Mathematics Coursework - Beyond Pythagoras

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Jenny Savage                

Mathematics Coursework - Beyond Pythagoras

The numbers 3,4 and 5 satisfy the condition

                3² + 4² = 5²

Because        3² = 3 x 3 = 9

                4² = 4 x 4 = 16

                5² = 5 x 5 = 25

And so

                3² + 4² = 9 + 16 = 25 = 5²

This condition also applies to the following sets of numbers

(smallest number)² + (middle number)² = (largest number)²

The numbers 5, 12 and 13 satisfy the condition

5² + 12² = 13²

Because        5² = 5 x 5 = 25

                12² = 12 x 12 = 144

                13² = 13 x 13 = 169

And so

                5² + 12² = 25 + 144 = 169 = 13²

This condition also applies to the following sets of numbers

(smallest number)² + (middle number)² = (largest number)²

The numbers 7, 24 and 25 satisfy the condition

7² + 24² = 25²

Because        7² = 7 x 7 = 49

                24² = 24 x 24 = 576

                25² = 25 x 25 = 625

And so

                7² + 24² = 49 + 576 = 625 = 25²

The numbers 3, 4 and 5 can be lengths of the sides of a right-angled triangle:

The perimeter and area of this triangle are:

Perimeter = 3 + 4 + 5 = 12 units

Area = ½ x 3 x 4 = 6 square units

                                                 

The table below shows the area and perimeter of the (5, 12, 13) and (7, 24, 25) right-angled triangles.

To work out the perimeter I added all the sides together, and to get the area I multiplied the shortest length with the middle length then divided it by 2.

(3, 4, 5); (5, 12, 13) and (7, 24, 25) are all called Pythagorean triples because they satisfy the condition:

a² + b² = c²

Below is a table, which shows the next results in the sequence:

The shortest length increases by two each time, as they are all odd numbers.

I have also noticed that the middle length increases by four and also the longest length increases by four.  

The area and the perimeter do not increase by a repetitious sequence so I then found out the difference between two values until I found the nth term, as you can see on the following page.

Also the difference between the middle side and the longest side is 1 of each right-angled triangle.

These are all triangles that have a shortest side, which is an odd number and also all the numbers are positive integers.

Shortest Length

The difference between each number is two, so it gives you the formula 2n.

I then worked out 2n and it gave me the sequence:

2, 4, 6, 8, 10

The sequence of 2n and the original sequence had a difference of 1 between each number so the final formula is:

2n + 1

Middle Length

The first line did not have the same number along the line so I then had to difference it.  This then came out to a difference of 4 between each number.  This gave you 2n² because you have to half the number on the second line then square it.

I then worked out the formula 2n² and it gave me the sequence:

2, 8, 18, 32, 50,

This then had a difference of 2, 4, 6, 8, and 10 between 2n² and the original sequence.  This difference then had a difference of 2 so the final part of the formula is 2n.  This meant the formula for the middle number is:

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2n² + 2n  

Longest Length

This again gave me the formula 2n², because it was all the same on the second line so I halved it and then squared it.  The formula, 2n² gave me the sequence 2, 8, 18, 32, 50.  The difference between this formula and the original sequence came to 3, 5, 7, 9, and 11.  This sequence had a difference of 2 between each number so you got 2n.  

This then gave you the formula 2n² + 2n.  I then worked this out and it came to:

4, ...

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