• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
20. 20
20
21. 21
21
22. 22
22
23. 23
23
24. 24
24
25. 25
25
• 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

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.

Or a2 + b2 = c2

Or 92 + 602 = 612

## Pythagorean Triples

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

a2 + b2 = c2

The numbers 3, 4 and 5 satisfy the condition

The numbers 3, 4 and 5 satisfy the condition

The numbers 3, 4 and 5 satisfy the condition

Because

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 are

Perimeter = 5 + 12 + 13 = 30 units

Area = ½ x 5 x 12 = 30units2

The perimeter and area of this triangle are

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

2n + 1

an + n

an + n + 1

(b + 1)

a + b + c

½ x a x b

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)

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

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

4n + 2

an + 2n

an + 2n + 2

(b + 2)

a + b + c

½ x a x b

4n + 2

4n2 + 4n

4n2 + 4n

+ 2

8n2 + 12n

+ 4

8n3 + 12n2

+ 4n

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

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

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Medicine and mathematics

x= 15:00 - 7:00 = (8) y=300*(0.69). = 5.08 Units Q7. 10 hours after administration the 1.81 as we can see on the graph . y=300*(0.610). We substitute x for 10, as the difference of time= 10. The answer is 1.813985, rounded up to 3.sf 1.81mg. Q8.

2. ## Geography Investigation: Residential Areas

higher in the CBD and gets better as you work your way out from the centre. In theory, the newer, planned houses are on the outskirts of the town, thus they will have a lower index of decay. However, the nineteenth century, older, unplanned houses in the more central part of town should have a higher index of decay.

1. ## Maths GCSE Courswork

360 x 264.57 / 2 = 95245.2 / 2 = 47622.6m2 OR Heron's Formula Area = = 500 (500 - 320)(500 - 360)(500 - 320) = 2268000000 = V2268000000 = 47622.6m2 The area of this isosceles triangle is 47622.6m2. The next triangle has two sides of length 330m and a

2. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

So on the actual trip I made sure that I took a photo of these things as well as other important buildings or objects e.g. a signpost that had previously read 'Barbican Leisure Park' but a local and very active member of the community had told us about her mission

1. ## Beyond Pythagoras

= 7 + 24 + 25 = 56cm I had to make ten more right angled triangles that follow the rule a2+b2=c2 The following triangles are what I made and I also found out the area and the perimeter of them.

2. ## Beyond pythagoras - First Number is odd.

n 8n2+12n+4 1 24 2 60 3 112 4 180 The formula to find the perimeter is 2(4n2+6n+2)= 8n2+12n+4. Area: The formula to find out the area is going to be 4 times as big as the formula for the odd numbers, because of 22.

1. ## Beyond Pythagoras

2 5 12 13 30 30 3 7 24 25 84 56 From this table I have observed the following patterns: * The shortest side length advances by two each time. * All the middle numbers are even. * The middle side is always a multiple of four.

2. ## Beyond Pythagoras

for the length of this side has to be 2n2 + ?n When n is 1, the number in the sequence is 4 2n2 + ?�1 = 4 2�12 + ?�1 = 4 2 + ?�1 = 4 ?=2 This means that the formula for b is 2n2 + 2n Longest side (c)

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to