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  • 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.

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