# For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

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Sarah Williams

## Introduction

For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world’s most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a²+b²=c². This is what the coursework is based on.

I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed.

### The coursework

The numbers 3, 4 and 5 satisfy the condition 3²+4²=5² because

3²=3x3=9

4²=4x4=16

5²=5x5=25

And so

3²+4²=9+16=25=5²

I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ²+(middle number) ²=(largest number) ²

a) 5, 12, 13

5²=5x5=25

12²=12x12=144

25+144=169

√169 = 13

This satisfies the condition as

5²+12²=25+144=169=13²

b) 7, 24, 25

7²=7x7=49

24²=24x24=576

49+576=625

√625=25

This satisfies the condition as

7²+24²=49+576=625=25²

The numbers 3,4 and 5 can be the lengths – in appropriate units – of the side of a right-angled triangle.

5

3

The perimeter and area of this triangle are:

Perimeter = 3+4+5=12 units

Area = ½x3x4=6 units ²

The numbers 5,12,13 can also be the lengths – in appropriate units – of a right-angled triangle.

### Perimeter = 5+12+13=30

Area=½x5x12=30

This is also true for the numbers 7,24,25

### Area=½x7x24=84

I have put these results into a table to see if I can work out any patterns.

I have noticed that there seems to be a recurring pattern between the length of the middle side and the length of the longest side. They are always consecutive numbers. I shall now investigate this.

I will assume that the hypotenuse has length b + 1 where b is the length of the middle side. (c=b+1)

b+1

a

b

Using Pythagoras:

a² + b² = (b+1) ²

Expanding this:

a² + b² = b² + 2b+ 1

therefore

a² = 2b +1

This means that a² must be odd because it equals 2b + 1.

Since a² is odd this means that a must be odd also.

(even x even = even)

(odd x odd = odd)

From the previous equation we can work out that

b = (a²-1)/2

The three sides are a, b and b +1 where a is an odd number. Now I will produce a table with the results.

To work out some more formulae I will draw this table again but using an n column.

I have worked out some formulae to find out any number in this sequence using n.

First of all I will work out a formula for the shortest side:

The simplest way to work this out would be to find a common difference between the numbers.

However 3 – 1 = 2

5 – 2 = 3

7 – 3 = 4

There is no common difference in the first difference so I will try the 2nd difference

3-2 = 1

4-3 = 1

This shows that there is a pattern between the numbers. So all I have to do now is find the formula. There might be a pattern if I multiply each number by the same number as in ...