# Beyond Pythagoras.

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Introduction

Beyond Pythagoras

For this piece of coursework I am trying to find Pythagorean triplets (these are whole numbers that’s satisfies Pythagoras theorem).

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

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

To find the perimeter we add all the sides together.

Perimeter = 3 + 4 + 5 = 12

Finally to find the areas we times the smallest and middle side and then divide by to.

Area = ½ x 3 x 4 = 6

I

Middle

5

11

60

61

132

330

6

13

84

85

182

545

7

15

112

113

240

840

8

17

144

145

306

1224

9

19

180

181

380

1710

10

21

220

221

462

2310

- s increases by +2 each term
- s is equal to the term number times 2 then add 1
- the square root of (m + l) = a
- (s x n) + n = m

I have worked out formulas for

- How to get s from n
- How to get m from n
- How to get l from n
- How to get the p from n
- How to get the a from n

To find the Smallest side (s) you use the formula 2n + 1

To find the Middle side (m) you use the formula 2n2 + 2n or (s x n) + n

To find the Longest side (l) you use the formula 2n2 + 2n + 1 or (n x s) + n + 1

To find the Perimeter (p) you use the formula 4n2 + 6n +2 or s + m + l

To find the Area (a) you use the formula 2n3+ 3n2

Conclusion

- The area = (s x m) divided by 2. Therefore I took my formula for ‘s’ (2n + 1) and my formula for ‘m’ (2n2 + 2n). I then did the following: -

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

Multiply this out to get

4n3 + 6n2 + 2n = area

2

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

2n3 + 3n2+ n = area (Also you can use the formula s x m divided by 2)

To prove my formulas for ‘s’, ‘m’ and ‘l’ are correct. I decided incorporate my formulas into s2 + m2 = l2

s2+ m2= l2

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

4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1

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

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

This proves that my ‘s’, ‘m’ and ‘l’ formulas are correct

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

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