• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 2107

Beyond Pythagoras

Extracts from this document...

Introduction

Beyond Pythagoras

Introduction.

A Pythagorean triple is a set of three integers a, b and c that specify the lengths of a right angled triangle - that is c2 = a2 + b2. The numbers 3, 4 and 5 is one example. Another way of writing this is (smallest number) 2 + (middle number) 2 = (largest number) 2  An example of the (3,4,5) triangle:

image00.png

The numbers 3,4 and 5 can be lengths, in appropriate units, of the sides of a right angled triangle.

In my investigation I will also look at perimeter and area.

The perimeter of this triangle is: 3+4+5=12 units.                                                                                             The area of the triangle is: ½ x 3 x 4= 6 units.

In my investigation I will observe patterns and work out formulae relating a side or measurement to the term number. I will see if each pattern has the same or different formulae. I will start with the shortest side being odd.

1. Shortest side being an odd integer.

Term number (N)

a

b

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

Formulae:

a= 2n+1

This is because there is a difference of two between each number in the pattern. When this happens as a general rule the formulae must have 2n then I found the difference between 2n and the actual length in each case, this happened to be 1.

Example of a=2n+1: The answer to a when n=6 is 13.The answer to the formula: 2x6+1=a

2x6=12

12+1=13 so the formula is correct.

b=2n2 +2n

...read more.

Middle

a

b

c

Perimeter

Area

1

4

3

5

12

6

2

6

8

10

24

24

3

8

15

17

40

60

4

10

24

26

60

120

5

12

35

37

84

210

Formulae:

a= 2n+2

This is because there is a difference of two between each number in the pattern. When this happens as a general rule the formulae must have 2n then I found the difference between 2n and the actual length in each case, this happened to be 2.

Example of a=2n+2: The answer to a when n=5 is 12.The answer to the formula: 2x5+2=a

2x5=10

10+2=12 so the formula is correct.

b=n2 +2n

This is because there is a second difference between each number. The first difference is not the same every time but the difference between this first difference was a constant of 2. As a general rule you halve this number and put n2 after it, the formula must have n2 in it. Then there is 2n because each time that was the difference between the answer for the first part and the actual answer.

Example of b=n2+2n: The answer to b when n=4 is 24. The answer to the formula:

4^2+2x4=b

4^2=16

2x4=8

16+8=24 So the formula is correct.

c=n2+2n+2

I noticed that the patterns in this length column were exactly the same as in b but they were two numbers more. So I used the same formula for the same reasons as in b but added 2.

Example of c=n2+2n+2: The answer to c when n=3 is 17.

...read more.

Conclusion

An example of s-L=(0.5s)2+1: The answer to L when s=4 is 5. The formula shows:

(0.5x4)2+1=s to L.

(0.5x4)2=4

4+1=5

So the formula is correct.

5. P=A

In each of the cases I have noticed that there is a clear link between each of the relevant perimeters and areas. In terms of n the area is always half n x perimeter. From this I can derive a way of finding the perimeter when I have the area and the area when I have the perimeter.

Area when I have perimeter:

Formula= A= n/2xP.

This is because in each of the cases there was a clear link between the two. For n=1 the area was always half the perimeter, for n=2 the area was always the same as the perimeter. This continues as so:

n

A

1

0.5*P

2

1*P

3

1.5*P

4

2*P

5

2.5*P

6

3*P

7

3.5*P

8

4*P

Etc…

Also each of the numbers to times the perimeter by were halve the term number.

Example of A= n/2xP: A perimeter of a 3,4,5 triangle is 12 and the perimeter is 6 also n=1. The formula shows:

A= 1/2x12

A=6 this is correct.

Perimeter when I have area

Formula= P=2A/n

I found this by making P the subject of the equation from above. By multiplying both sides by two and dividing by n I got this formula.

Example of P=2A/n: The area of an 8,15,17 triangle is 60, the term number is three and the perimeter is 40. The formula shows:

P=2x60/3

2x60=120

120/3=40 so the formula is correct.

...read more.

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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Geography Investigation: Residential Areas

    (Figure 12 - next page) Figure 12 Stacked Bar Chart On the route out of Basingstoke's CBD along Winchester Road, you can see above that from Sarum Hill to Cumberland Avenue the amount of residents not wanting to move increases.

  2. Beyond Pythagoras

    cm 24 cm 25 cm 56 cm 84 cm� 9 cm 40 cm 41 cm 90 cm 180 cm� 11 cm 60 cm 61 cm 132 cm 330 cm� 13 cm 84 cm 85 cm 182 cm 546 cm� 15 cm 112 cm 113 cm 240 cm 840 cm� 17

  1. Beyond pythagoras - First Number is odd.

    2 x 2� + 2 x 2 = 12 2 x 4 + 4 = 12 8 + 4 = 12 12 = 12 My formula also works for the 2nd term. If it works for the 3rd term I can safely say that 2n2 + 2n is the correct formula.

  2. Beyond Pythagoras

    To get from 5 to 13, I have to add 8; then from 13 to 25, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further.

  1. Beyond Pythagoras

    The longest side of the triangle (c) is always 2 more than b because I am investigating triangles where c = b + 2. Therefore, the formula must be n2 + 1. Area (A) The formula for the area (A)

  2. Beyond Pythagoras.

    The result can be written as f(x) = 2n3 - n2 - 5n - 2= (n - 2)(2n2 + 3n + 1) If 2 was substituted for the x in this identity so that n - 2 = 0, the quotient is eliminated giving f (2)

  1. Beyond Pythagoras.

    , 40 (M) , and 41 (L) S + M + L = P 9 + 40 + 41 = 90 This proves the formula works. I will now try and use the same formula but modify it in some ways to see if I can find a formula for area.

  2. Beyond Pythagoras.

    = a = 4 2 = a = 2 = 3a + b = 1st diff = 3 � 2 + b = 8 = 6 + b = 8 = b = 8 - 6 = b = 2 term: = a + b + c = 1st term

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