Maths Coursework: Equable Shapes

2a + 6a = a * 2a - Now I have to simplify.

8a = 3a ^2 - Change the formula around to make ...= 0

-3a ^2 + 8a =0 - Now I need to factorise.

-a (3a - 8 ) = 0 - Now I work out what "a ="

-a = 0 ( a = 0 )

3a - 8 = 0 ( a = 2 2/3 )

Therefore a = 2 2/3

Now after doing this same method for several more rectangles the following pattern emerged...

"n"

"a"

4

2

3

3

2 2/3

4

2.5

5

2.4

6

2 1/3

7

2 2/7

The "a" number decreases in a familiar pattern:

-1/3 -1/6 -1/10 -1/15 -1/21 -1/28 ...

After finding the difference between these numbers, I carried on going but there were no similarities between the numbers so I decided to see if I could come to a formula using the original method as of the beginning.

The drops between each ratio get smaller each time.

The First Pattern was this: (for 2 : 1)

2(a + 2a) = a * 2a

2a + 4a = a * 2a

6a = 2a ^2

-2a ^2 + 6a =0

-2a ( a - 3 ) = 0

If we call "2" n we can see that with all the methods on the first line ( 2(a + 2a) = a * 2a ) the a * 2a on the end can be changed to "a * na".

Also, the "( a + 2a )" can be changed to "a + na"

With all the formulas we can see that the 2 at the front of the formula stays there each time.

So now we have a formula which works each time :

2( a + na ) = a * na

So from here we do the rest of the method...

2( a + na ) = a * na

2a + 2na = na ^2

-na ^2 + 2a + 2na = 0

Now we need to separate "a" from "n" by rearranging the formula.

-a(na - 2 - 2n) = 0 (which expands to the formula above)

So now we have these possibilities:

-a = 0 ( a = 0 )

na - 2 - 2n = 0

From this formula we can create an equation that can find the length of "1" when "n" is input.

-2 - 2n = -na Now we get rid of the "-n" to leave it as "a ="

a = ( - 2 - 2n ) / -n We can simplify this even more:

a = ( 2 + 2n ) / n

To prove this formula works we can try a 1:1 rectangle...

a = ( 2 + 2 ) / 1

Which gives us the answer 4 (both sides are 4)

Area = 4 * 4 = 16

Perimeter = 4 + 4 + 4 + 4 = 16

As the perimeter and the area match, the formula must be correct.

I can use excel or visual basic to create a table of equable rectangles.

Investigating Equable Polygons

After investigating Equable Rectangles I decided to investigate Polygons in the same way.

"n" = Number of sides

x

To work out the area of a 6 sided polygon I have to use Trigonometry. I have to disect the Polygon into 6 triangles and find out the area of the triangle (all are equal) and times it by how many sides there are (6)

For this polygon I will substitute "a" as 5.

We have tomake 6a = The area. To work out the area we do this:

360 / 6 (This Gives us the inside angle of each triangle.)

80 - (360 / 6) (After getting this we divide it by 2 to get the angles for the bottom of the triangle (both equal)

(180 - (360 / 6 )) / 2 (Now we have the bottom angles. We use "tan" now to work out the height of the triangles.

Tan ((180 - (360 / 6)) / 2) (We times this by half "a" to get the height)

To get the area of the triangle it is half base times height. In this case "a" is 5 so:

2.5 *(2.5a *(tan ((180 - (360 / 6)) / 2) = the area of one triangle. To get the area of the polygon I need to times this by how many sides there are (6)

6 * (2.5a *(2.5a *(tan ((180 - (360 / 6)) / 2) = area of polygon

Now that we have the area we can substitute some of the values as "a" and "n"

So we start from the beginning substituting side lengths as "a" and number of sides as "n"

360 / n (This Gives us the inside angle of each triangle.)

80 - (360 / n) (After getting this we divide it by 2 to get the angles for the bottom of the triangle (both equal)

(180 - (360 / n )) / 2 (Now we have the bottom angles. We use "tan" now to work out the height of the triangles.

Tan ((180 - (360 / n)) / 2) (We times this by half "a" to get the height)

To get the area of the triangle it is half base times height.

/2a *(1/2a *(tan ((180 - (360 / n)) / 2) = the area of one triangle. To get the area of the polygon I need to times this by how many sides there are (n)

n * (1/2a *(1/2a *(tan ((180 - (360 / n)) / 2) = area of polygon.

n * 1/4 a^2 *(tan ((180 - (360 / n)) / 2) Now I have to rearrange the formula to make the perimeter = area.

na = n * 1/4 a^2 *(tan ((180 - (360 / n)) / 2)

The two n's at the beginning of the formula cancel each other out so I am now left with:

a = 1/4 a ^2*(tan ((180 - (360 / n)) / 2) I can now rearrange the front "a ^2" to get this:

a/a ^2 = 1/4 * tan ((180 - (360 / n)) / 2) I can now also take away the "1/4"

4 (a/a ^2) = tan ((180 - (360 / n)) / 2)

a/a ^2 can be simplified to 1 / a so now we can get this:

4(1/a) = tan ((180 - (360 / n)) / 2)

Now that we have 4 (1/a) the rest of the formula is quite simple.

4/a = tan ((180 - (360 / n)) / 2)

Now the last thing to do is to make the formula "a ="

a = 4 / (tan ((180 - (360 / n))) / 2)

To prove this formula works we can use it on a rectangle (scale 1:1 as in regular polygons, all sides and angles are the same...)

So I will substitute 4 as the sides and lengths default and the equation should return 4.

4 = 4 / (tan ((180 - (360 / 4))) / 2)

And the answer came up with 4 proving that formula is valid. I can now make a graph showing the different length sides needed for certain polygons.

"n"

"a"

4

4

5

2.906170112

6

2.309401077

7

1.926298475

8

1.656854249

9

1.455880937

0

1.299678785

1

1.174505972

2

1.07179677

3

0.985911452

4

0.912973898

5

0.850226247

6

0.79564947

7

0.747729588

8

0.705307923

9

0.667481945

20

0.633537761

With this graph we can see that the equable side length needed for the polygons has a clear pattern. The more sides the polygon has the shorter the length of the sides. Also, as the number of sides increases the drop between the side lengths decreases like for the equable rectangles.

Analysis

If we put the two graphs next to each other we can see the similarities between them quite clearly.

As we can see with these two graphs there is a definate pattern. The side lengths both decrease and the amount the side length drops by decreases the higher "n" reaches. As both formulas are very different we can not compare them easily.

One uses Trigonometry while another uses simple algebra. They are similar in the way I found them because I found out how to do one of the shapes (e.g. 2:1 and 6 sides) and then substituted "a" and "n" in the formulas. I then kept working on the formulas making them smaller until the formula had all the a's on one side and the n's on the other (a=...)

For my equable rectangles there is a steep drop towards the beginning of the graph (1:1, 2:1...) This is also evident for equable polygons although the drop is not quite as steep. The side length drops because the more sides there are on the shape the more you have to times the side length by. If the number of sides were 209 you would have to times the side length by it.

The reason my formula work is because I took the formula from a real example. For the rectangles I used the scale 2:1 and substituted "n" and "a" in. I then did not change the formula but rearranged it to make "a=". This formula should then work with all equable rectangles because they are all formed in the same way. It also works like this for my polygons because I took a real example (6 sides) and used substitution. I used the exactly same method for both formulas.

There are many ways of approaching the problem but I believe my method was the easiest. By using substitution there are not many margins for error and it uses simple maths. If you were to create a list of all the possible results and ten lay them out to find differences between them etc. this would take a long time and could easily go wrong. I did chose this method originally for my rectangles but after several hours the difference between each of the side lengths became ridiculous.

I think that I have completed the task well. I have find out an equation for each of the problems I have been working and they both work effectively for what they are designed to do.

If I were to redo these problems with more time I would use different methods of finding formulas to see which one was quickest and most efficient. Also, I would investigate trapeziums and rhombuses etc. to see if they are similar to the results I have compared against each other. I would also ask if they followed the same pattern as mine.