# Beyond Pythagoras

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Introduction

## Daniel Cook 10R

## Beyond Pythagoras: Year 10 GCSE Coursework

I am going to study Pythagoras’ theorem. Pythagoras Theorem is a2 + b2 = c2. ‘a’ being the shortest side, ‘b’ being the middle side and ‘c’ being the longest side (hypotenuse) of a right angled triangle.

For example, I will use 32 x 42 = 52 . This is because:

32 = 3 x 3 = 9

42 = 4 x 4 = 16

52 = 5 x 5 = 25

So.. 9 +16 = 25

For this table, I am using the term a, b, b + 1

Triangle Number (n) | Length of shortest side | Length of middle side | Length of longest side | 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 |

6 | 13 | 84 | 85 | 183 | 546 |

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

8 | 17 | 144 | 145 | 296 | 1224 |

## Formulas

Shortest side = 2n + 1, n being the triangle number

Middle side = 2n2 + 2n. This is because:

Triangle Number

1 = 2 x 2

2 = 3 x 4

3 = 4 x 6

Middle

4n³

2n²

2n

+1

2n²

2n

1

= 4n4 + 8n³ + 8n² +4n + 1

A + B = C

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

My results show that shortest² + middle² = longest². I have Pythagoras’ theorem.

For this table, I am using the term a, b, b + 2

Triangle Number (n) | Length of shortest side | Length of middle side | Length of longest side | Perimeter | Area |

1 | 6 | 8 | 10 | 24 | 48 |

2 | 8 | 15 | 17 | 40 | 120 |

3 | 10 | 24 | 26 | 60 | 240 |

4 | 12 | 35 | 37 | 84 | 420 |

5 | 14 | 48 | 50 | 112 | 672 |

6 | 16 | 63 | 65 | 144 | 1008 |

7 | 18 | 80 | 82 | 180 | 1440 |

8 | 20 | 99 | 101 | 220 | 1980 |

## Formulas

Shortest side = 2n + 4. This is because:

2 x 1 + 4 = 6

2 x 2 + 4 = 8

2 x 3 + 4 = 10 etc

Middle side = n² + 4n + 3

### Longest side = n² + 4n + 5

## Box Methods

Shortest² = (2n + 4)²

2n | +4 | |

2n | 4n² | 8n |

+4 | 8n | 16 |

= 4n² + 16n + 16

Middle² = (n² + 4n + 3)²

n² | +4n | +3 | |

n² | n4 | 4n³ | 3n² |

+4n | 4n³ | 16n² |

Conclusion

Length of longest side

Perimeter

Area

1

12

16

20

48

96

2

16

30

34

80

240

3

20

48

52

120

480

4

24

70

74

168

840

5

28

96

100

224

1344

6

32

126

130

288

2016

7

36

160

164

360

2880

8

40

198

202

440

3960

## Formulas

Shortest side = 4n + 8

Middle side = 2n² + 8n + 6

Longest side = 2n² + 8n + 10

## Box Methods

Shortest² = (4n + 8)²

4n | +8 | |

4n | 16n² | 32n |

+8 | 32n | 64 |

= 16n² + 64n + 64

Middle² = (2n² + 8n + 6)²

2n² | +8n | +6 | |

2n² | 4n4 | 16n³ | 12n² |

+8n | 16n³ | 64n² | 48n |

+6 | 12n² | 48n | 36 |

= 4n4 + 32n³ + 88n² + 96n + 36

Longest² = (2n² + 8n + 10)²

2n² | +8n | +10 | |

2n² | 4n4 | 16n³ | 20n² |

+8n | 16n³ | 64n² | 80n |

+10 | 20n² | 80n | 100 |

= 4n4 + 32n³ + 104n² + 160n + 100

A + B = C

16n² + 64n + 64 + 4n4 + 32n³ + 88n² + 96n + 36 = 4n4 + 32n³ + 104n² + 160n + 100

All of my research has proved that Pythagoras’ theorem is correct, even if you come at the theorem at different angles; you’re sure to find out that Pythagoras is 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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