Beyond Pythagoras.

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Saneeta Mandil 10R

Maths Coursework

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.

Pythagorean triples are a family of right-angled triangles for which all the positive integers and the shortest side is an odd number.

The first part of my investigation is to study the family of Pythagorean triples where the shortest side is an odd number and all three sides are positive integers.

The numbers 3, 4 and 5 satisfy the condition

32 + 42 = 52

(Smallest number) 2 + (middle number) 2 = (largest number) 2

The numbers 3, 4 and 5 can be the lengths - in appropriate sides units - of the sides of a right-angled triangle.

5

3

4

32 = 3 x 3 = 9

42 = 4 x 4 =16

52 = 5 x 5 =25

and so 32+ 42 = 9 + 16 =25

The numbers 5, 12 and 13 can also be the length in appropriate units of a right-angled triangle.

3

5

12

52 = 5 x 5 = 25

22 = 12 x12 =144

32 = 13 x 13 =169

And so 52 + 42 = 25 + 144 =169

This is also true for the numbers 7, 24 and 25.

25

7

24

7 = 7 x 7 = 49

24 =24 x 24 = 576

25 = 25 x 25 = 625

and so 7 + 24 = 49 + 576 =25

(3,4,5), (5,12,13) and (7,24,25) are called Pythagorean triples because they satisfy the condition.

The perimeter and area of the triangle with lengths 3, 4,5 are

Perimeter = 3 + 4 + 5 = 12 units

Area = 1/2 x 3 x 4 = 6 square units

I have put these Pythagorean triples (3,4,5), (5,12,13) and (7,24,25) into a table in which I have noticed many patterns occurring.

n

a

b

c

Perimeter

Area

3

4

5

2

6

2

5

2

3

30

30

3

7

24

25

56

84

Patterns

* 'a' increases by + 2 each term.

* 'a' is equal to term number times 2 then add 1.

* The last digit of 'b' is in the pattern of 4,2,4.

* The last digit of 'c' is in the pattern of 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 the first three Pythagorean triples, I have found formulae for the sides of a, b, c, and the perimeter and area.
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Side 'a'

To get a from n = 2n + 1

I obtained this rule because I realised from the pattern there was a difference of 2. Consequently using the nth term, I knew that 2n would be involved. I did 2 x 1 = 2, (1 being the first term of n). The first number of 'a' is 3, which meant there was only a difference of 1. Therefore the formula was 2n + 1. This rule worked for the next two numbers in side 'a' (5,7).

Side 'b'

To get 'b' from ...

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