The area inside the curve is given by the area of the Stage 0 triangle plus the sum of all the areas of the smaller triangles that we would add on at all the stages. Let us just consider the triangles added to ONE edge and, for simplicity, let's take the area of the triangles added at Stage 1 to be A. (We'll compute A later.) At each stage we add on triangles whose area is $\frac{1}{9}$th of the area at that stage, and the number of segments is multiplied by 4 so we add on 1, 4, 16, 64 .. . triangles (remember we are only considering ONE edge). So the sum of the areas added is:

${\displaystyle A+(\frac{4}{9})A+(\frac{4}{9}{)}^{2}A+(\frac{4}{9}{)}^{3}A+\dots }$

The sum of this infinite geometric series is

${\displaystyle A\frac{1}{1-\frac{4}{9}}=\frac{9A}{5}.}$

Now $A$ is $\frac{1}{9}$th of the area of the triangle at Stage 0 so the total area added to all three edges is

${\displaystyle 3\times \frac{9A}{5}}$

which is $\frac{3}{5}$th of the area of the original triangle. With side 1 unit the area of the Stage 0 triangle is $\frac{\sqrt{3}}{4}$ so the area inside the Von Koch curve is

${\displaystyle \frac{2\sqrt{3}}{5}.}$

The magnification factor is 3 and the generator contains 4 copies of the line segment it replaces so the dimension is given by

${\displaystyle 3^d=4}$

which gives $d=\frac{\log 4}{\log 3}\approx 1.262$.