# Beyond Pythagoras.

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Introduction

The Pythagoras tethered was found by a mathematician from Italy called Pythagoras who is said to be the first ever true mathematician. He was also the first to teach that the Earth is a sphere that revolves around the Sun.

This how the Pythagoras thorium is defend:

A thorium stating that a square of a hypotenuse of a right angled triangle is equal to the sum of the squares of the other two sides.

Formula that Pythagoras discovered:

A² + B² = H²

H

To Check that each of the following sets of numbers satisfy a

Similar condition of the

(Smallest)²+ (middle)² = (largest number)²

1) a) 5,12,13

a²+b²=h²

5²+12²=h²

25+144=169

169= 13

I put these figures in the formula for the Pythagoras theorem to

find that these numbers do satisfy the condition

(Smallest)²+ (middle)² = (largest number)²

l) b) 7,24,25

a²+b²=h²

7²+24²=h²

49+576=625

625=25

After finding that they are Pythagoras triples, I found the perimeter and area.

2)

(5,12,13)

Perimeter = 5 + 12 + 13

= 30

Area = 5 × 12

2

= 60

2

= 30

(7,24,25)

Perimeter = 7 + 24 + 25

= 56

Area = 7 × 24

2

= 168

2

= 84

I put everything that I did in to a table showing the unit of each side and the perimeter and area.

Shortest side ‘a’ | Middle side ‘b’ | Longest side (Hypotenuse) ‘h’ | Perimeter | Area |

3 | 4 | 5 | 12 | 6 |

5 | 12 | 13 | 30 | 30 |

7 | 24 | 25 | 56 | 84 |

To find the Perimeter I simply had to find the total of all three sides. For the area I had to

P = a + b + h

find multiply side ‘a’ with side ‘b’ and half the total

A = ab

2

Bellow is have shown in detail how I found the perimeter and area for each triple.

2)

(5,12,13)

Perimeter = 5 + 12 + 13

= 30

Area = 5 × 12

2

= 60

2

= 30

(7,24,25)

Perimeter = 7 + 24 + 25

= 56

Middle

= 2 × 4² + 2 × 4

= 2 × 16 + 2 × 4

= 32 + 8

= 40

Term 5:

‘b’ = 2n2 + 2n

= 2 × 5² + 2 × 5

= 2 × 25 + 2 × 5

= 50 + 10

= 60

Side ‘h’

It is obviouse that the set of results found for side ‘h’ are just one plus to the results of side ‘b’

Middle side ‘b’ | Longest side (Hypotenuse) ‘h’ |

4 + 1 | 5 |

12 + 1 | 13 |

24 + 1 | 25 |

40 + 1 | 41 |

60 + 1 | 61 |

84 + 1 | 85 |

So I took the formula for side ‘b’ and simply added one onto it:

2n2 + 2n + 1

I tested this formula by randomly selecting 2 terms.

Term 5:

‘h’ = 2n2 + 2n + 1

= 2 × 5² + 2 × 5 + 1

= 2 × 25 + 10 + 1

= 50 + 11

= 61

Term 4:

‘h’ = 2n2 + 2n + 1

= 2 × 4² + 2 × 4 + 1

= 2 × 16 + 8 + 1

= 32 + 9

= 41

Perimeter

To find the perimeter for any shape you must find the total of all the sides so I would do

a + b + c. So I took the formulas that I fount for all 3 sides and added them together to make the formula for the perimeter.

2n+1 + 2n2 + 2n + 2n2 + 2n + 1

4n2 + 6n +2

I then tested this formula

Term 2:

‘p’ = 4n2 + 6n + 2

= 4×22 + 6 × 2 + 2

= 4×4 + 12 + 2

= 16 + 14

= 30

Term 5:

‘p’ = 4n2 + 6n + 2

= 4×52 + 6 × 5 + 2

= 4×25 + 30 + 2

= 100 + 32

= 132

Area

To find the area of a right angled triangle I would have to multiply the base and the height and half the answer as one would to find the area for a simple right angled triangles. So I took both formulas from the base and height (‘a’’b’) and put them into the following formula:

½ (2n+1) (2n2 + 2n)

Then multiplied it out to get:

4n3 + 6n2 + 2n

2

Then divide 4n3 + 6n2 + 2n by 2 to get

2n3 + 3n2 + n

I then looked at the relationship between the perimeter and area.

P = 4n2 + 6n + 2

A = 2n3 + 3n2 + n

Conclusion

b = 4

c = 0

I then used this to replace the letters in the formula to find the nth term

an² + bn + c = 0

When replaced

4n² + 4n + 0

So the nth term formula would be : 2n² + 2n

It could see that everything double and I ended up with a double nth term equation

I then tested this formula for 2 randomly selected terms.

Term 4:

‘b’ = 4n2 + 4n

= 4 × 4² + 4 × 4

= 4 × 16 + 4 × 4

= 64 + 16

= 80

Term 5:

‘b’ = 4n2 + 4n

= 4 × 5² + 4 × 5

= 4 × 25 + 4 × 5

= 100 + 20

= 120

Side ‘h’ even

I only had to double the formula for for the hypotenuse

2n2 + 2n + 2

4n2 + 4n + 2

But as the results have doubled, instead of there being a difference of 1 between side ‘a’ and ‘h’ there is a difference of 2

Middle side ‘b’ | Longest side (Hypotenuse) ‘h’ |

8 + 2 | 10 |

24 + 2 | 26 |

48 + 2 | 50 |

80 + 2 | 82 |

60 + 2 | 122 |

168 + 2 | 170 |

I then tested this formula

Term 5:

‘h’ = 4n2 + 4n + 2

= 4 × 5² + 4 × 5 + 2

= 4 × 25 + 20 + 2

= 100 + 22

= 122

Term 4:

‘h’ = 4n2 + 4n + 2

= 4 × 4² + 4 × 4 + 2

= 4 × 16 + 16 + 2

= 64 + 18

= 82

Perimeter even

There are 2 ways to find the perimeter for even. I can add all the formulas I have found together or I can multiply the whole equation by 2. Either way this is the formula I got:

4n+2 + 4n2 + 4n + 4n2 + 4n + 2

8n2 + 12n +4

I then tested this formula

Term 2:

‘p’ = 8n2 + 12n + 4

= 8×22 + 12 × 2 + 4

= 8×4 + 24 + 4

= 32 + 28

= 60

Term 5:

‘p’ = 8n2 + 12n + 4

= 8×52 + 12 × 5 + 4

= 8×25 + 60 + 4

= 200 + 64

= 264

Area even

The area for the even results would be

A = pn

If I multiplied the perimeter by n I should get the area.

P = 8n2 + 12n + 4

I will now see if this is correct by using the formulas for side a and b (simple area formula)

Area = ½ (4n + 1) (4n2 + 4n)

= 16n³ + 10n + 8n2 + 8n

2

= 16n³ + 12n2 + 8n

= 8n + 12n2 + 4n

Area = p × n

This shows that if the perimeter is multiplied by n it well equal to the answer.

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

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