## Area Bounded by the Koch Curve

Suppose that the length of the initial horizontal chord is 1. Let \(A\) be the area bounded by the Koch curve and this initial chord.

The side length of the equilateral triangle inside the bounded region is 1/3, so the area of this equilateral triangle is \(\displaystyle \frac{\sqrt{3}}{4}\cdot\left(\frac{1}{3}\right)^2 = \frac{\sqrt{3}}{36}\). Each of the four smaller regions outside this equilateral triangle is bounded by a copy of the Koch curve that has been scaled by 1/3, so each of them has area A/9. Therefore
\[A = \frac{\sqrt{3}}{36} + 4\cdot\frac{A}{9} \Rightarrow \frac{5}{9}A = \frac{\sqrt{3}}{36} \Rightarrow A = \frac{\sqrt{3}}{36}\cdot \frac{9}{5}= \frac{\sqrt{3}}{20}.\]

More generally, if the length of the initial chord is \(L\), then the area bounded by the Koch curve is \(\displaystyle \frac{\sqrt{3}}{20}L^2\).