Area of the Koch Snowflake

The first observation is that the area of a general equilateral triangle with side length a is

\[\frac{1}{2} \cdot a \cdot \frac{{\sqrt 3 }}{2}a = \frac{{\sqrt 3 }}{4}{a^2}\]

as we can determine from the following picture

For our construction, the length of the side of the initial triangle is given by the value of s. By the result above, using a = s, the area of the initial triangle S(0) is therefore \(\dfrac{{\sqrt 3 }}{4}{s^2}\).

Area after first iteration: (using a = s/3)

\[{\text{Area: }}\frac{{\sqrt 3 }}{4}{s^2} + 3 \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{3}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{3}{9}} \right)\]

Area after second iteration: (using a = s/3²)

\[{\text{Area: }}\frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{3}{9}} \right) + 3 \cdot 4 \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{9}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{3}{9} + \frac{{3 \cdot 4}}{{{9^2}}}} \right)\]

Area after third iteration: (using a = s/3³)

\[{\text{Area: }}\frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{3}{9} + \frac{{3 \cdot 4}}{{{9^2}}}} \right) + 3 \cdot 4 \cdot 4 \cdot \frac{{\sqrt 3 }}{4}\left( {\frac{s}{{{3^3}}}} \right) = \frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{3}{9} + \frac{{3 \cdot 4}}{{{9^2}}} + \frac{{3 \cdot {4^2}}}{{{9^3}}}} \right)\]

By now the pattern should be clear. At the kth iteration we add 3×4^k-1 additional triangles of area \(\frac{{\sqrt 3 }}{4}{\left( {\frac{s}{{{3^k}}}} \right)^2}\). This means we add a total area of

\[3 \cdot {4^{k - 1}} \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{{{3^k}}}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} \right)\]

to the area S(k-1) to get the area of S(k). Hence after n iterations we get the area of S(n) to be

\[\frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \sum\limits_{k = 1}^n {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} } \right)\]

The sum inside the parentheses is the partial sum of a geometric series with ratio r = 4/9. Therefore the sum converges as n goes to infinity, so we see that the area of the Koch snowflake is

\[\begin{align} \frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \sum\limits_{k = 1}^\infty {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} } \right) &= \frac{{\sqrt 3 }}{4}{s^2}\left( {1 + \frac{{3/9}}{{1 - 4/9}}} \right) \\ &= \frac{{\sqrt 3 }}{4}{s^2}\left( {\frac{8}{5}} \right) \\ &= \frac{{2\sqrt 3 }}{5}{s^2} \\ \end{align}\]