# Beyond Pythagoras - Year 10 Maths Coursework

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Introduction

Beyond Pythagoras - Year 10 Maths Coursework

By Nadia House 10G

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 now have to find out if the following sets of numbers satisfy a similar condition of:

Smallest number² + Middle number² = Largest number²

(S² + M = L²)

a) 5, 12, 13

5² + 12² = 13² ~ (25 + 144 = 169)

b) 7, 24, 25

7²+24² = 25² ~ (49 + 576 = 625)

Here is a table containing the results:

Short Side | Middle Side | Long Side |

3 | 4 | 5 |

5 | 12 | 1 |

7 | 24 | 25 |

I looked at this table 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 number² + Middle number² = Largest number². So I know that there will be a connection between the numbers written above. The only problem is that it is obviously not:

Middle Side² + Large Side² = Small Side²

Because, 12² + 13² = 144 + 169 = 313

and 5² = 25

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

I will now try 2 sides squared.

M² + L = S²

= 12² + 13 = 5²

= 144 + 13 = 25

= 157 = 25

This does not work and neither will 13², because it is larger than 12². 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 squared.

12² + 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 = 5²

25 = 25

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

4 +5 = 3²

9 = 9

And…

24 + 25 = 7²

49 = 49

It works with both of my other triangles. So…

M + L = S²

Middle

4n2 - 4(n - 1)² = Middle side.

This should in theory give me my middle side. I will test my theory with the first term.

4 x 12 - 4(1-1)2 = 4

4 x 1 - 4 x 02 = 4

4 - 4 x 0 = 4

4 - 0 = 4

4 = 4

My formula works for the first term. I will now check if it works using the 2nd term.

4 x 22 - 4(2-1)2 = 12

4 x 4 - 4 x 12 = 12

16 - 4 x 1 = 12

16 - 4 = 12

12 = 12

My formula also works for the 2nd term. It’s looking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.

4 x 32 - 4(3-1)2 = 24

4 x 4 - 4 x 22 = 24

36 - 4 x 4 = 24

36 – 16 = 24

20 = 24

My formula doesn’t work for the 3rd term. It now looks as if “4n² - 4(n - 1)2” is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.

4 x 42 - 4(4-1)2 = 40

4 x 16 - 4 x 32 = 40

64 - 4 x 9 = 40

64 – 36 = 40

28 = 40

My formula doesn’t work for the 4th term either. I can now safely say that 4n² - 4(n-1)² is definitely not the correct formula for the middle side.

I believe the problem with 4n² - 4(n-1)2 was that 4n², once you start using larger numbers, becomes far to high to bring it back down to the number that I want for the middle side. Also, 4(n-1)² is not as small when it gets larger so it doesn’t bring the 4n² down enough, to equal the middle side.

I know that the final formula will have something to do with 4 and have to be n². I will now try n² + 4.

Conclusion

Perimeter - Shortest Side = Relationship.

4n² + 6n + 2 – 2n + 1 = Relationship.

4n² + 4n + 1 = Relationship.

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

Relationship between the Perimeter and Shortest side = 4n² + 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 = 2n² +2n

2

Here is the final table.

S | M | L | P | A |

3 | 4 | 5 | 12 | 6 |

5 | 12 | 13 | 30 | 30 |

7 | 24 | 25 | 56 | 984 |

9 | 40 | 41 | 90 | 180 |

11 | 60 | 61 | 132 | 330 |

13 | 84 | 85 | 182 | 546 |

15 | 112 | 113 | 240 | 840 |

17 | 144 | 145 | 306 | 1224 |

19 | 180 | 181 | 380 | 1710 |

21 | 220 | 221 | 462 | 2310 |

I would like to know whether or not the Pythagorean triple 3, 4, 5 is the basis of all triples just some of them.

To find this out I have been to the library and looked at some A-level textbooks and also searched on the internet. I have learnt ‘Arithmetic Progression’

3, 4, 5 is a Pythagorean triple

The pattern is plus one

f a = 3 and d = difference (which is +1) then

3 = a

4 = a + d

5 = a +2d

a, a +d, a + 2d

Therefore if you incorporate this into Pythagoras theorem

a2 + (a + d)2 = (a + 2d)2

a2 + (a + d)(a + d) = (a + 2d)2

a2 + a2 + ad + ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)(a + 2d)

2a2 + 2ad + d2 = a2 + 2ad + 2ad + 4d2

2a2 + 2ad + d2 = 4d2 + a2 + 4ad

If you equate these equations to 0 you get

a2 – 3d2 – 2ad = 0

Change a to x

x2 – 3d2 – 2dx = 0

Factorise this equation to get

(x + d)(x – 3d)

Therefore

x = -d

x = 3d

x = -d is impossible as you cannot have a negative dimension

a, a + d, a + 2d

Is the same as

3d, 4d, 5d

This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.

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

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