# Pythagoras Theorem is a2 + b2 = c2. 'a' being the shortest side, 'b' being the middle side and 'c' being the longest side (hypotenuse) of a right angled triangle.

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Introduction

Pythagoras Theorem is a2 + b2 = c2. ‘a’ being the shortest side, ‘b’ being the middle side and ‘c’ being the longest side (hypotenuse) of a right angled triangle.

The numbers 3, 4 and 5 satisfy this condition

32 + 42 = 52

because 32 = 3 x 3 = 9

42 = 4 x 4 = 16

52 = 5 x 5 = 25

and so 32 + 42 = 9 + 16 = 25 = 52

The numbers 5, 12, 13 and 7, 24, 25 also work for this theorem

52 + 122 = 132

because 52 = 5 x 5 = 25

122 = 12 x 12 = 144

132 = 13 x 13 = 169

and so 52 + 122 = 25 + 144 = 169 = 132

72 + 242 = 252

because 72 = 7 x 7 = 49

242 = 24 x 24 = 576

252 = 25 x 25 = 625

and so 72 + 242 = 49 + 576 = 625 = 252

3 , 4, 5

Perimeter = 3 + 4 + 5 = 12

Area = ½ x 3 x 4 = 6

5, 12, 13

Perimeter = 5 + 12 + 13 = 30

Area = ½ x 5 x 12 = 30

7, 24, 25

Perimeter = 7 + 24 + 25 = 56

Area = ½ x 7 x 24 = 84

From the first three terms I have noticed the following: -

- ‘a’ increases by +2 each term
- ‘a’ is equal to the term number times 2 then add 1
- the last digit of ‘b’ is in a pattern 4, 2, 4
- the last digit of ‘c’ is in a pattern 5, 3, 5
- the square root of (‘b’ + ‘c’) = ‘a’
- ‘c’ is always +1 to ‘b’
- ‘b’ increases by +4 each term
- (‘a’ x ‘n’) + n = ‘b’

Middle

90

180

5

11

60

61

132

330

I have worked out formulas for

- How to get ‘a’ from ‘n’
- How to get ‘b’ from ‘n’
- How to get ‘c’ from ‘n’
- How to get the perimeter from ‘n’
- How to get the area from ‘n’

My formulas are

- 2n + 1
- 2n2 + 2n
- 2n2 + 2n + 1
- 4n2 + 6n + 2
- 2n3 + 3n2 + n

To get these formulas I did the following

- Take side ‘a’ for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph

- From looking at my table of results, I noticed that ‘an + n = b’. So I took my formula for ‘a’ (2n + 1) multiplied it by ‘n’ to get ‘2n2 + n’. I then added my other ‘n’ to get ‘2n2 + 2n’. This is a parabola as you can see from the equation and also the graph

- Side ‘c’ is just the formula for side ‘b’ +1
- The perimeter = a + b + c.

Conclusion

4n2 + 6n + 2

4(n +3)2 – 9 + 2

4(n +3)2 – 7

4(n +3)2 = 7

(n +3)2 = 1.75

n + 3 = 1.322875656

n + 3 = -1.322875656

n = -1.677124344

n = -4.322875656

## Arithmatic Progression

I would like to know whether or not the Pythagorean triple 3,4,5 is the basis of all triples just some of them.

To find this out I have been to the library and looked at some A-level textbooks and learnt ‘Arithmatic Progression’

3, 4, 5 is a Pythagorean triple

The pattern is plus one

If a = 3 and d = difference (which is +1) then

3 = a

4 = a + d

5 = a +2d

a, a +d, a + 2d

Therefore if you incorporate this into Pythagoras theorem

a2 + (a + d)2 = (a + 2d)2

a2 + (a + d)(a + d) = (a + 2d)2

a2 + a2 + ad + ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)(a + 2d)

2a2 + 2ad + d2 = a2 + 2ad + 2ad + 4d2

2a2 + 2ad + d2 = 4d2 + a2 + 4ad

If you equate these equations to 0 you get

a2 – 3d2 – 2ad = 0

Change a to x

x2 – 3d2 – 2dx = 0

Factorise this equation to get

(x + d)(x – 3d)

Therefore

x = -d

x = 3d

x = -d is impossible as you cannot have a negative dimension

a, a+d, a + 2d

Is the same as

3d, 4d, 5d

This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.

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

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