• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  • Level: GCSE
  • Subject: Maths
  • Word count: 2286

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

Extracts from this document...

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                image00.png

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 = 13²image01.png

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 = 25²image08.png

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 = 330image09.png

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

...read more.

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.

...read more.

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.image07.png

Wednesday, 02 May 2007,  15:57:22

        -  -

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Beyond Pythagoras

    Shortest side (a) The length of this side increases by 2 for each triangle. It is also 1 more than double the triangle number (n). This means that the formula is (2�n) + 1, which is more commonly written as 2n + 1 Middle side (b)

  2. Beyond Pythagoras.

    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 I will now test it using the first three terms.

  1. Beyond Pythagoras

    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.

  2. Beyond Pythagoras.

    4 + 1 = 5 Term 3 'a' = 2n + 1 = 2 � 3 + 1 = 6 + 1 = 7 Side 'b' I used quadratic rule to find the formula for this side: 1 diff = 8 The first difference would be: difference of 1st sequence

  1. Beyond Pythagoras.

    Now I would need to find a formula to help me work out each side. To work out the shortest length I would need to see what I need to do to 'n' to make it equal the first length.

  2. Beyond Pythagoras

    Now I will try this out on some odd numbers, plus find there area and perimeter: 9� = 81 = 40.5 2 Lower bound = 40, Upper bound = 41. Middle side = 40, Largest side = 41. Perimeter= 9+40+41= 90 Area= 9 x 40 � 2 = 180 11�

  1. Beyond Pythagoras

    So I used the same formula for the same reasons as in b but added 1. Example of c=2n2 +2n+1: The answer to c when n=3 is 25. The answer to the formula is: 2x3^2+2x3+1=c 2x3^2=18 2x3=6 18+6+1=25 so the formula is correct.

  2. Beyond Pythagoras.

    + n = 'b' From these observations I have worked out the next two terms. I will now put the first five terms in a table format. Term Number 'n' Shortest Side 'a' Middle Side 'b' Longest Side 'c' Perimeter Area 1 3 4 5 12 6 2 5 12

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work