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Introduction

DOTTY PATTERNS

AIM:         I have been set a task for my coursework to find out the relationship of the dots inside a shape of different sizes.

PLAN: I have planned to use a specific quadrilateral shape for my investigation in which lines will be 45o         (diagonal), one dot to the other; touching each others ends and being closed from all sides. I will be using         the following technique for my investigation. First of all I will commence with the shape-size being 1 cm2         increasing it every step by another 1 cm2. At the same time I will be counting the dots inside that         particular         shape. I will be using this method until I find a pattern; thereafter I will generate a suitable formula from that         pattern.

METHOD: I will be using more or less 5 diagrams and possibly the 6th one for my prediction.

DIAGRAM 1 AREA DOTS PERIMETER 1 cm2 1 4 DIAGRAM 2 AREA DOTS PERIMETER 2 cm2 5 8

DIAGRAM 3 AREA DOTS PERIMETER 3 cm2 13 12

It seems that a pattern is forming for both. Firstly for the dots and area you add 4, there after you double the number, secondly for the perimeter you just add four at each level.

DIAGRAM 4 AREA DOTS PERIMETER 4 cm2 25 16

DIAGRAM 5 AREA DOTS PERIMETER 5 cm2 41 20

Middle

2n2) + (-2n) + 1

(2 x 42) + (-2 x 4) + 1

32 – 8 + 1

32 – 7 = 25

FORMULA CORRECT

TEST PREDICTION:

((2n2) + (-2n)) + 1

(2 x 62) + (-2 x 6) + 1

(72 – 12) + 1

60 + 1 = 61

My prediction was correct.

I can change the formula from ((2n2) + (-2n)) + 1 to ((2n2 - 2n)) + 1 because it makes no difference;

+ x - = -

 AREA DOTS in perimeter 1 CM2 – 04                > 042 CM2 – 08                > 043 CM2 – 12                > 044 CM2 – 16                > 045 CM2 – 20

FORMULA FOR PERIMETER

As the number is doubling by four and there is only one difference I think that linear formula will be more suitable. So;

bn + c

b= first difference

n= any number of sequence

c= output when input is zero

4n + 0

Test formula:

4x2 = 8           Formula correct

Test prediction:

4x6 = 24        Prediction correct

SECOND INVESTIGATION:

The method I will be implying for my second investigation is that I will add 1cm on the length as well as on the width. Diagram 1 AREA cm DOTS PERIMETER 1x2 2 6

Diagram 2 AREA cm DOTS PERIMETER 2x3 8 10

Diagram 3 AREA cm DOTS PERIMETER 3x4 18 14

It looks like that after starting with 6, four is added each time. As for the perimeter 4 is added each time like you can see clearly.

Diagram 4 AREA cm DOTS PERIMETER 4x5 32 18

Diagram 5

 AREA cm DOTS PERIMETER 5x6 50 22

The comment I made before for the shape and area was wrong.

Conclusion

Test prediction:

2n2

2x62

2x36 = 72         prediction correct

Formula for perimeter

 AREAcm DOTS in perimeter 0x1 – 02              > 041x2 – 06             > 042x3 – 10             > 043x4 – 14             > 044x5 – 18             > 045x6 – 22

As the number is doubling by four and there is only one difference I think that linear formula will be more suitable. So;

bn + c

b= first difference

n= any number of sequence

c= output when input is zero

4n + 2

Test formula:

4n + 2

4x3 + 2 = 14                formula correct

Test prediction:

4n + 2

4x6 + 2 = 26                prediction correct

EVALUATION:

I had a bit of problems in finding the formula for my second investigation, but I realised that the first part of the investigation appears to give the answer (2n2). I tested the formula and fortunately it worked. I was very pleased with it and it also gave me a tip for the next time I had to do the job; look closely at the formula.

CONCLUSION:

I came to the conclusion that the formulae are as follows:

1stinvestigation:

2n2 + 2n – 1                 for dots & area 4n                        for dots in perimeter 2ndinvestigation:

2n2for dots & area 4n + 2                        for dots in perimeter Note:

In the second investigation you can use the following formula to make things a bit                 simpler;

2d2

As d equals to the width of a shape, so;

2d2

2x32 = 18

Same goes for the formula for perimeter:

4d + 2

-END-

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