# Mathematics Coursework - Beyond Pythagoras

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Introduction

Francis Musonda

Mathematics Coursework

Beyond Pythagoras

The numbers 3, 4 and 5 satisfy the condition

3 +4 =5

Because

3

= 3x3

=9

4

= 4x4

=16

5

= 5x5

=25

So

3 +4

=9+16

=25

=5

I now have to find out whether the following sets of numbers satisfy a similar condition of

(smallest number) + (middle number) = (largest number)

a) 5, 12, 13

5 +12

= 25+144

= 169

= 13

b) 7, 24, 25

7 +24

= 49+576

= 625

=25

I looked at the table of results and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.

I already know that the (smallest number) + (middle number) = (largest number) . So I think that there might be a connection between the numbers. The problem is that this is not completely correct.

(Middle number) + (largest number) = (smallest number)

Because

12 + 13

= 144+169

= 313

5 = 25

The difference between 25 and 313 is 288 which is far to big, so this means that the equation I need and want has nothing to do with 3 sides being squared. So I shall now try 2 sides being squared.

(middle number ) + largest number = (smallest number)

= 12 + 13 = 52

= 144 + 13 = 25

= 157

= 25

This does not work. There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side being squared.

12 + 13 = 5

This couldn’t work because 122 is already larger than 5, this also goes for 132. The only number now I can try squaring is the smallest number.

12 + 13 = 5

25 = 25

=5

This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles that I know.

4 +5 = 32

9 = 9

And…

24 + 25 = 7

49 = 49

It works with both of my other triangles. So this means that the

Middle

2

Lower bound = 220, Upper bound = 221.

Middle Side = 220, Largest Side =221.

I now have 10 different triangles, which I think makes life easier when it comes to finding a relationship between each side.

Shortest side = Middle side + Longest side

The way I mentioned above and on the previous pages, is quite a good way of finding the middle and longest sides. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest, middle and longest).

The formula I will work out first is for the shortest side.

3 , 5 , 7 , 9 , 11

2 2 2 2

The difference between the lengths of the shortest side is 2. This means the equation must have something to do with 2n. There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct, I will now test this formula.

The Shortest side and 2n+1 columns match meaning that:

Shortest Side = 2n+1

The next formula I need to work out is the formula for the middle side.

4 , 12 , 24 , 40 , 60

8 12 16 20

4 4 4

The difference of the difference of the lengths of the middle sides is 4. This means that the formula has to have something to do with 4, most likely 4n. However, because 4 are the difference of the difference, the formula must be n . I now believe that the answer will have something to do with 4n . So, I will now write out the answers for 4n .

Conclusion

2n - 1 and the Relationship between the shortest and middle side both match. So…

Relationship between Shortest and Middle sides = 2n – 1

The next relationship I’m going to work out is the relationship between the shortest and longest sides.

Longest Side – Short Side = Relationship.

2n + 2n + 1 – 2n + 1 = Relationship.

2n + 1 = Relationship.

To check if this formula is right, I am going to write out a table containing 2n + 1 and the relationship between the shortest and Longest side.

2n + 1 and the Relationship between the shortest and longest sides both match. So…

Relationship between the shortest and longest sides = 2n + 1

The next relationship I’m going to find is between the middle and longest side.

Longest Side – middle Side = Relationship.

2n + 2n + 1 – 2n + 2n = Relationship.

1 = Relationship.

I am certain that the relationship between the longest and middle side is 1, so I have a table to prove it.

1 and the Relationship between the middle and longest sides both match. So…

Relationship between the middle and longest sides = 1

Using these same principles, I can work out any relationship, to prove this, I will work out the relationship between the shortest side and the perimeter and then the area.

Perimeter - Shortest Side = Relationship.

4n + 6n + 2 – 2n + 1 = Relationship.

4n + 4n + 1 = Relationship.

I am certain that this is the write answer. So…

Relationship between the Perimeter and Shortest side = 4n + 4n + 1

The area is even simpler because, all you have to do is knock the 2n + 1 out of the equation. So…

Relationship between the Area and Shortest side = 2n +2n

2

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.

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