# Beyond Pythagoras

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Introduction

TO: Mr. Fray

FROM: Nadia Falcon (12 – GH)

DATE: 11th March, 2002

TITLE: Beyond Pythagoras

1. INTRODUCTION:

In this coursework I am required to find out the area and perimeter of the following sets of numbers 3,4,5; 5,12,13 and 7,24,25 of right angle triangles and whether they satisfy a similar condition by using the formula of the Pythagoras Theorem, which is,

(shortest side) ² + (middle side) ² = (longest side) ²

2. PROCEDURE:

I am going to work out the area and perimeter of the following three triangles with measurements, 3,4,5; 5,12,13 and 7,24,25 to check if they satisfy the condition

(Shortest side) ² + (Middle side) ² = (Longest side) ² of the Pythagoras Theorem.

1) The numbers 3,4,5 satisfy this condition because -

3² + 4² = 5²

So shortest side 3² = 3 x 3 = 9

Middle side 4² = 4 x 4 = + 16

--------

25

And Longest side 5² = 5 x 5 = 25

2) 5, 12, 13

5² + 12² = 13²

Shortest side 5² = 5 x 5 = 25

Middle side 12² = 12 x 12 = + 144

169

Longest side 13² = 13 x 13 = 169

3) 7, 24, 25

7² + 24² = 25²

Shortest side 7² = 7 x 7 = 49

Middle side 24² = 24 x 24 = + 576

625

Longest side 25² = 25 x 25 = 625

FAMILIES OF TRIPLES:

With the help of the above set of triples, I am now going to prove that all the three sets of numbers satisfy the Pythagoras Theorem and obey the rule, i.e. (Shortest side) ² + (Middle side) ² = (Longest side) ²

This I have down by multiplying the triples by 2 and 3. This is shown as below:

TRIPLES |

3, 4, 5 X 2 = 6, 8, 10 X 3 = 9, 12, 15 |

5, 12, 13 X 2 = 10, 24, 26 X 3 = 15, 36, 39 |

7, 24, 25 X 2 = 14, 48, 50 X 3 = 21, 72, 75 |

Therefore, 6² + 8² = 10² and 9² + 12² = 15²

36 + 64 = 100 81+144 = 225

100 = 100 225 = 225

They can also be called as Pythagorean triples.

Now I am going to work out the Area and the Perimeter of each triangle.

1)3,4,5

The formula for Area of a triangle is: ½ x b x h

Area = ½ x 4 x 3

= 4 x 3 = 2 x 3 = 6

2

Middle

Note: The sides of the new three Pythagorean triples are highlighted in red.

PATTERNS IN LENGTH | |||

Term | Shortest Side | Middle Side | Longest Side |

1 | 3 | 4 | 5 |

2 | 5 | 12 | 13 |

3 | 7 | 24 | 25 |

4 | 9 | 40 | 41 |

5 | 11 | 60 | 61 |

6 | 13 | 84 | 85 |

The table above explains that for the terms 1,2,3, the difference between the lengths of the shortest sides 3,5,7 is 2. This means that to get the new shortest sides for the new triples, I will add 2 each time. By doing so I have got: 7 + 2 = 9; 9 + 2 = 11 and 11 + 2 = 13.

Similarly, looking at the lengths of the Middle sides 4,12,24, I can see that to get from 4 to 12, I have to add 8; then from 12 to 24, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further. Hence, the next lengths of the three middle sides would be: 24 + 16 = 40; then 40 + 20 = 60 and 60 + 24 = 84.

The same pattern used to find out the middle sides was used to find the longest sides. To get from 5 to 13, I have to add 8; then from 13 to 25, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further. Hence, the next lengths of the three longest sides would be: 25 + 16 = 41; then 41 + 20 = 61 and 61+ 24 = 85.

Also, we can see that there is another way to find the longest sides besides as mentioned above. We can see that the longest side for the first 3 terms is 1 more than the middle side.

Conclusion

- I can calculate ‘b’ by substituting ‘a’ and ‘c’ for the values I have found.

Substituting it - an² + bn + c

Input valueoutput value

2 x (2) ² + b(2) + 0 = 12

2 x 4 + 2b = 12

8 + 2b = 12

2b = 12 – 8

2b = 4

b = 4/2 = 2

Hence b = 2

Now I am going to find the expression for the size of the longest side, using the same procedures as I have used to find the expression for the middle side.

Term ‘n’ | Longest sides ‘LS’ | First Difference | Second Difference |

1 | 5 | 8 | 4 |

2 | 13 | 12 | 4 |

3 | 25 | 16 | 4 |

4 | 41 | 20 | 4 |

5 | 61 | 24 | 4 |

6 | 85 |

However, this is straightforward, because longest side is just the formula for the middle side, which is ‘MS’ +1. However, looking at the table of results, I noticed that 2n² + 2n + 1= LS

Hence the expression for the longest side will be: LS = 2n² + 2n + 1

‘LS’ meaning length of the longest side.

‘n’ meaning number of terms.

I can prove that this works by testing it against any one length of the LS and the ‘n’ term.

The triangle for the second term has a LS of 13 (output value) and the ‘n’ term against it is 2 (input value).

2n² + 2n + 1 = LS

Therefore, 2 x 2² + 2 x 2 + 1 = 13

8 + 4 + 1 = 13

So, 13 = 13

Finally, to prove my formulas for ‘SS’, ‘MS’ and ‘LS’ are correct, I decided to incorporate them formulas into 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

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

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

I now have the nth term for each of the 3 sides of a right-angled triangle. This proves that my ‘SS’, ‘MS’ and ‘LS’ formulas are correct.

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

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