• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1924

Mathematics Coursework - Beyond Pythagoras

Extracts from this document...

Introduction

Edexcel 2002        BEYOND PYTHAGORAS

Mathematics GCSE Coursework         Ms Pathak

image22.jpgimage00.png

image22.jpgimage01.png

1)

The numbers 3, 4, and 5 satisfy the condition

                        3² + 4² = 5²

because                3² = 3x3 =9

                        4² = 4x4 = 16

                        5² = 5x5 = 25

and so

                3² + 4² = 9 + 16 = 25 = 5²

I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) ² + (middle number) ² = (largest number) ².

a) 5, 12, 13

5² + 12² = 25 + 144 = 169 = 13²

b) 7, 24, 25

7² + 24² = 49 + 576 = 625 = 25²

2)        Perimeter

image08.png

image14.png

image16.png

image17.png

image18.png

image19.png

image20.png

image21.png

image02.png

image03.png

b)

...read more.

Middle

15

112

113

240

840

8image06.png

17

144

145

306

1224

9

19

180

181

380

1710

10

21

220

221

462

2310

I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest side. And also if you can see in the shortest side column, it goes up by 2. I have also noticed that the area is

½ (shortest side) x (middle side).  

3)

In this section I will be working out and finding out the formulas for:

  • Shortest side
  • Middle side
  • Longest side

In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I don’t know the formula so I will have to work that out.

Firstly I will be finding out the formula for the shortest side.

3    5    7    9    11

   2    2    2    2

The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n.

Let’s see…

Nth term

Length of shortest side

1

3

2n

2 x 1 = 2 (wrong)image10.png

There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct I will now test this formula.

2n+1

Nth term

Length of shortest side

1

3

2x1+1=3 (correct)image11.png

Just in case I will test this formula in the next term:

2n+1

Nth term

Length of shortest side

2

5

2x2=4

4+1=5 (correct)image12.png

I now have to work out the formula for the middle side.

...read more.

Conclusion

So 12 is the perimeter for the first term

2nd term 5 + 12 + 13 = 30

3rd term 7 + 24 + 25 = 56

And so on. All I have to do is put all the 3 formulas together.

Perimeter = (shortest side) + (middle side) + (longest side)

 = 2n + 1 + 2n² + 2n + 2n² + 2n + 1

 = 2n² + 2n² + 2n + 2n + 2n + 1 + 1

 = 4n² + 6n + 2

If I have done my calculations properly then I should have the right answer. To check this I am going use the 4th, 5th and 6th terms.

4th term:

4n² + 6n² + 2 = perimeter

4 x 4² + 6 x 4 + 2 = 9 + 40 + 41

64 + 24 + 2 = 90

90 = 90

It works for the 4th term

Let’s see if it works for the 5th term:

4n² + 6n² + 2 = perimeter

4 x 5² + 6 x 2 = 11 + 60 + 61

100 + 30 + 2 = 132

132 = 132

And it works for the 5th term

And finally the 6th term:

4n² + 6n² + 2 = perimeter

4 x 6² + 6 x 6 + 2 = 13 + 84 + 85

144 + 36 + 2 = 182

182 = 182

It works for all the terms so:

Perimeter = 4n² + 6n + 2

Like the area I know that the area of a triangle is found by:

Area = ½ (b x h)

b = base

h = height

Depending on which way the right angled triangle is the shortest or middle side can either be the base or height because it doesn’t really matter which way round they go, as I’ll get the same answer either way.

Area = ½ (shortest side) X (middle side)

 = ½ (2n + 1) x (2n² + 2n)

 = (2n + 1)(2n² + 2n)

I will check this formula on the first two terms:

(2n + 1)(2n² + 2n) = ½ (b x h)

(2 x 1 + 1)(2 x 1² + 2 x 1) = ½ x 3 x 4

3 x 4 = ½ x 12

12 = 6

6 = 6

2nd term:

(2n + 1)(2n² + 2n) = ½ b h

(2 x 2 + 1)(2 x 2² + 2 x 2) = ½ x 5 x 12

5 x 12 = ½ x 60

60 = 30

30 = 30

It works for both of the terms. This means:

Area = (2n + 1)(2n² + 2n)

By Asif Azam

10L4

Mathematics Coursework

...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. Medicine and mathematics

    that every hour after the administration of the original dose, only 60% of insulin will remain active. This basically is means that every successive hour, the next value will be 60% of the value before it. The insulin was administrated at 07:00, and amount of penicillin active at that moment was 300.00.

  2. Fencing - maths coursework

    Area = 0.5 x 288.6751346m x 333.33333333333m = 48112.52243m2 333.33333333333m 250m = Area = 48112.52243m2 Area = 62500 m2 Perimeter = 1000m Which one has the greatest area? The square has the greatest area = 62500 m The difference area between the square and the triangle was by = 62500

  1. Fencing problem.

    Firstly I shall find the area of the rectangle then the trapezium that the rectangle has been placed upon. Area of rectangle = Base � Height Area of rectangle = 325 � 25 Area of rectangle = 9375m2 Now I shall find the area of the trapezium.

  2. Maths GCSE Courswork

    base of 340m: Base of one triangle = 340 / 2 = 170m Length A = c2 - b2 = a2 = 3302 - 1702 = 800002 = V80000 = 282.84m Therefore, the height of the triangle is 282.843m. Area = base x height / 2 = 340 x 282.843

  1. Beyond Pythagoras

    7 2 4 41-32 9 2 5 61-50 11 2 Rule = 2n+1 So the rule that the hypotenuse line column follows is = 2n�+2n+1 This is how I found out the rule the Fourth column follows (perimeter):- A Input Output 1st Difference 2nd Difference 1 12 18 8 2

  2. Beyond pythagoras - First Number is odd.

    Length of Middle Side (b) Length of Longest Side (c) Perimeter in Units Area in Square Units 1 6 8 10 24 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 n 4n2+4n+2 1 10 2 26 3 50 4 82 As I expected my formula works.

  1. Beyond Pythagoras

    = 180 2 Perimeter of a triangle = side + side + side = 9 + 40 + 41 = 90 2) 11, 60, 61 The formula for Area of a triangle is: 1/2 x b x h Area = 1/2 x 60 x 11 = 60 x 11 =

  2. Maths Coursework - Cables: For this assignment I have been requested to study a ...

    We can see that the length (y) is equivalent to twice this radius i.e. Now we need to know the value of (x) to conclude. We can do this using this formula: Arc Length = (angle/360) x (total circumference, ? x diameter) Because we know that (x)

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