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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

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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

  1. How to get s from n
  2. How to get m from n
  3. How to get l from n
  4. How to get the p from n
  5. 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

...read more.

Conclusion

  1. 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

...read more.

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