Equable shapes

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Section one

Introduction

During this project I will be investigating equable shapes. An equable shape is a shape which has an equal perimeter and area. An example of this is a rectangle of sides 6x3.

Section two

Squares

First I shall investigate squares, trying to find equable squares. The way I will be doing this is by setting out a table.  The table below shows the perimeter and area of each square with a width and length of one to six.

Now I will use my table of results to produce a graph.

Section Three

Rectangles

I started investigating rectangles by filling in different tables. The first table I did kept a fixed width of 1 (shown below is the table showing my results).

By looking at this graph we can see that the linear graphs for area and perimeter are parallel. This suggests that we can’t have an equable rectangle with a width of 2. Now I will try to prove this algebraically.

 

Area = 2L

Perimeter = 2L+4

      2L=2L+4

2L-2L=4

      0L=4

        L=

Clearly, you can’t have a rectangle that is infinitely long. I was right in thinking an equable rectangle with a fixed width of 2 doesn’t exist.

Now I will be looking at Rectangles with a fixed width of three.

I produced a graph to show these results.

As you can see the Lines do not touch but if I had carried my investigation into higher numbers there would be an equable shape. I will produce a formula to find out the equable shape. The formula is:

Join now!

P=2L+6

A=3L

      3L=2L+6

3L-2L=6

       L=6

This shows that the length must be six for a rectangle with a fixed width of three to be equable.

Now I will look at the general rectangle width Length L and Width W.

Area = LW

Perimeter = 2(L+W)

      LW = 2(L+W)

      LW = 2L + 2W

LW-2L = 2W

L(W-2) = 2W

L =      This is the formula to find L (the length) given W (the width)

Test the formula

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