# Beyond Pythagoras

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Introduction

Beyond Pythagoras 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 32 = 3 x 3 = 9 42 = 4 x 4 = 16 52 = 5 x 5 = 25 Therefore, 32 + 42 = 9 + 16 = 25 = 52 For the numbers 3, 4 and 5, the perimeter and area are as follows: Perimeter = 3 + 4 + 5 = 12 Area = 1/2 x 3 x 4 = 6 The numbers 5, 12 and 13 also satisfy this condition. 52 + 122 = 132 52 = 5 x 5 = 25 122 = 12 x 12 = 144 132 = 13 x 13 = 169 Therefore, 52 + 122 = 25 + 144 = 169 = 132 5, 12, 13 Perimeter = 5 + 12 + 13 = 30 Area = 1/2 x 5 x 12 = 30 The numbers 7, 24 and 25 also satisfy this condition. ...read more.

Middle

I have worked out formulas for Term Number 'n' Shortest Side 'a' Middle Side 'b' 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 7 15 I have worked out formulas for: 1. How to get 'a' from 'n' 2. How to get 'b' from 'n' 3. How to get 'c' from 'n' 4. How to get the perimeter from 'n' 5. How to get the area from 'n' My formulas are 1. 2n + 1 2. 2n2 + 2n 3. 2n2 + 2n + 1 4. 4n2 + 6n + 2 5. 2n3 + 3n2 + n To get these formulas I did the following 1. 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) ...read more.

Conclusion

Therefore I took my formula for 'a' (2n + 1) and my formula for 'b' (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 To prove my formulas for 'a', 'b' and 'c' are correct. I decided incorporate my formulas into a2 + b2 = c2: - a2 + b2= c2 (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 'a', 'b' and 'c' formulas are correct Luke Heywood 10B Maths Coursework ...read more.

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