Larry Riddle, Agnes Scott College

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*/3^{2})

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

**Area after third iteration**: (using **a** = *s*/3^{3})

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

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

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

Suppose that the area of the center equilateral triangle in the Koch snowflake is 1. Let S be the area outside this triangle, but inside the part of the snowflake outlined in green in the following image.

The area of the equilateral triangle inside the green section is the center equilateral triangle scaled by 1/3, so its area is 1/9. Each of the four smaller copies of the green section are a 1/3-scaled version of the entire green section, so each of them has area S/9. Therefore \[S = \frac{1}{9} + 4\cdot\frac{S}{9} \Rightarrow S = \frac{1}{5}.\] The blue and red sections outside the center equilateral triangle but inside the snowflake have the same area as the green section. Thus the area of the Koch snowflake is 1 + 3(1/5) = 8/5.

More generally, the area of the snowflake would be 8/5 times the area of the original equilateral triangle. If the length of a side of the triangle is \(s\), then the area of the triangle is \(\displaystyle \frac{\sqrt{3}}{4}s^2\) and therefore the area of the snowflake would be \[\left( {\frac{8}{5}} \right)\frac{{\sqrt 3 }}{4}{s^2} = \frac{{2\sqrt 3 }}{5}{s^2}. \]

Here is an animation that uses the same idea of self-similarity to find the area bounded by the Koch curve and its initial line segment, and then applies the result to find the area bounded by the Koch snowflake (and the Koch anti-snowflake).

- McWorter, William A. Personal correspondence, December 5, 1999.
- Sandefur, James T., "Using Self-Similarity to Find Length, Area, and Dimension", The American Mathematical Monthly, Vol. 103, No. 2 (Feb., 1996), pp. 107-120 [Available from JSTOR (subscription required)].