# Pythagorean Theorem Coursework

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Introduction

Jaymin Patel 10Q Page of

Pythagoras’ theorem states that a²+b²=c²

As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number.

The numbers 3, 4 and 5 work in Pythagoras’ theorem,

3²+4²=5²

because 3²=3×3=9

4²=4×4=16

5²=5×5=25

and so 3²+4²=9+16=25=5²

The numbers 5, 12 and 13 also work,

5²+12²=13²

because 5²=5×5=25

12²=12×12=144

13²=13×13=169

and so 5²+12²=25+144=169=13²

The numbers 7, 24 and 25 also work,

7²+24²=25²

because 7²=7×7=49

24²=24×24=576

25²=25×25=625

and so 7²+24²=49+576=625=25²

3, 4 and 5

Perimeter= 3+4+5=12

Area= ½ ×3×4=6

5, 12 and 13

Perimeter= 5+12+13=30

Area= ½ ×5×12=30

7, 24 and 25

Perimeter= 7+24+25=56

Area= ½ ×7×24=84

From the first three terms I have realised that: -

- a increases by 2 each time

- a is equal to the formula ×2+1 from n

- b is always even

- c is always odd

- c is always +1 of b

- b=(a×n) + n

I have also added 2 more terms using what I think are my formulae.

‘n’ | ‘a’ | ‘b’ | ‘c’ | Perimeter | Area |

1 | 3 | 4 | 5 | 12 | 6 |

2 | 5 | 12 | 13 | 30 | 30 |

3 | 7 | 24 | 25 | 56 | 84 |

4 | 9 | 40 | 41 | 90 | 180 |

5 | 11 | 60 | 61 | 132 | 330 |

Here are my formulas.

Middle

‘n’ | ‘a’ | ‘b’ | ‘c’ | area | perimeter |

1 | 3 | 4 | 5 | 6 | 12 |

2 | 5 | 12 | 13 | 30 | 30 |

3 | 7 | 24 | 25 | 84 | 56 |

4 | 9 | 40 | 41 | 180 | 90 |

5 | 11 | 60 | 61 | 330 | 132 |

6 | 13 | 84 | 85 | 546 | 182 |

7 | 15 | 112 | 113 | 840 | 240 |

8 | 17 | 144 | 145 | 1224 | 306 |

9 | 19 | 180 | 181 | 1710 | 360 |

10 | 21 | 220 | 221 | 2310 | 462 |

Now I have to prove that my formulae for a, b and c work. To do this I must incorporate a²+b²+c²: -

a²+b²=c²

(2n+1)²+(2n²+2n) ²=(2n²+2n+1) ²

(2n+1)(2n+1) + (2n²+2n)(2n²+2n) = (2n²+2n+1)(2n²+2n+1)

4n²+2n+2n+1+4n⁴+4n²+4n³+4n³ = 4n⁴+8n³+8n²+4n+1

4n⁴+8n³+8n²+4n+1 = 4n⁴+8n³+8n²+4n+1

This proves that my formulae for a, b and c are all correct!

Now for part 2, I will try to solve the problem of Pythagorean triples with even numbers for a.

The numbers 6, 8 and 10 work for Pythagoras’ theorem,

6²+8²=10²

because 6²=6×6=36

8²=8×8=64

10²=10×10=100

and so 6²+8²=36+64=100=10²

Conclusion

So 2n+n²-1+n²+1

= 2n+2n²

= 2n²+2n

Here is how I got to my b and c formulae.

Getting b

a²+b²=c²

=c²= (b+2) ²

=(2n) ²+b²=(b+2) ²

=4n²+b²=b²+4b+4

=4n²-4=4b

=n²-1=b

=b=n²-1

Getting c

By looking at b you can see that c is always +2 than b.

So c is n²-1+2

=n²+1

Here is a table with all of the results for even numbers.

‘n’ | ‘a’ | ‘b’ | ‘c’ | Perimeter | Area |

1 | 2 | 0 | 2 | 4 | 0 |

2 | 4 | 3 | 5 | 12 | 6 |

3 | 6 | 8 | 10 | 24 | 24 |

4 | 8 | 15 | 17 | 40 | 60 |

5 | 10 | 24 | 26 | 60 | 120 |

6 | 12 | 35 | 37 | 84 | 210 |

7 | 14 | 48 | 50 | 112 | 336 |

8 | 16 | 63 | 65 | 144 | 504 |

9 | 18 | 80 | 82 | 180 | 720 |

10 | 20 | 99 | 101 | 220 | 990 |

Now I have to prove my formulae for a, b and c. To do this I have to incorporate a²+b²=c²

a²+b²=c²

= (2n)²+(n²-1) ²=(n²+1) ²

= (2n)(2n)=4n²

(n²-1)(n²-1)=n⁴-n²-n²+1

(n²+1)(n²+1)=n⁴+n²+n²+1

= 4n²+n⁴-n²-n²+1=n⁴+n²+n²+1

= n⁴+2n³+1=n⁴+2n³+1

This proves that my formulas for a, b and c are all correct.

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

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