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Koch’s Snowflake Investigation

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Introduction

David Paton 10PA        Koch’s Snowflake Investigation

Koch’s Snowflake Investigation

Introduction

I am conducting an investigation into how Koch’s Snowflake is built up.  I will investigate how the perimeter and area change with each new shape.  On every new snowflake the middle third of every edge is taken out and a smaller triangle is put in its place ( see diagrams at back of project).  First, I will investigate how the perimeter increases with every shape.

Perimeter

I will not count manually the length of each shape’s perimeter as this would be hard and inefficient.  Instead, I will calculate the perimeter by multiplying the number of edges on the shape by the length of each edge.

I have noticed that the number of edges on each shape is multiplied by 4 each time.  This is because every time a triangle is added, the side is split into 4 separate edges (see diagram above).  I have also noticed that the length of each edge is divided by 3 every time (see diagram).

Now, by multiplying the number of edges by the length of each edge on a shape, I can find its perimeter.

Shape Number

1

2

3

4

No. Of Edges

3

×4

12

×4

48

×4

192

Length of Each Edge

9

÷3

3

÷3

1

÷3

Perimeter

27

36

48

64

I have now noticed that the perimeter is increasing in a geometric sequence, by ⅓ of its value each time.  It is also increasing by a larger amount each time, and so is a divergent series.

For example:          27 × 1⅓ = 36

36 × 1⅓ = 48

48 × 1⅓ = 64   and so on…

After considering this, I have discovered that this is connected to how I first calculated the perimeter.  It is evident on the table above that I multiplied the previous perimeter by 4, then divided by 3 to find the next perimeter.  So, the perimeter can be calculated by multiplying the previous one by image11.png.

27 × image11.png = 36

36 × image11.png = 48

48 × image11.png = 64

let n = shape number   let p = perimeter

n

1

2

3

4

p

27

× image14.png =

36

× image14.png =

48

× image14.png =

64

...read more.

Middle

27 × image14.png = 36, which is shape no. 2image01.png

27 × image14.png = 48, which is shape no. 3image03.png

27 × image14.png = 64, which is shape no. 4

image12.png the general rule for the perimeter is:image04.png

Area

I will now investigate how the area increases with each shape.  Instead of working out the area of each shape in cm³, I will count the number of triangles in each shape, as this is more efficient.

In the first snowflake (see back of project) there were 81 triangles.  The sides of the whole triangular shape are each 9 centimetres long, the area of one of the triangles can be found by squaring the length of one side ( 9 ² = 81).

I will calculate the area of each snowflake by breaking it down into individual triangle shapes.  For example, shape 2 consists of one large triangle, from shape 1, and three new, smaller triangles:

The area of shape 2 can therefore be calculated by:

area of large triangle + ( 3 × area of small triangle).

In order to work out the area in this manner, I will have to first work out the area of each triangle added on and the amount of triangles added on with each shape.

This table shows how the area of each small triangle decreases with each shape:

Shape Number

Area of Smallest Triangle

1

81

2

9

3

1

4

1/9

5

1/81

So, I now know that the area of the smallest triangle is divided by 9 with each new shape.

This table shows how the number of new triangles in each shape increases:

Shape Number

Number of Small Triangles

1

1

2

3

3

12

4

48

5

192

...read more.

Conclusion

image13.png, as I worked out earlier.  I substituted these figures into the equation:image06.png

I then tested this equation:

n

Area using equation

Real Area

1

27

81

2

39

108

3

44⅓

120

4

46.7

125⅓

5

47.8

127.7

This equation is not correct.  I will now try adding on 81 to my equation, as previously I had totally discarded it for being an exception to the pattern.  As 27 is really the second term and not the first, I will also change ‘n’ in my equation to ‘n-1’:

image07.png

This equation works.  It can be used to find the area of any number of snowflake.

As the area series is convergent, I can also find the sum to infinity.

Using the Foundation A-Level book again, I have found the following equation:image08.png

image09.png

I can now adapt this equation to suit my sequence:


This answer of 129.6 is consistent with my previous results as the area gets closer and closer to this figure, without ever reaching it.  This is evident on the graph I have drawn (see back of project).image10.pngimage02.png

129.6 is 1.6 times the original area of the first snowflake.

...read more.

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