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  • Level: GCSE
  • Subject: Maths
  • Word count: 1235

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 1image03.png

AREA

DOTS

PERIMETER

1 cm2

1

4

image04.png

DIAGRAM 2

image07.png

AREA

DOTS

PERIMETER

2 cm2

5

8

DIAGRAM 3

image08.png

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

image09.png

AREA

DOTS

PERIMETER

4 cm2

25

16

DIAGRAM 5

image10.png

AREA

DOTS

PERIMETER

5 cm2

41

20

...read more.

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

                > 04

2 CM2 – 08

                > 04

3 CM2 – 12

                > 04

4 CM2 – 16

                > 04

5 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.

image11.png

        Diagram 1

image12.png

AREA cm

DOTS

PERIMETER

1x2

2

6

Diagram 2

image13.png

AREA cm

DOTS

PERIMETER

2x3

8

10

Diagram 3

image14.png

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

image05.png

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.

...read more.

Conclusion

Test prediction:

2n2

2x62

2x36 = 72         prediction correct

Formula for perimeter

AREA

cm

DOTS in

perimeter

0x1 – 02

             > 04

1x2 – 06

             > 04

2x3 – 10

             > 04

3x4 – 14

             > 04

4x5 – 18

             > 04

5x6 – 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 & areaimage00.png

                4n                        for dots in perimeterimage01.png

        2ndinvestigation:

                2n2for dots & areaimage01.png

                4n + 2                        for dots in perimeterimage02.png

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-

...read more.

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