Beyond Pythagoras

72 + 242 = 252

72 = 7 x 7 = 49

242 = 24 x 24 = 576

252 = 25 x 25 = 625

Therefore, 72 + 242 = 49 + 576 = 625 = 252

7, 24, 25

Perimeter = 7 + 24 + 25 = 56

Area = 1/2 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'

From these observations I have worked out the next two terms.

I will now put the first seven terms in a table format.

I have worked out formulas for

Term Number 'n'

Shortest Side 'a'

Middle Side 'b'

Longest Side 'c'

Perimeter

Area

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

7

5

I have worked out formulas for:

. 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

. 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

. 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

2. 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

3. Side 'c' is just the formula for side 'b' +1

4. The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n2 + 2n) and my formula for 'c' (2n2 + 2n + 1). I then did the following: -

2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter

Rearranges to equal

4n2 + 6n + 2 = perimeter

5. The area = (a x b) divided by 2. 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