# Beyond Pythagoras

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Introduction

Mathematics Coursework

## Beyond Pythagoras

Introduction

There once lived a Greek Mathematician/Philosopher called Pythagoras. He discovered many interesting theories and I will be looking at one in particular. Pythagoras’ theorem of right-angled triangles, which states one clear fact:

a2 + b2 = c2 or "The square of the hypotenuse is equal to the sum of the squares of the other two sides". Other mathematical formulas I will be using to aid me in my tasks are: Perimeter = a + b + cand

Area = ½ x a x b

I have decided to investigate the relationship between different aspects of right-angled triangles. I will study the connections between all three sides, the perimeter and the area of the triangles. To do this, I will begin by constructing a table of various measurements taken from some right-angled triangles. The triangles I have chosen, all increase by one unit on the shortest side. In my table, s=a, m=b and l=c:

Term Number (n) | Length of shortest side (s) | Length of middle side (m) | Length of longest side (l) | 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 |

Initially, I calculated the relationship between the shortest side and the other two sides until

Middle

By constructing another table, I discovered that the expression is:

m = 2n2 + 2n

To check this, I can use the formula (n x s) + n.

## Length of Longest Side

By referring back to my table, I noticed that the longest side is 1 unit larger than the length of the middle side, which makes them consecutive. This means that the formula for the length of the longest side would be exactly the same as the formula for the length of the middle side, with an addition of 1 unit. So my new formula is as follows: l = 2n2 + 2n + 1.

The reason for this is that the difference between the length of the shortest side and the length of the longest side, and the actual length of the shortest side, are equal to the length of the longest side.

To check that my formula worked, I decided to find the formula using a different method. I realized that the difference between s and l was 2n2 by calculating that the third difference was 4 and then added this to my formula for s, and, to prove my formula correct, the expression was the same.

#### Proof

To prove my formulas are correct, I referred back to the initial formula: a2 + b2 = c2

Conclusion

Area: a = (n + 3n2 + 2n3) x 2

a = 2n + 6n2 + 4n3

After applying my new formulas to the new table, I discovered that my prediction is correct.

Investigating Other Pythagorean Triples

I decided to delve deeper into Pythagorean triples (sides of a right-angled triangle that are integers). I began by drawing a table with s = 2 and used my formulas to calculate the lengths of the other sides, to check if they were integers.

S | m | l |

2 | 2 | 2.93 |

2 | 3 | 3.61 |

2 | 4 | 4.47 |

2 | 5 | 5.39 |

2 | 6 | 6.32 |

2 | 7 | 7.28 |

After looking at the table, I realised that there could not possibly be a Pythagorean triple with shortest side 2, because, the longest side would need to be an integer and it would obviously never reach a whole number. We can see this because, in the l column the number is always l = m + 0.x. Because of this fact, and the fact that in the first possible s and m row, x is only 93 and from then on x decreases, we know that l can never reach a whole number that is larger than m.

I then tried a table for s = 4 but achieved the same outcome. As it would be impractical and extremely time consuming to conduct every possible table for every possible s, I believe that there are no Pythagorean triples where s is an even number, except those that are enlargements of the initial Pythagorean triple.

Asha Lancaster Thomas 11JL

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

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