Area of a Koch Snowflake

Question: A Koch Snowflake is a fractal which can be built by starting with an equilateral triangle, removing the inner third of each side, building another equilateral triangle at the location where the side was removed, and then repeating the process indefinitely.

Find the area of a Koch Snowflake when the sides of the starting equilateral triangle has the length x.

Answer: The easiest way to solve this problem is to calculate the area added to the Koch Snowflake after each iteration.

Area of the first iteration can easily be calculated by using Pythagoras.

x^2 = (\frac{x}{2})^2 + h^2
h^2 = \frac{3}{4} x^2
h = \frac{\sqrt{3}}{2} x

\therefore area = (\frac{\sqrt{3}}{2} x)(\frac{x}{2}) = \frac{\sqrt{3}}{4} x^2

Using the solution for the first iteration, we can easily calculate the area of the second iteration. The area will be the area of the original equilateral triangle plus the area of 3 smaller equilateral triangles.

area =  \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2

The area of the third iteration would, therefore, be:

area =  \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2 + 3 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^2})^2

If we continue doing this, we will get the following summation.

area =  \frac{\sqrt{3}}{4} x^2 + 3 \frac{\sqrt{3}}{4} (\frac{x}{3})^2 + 3 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^2})^2 + 3 \cdot 4 \cdot 4 \frac{\sqrt{3}}{4} (\frac{x}{3^3})^2 \ldots
area =  \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty}3 \cdot 4^i \frac{\sqrt{3}}{4} \frac{x^2}{3^{2(i+1)}}
area =  \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty} \frac{3^{\frac{3}{2}}}{4} x^2 \frac{4^i}{9^{i+1}}
area =  \frac{\sqrt{3}}{4} x^2 + \sum_{i=0}^{\infty} \frac{1}{4 \sqrt{3}} x^2 (\frac{4}{9})^i

Since the summation is a geometric series, we know that \sum_{i=0}^{\infty} a r^i = \frac{a}{1-r} when r < 1.

Therefore,

area =  \frac{\sqrt{3}}{4} x^2 + \frac{1}{4 \sqrt{3}} x^2 \cdot \frac{1}{1 - \frac{4}{9}}
area =  \frac{\sqrt{3}}{4} x^2 + \frac{1}{4 \sqrt{3}} x^2 \cdot \frac{9}{5}
area =  \frac{2 \sqrt{3}}{5} x^2