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  • Level: GCSE
  • Subject: Maths
  • Word count: 1741

Beyond Pythagoras

Extracts from this document...

Introduction

William Kerslake        Mathematics GCSE Coursework

Beyond Pythagoras

Aim

I am conducting this investigation to discover formulae that will allow me to calculate many Pythagorean triples. I will first find formulae for odd Pythagorean triples and then even ones. For each, I will find formulae for ‘a’, ‘b’, ‘c’, perimeter and area, all in terms of ‘n’. After this I hope to discover a general formula for all Pythagorean triples, although it is unlikely that I will.

Contents

Page Number

1

Aim and Contents

2             Odd and Even

Tables of Pythagorean triples

3             Odd

Formula for ‘a’ in terms of ‘n’

4             Odd

Formula for ‘b’ and ‘c’ in terms of ‘a’

5             Odd

Formula for ‘b’ and ‘c’ in terms of ‘n’

6             Odd

Perimeter and Area in terms of ‘n’

7             Even

Formula for ‘a’ in terms of ‘n’

8             Even

Formula for ‘b’ and ‘c’ in terms of ‘a’

9             Even

Formula for ‘b’ and ‘c’ in terms of ‘n’

10           Even

Perimeter and Area in terms of ‘n’

11           Odd and Even

Summary – All Formulae


Odd Pythagorean Triples

Table of first 10 odd triples

n

a

b

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

...read more.

Middle

Difference

1

1

3

2

1

2

5

2

1

3

7

2

1

4

9

2

The differences between the numbers show that ‘a’ is changing twice as fast as ‘n’. Therefore, 2n must be part of the formula.

Difference

2n

Difference between n and a

a

Difference

2

2

1

3

2

2

4

1

5

2

2

6

1

7

2

2

8

1

9

2

The differences show that ‘a’ and ‘n’ are changing at the same rate – increasing by 2 each time, thus proving that the formula includes 2n. The difference between ‘a’ and ‘n’ is always 1; therefore, our formula needs to have a +1 included.

From this, I know the formula for ‘a’, in terms of ‘n’, to be;

a  =  2n + 1

I personally cannot see an obvious link between ‘n’ and ‘b’ or ‘c’, so I will use the known formula, a² +b² = c² to discover ‘b’ and ‘c’ in terms of ‘n’.

3² + 4² = 5² and 5² + 12² = 13² both have a hypotenuse one greater than one side (in this case known as ‘b’). Looking at my table of triples I can see that there are more odd triples that follow this pattern. Using this, I should be able to discover ‘b’ and ‘c’ in terms of ‘a’.

3² + 4²        = 5²

a² + b²        = c²

a² + b²        = (b + 1)²

a² + b²        = b² + 1 + 2b

a²         = 2b + 1

2b        = (a² - 1)

b         = (a² - 1)

                2

Remove extra variable ‘c’ from formula

Multiply out brackets

Simplify formula

Rearrange formula

Rearrange formula

From this, I know the formula for ‘b’, in terms of ‘a’, to be;

b = (a² - 1)

     2

From this formula for ‘b’ in terms of ‘a’ I can discover ‘c’ in terms of ‘a’;

a

b

(a² - 1)/2

a²-1

a²+1

(a² + 1)/2

c

3

4

4

8

9

10

5

5

5

12

12

24

25

26

13

13

7

24

24

48

49

50

25

25

9

40

40

80

81

82

41

41

...read more.

Conclusion

((2n + 4) ² - 4)

                        4

b         = 4n² + 8n + 8n +16 - 4

 4

b         = 4n² + 16n + 12

                4

b        = n² + 4n + 3

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

And the same with ‘c’;

c         = (a² + 4)

                4

c         = ((2n + 4) ² + 4)

                        4

c         = 4n² + 8n + 8n +16 + 4

 4

c         = 4n² + 16n + 20

                4

c        = n² + 4n + 5

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

b = 2n² + 2n

c = 2n² + 2n + 1


Now that I have formulae for ‘a’, ‘b’ and ‘c’ in terms of ‘n’; I can put these into my simple area and perimeter formulae and will then be able to calculate both area and perimeter in terms of ‘n’.

Perimeter

Perimeter = a + b + c

P = (2n + 4) + (n² + 4n +3) + (n² + 4n + 5)

P = 2n + 4 + n² + 4n +3 + n² + 4n + 5

P = 10n + 12 + 2n²

Perimeter = 2n² + 10n + 12

Area

Area = a x b

           2

A = (2n+4) x (n² + 4n +3)

                2

A = 2n³ + 8n² + 6n + 12 + 4n² + 16n

                2

A = 2n³ + 12n² + 22n + 12

        2

A = n³ + 6n² + 11n + 6

Area = n³ + 6n² + 11n + 6


Formulae Summary

These are all the formulae I have discovered in my investigation. I can now workout any odd or even Pythagorean triple and its perimeter and area knowing only the number, n. For example, the 26 odd triple is: a = 53, b = 1404, c = 1405, perimeter = 2862. area = 37206

Odd Pythagorean Triples

a                =        2n + 1

b                =        2n² + 2n

c                =        2n² + 2n + 1

Perimeter        =        4n² + 6n + 2

Area                =        2n³ + 3n² + n

Even Pythagorean Triples

a                =        2n + 4

b                =        n² + 4n + 3

c                =        n² + 4n + 5

Perimeter        =        2n² + 10n + 12

Area                =        n³ + 6n² + 11n + 6

/

...read more.

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