Larry Riddle, Agnes Scott College

## 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 3×4k-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}