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Introduction

## 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

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

## Box Methods

Shortest² = (2n + 4)²

 2n +4 2n 4n² 8n +4 8n 16

= 4n² + 16n + 16

Middle² =  (n² + 4n + 3)²

 n² +4n +3 n² n4 4n³ 3n² +4n 4n³ 16n²

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.

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

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