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# Borders - Fencing problem.

Extracts from this document...

Introduction

GCSE Coursework

Borders

Here are the first four examples of the shapes in this sequence:

The sequence begins with a single white square, which is then surrounded by black squares to form the second shape. Each new cross is then formed by completely surrounding the previous cross with a border of black squares. In each new cross, the previous cross can be seen as the area of white squares in the centre.

In this investigation I will try to get algebraic formulas from the sequence, each expressing one property in terms of another (e.g. defining the area as the diameter squared). These formulas can then be checked and, proven, using a variety of maths skills.

## Defining N

The variable that I will be using to refer to terms in this investigation will be n.

Middle

0

0

1

4

2

8

3

12

4

16

5

20

From this table it becomes obvious to see that the formula for the perimeter of the shape is P = 4n

If you divide the perimeter of a cross by four to obtain a single side of the cross then you can see why this works. The length of one side is exactly the same as n. The length of one side must be multiplied by 4 to give the perimeter and for this reason P = 4n.

## Area

To work out this formula I will draw a diagram that finds out the differences between consecutive terms, and then the differences between those differences.

N:  0   1    2     3

Area: 1    5   13   25

4     8  12

44

Because the diagram needs to go down two levels the find

Conclusion

of the cross. Therefore the 2n2 must be the rest of the circle. Because it is 2n2 then it must be possible to split in to two equal parts each part being n2. n2 must equal the part of the cross above or below the row of squares worked out by 2n+1.

I will also try the formula out on the next cross to see if it works.

The n value of this cross is 6.

I have counted the squares manually in the cross and there were 85 squares altogether.

So:

2n2 + 2n +1

2(6)2 + 2(6) + 1

72 + 12 + 1

= 85

Again the formula works. I wont draw out any more of the crosses because I think we have drawn enough. Now I will draw a table showing the area of the next few crosses in the sequence by using the formula to work it out.

 n 2n2+2n+1 7 113 8 145 9 181 10 221

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