Agnes Scott College
Larry Riddle, Agnes Scott College
KochSnowflake100

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

height

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)

S1area

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

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

S2area

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

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

S3area

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

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

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