# Fractal Geometry

### Area of the Koch Snowflake

For any equilateral triangle with side s, $\text{Area}=\frac{\sqrt{3}}{4}{s}^{2}.$ We will use this to find the area of the Koch snowflake curve.

For curve 1,

${A}_{1}=\frac{\sqrt{3}}{4}$

For curve 2,

$\begin{array}{l}{A}_{2}={A}_{1}+\frac{\sqrt{3}}{4}{\left(\frac{1}{3}\right)}^{2}\left(3\right)\\ {A}_{2}=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}{\left(\frac{1}{3}\right)}^{2}\left(3\right)\end{array}$

Similarly,

$\begin{array}{l}{A}_{3}={A}_{2}+\frac{\sqrt{3}}{4}{\left(\frac{1}{9}\right)}^{2}\left(3\right)\left(4\right)\\ {A}_{3}=\left[\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\left(\frac{1}{9}\right)\left(3\right)\right]+\frac{\sqrt{3}}{4}{\left(\frac{1}{9}\right)}^{2}\left(3\right)\left(4\right)\end{array}$

You should be able to see a pattern developing.

$\begin{array}{l}{A}_{4}={A}_{3}+\frac{\sqrt{3}}{4}{\left(\frac{1}{27}\right)}^{2}\left(3\right)\left(16\right)\\ {A}_{4}=\left[\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\left(\frac{1}{9}\right)\left(3\right)+\frac{\sqrt{3}}{4}{\left(\frac{1}{9}\right)}^{2}\left(3\right)\left(4\right)\right]+\frac{\sqrt{3}}{4}{\left(\frac{1}{27}\right)}^{2}\left(3\right){\left(4\right)}^{2}\end{array}$

In general,

The expression in parentheses is a convergent geometric series with $a=\frac{1}{3}$ and $r=\frac{4}{9}.$ The sum of the series is given by the formula

$Sum=\frac{a}{1-r}$

Thus, the area of the Koch snowflake is

$\begin{array}{l}{A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{\frac{1}{3}}{1-\frac{4}{9}}\right]\\ {A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{\frac{1}{3}}{\frac{5}{9}}\right]\\ {A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{1}{3}\left(\frac{9}{5}\right)\right]\\ {A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{3}{5}\right]\\ {A}_{n}=\frac{\sqrt{3}}{4}\left[\frac{8}{5}\right]\end{array}$

So we have,

$\underset{n\to \infty }{\mathrm{lim}}{A}_{n}=\frac{\sqrt{3}}{4}\left[\frac{8}{5}\right]$

In other words, the area of the Koch snowflake curve approaches a limit that is 8/5 the area of the original triangle. Recall, as $n\to \infty ,$ the perimeter of the Koch snowflake becomes infinite. Yet, we have just seen that the area of the snowflake curve is finite. So what we have is a curve of infinite length enclosing only a (small) finite area! This remarkable property of the Koch snowflake may seem paradoxical. This is just one of many fascinating and unexpected results one finds when studying fractal geometry.

The terms fascinating and unexpected are general descriptors of fractal geometry. For instance, consider our next topic, fractal dimension vs. ordinary dimension.

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