# Beyond Pythagoras

Extracts from this document...

Introduction

## BEYOND PYTHAGORAS

Pythagoras theorem states that in a right angled triangle, a² + b² = c², this can also be interpreted as (shortest side) ² + (middle side) ²

= (longest side) ².

Two sets of numbers that satisfy the theorem are 5, 12, 13 and

7, 24, 25.

5² + 12² = 13² 7² + 24² = 25²

25 + 144 = 169 49 + 576 = 625

The perimeter of a triangle is calculated using a + b + c. The area of a triangle is calculated using ½ x a x b. The perimeters and areas for the triangles above are:

PERIMETER –

5 + 12 + 13 = 30

7 + 24 + 25 = 56

AREA -

½ x 5 x 12 = 30

½ x 7 x 24 = 84

Below is a table of side length, area and perimeter for Pythagorean triples that start with ODD numbers.

TERM | LENGTH OF | LENGTH OF | LENGTH OF | PERIMETER | AREA |

SHORTEST SIDE | MIDDLE SIDE | LONGEST SIDE | |||

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 |

Using the sequence formula :

nth term = a + (n-1)d + ½ (n-1) (n-2)c

a = first number

d = first difference

c = second difference

Middle

20

25

60

150

nth

3n

4n

5n

12n

6n²

## TRIPLE: 5,12,13

TERM | LENGTH OF | LENGTH OF | LENGTH OF | PERIMETER | AREA |

SHORTEST SIDE | MIDDLE SIDE | LONGEST SIDE | PERIMETER | AREA | |

1 | 5 | 12 | 13 | 30 | 30 |

2 | 10 | 24 | 26 | 60 | 120 |

3 | 15 | 36 | 39 | 90 | 270 |

4 | 20 | 48 | 52 | 120 | 480 |

5 | 25 | 60 | 65 | 150 | 750 |

nth | 5n | 12n | 13n | 30n | 30n² |

TRIPLE: 7,24,25

TERM | LENGTH OF | LENGTH OF | LENGTH OF | PERIMETER | AREA |

SHORTEST SIDE | MIDDLE SIDE | LONGEST SIDE | |||

1 | 7 | 24 | 25 | 56 | 84 |

2 | 14 | 48 | 50 | 112 | 336 |

3 | 21 | 72 | 75 | 168 | 756 |

4 | 28 | 96 | 100 | 224 | 1344 |

5 | 35 | 120 | 125 | 280 | 2100 |

nth | 7n | 24n | 25n | 56n | 84n² |

I put the nth terms in a table against the number triple it was (t). I was then able to work

Conclusion

Formula for Area:

(n² - m²) 2mn

2

## Perimeter = Area:

(n² - m²) + 2mn + n² + m² = (n² - m²) 2mn

2

I simplified the formula to give so that I was able to work out values for n and m.

(n² - m²) + 2mn + n² + m² = (n² - m²) 2mn

2

2mn + 2n² = (n + m)(n -m)mn

2n(m + n) = (n + m)(n – m)mn

2 = m x (n - m)

m = 2 (n – m) = 1 n = 3

or m = 1 (n – m) = 2 n = 3

When these two solutions for n and m are inserted into the general formulas the two Pythagorean triples with perimeter = area are generated.

m = 2 n = 3

n² - m ² 2mn n² + m²

3² - 2² 2x2x3 = 12 3² + 2²

9 – 4 = 59 + 4 = 13

Pythagorean triple in which area = perimeter : 5, 12, 13

m = 1 n = 3

n² - m² 2mn n² + m²

3² - 1² 2x1x3 = 6 3² + 1²

9 - 1 = 8 9 + 1 = 10

Pythagorean triple in which area = perimeter: 6, 8 , 10

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

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