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  • Level: GCSE
  • Subject: Maths
  • Word count: 2399

Beyond Pythagoras

Extracts from this document...

Introduction

Mahmoud ElsherifBeyond PythagorasP.1

Pythagoras Theorem is a2+b2= c2 ‘a’ is being the shortest side, ‘b’ being the middle side and ‘c’ being the longest side (hypotenuse) of a right angled triangle.

The numbers 3,4,5 satisfy this condition and so

32+ 42=52

Because 32= 3*3=9

42=4*4=16

52=5*5=25

32+ 42=52

9+16=25

25=25

image00.png

This proves Pythagoras Theorem goes with the right angled triangle with the numbers 3,4,5. Next I shall prove that Pythagoras’s Theorem applies to 5,12,13 right angled triangle.

52+122=132

Because 52= 5*5=25

122= 12*12=144

132= 13*13=169

image01.png

Mahmoud ElsherifBeyond PythagorasP.2

This satisfies the Theorem of Pythagoras’s goes with these numbers 5,12,13. Finally I shall prove that Pythagoras’s Theorem applies to 7,24,25 right angled triangle.

72+ 242=252

Because 72= 7*7=49

242= 24*24= 576

252=25*25=625.

So

a2+b2=c2

72+242=252

49+576=625

image02.png

This proves Pythagoras Theorem goes with the right angle triangle with the sides 7,24,25

Shortest Side

Middle Side

Longest Side

3

4

5

5

12

13

7

24

25

9

40

41

11

60

61

13

84

85

Mahmoud ElsherifBeyond PythagorasP.3

I shall find the prediction of the shortest side first.

3,5,7

It goes up in 2 so in my conclusion so it will become

3,5,7,9,11,13

Now I will find the difference between them.

The difference is 2

Next I shall find the prediction of the middle side next.

4,12,24

It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20,24

So it will be 4,12,24, 40, 60 84.

The difference is 4,8,12. Now I shall find the difference and it is n*4

...read more.

Middle

+4n+1)- (4n+ 4n2+1)


Finally I will investigate the shortest Term.

Mahmoud ElsherifBeyond PythagorasP.7

Shortest Term2= Longest Term2- Middle Term2

(nth term)2=  (nth term)2     +  (nth term)2

(2n2+1)2 =(2n2+2n+1) 2 - (2n+2n2)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

(4n+ 4n2+1)= (4n4+8n3+8n2+4n+1)-(4n2 +8n3+4n4)

Now I have finished that I will start having even numbers, to see if Pythagoras’s Theorem works. I shall do this all over again but with an even short side.

Shortest Side

Middle Side

Longest Side

6

8

10

10

24

26

14

48

50

18

80

82

22

120

122

26

168

170

Mahmoud Elsherif        Beyond Pythagoras        P.8

I shall find the prediction of the shortest side first.

6,10,14

It goes up in 4 so in my conclusion so it will become

6,10,14,18,22,26.

Now I will find the difference between them.

The difference is 4

Next I shall find the prediction of the middle side next.

8,24,48

It goes up by 8,16,24. So in my conclusion I think it will become 8,16,24,32,40,48

So it will be 8,24,48, 80, 120, 168.

The difference is 8,16,24. Now I shall find the difference and it is n*8

...read more.

Conclusion

9

12

15

15

36

39

21

72

75

27

120

123

33

180

183

39

252

255

Finally to get the fourth part you have to times the first part by four.

Shortest Side

Middle Side

Longest Side

12

16

20

20

48

52

28

96

100

36

160

164

44

240

244

52

336

340

So my theory is correct with the times.
For the shortest formula it is x (2n+1)

For the middle formula it is x (2n2+2n)

For the longest formula it is x (2n2+2n+1)

1(2n+1) 2* 1 =2 +1=3

2 (2n+1) 4n +2 = 4*1= 4+2=6

Mahmoud ElsherifBeyond PythagorasP.14

3 (2n+1) 6n+3= 6*1 + 3= 9

4 (2n+1) 8n+4= 8*1+4= 12

M(2n +1)

1 (2n2+2n) 2*12= 2*1=2, 2+2=4

2(2n2+2n) 4n2+4n = 4*12= 4, 4*1=4. 4+4=8

3(2n2+2n) 6n2+6n= 6*12= 6, 6*1=6. 6+6= 12

4(2n2+2n) 8n2+8n= 8*12= 8, 8*1=8. 8+8= 16

M(2n2+2n)

1 (2n2+2n+1) 2*12= 2*1=2, 2+2+1=5

2 (2n2+2n+1) 4n2+4n = 4*12= 4, 4*1=4. 4+4+2=10

3 (2n2+2n+1) 6n2+6n = 6*12= 6, 6*1=6. 6+6+3=15

4 (2n2+2n+1) 8n2+8n = 8*12= 8, 8*1=8. 8+8+4=20

M (2n2+2n+1)

N=1

Now to investigate the Pythagoreans Triples

1st Family= c, b, b+1 generate this family using (b+1)2 = b2+c2

   (4+1)2= 42+ 32

                                                    52=42+ 32

                                                  25=16+ 9

2nd Family=c, b, b+2 generate this family using (b+2)2 = b2+c2

   (8+2)2= 82+ 62

                                                    102=82+ 62

                                                   100=64+ 36

3rd Family= c, b, b+3 generate this family using (b+3)2 = b2+c2

   (12+3)2= 122+ 92

                                                    152=122+ 92

                                                    225= 144+81

Mahmoud ElsherifBeyond PythagorasP.15

4th Family= c,b,b+4 generate this family using(b+4)2 = b2+c2

   (16+4)2= 162+ 122

                                                    202=162+ 122

                                                    400= 256+144

I have noticed you have to make b the subject and the formula is this.

Xth family= c, b, b,+ x generates this rule using (b+ x)2= b2+c2

(B+X)2= b2+c2

  X2= b2+c2

                                                   X= b+c

                                                 
This is the formula I have proven this by my methods by using the families of the Pythagorean triples.

...read more.

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