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Part II - Snowflake Island 0r The fractal called snowflake island (or Kochs snowflake) is constructed as follows be You will
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a) The table data will be

Step (n) Length of one side (Sn) Number of sides Perimeter (Pn) Area of one new triangle Number of new triangles Area of snowflake (An)
0 1 3 3 n/a 0 A
1 1/3 12 4 1/9A 3 A + 1/3A
2 1/9 48 16/3 1/81A 12 A + 1/3A + 4/27A
3 1/27 192 64/9 1/729A 48 A + 1/3A + 4/27A + 16/243 A
4 1/81 768 256/27 1/6561A 192 A + 1/3A + 4/27A + 16/243A + 64/2187A
5 1/243 3072 1024/81 1/59049 768 A + 1/3A + 4/27A + 16/243A + 64/2187A + 256/19683A
n 1/3^n 3(4)^n 3(4/3)^n 1/2^(2n)A 3(4)^(n-1) A(n-1) + 1/3*(4/9)^n-1 * A

b) The perimeter of Koch flake is given by the formula

P_n = 3(\frac{4}{3})^n

So, when n tends to infinity, hence the perimeter will tend to infinity

c) The area of the Koch flake will be

A_n = A_{n-1} + \frac{1}{3}(\frac{4}{9})^{n-1}A

where A is the area of the initial triangle

Solving this recurrence relation we get

9 3 3

So, when n tends to infinity, the area will be tending to 8/5 * A

Note - Post any doubts/queries in comments section.

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