# Beyond Pythagoras

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Introduction

## Beyond Pythagoras

Introduction: -

This investigation is to get me to find out about Pythagorean Triple. I know some of these and I will try and find a pattern. This investigation is also trying to find a formula for Pythagorean Triple. I will generate my formula by using examples and I will justify by examples.

## Searching for Pythagorean Triple: -

This is an example to show you that two specific whole numbers squared will add up to get another specific number that when you square root it you will get a square number which is a whole number. These numbers are unique.

Example: - Numbers 3,4 and

Left Hand Side = 3 + 4

= 9+16

= 25

Right Hand Side= 5

= 25

Left Hand Side = Right Hand Side

Now I am going to justify this with more examples.

A) Numbers 5,12 and 13

Left Hand Side = 5 + 12

= 25 + 144

= 169

Right Hand Side = 13

= 169

Left Hand Side =Right Hand Side

B) Numbers 7,24 and 25

Left Hand Side = 7 + 24

= 49 + 576

= 625

Right Hand Side = 25

= 625

Left Hand Side = Right Hand Side

The numbers (3,4,5), (5,12,13) and (7,24,25) can be set out in a formula.

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

Large

Small

Middle

To find the perimeter of this right – angled triangle I am have to know the smallest side, the middle side and the largest number.

Smallest number + middle number + largest number = Perimeter

Example: -

Perimeter = 3 + 4 + 5 = 12 units

Middle

40

9

41

## Area = ½ * A * B

Area = ½ * 9 * 40

½ * 9 * 40 = 180 units

The perimeter of the numbers 9,40 and 41 is 90 units.

The area of the numbers 9,40 and 41 is 180 units.

All of these numbers (3,4,5), (5,12,13), (7,24,25) and (9,40,41) are all called Pythagorean Triple because they all satisfy the condition.

A + B = C or 3 + 4 = 5

3 5

4

I am now going to investigate the relationships between the lengths of all three sides of the right – angled triangles. I am now going to find a formula for A.

N | A |

1 | 3 |

2 | 5 |

3 | 7 |

4 | 9 |

The difference between the lengths of the shortest side is 2. This means the equation must be something with 2n. So, there is only a difference of +1 between 2n and the shortest side, s this means the formula should be 2n+1

A=2n+1

I will now test this formula: -

N=1

A=2*1+1

A=3

This means that my first term is right.

N=2

A=2*2+1

A=5

This shows that my second term is right.

N=3

A=2*3+1

A=7

This shows that my third term is right.

I will now try and find a method for B.

N | A | B |

1 | 3 | 4 |

2 | 5 | 12 |

3 | 7 | 24 |

4 | 9 | 40 |

From this table I can see if I *1 and then +1 from A, I get B.

### Example

N=1

A=3

## B=3*1+1

B=4

This means this way works for the first term. Now I will try this for the next line.

N=2

A=5

B=5*1+1

B=6

Conclusion

A=2n+1

B=2n +2n

C=2n +2n+1

Left Hand Side

A=(2n+1)

I have to factorise this formula to get rid of the brackets.

A=(2n+1) (2n+1)

A=2n*2n=4n

A=2n*1=2n

A=1*2n=2n

A=1*1=1

A=4n +2n+2n+1

I can simplify this to,

A=4n +4n+1

I now have to do the same to the formula of B.

B=(2n +2n)

B=(2n +2n) (2n +2n)

B=2n +2n =4n

B=2n +2n=4n

B=2n+2n =4n

B=2n+2n=4n

B=4n +4n +4n +4n

This can be simplified to,

B=8n +8n

I now know the left hand side equation is: -

A and B=4n +4n+1+8n +8n

Now that I found the left Hand Side of the formula, I must find the right hand side of the formula. I have to do to C’s formula exactly what B’s formulas is and what A formula is.

C=2n +2n+1

I now have to factorise this.

C=(2n +2n+1) (2n +2n+1)

C=2n *2n =4n

C=2n *2n=4n

C=2n +1=2n

C=2n*2n =4n

C=2n*2n=4n

C=2n*1=2n

C=1*2n =2n

C=1*2n=2n

C=1*1=1

C=4n +4n +2n +4n +4n +2n+2n +2n+1

This can be simplified to,

C=4n +4n+1+8n +8n

I got this answer by adding the two 4n together and then the other two 4n together.

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

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