# Pythagoras Theorem.

Extracts from this document...

Introduction

Chetan Mandlia Mathematics 10 White

## Mathematics GCSE Coursework: Beyond Pythagoras

### Pythagoras

Pythagoras was a Greek mathematician and philosopher who was born around 582BC and died around 500BC. The theorem that carries his name, Pythagoras’ Theorem, is perhaps the best-known theorem in the whole of mathematics. The theorem is about finding the lengths of sides in right-angled triangles.

## Pythagoras Theorem

The theorem states that for any right-angled triangle, the area of the square formed on the hypotenuse is equal to the sum of the squares formed on the other two sides.

Or a2 + b2 = c2

Or 92 + 602 = 612

## Pythagorean Triples

(3, 4, 5); (5, 12, 13); (7, 24, 25) and etc are all called Pythagorean triples because the satisfy the condition

a2 + b2 = c2

The numbers 3, 4 and 5 satisfy the condition

The numbers 3, 4 and 5 satisfy the condition

The numbers 3, 4 and 5 satisfy the condition

Because

The perimeter and area of this triangle are

Perimeter = 3 + 4 + 5 = 12 units

Area = ½ x 3 x 4 = 6units2

The perimeter and area of this triangle are

Perimeter = 5 + 12 + 13 = 30 units

Area = ½ x 5 x 12 = 30units2

The perimeter and area of this triangle are

Perimeter = 7 + 24 + 25 = 56 units

Area = ½ x 7 x 24 = 84units2

## Results Table

Odd numbers for Length of Shortest Side (a).

n | Length of Shortest Side (a) | Length of Middle Side (b) | Length of Longest Side (c) | Perimeter | Area |

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 |

6 | 13 | 84 | 85 | 182 | 546 |

7 | 15 | 112 | 113 | 240 | 840 |

8 | 17 | 144 | 145 | 306 | 1224 |

9 | 19 | 180 | 181 | 380 | 1710 |

10 | 21 | 220 | 221 | 462 | 2310 |

11 | 23 | 264 | 265 | 552 | 3036 |

## Relationships | 2n + 1 | an + n | an + n + 1 (b + 1) | a + b + c | ½ x a x b |

## nth Term | 2n + 1 | 2n2 + 2n | 2n2 + 2n + 1 | 4n2 + 6n + 2 | 2n3 + 3n2 + n |

## nth Term factorised | N/A | 2(n2 + n) | N/A | 2(2n2 + 3n + 1) | n(2n2 + 3n +1) |

Middle

8

16

14

26

6

3

56

8

36

20

34

6

4

90

8

64

26

42

6

5

132

8

100

32

50

6

6

182

144

38

New sequence = Original sequence – 4n2

Quadratic equation = an2 + bn + c

a = 2nd difference/2 = 8/2 = 4

b = 1st difference = 6

c = 1st no. in New sequence – 1st difference = 8 – 6 = 2

∴nth term for perimeter =

4n2 + 6n + 2

4n2 + 6n + 2 =

2(2n2 + 3n + 1)

### Nth Term for Area

area = ab/2

a = 2n + 1

b = 2n2 + 2n

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

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

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

4n3 + 4n2 + 2n2 + 2n =

4n3 + 6n2 + 2n

(4n3 + 6n2 + 2n)/2 =

2n3 + 3n2 + n

2n3 +3n2 + n =

n(2n2 +3n + 1)

##### OR

nth term | Orig. seq. | 1st diff. | 2nd diff. | 3rd diff | 2n3 | New seq. | 1st diff. | 2nd diff. | 3n2 | New seq. | 1st diff. |

0 | |||||||||||

1 | 6 | 2 | 4 | 3 | 1 | ||||||

24 | 10 | 1 | |||||||||

2 | 30 | 30 | 16 | 14 | 6 | 12 | 2 | ||||

54 | 12 | 16 | 1 | ||||||||

3 | 84 | 42 | 54 | 30 | 6 | 27 | 3 | ||||

96 | 12 | 22 | 1 | ||||||||

4 | 180 | 54 | 128 | 52 | 6 | 48 | 4 | ||||

150 | 12 | 28 | 1 | ||||||||

5 | 330 | 66 | 250 | 80 | 6 | 75 | 5 | ||||

216 | 34 | 1 | |||||||||

6 | 546 | 432 | 114 | 108 | 6 |

New sequence = Original sequence – 2n3

New sequence = New sequence – 3n2

Cubic equation = an3 + bn2 + cn + d

a = 3rd difference/6 = 12/6 = 2

b = 2nd difference/2 = 6/2 = 3

c = 1st difference = 1

d = 1st no. in New sequence – 1st difference = 1 – 1 = 0

## ∴nth term for area =

## 2n3 + 3n2 + 1n + 0 =

## 2n3 +3n2 + n

2n3 +3n2 + n =

n(2n2 +3n + 1)

## Results Table

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

n | Length of Shortest Side (a) | Length of Middle Side (b) | Length of Longest Side (c) | Perimeter | Area |

1 | 6 | 8 | 10 | 12 | 24 |

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

3 | 14 | 48 | 50 | 112 | 336 |

4 | 18 | 80 | 82 | 180 | 720 |

5 | 22 | 120 | 122 | 264 | 1320 |

6 | 26 | 168 | 170 | 364 | 2184 |

7 | 30 | 224 | 226 | 480 | 3360 |

8 | 34 | 288 | 290 | 612 | 4896 |

9 | 38 | 360 | 362 | 760 | 6840 |

10 | 42 | 440 | 442 | 924 | 9240 |

11 | 46 | 528 | 530 | 1104 | 12144 |

## Relationships | 4n + 2 | an + 2n | an + 2n + 2 (b + 2) | a + b + c | ½ x a x b |

## nth Term | 4n + 2 | 4n2 + 4n | 4n2 + 4n + 2 | 8n2 + 12n + 4 | 8n3 + 12n2 + 4n |

## nth Term factorised | 2(2n + 1) | 4(n + n) | 2(2n + 2n + 1) | 4(2n + 3n + 1) | 4n(2n + 3n + 1) |

Fig.2

## Relationships

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

The Length of The Shortest Side (a)

The length of the shortest side (a) can be calculated using n.

The relationship between a and n is that: -

a = 4n + 2

For example, if n = 6

a = 4n +2

a = (4 x 6) + 2

a = 24 + 2

∴a = 26

The Length of The Middle Side (b)

The length of the middle side (b) can be calculated using n and a.

The relationship between b, a and n is that: -

b = an + 2n

For example, if a = 18 and n = 4

b = an + 2n

b = (18 x 4) + (2 x 4)

b = 72 + (2 x 4)

b = 72 + 8

∴b = 80

The Length of The Longest Side (c)

The length of the longest side (c) can be calculated using a, n and b.

The relationship between c, n and a is that: -

c = an + 2n + 2

For example, if a = 18 and n = 4

c = an + 2n + 2

c = [(18 x 4) + (2 x 4] +2 = [72 + (2 x 4] + 2 = [72 + 8] + 2

∴c = 80 + 2 = 82

The relationship between c and b is that: -

c = b + 2

For example, if b = 80

c = b + 2

c = 80 + 2

∴c = 82

The Perimeter

The perimeter can be calculated using a, b and c.

The relationship between the perimeter, a, b and c is that: -

perimeter = a + b + c

For example, if a = 34, b = 288 and c = 290

perimeter = a + b + c

perimeter = 34 + 288 + 290

∴perimeter = 612

The Area

The area can be calculated using a and b.

The relationship between the area, a and b is that: -

area = ½ab

For example, if a = 30 and b = 224

area = ½ab

area = ½(30 x 224)

area = ½ x 6720

∴area = 3360

## Generalisations

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

Nth Term for Shortest Side (a)

nth term | Sequence | 1st difference |

0 | ||

1 | 6 | |

4 | ||

2 | 10 | |

4 | ||

3 | 14 | |

4 | ||

4 | 18 | |

4 | ||

5 | 22 | |

4 | ||

6 | 26 |

Conclusion

Below is the proof that the nth terms I have produced satisfy the condition a2 + b2 = c2.

Odd numbers for Length of Shortest Side (a).

nth term for a = 2n + 1

nth term for b = 2n2 + 2n

nth term for c = 2n2 + 2n + 1

a2 + b2 = c2

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

a = (2n + 1) 2 =

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

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

4n2 + 2n + 2n + 1=

4n2 + 4n + 1

b = (2n2 + 2n) 2 =

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

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

4n4 + 4n3 + 4n3 + 4n2 =

4n4 + 8n3 + 4n2

a2 + b2 =

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

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

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

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

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

4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1=

4n4 + 8n3 + 8n2 + 4n + 1 = a2 + b2 = c2

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

nth term for a = 4n + 2

nth term for b = 4n2 + 4n

nth term for c = 4n2 + 4n + 2

a2 + b2 = c2

(4n + 2) 2 + (4n2 + 4n) 2 + (4n2 + 4n + 2) 2

a = (4n + 2) 2 =

(4n + 2)(4n + 2) =

4n(4n + 2) + 2(4n + 2) =

16n2 + 8n + 8n + 4=

16n2 + 16n + 4

b = (4n2 + 4n) 2 =

(4n2 + 4n)(4n2 + 4n) =

4n2 (4n2 + 4n) + 4n(4n2 + 4n) =

16n4 + 16n3 + 16n3 + 16n2 =

16n4 + 32n3 + 16n2

a2 + b2 =

(16n2 + 16n + 4) + (16n4 + 32n3 + 16n2) =

16n4 + 32n3 + 32n2 + 16n + 4

c = (4n2 + 4n + 2) 2 =

(4n2 + 4n + 2)(4n2 + 4n + 2) =

4n2 (4n2 + 4n + 2) + 4n(4n2 + 4n + 2) + 2(4n2 + 4n + 2) =

16n4 + 16n3 + 8n2 + 16n3 + 16n2 + 8n + 8n2 + 8n + 4=

16n4 + 32n3 + 32n2 + 16n + 4 = a2 + b2 = c2

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month