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• Level: GCSE
• Subject: Maths
• Word count: 2286

# Beyond pythagoras - This investigation is about Pythagorean triples and the similarities between them.

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Introduction

KELVIN STAPLETON 10G4                31208.doc

BEYOND PYTHAGORAS

This investigation is about Pythagorean triples and the similarities between them. A Pythagorean triple is three lengths on a triangle which satisfy the equation a² + b² = c². An example of this is the lengths 3, 4, 5 which is because:

3² + 4² = 5²

3² = 3 x 3 = 9

4² = 4 x 4 = 16

5² = 5 x 5 = 25

9 + 16 = 25

So the numbers 3, 4, 5 are Pythagorean triples. The numbers chosen need to be integers and need to get longer with each number.

The perimeter of the triangle with sides 3, 4, 5 is 3 + 4 + 5 = 12.

The area of this triangle is ½ x 3 x 4 = 6

The numbers 5, 12, 13 can be the lengths of a Pythagorean triple as well:

5² + 12² = 13²

5² = 5 x 5 = 25

12² = 12 x 12 = 144

13² = 13 x 13 = 169

144 + 25 = 169

The perimeter of the triangle with sides 5, 12, 13 is 5 + 12 + 13 = 30

The area of this triangle is ½ x 5 x 12 = 30

The numbers 7, 24, 25 are another set of Pythagorean triples as well:

7² + 24² = 25²

7² = 7 x 7 = 49

24² = 24 x 24 = 576

25² = 25 x 25 = 625

576 + 49 =625

The perimeter of the triangle with sides 7, 24, 25 is 7 + 24 + 25 = 56

The area of this triangle is ½ x 7 x 24 = 84

 Length of the shortest side Length of the middle side Length of the longest side Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180 11 60 61 132 330

The triangle with sides 9, 40, 41 has a perimeter of 9 + 40 + 41 = 90

This triangle has an area of ½ x 9 x 40 = 180

The triangle with sides 11, 60, 61 has a perimeter of 11 + 60 + 61 = 132

This triangle has an area of ½ x 11 x 60 = 330

This is how I found the formula for the shortest side:

I think the formula for the shortest side is 2n +1

I will now test this formula:

For the 1st term it is 2 x 1 + 1= 3

For the 2nd term it is 2 x 2 + 1 = 5

Middle

Length of shortest side

Length of middle side

Length of longest side

Perimeter

Area

3

4

5

12

6

5

12

13

30

30

7

24

25

56

84

9

40

41

90

180

11

60

61

132

330

13

84

85

182

546

15

112

113

240

840

I am now going to prove my formulas are correct .

I will do this by using pythagoras’s theoreom which is: a² + b² = c²

Which in other words means the shortest side² + the middle side² = the longest side²

So if I put the first set of numbers into the formula they come out as:

3² + 4² = 5²

9 + 16 = 25

The first sequence works so now I will try the the third set of numbers into the formula to see if they work and they come out as:

7² + 24² = 25²

49 + 576 = 625

So this one works as well I think I will now try it on the last sequence I found to make sure it still works at the end of the sequences:

15² + 112² = 113²

225 + 12544 = 12769

So I think my formulas are correct as this one works as well.

I am now going to predict the 100th sequence to see if my formulas work when you get further on in the triangle sequences.

The shortest side is: 2 x 100 + 1 = 201

The middle side is: 2 x 100² + 2 x 100 = 20,200

The longest side is: 2 x 100² + 2 x 100 + 1 = 20,201

I am now going to use pythagoras’s theorem again to prove my prediction of the 100th sequence is correct:

a² + b² = c²

201² + 20,200² = 20,201²

40,401 + 408,040,000 = 408,080,401

So my formulas are correct and work later on as it worked for the 100th term.

Conclusion

The first odd triangle with sides 3, 4, 5 with a perimeter of 12 and an area of 6 is half of the second even triangle with sides 6, 8, 10 with a perimeter of 24 which is the first triangles numbers doubled. Also I had to prove that my formulas worked by using Pythagoras’s theorem (a² + b² = c²) and I did this with the numbers I got from the calculations I did with my formulas and so proved my formulas were correct and worked. Also I had to prove my formulas worked when I put them into a algebraic format using the formulas I used to get my answers. This I did and it was successful and worked quite well but I did make a few mistakes which I had to correct so was quite diffucult. To work out what my formulas were when they were squared I had to multiply them out of their brackets and collect the like terms together. It took me quite a while to figure out what the formulas were but when I did it was easy. After that it was quite simple for me to put them together and work them out.

Now I am going to put all of my results into a single table so it is easier to see what I mean by every other even term being double an odd one:

This table shows the way the even triangles have a relationship with the odd triangles.

Wednesday, 02 May 2007,  15:57:22

-  -

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