Beyond Pythagoras

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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 = 1/2 x 3 x 4 = 6

Odd Triples

I will now put the first term in to a table and try to find the next terms up to 10.

Term Number (n)

Shortest Side (s)

Middle Side (m)

Longest Side (l)

Perimeter (p)

Area (a)

3

4

5

2

6

2

5

2

3

30

30

3

7

24

25

56

84

4

9

40

41

90

80

5

1

60

61

32

330

6

3

84

85

82

545

7

5

12

13

240

840

8

7

44

45

306

224

9

9

80

81

380

710

0

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

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
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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 + n or s x m divided by 2

To get these formulas I did the following

. Take side 's' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula ...

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