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  • Level: GCSE
  • Subject: Maths
  • Word count: 2681

Pythagoras Theorem.

Extracts from this document...

Introduction

Chetan Mandlia        Mathematics        10 White

Mathematics GCSE Coursework: Beyond Pythagoras

Pythagoras

image15.jpg

Pythagoras was a Greek mathematician and philosopher who was born around 582BC and died around 500BC. The theorem that carries his name, Pythagoras’ Theorem, is perhaps the best-known theorem in the whole of mathematics. The theorem is about finding the lengths of sides in right-angled triangles.

Pythagoras Theorem

The theorem states that for any right-angled triangle, the area of the square formed on the hypotenuse is equal to the sum of the squares formed on the other two sides.

image00.png

image07.pngimage01.png

image09.pngimage08.png

Or a2 + b2 = c2

image10.png

Or 92 + 602 = 612image11.png

Pythagorean Triples

(3, 4, 5); (5, 12, 13); (7, 24, 25) and etc are all called Pythagorean triples because the satisfy the conditionimage12.png

a2 + b2 = c2

The numbers 3, 4 and 5 satisfy the conditionimage13.pngimage14.png

The numbers 3, 4 and 5 satisfy the conditionimage03.pngimage02.png

The numbers 3, 4 and 5 satisfy the conditionimage04.pngimage05.png

Because  

image16.png

The perimeter and area of this triangle are

Perimeter = 3 + 4 + 5 = 12 units

Area = ½ x 3 x 4 = 6units2

The perimeter and area of this triangle areimage03.png

Perimeter = 5 + 12 + 13 = 30 units

Area = ½ x 5 x 12 = 30units2

The perimeter and area of this triangle areimage06.png

Perimeter = 7 + 24 + 25 = 56 units

Area = ½ x 7 x 24 = 84units2

Results Table

Odd numbers for Length of Shortest Side (a).

n

Length of Shortest Side (a)

Length of Middle Side (b)

Length of Longest Side (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

6

13

84

85

182

546

7

15

112

113

240

840

8

17

144

145

306

1224

9

19

180

181

380

1710

10

21

220

221

462

2310

11

23

264

265

552

3036

Relationships

2n + 1

an + n

an + n + 1

(b + 1)

a + b + c

½ x a x b

nth Term

2n + 1

2n2 + 2n

2n2 + 2n

+ 1

4n2 + 6n

+ 2

2n3 + 3n2

+ n

nth Term factorised

N/A

2(n2 + n)

N/A

2(2n2 + 3n

+ 1)

n(2n2 + 3n

+1)

...read more.

Middle

8

16

14

26

6

3

56

8

36

20

34

6

4

90

8

64

26

42

6

5

132

8

100

32

50

6

6

182

144

38

New sequence = Original sequence – 4n2

Quadratic equation = an2 + bn + c

a = 2nd difference/2 = 8/2 = 4

b = 1st difference = 6

c = 1st no. in New sequence – 1st difference = 8 – 6 = 2

nth term for perimeter =

4n2 + 6n + 2

4n2 + 6n + 2 =

2(2n2 + 3n + 1)

Nth Term for Area

area = ab/2

a = 2n + 1

b = 2n2 + 2n

area = (2n + 1)(2n2 + 2n)/2

(2n + 1)(2n2 + 2n) =

2n(2n2 + 2n)+1(2n2 + 2n) =

4n3 + 4n2 + 2n2 + 2n =

4n3 + 6n2 + 2n

(4n3 + 6n2 + 2n)/2 =

2n3 + 3n2 + n

2n3 +3n2 + n =

n(2n2 +3n + 1)

OR

nth term

Orig. seq.

1st diff.

2nd diff.

3rd diff

2n3

New seq.

1st diff.

2nd diff.

3n2

New seq.

1st diff.

0

1

6

2

4

3

1

24

10

1

2

30

30

16

14

6

12

2

54

12

16

1

3

84

42

54

30

6

27

3

96

12

22

1

4

180

54

128

52

6

48

4

150

12

28

1

5

330

66

250

80

6

75

5

216

34

1

6

546

432

114

108

6

New sequence = Original sequence – 2n3

New sequence = New sequence – 3n2

Cubic equation = an3 + bn2 + cn + d

a = 3rd difference/6 = 12/6 = 2

b = 2nd difference/2 = 6/2 = 3

c = 1st difference = 1

d = 1st no. in New sequence – 1st difference = 1 – 1 = 0

nth term for area =

2n3 + 3n2 + 1n + 0 =

2n3 +3n2 + n

2n3 +3n2 + n =

n(2n2 +3n + 1)

Results Table

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

n

Length of Shortest Side (a)

Length of Middle Side (b)

Length of Longest Side (c)

Perimeter

Area

1

6

8

10

12

24

2

10

24

26

60

120

3

14

48

50

112

336

4

18

80

82

180

720

5

22

120

122

264

1320

6

26

168

170

364

2184

7

30

224

226

480

3360

8

34

288

290

612

4896

9

38

360

362

760

6840

10

42

440

442

924

9240

11

46

528

530

1104

12144

Relationships

4n + 2

an + 2n

an + 2n + 2

(b + 2)

a + b + c

½ x a x b

nth Term

4n + 2

4n2 + 4n

4n2 + 4n

+ 2

8n2 + 12n

+ 4

8n3 + 12n2

+ 4n

nth Term factorised

2(2n + 1)

4(n + n)

2(2n + 2n

+ 1)

4(2n + 3n

+ 1)

4n(2n + 3n + 1)

Fig.2

Relationships

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

The Length of The Shortest Side (a)

The length of the shortest side (a) can be calculated using n.

The relationship between a and n is that: -

a = 4n + 2

For example, if n = 6

a = 4n +2

a = (4 x 6) + 2

a = 24 + 2

a = 26

The Length of The Middle Side (b)

The length of the middle side (b) can be calculated using n and a.

The relationship between b, a and n is that: -

b = an + 2n

For example, if a = 18 and n = 4

b = an + 2n

b = (18 x 4) + (2 x 4)

b = 72 + (2 x 4)

b = 72 + 8

b = 80

The Length of The Longest Side (c)

The length of the longest side (c) can be calculated using a, n and b.

The relationship between c, n and a is that: -

c = an + 2n + 2

For example, if a = 18 and n = 4

c = an + 2n + 2

c = [(18 x 4) + (2 x 4] +2 = [72 + (2 x 4] + 2 = [72 + 8] + 2

c = 80 + 2 = 82

The relationship between c and b is that: -

c = b + 2

For example, if b = 80

c = b + 2

c = 80 + 2

c = 82

The Perimeter

The perimeter can be calculated using a, b and c.

The relationship between the perimeter, a, b and c is that: -

perimeter = a + b + c

For example, if a = 34, b = 288 and c = 290

perimeter = a + b + c

perimeter = 34 + 288 + 290

perimeter = 612

The Area

The area can be calculated using a and b.

The relationship between the area, a and b is that: -

area = ½ab

For example, if a = 30 and b = 224

area = ½ab

area = ½(30 x 224)

area = ½ x 6720

area = 3360

Generalisations

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

Nth Term for Shortest Side (a)

nth term

Sequence

1st difference

0

1

6

4

2

10

4

3

14

4

4

18

4

5

22

4

6

26

...read more.

Conclusion

Below is the proof that the nth terms I have produced satisfy the condition a2 + b2 = c2.

Odd numbers for Length of Shortest Side (a).

nth term for a = 2n + 1

nth term for b = 2n2 + 2n

nth term for c = 2n2 + 2n + 1

a2 + b2 = c2

(2n + 1) 2 + (2n2 + 2n) 2 + (2n2 + 2n + 1) 2

a = (2n + 1) 2 =

(2n + 1)(2n + 1) =

2n(2n + 1) + 1(2n + 1) =

4n2 + 2n + 2n + 1=

4n2 + 4n + 1

b = (2n2 + 2n) 2 =

(2n2 + 2n)(2n2 + 2n) =

2n2 (2n2 + 2n) + 2n(2n2 + 2n) =

4n4 + 4n3 + 4n3 + 4n2 =

4n4 + 8n3 + 4n2

a2 + b2 =

(4n2 + 4n + 1) + (4n4 + 8n3 + 4n2) =

4n4 + 8n3 + 8n2 + 4n + 1

c = (2n2 + 2n + 1) 2 =

(2n2 + 2n + 1)(2n2 + 2n + 1) =

2n2 (2n2 + 2n + 1) + 2n(2n2 + 2n + 1) + 1(2n2 + 2n + 1) =

4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1=

4n4 + 8n3 + 8n2 + 4n + 1 = a2 + b2 = c2

Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).

nth term for a = 4n + 2

nth term for b = 4n2 + 4n

nth term for c = 4n2 + 4n + 2

a2 + b2 = c2

(4n + 2) 2 + (4n2 + 4n) 2 + (4n2 + 4n + 2) 2

a = (4n + 2) 2 =

(4n + 2)(4n + 2) =

4n(4n + 2) + 2(4n + 2) =

16n2 + 8n + 8n + 4=

16n2 + 16n + 4

b = (4n2 + 4n) 2 =

(4n2 + 4n)(4n2 + 4n) =

4n2 (4n2 + 4n) + 4n(4n2 + 4n) =

16n4 + 16n3 + 16n3 + 16n2 =

16n4 + 32n3 + 16n2

a2 + b2 =

(16n2 + 16n + 4) + (16n4 + 32n3 + 16n2) =

16n4 + 32n3 + 32n2 + 16n + 4

c = (4n2 + 4n + 2) 2 =

(4n2 + 4n + 2)(4n2 + 4n + 2) =

4n2 (4n2 + 4n + 2) + 4n(4n2 + 4n + 2) + 2(4n2 + 4n + 2) =

16n4 + 16n3 + 8n2 + 16n3 + 16n2 + 8n + 8n2 + 8n + 4=

16n4 + 32n3 + 32n2 + 16n + 4 = a2 + b2 = c2

...read more.

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