Height (h) = cos -1 (½ / 1) = 60º = ½
= tan 60 () x ½ =
Area = =
2
Area of Stage 1 = Area of Stage 0 + Area of 3 shaded triangles
Area of shaded small triangle = base x height
2
Height = cos -1 (() / ()) = 60º = ½
= tan 60 () x =
Area of shaded small triangle
= x
2
=
Total area of Stage 1 =
=
=
Area of Stage 2 = Area of Stage 1 + Area of 12 small shaded triangles
Height= cos -1 (() / ()) = 60º = ½
= tan 60 () x =
Area of shaded small triangle
= x
2
=
Total area of Stage 2 = +
=
=
Area of Stage 3 = Area of Stage 2 + Area of 48 small shaded triangles
Height= cos -1 (() / ()) = 60º = ½
= tan 60 () x =
Area of shaded small triangle
= x
2
=
Total area of Stage 3 = +
=
=
N n ) The number of sides increases from stage to stage by a factor of 4. This is because we remove the inner third of each stage and we then build an equilateral triangle where it was removed, therefore the number of sides of each stage is equal to the number of sides of the previous stage, plus 3 (number of sides of the equilateral triangle) times the number of sides of the previous stage. This is equal to 4 times the number of sides of the previous stage.
L n ) The length of a single side decreases by a factor of 3 because the inner third of each side is removed therefore if the original length is 1 and you decrease it by a factor of 3 (multiply by 1/3) you’ll get the following length to be 1/3
P n ) The length of the perimeter is simply N n times L n and since N n is increasing by a factor of 4 and L n is decreasing by a factor of 3, then P n is increasing by a factor of 4/3
A n ) The area of each stage was found by adding the area of the small triangles to the area of the shape at the previous stage.
Part 2 :
Graph 1
Graph 2
Graph 3
Graph 4
Part 3:
In graph 1, which shows the number of sides of each stage, we can conclude that the formula for calculating the number of sides at the nth stage is 3 x 4n (where n is the number of the stage) since the gradient of the graph is 4 and the first term is 3.
In graph 2, which shows the length of a single side of each stage, we can conclude that the formula for calculating the number of sides at the nth stage is 1 x (1/3)n (where n is the number of the stage) since the gradient of the graph is -3 (1/3) and the first term is 1
In graph 3, which shows the perimeter of the different stages, the formula for calculating the perimeter at the nth stage is Nn x Ln . Nn = 3 x 4n and Ln= 1 x (1/3)n for the reasons previously explained, therefore the formula for the perimeter is 3 x (4/3)n .
In graph 4, which shows the area of each shape, we can see that as we progress in stage the area increases. To find the formula in this case we’ll need to look at the similarities between each stage and perform some calculations as shown below:
Area Stage 0 = .
Area Stage 1 = +
Area Stage 2 = + +
Area Stage 3 = + + +
An = An-1 +
An = An-1 +
An = An-1 +
Part 4:
Nn = 3 x 4n
= 3 x 44
= 768
Ln = 1 x (1/3)n
= 1 x (1/3) 4
=
Pn = 3 x (4/3)n
= 3 x (4/3) 4
= 3 x
=
= 9
An = An-1 +
= + (x 768 x )
16
= +
16
= +
= +
=
Drawing of the Koch snowflake at the 4th stage:
Part 5:
The value of n where An+1 is equal to An to six places of decimals is 17. This was found using a spreadsheet on Microsoft Excel.
Part 6:
The formula for calculating the perimeter is 3 x (4/3) n. As n gets very large or in other words as n , the perimeter becomes larger and larger (we can prove that using the spreadsheet above) therefore we can conclude that the perimeter is infinite:
The formula for calculating the area is An-1 + if we look back to the
end of part 3 we can see that the Area of the nth stage is equal to the area of the previous stage plus the area of the smaller triangles added at each stage. This new area added, as, becomes smaller and smaller and will be converging to zero therefore the Area for the nth stage must be converging to a certain number so it must be finite.
In order to find the number to which it converges we can simply use the spreadsheet and from looking at a part of the spreadsheet below we can see that as n is getting larger the value for An is remaining the same therefore it must be converging to that number.
We can see that as n becomes larger than 25 (as n ) the area remains to be 0.692820323 therefore it’s converging to that value. To find an exact value for this I can simply use my trigonometry knowledge…= 0.7071067812, therefore by trial and improvement I discovered that 0.692820323 = which in turn is equal to.
In conclusion we can state that the Koch Snowflake has an infinite perimeter and a finite area as n . We can also see that as n , the length of each side (Ln ) decreases and the number of sides Nn increases.
