# Mathematics Coursework - Beyond Pythagoras

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Introduction

Edexcel 2002 BEYOND PYTHAGORAS

Mathematics GCSE Coursework Ms Pathak

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

b)

Middle

15

112

113

240

840

8

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)

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)

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)

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

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

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