• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # Find out the relationship of the dots inside a shape of different sizes.

Extracts from this document...

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-

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

# Related GCSE Fencing Problem essays

1. ## t shape t toal

+ (T+2) + (T+8) + (T+1) = T-total 5T + 7 = T-total Does this formula work for other t-shapes in the grid. 5T + 7 = T-total 5 ? 26 + 7 = T-total 137 = 137 The formula 5T + 7 works for all t-shapes regardless of what grid size the t-shape has come from.

2. ## t shape t toal

the formula for a 90� to the right T-Shape on any size grid. Although I should solve the full working out, I found that the formula 5T+7 could be worked for all grid sizes. Here is an example on a 3 by 3 grid size, which you could use: 1

1. ## t shape t toal

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

2. ## Investigating different shapes to see which gives the biggest perimeter

x 137.64m = 100 x 137.64 = 13763.8m� So the area of one isosceles triangle within the pentagon is 13763.8m�. The area of the pentagon is 5 times the value of one isosceles triangle. Therefore the value of the pentagon is 13763.8 x 5 = 68819.1m�.

1. ## Geography Investigation: Residential Areas

< 30 0 < 30 6 30 < 35 2 < 35 8 35 < 40 0 < 40 8 40 < 45 1 < 45 9 45 < 50 0 < 50 9 50 < 55 1 < 55 10 I will now plot the 'Cumulative Frequency' data from

2. ## Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping ...

Then, by a series of calculations, two Mann-Whitney U values are produced. As long as the smaller of these two values is above that needed for amount of values, H0 can be rejected. H0 = There is no significant difference in grass heights between the wooded, and grassland areas of Rushey Plain, Epping Forest.

1. ## The best shape of guttering

Since the polygons will be halved I shall only investigate even number sided shapes since the odd numbered ones would only be irregular variants of these. To calculate the area of each polygon I first find the area of a single triangle inside the shape, ie in a hexagon: I use trigonometry to do this - A B SinC 0.5.

2. ## Geographical Inquiry into the proposed redevelopment plan of the Elephant and Castle.

The increased retailing will allow more opportunities for jobs to spend some of their spare time. The Elephant and Castle's Holistic approach is informed by the people's vision of providing individual opportunity, creating employment, improving health and education and enhancing living standards. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 