# Find a formula to enable the perimeter to be found for any odd Pythagorean triple.

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Introduction

Paul Fudio

Perimeter:

I want to find a formula to enable me to find the perimeter for any odd Pythagorean triple. I know that perimeter is a + b + c so I know the formula for the shortest, middle and longest side so I can assume that if I substitute the formulas I know, I can presume I will find the formulas for the perimeter.

The perimeter = a + b + c. Therefore I took my formula for ‘a’ (2n + 1), my formula for ‘b’ (2n² + 2n) and my formula for ‘c’ (2n² + 2n + 1). I then did the following:

a + b + c

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

2n + 1 + 2n + 2n + 1 + 4n² =

4n² + 6n + 1 + 1 =

4n² + 6n + 2 = formula for perimeter

I will further prove this formula by using sequences.

3,4,5 5,12,13 7,24,25 9,40,41 11,60,61

12 30 56 90 132

+18 +26 +34 +42

+8 +8 +8

Here the first differences aren’t the same, but the second differences are. If the second difference is a constant, then the formula for the nth term contains n². The number in front of n² is half the constant difference. Therefore the first part of the formula is 4n².

Middle

+10 +16 +22 +28

+6 +6 +6

Value of 3n²: 3 12 27 48 75

I have found a similarity between these two parts (difference and 3n²). I will make this easier by moving the constants out of the way.

Term number: 1 2 3 4 5

Difference: +4 +14 +30 +52 +80

+1 +2 +3 +4 +5

Value of 3n²: 3 12 27 48 75

In this table it is clear to see that you add the term number every time so therefore the last part of the formula is +n, this makes up all of the formula to give evidence that my formula was correct in my first statement and is 2n³ + 3n²+ n.

Here you can notice a similar pattern between the area formula and the perimeter formula. First I factorised the area formula and I got:

2 (4n² + 6n + 2) which means, 2 x area = perimeter, as you can see easier

N N

in the table I am going to draw below.

N | Perimeter A + b + c | Area (h x b) 2 | Formula 2 x area = perimeter N | Check |

1 | 12 | 6 | 2/1 x 6 = | 12 |

2 | 30 | 30 | 2/2 x 30 = | 30 |

3 | 56 | 84 | 2/3 x 84 = | 56 |

4 | 90 | 180 | 2/4 x 180 = | 90 |

5 | 132 | 330 | 2/5 x 330 = | 132 |

6 | 182 | 546 | 2/6 x 546 = | 182 |

7 | 240 | 840 | 2/7 x 840 = | 240 |

8 | 306 | 1224 | 2/8 x 1224 = | 306 |

9 | 380 | 1710 | 2/9 x 1710 = | 380 |

10 | 462 | 4620 | 2/10 x 4620 = | 462 |

This proves in my formulae are similar to each other. Therefore another statement I have made is correct.

Proving:

Conclusion

## c = cos –1 = 40

## 41

## = 12.68

## = 12.7˚ (1 dp)

b = 180 – 90 – 12.7 = 77.3˚

a = 90˚

## c = cos –1 = 60

## 61

## = 10.38

## = 10.4˚ (1 dp)

b = 180 – 90 – 10.4 = 79.6˚

a = 90˚

n | Shortest side 2n + 1 | Middle side 2n² + 2n | Longest side 2n² + 2n +1 | a˚ | b˚ | c˚ |

1 | 3 | 4 | 5 | 90 | 53.1 | 36.9 |

2 | 5 | 12 | 13 | 90 | 67.4 | 22.6 |

3 | 7 | 24 | 25 | 90 | 73.7 | 16.3 |

4 | 9 | 40 | 41 | 90 | 77.3 | 12.7 |

5 | 11 | 60 | 61 | 90 | 79.6 | 10.4 |

After I looked at the graph there were no sequences or patterns apart from they all add up to 180˚ and b + c add up to 90˚. Also angle b increases when the sides are longer while c angle decreases when the sides are longer.

Conclusion:

Throughout my whole investigation I have only used odd triples and numbers, but there are also even numbers in triples and Pythagoras’ work. I would expect the same result and formulae as those of odd numbers, and maybe in the future I can investigate this.

I am also amazed at the formulae we had to find and how well it fitted together with our table and substitutions. We found the formulae for finding each side of a right-angled triangle that conforms to Pythagoras’ theorem.

I also came to a conclusion that the formula for area and the formula for perimeter are extremely similar. For some extra work I looked at trigonometry and angles and found out that angles didn’t have any correlation apart from having 180˚ between them.

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