• 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: 1272

Pyhtagorean Theorem

Extracts from this document...

Introduction

Jaymin Patel 10Q                                                               Page  of

image00.png

Pythagoras’ theorem states that a²+b²=c²                image01.png

As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number.

The numbers 3, 4 and 5 work in Pythagoras’ theorem,

3²+4²=5²

because        3²=3×3=9

                4²=4×4=16

                5²=5×5=25

and so                3²+4²=9+16=25=5²

The numbers 5, 12 and 13 also work,

5²+12²=13²

because        5²=5×5=25

                12²=12×12=144

                13²=13×13=169

and so                5²+12²=25+144=169=13²

The numbers 7, 24 and 25 also work,

                7²+24²=25²

because        7²=7×7=49

                24²=24×24=576

                25²=25×25=625

and so                7²+24²=49+576=625=25²

3, 4 and 5

Perimeter= 3+4+5=12

Area= ½ ×3×4=6

5, 12 and 13

Perimeter= 5+12+13=30

Area= ½ ×5×12=30

7, 24 and 25

Perimeter= 7+24+25=56

Area= ½ ×7×24=84

From the first three terms I have realised that: -

  • a increases by 2 each time
  • a is equal to the formula ×2+1 from n
  • b is always even
  • c is always odd
  • c is always +1 of b
  • b=(a×n) + n

I have also added 2 more terms using what I think are my formulae.

 ‘n’

‘a’

‘b’

‘c’

Perimeter

Area

1

3

4

5

12

6

2

5

12

13

30

30

3

7

24

25

56

84

4

9

40

41

90

180

5

11

60

61

132

330

Here are my formulas.

...read more.

Middle

‘n’

‘a’

‘b’

‘c’

area

perimeter

1

3

4

5

6

12

2

5

12

13

30

30

3

7

24

25

84

56

4

9

40

41

180

90

5

11

60

61

330

132

6

13

84

85

546

182

7

15

112

113

840

240

8

17

144

145

1224

306

9

19

180

181

1710

360

10

21

220

221

2310

462

Now I have to prove that my formulae for a, b and c work. To do this I must incorporate 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

4n⁴+8n³+8n²+4n+1 = 4n⁴+8n³+8n²+4n+1

This proves that my formulae for a, b and c are all correct!

Now for part 2, I will try to solve the problem of Pythagorean triples with even numbers for a.

The numbers 6, 8 and 10 work for Pythagoras’ theorem,

                6²+8²=10²

because        6²=6×6=36

                8²=8×8=64

                10²=10×10=100

and so                6²+8²=36+64=100=10²

The numbers 10, 24 and 26 also work,

                10²+24²=26²

because        10²=10×10=100

                24²=24×24=576

                26²=26×26=676

and so                10²+24²=100+576=676=26²

...read more.

Conclusion

So                2n+n²-1+n²+1

=                2n+2n²

=                2n²+2n

Here is how I got to my b and c formulae.

Getting b

a²+b²=c²

=c²= (b+2) ²

=(2n) ²+b²=(b+2) ²

=4n²+b²=b²+4b+4

=4n²-4=4b

=n²-1=b

=b=n²-1

Getting c

By looking at b you can see that c is always +2 than b.

So c is n²-1+2

=n²+1

Here is a table with all of the results for even numbers.

‘n’

‘a’

‘b’

‘c’

Perimeter

Area

1

2

0

2

4

0

2

4

3

5

12

6

3

6

8

10

24

24

4

8

15

17

40

60

5

10

24

26

60

120

6

12

35

37

84

210

7

14

48

50

112

336

8

16

63

65

144

504

9

18

80

82

180

720

10

20

99

101

220

990

Now I have to prove my formulae for a, b and c. To do this I have to incorporate a²+b²=c²

a²+b²=c²

=        (2n)²+(n²-1) ²=(n²+1) ²

=        (2n)(2n)=4n²

(n²-1)(n²-1)=n⁴-n²-n²+1

        (n²+1)(n²+1)=n⁴+n²+n²+1

=        4n²+n⁴-n²-n²+1=n⁴+n²+n²+1

=        n⁴+2n³+1=n⁴+2n³+1

This proves that my formulas for a, b and c are all correct.

...read more.

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras 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 Beyond Pythagoras essays

  1. Beyond Pythagoras

    1: 4 x 3 = 6 2 Triangle 2: 12 x 5 = 30 2 Triangle 3: 24 x 7 = 84 2 Triangle 4: 40 x 9 = 180 2 Triangle 5: 60 x 11 = 330 2 Triangle 6: 84 x 13 = 546 2 nth term Next,

  2. Beyond Pythagoras

    If the formulas are correct they will suit the Pythagoras equation. a (smallest side) = 2n + 1 b (middle side) = 2n + 2n2 c (largest side) = 2n + 2n2 + 1 a2 = (2n+1) x (2n+1) = 4n2 + 1 + 2n + 2n = 1 + 4n + 4n2 b2 = (2n + 2n2)

  1. Pythagorean Theorem Coursework

    n to b- 2n�+2n 3. n to c- 2n�+2n+1 4. n to perimeter- 4n�+6n+2 5. n to area- 2n�+3n�+n Here is how I got to the area and perimeter formulae. We all know that the area of a triangle has a formula of area= 1/2 �a�b.

  2. Research on Pythagoras and his work.

    However to Pythagoras numbers had personalities which we hardly recognise as mathematics today [3]:- Each number had its own personality - masculine or feminine, perfect or incomplete, beautiful or ugly. This feeling modern mathematics has deliberately eliminated, but we still find overtones of it in fiction and poetry.

  1. Dice Maths Investigation

    x (5/18)n-1 Test: 1st round - 1/6 (1/6) x (5/18)0 = 1/6 - Correct 2nd round - 5/108 (1/6) x (5/18)1 = 5/108 - Correct 3rd round - 25/1944 (1/6) x (5/18)2 = 25/1944 - Correct 4th round - 125/34992 (1/6) x (5/18)3 = 125/34992 - Correct The formula appears to work. We now have the following formulas.

  2. Beyond Pythagoras - I am investigating the relationships between the lengths of the three ...

    =12 2 x 3(3+1) =24 2 x 4(4+1) =40 2 x 5(5+1) =60 2 x 6(6+1) =84 2 x 7(7+1) =112 2 x 8(8+1) =144 2 x 9(9+1) =180 2 x 10(10+1) =220 etc. nth term for 'Length of longest side' 5, 13, 25, 41, 61, 85, 113, 145, 181, 221...

  1. Beyond Pythagoras

    * (2n2+2n) 4n4+4n3 4n3+ 4n2 4n2 +8n3+4n4 Longest (2n2+2n+1) * (2n2+2n+1) 4n4+4n3+2n2 4n3+ 4n2+2n 2n2+2n+1 4n4+8n3+8n2+4n+1 (4n4+8n3+8n2+4n+1)= (4n2 +8n3+4n4)+ (4n+ 4n2+1) Mahmoud Elsherif Beyond Pythagoras P.6 Next I will investigate the Middle Term Middle Term2= Longest Term2- ShortestTerm2 (nth term)2= (nth term)2 + (nth term)2 (2n2+2n)2 = (2n2+2n+1) 2 - (2n+1)2 Shortest (2n+1)* (2n+1)

  2. Beyond Pythagoras ...

    I will start with shortest term first. Longest Term2= Middle Term2+ ShortestTerm2 (nth term)2= (nth term)2 + (nth term)2 (2n2+2n+1)2 (2n2+2n) 2 + (2n+1)2 Shortest 2n+1+ 2n+1 4n2+ 2n 2n + 1 4n+ 4n2+1 Middle 2n2+2n + 2n2+2n 4n4+4n3 4n3+ 4n2 4n2 +8n3+4n4 Longest 2n2+2n+1 + 2n2+2n+1 4n4+4n3+2n2 4n3+ 4n2+2n 2n2+2n+1 4n4+8n3+8n2+4n+1 (4n4+8n3+8n2+4n+1)= (4n2 +8n3+4n4)+ (4n+ 4n2+1)

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