extension, I realized that you can simply find the area of the original
square by multiplying the base of the triangle times the height of the triangle. This
would work because, as you can see in the illustration the measurements of AD and DC
have not changed and remain at 4.50cm. Since I did not state the extension to state
that the area had to be found using the perimeter of the triangle then, the easiest way to
find the area would simply be to multiply the base of the triangle times the height of the
triangle. As you can see the areas the calculations of the area for the square matched
up for the square. If I would have stated the problem where I would have to find the area
of the original square given the perimeter of x, then I could go about solving the problem
using the pythagorem theorem.
Author & Contact
Carla McNeely, Middle Grades Education student, concentrating in English/Language
Arts and Math. I am currently a junior at Georgia College and State University.
carlalynnmc@yahoo.com