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Beyond Pythagoras

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Introduction

Daniel Cook 10R

Beyond Pythagoras: Year 10 GCSE Coursework

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

         For example, I will use 32 x 42 = 52 . This is because:

32 = 3 x 3 = 9

42 = 4 x 4 = 16

52 = 5 x 5 = 25

        So.. 9 +16 = 25

        For this table, I am using the term a, b, b + 1

Triangle Number (n)

Length of shortest side

Length of middle side

Length of longest side

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

6

13

84

85

183

546

7

15

112

113

240

840

8

17

144

145

296

1224

Formulas

Shortest side = 2n + 1, n being the triangle number

Middle side = 2n2 + 2n. This is because:

Triangle Number

        1                =        2 x 2

        2                =        3 x 4

        3                =        4 x 6

...read more.

Middle

+2n

4n³

2n²

2n

+1

2n²

2n

1

= 4n4  + 8n³ + 8n² +4n + 1

A                +                B                    =                   C

4n² + 4n + 1       +       4n4  + 8n³ + 4n²        =      4n4+ 8n³ + 8n² +4n + 1

        My results show that shortest² + middle² = longest². I have Pythagoras’ theorem.

        For this table, I am using the term a, b, b + 2

Triangle Number (n)

Length of shortest side

Length of middle side

Length of longest side

Perimeter

Area

1

6

8

10

24

48

2

8

15

17

40

120

3

10

24

26

60

240

4

12

35

37

84

420

5

14

48

50

112

672

6

16

63

65

144

1008

7

18

80

82

180

1440

8

20

99

101

220

1980

Formulas

Shortest side = 2n + 4. This is because:

2 x 1 + 4 = 6

2 x 2 + 4 = 8

2 x 3 + 4 = 10  etc

Middle side = n² + 4n + 3

Longest side = n² + 4n + 5

Box Methods

Shortest² = (2n + 4)²

2n

+4

2n

4n²

8n

+4

8n

16

= 4n² + 16n + 16

Middle² =  (n² + 4n + 3)²

+4n

+3

n4

4n³

3n²

+4n

4n³

16n²

...read more.

Conclusion

Length of longest side

Perimeter

Area

1

12

16

20

48

96

2

16

30

34

80

240

3

20

48

52

120

480

4

24

70

74

168

840

5

28

96

100

224

1344

6

32

126

130

288

2016

7

36

160

164

360

2880

8

40

198

202

440

3960

Formulas

Shortest side = 4n + 8

Middle side = 2n² + 8n + 6

Longest side = 2n² + 8n + 10

Box Methods

Shortest² = (4n + 8)²

4n

+8

4n

16n²

32n

+8

32n

64

= 16n² + 64n + 64

Middle² = (2n² + 8n + 6)²

2n²

+8n

+6

2n²

4n4

16n³

12n²

+8n

16n³

64n²

48n

+6

12n²

48n

36

= 4n4 + 32n³ + 88n² + 96n + 36

Longest² = (2n² + 8n + 10)²

2n²

+8n

+10

2n²

4n4

16n³

20n²

+8n

16n³

64n²

80n

+10

20n²

80n

100

= 4n4 + 32n³ + 104n² + 160n + 100

A               +                       B =                         C

16n² + 64n + 64    +    4n4 + 32n³ + 88n² + 96n + 36   =   4n4 + 32n³ + 104n² + 160n + 100

        All of my research has proved that Pythagoras’ theorem is correct, even if you come at the theorem at different angles; you’re sure to find out that Pythagoras is correct.

...read more.

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