Von Koch Curve
At each stage the length of the curve is increased by a factor
${4\over 3}$. So at Stage $n$ the length is $3 \times ({4\over
3})^n$. So the length of the curve keeps increasing without any
bounds. The length of the curve is infinite.
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
${1\over 9}$th of the area at that stage, and the number of
segments is multiplied by $4$ so we add on $1, 4, 16, 64, \ldots$
triangles (remember we are only considering ONE edge). So the sum
of the areas added is: $$A + \left({4\over 9}\right)A+
\left({4\over 9}\right)^2A +\left({4\over 9}\right)^3A+ \cdots$$
The sum of this infinite geometric series is $$A{1\over 1-{4\over
9}}= {9A\over 5}$$ Now $A$ is ${1\over 9}$th of the area of the
triangle at Stage $0$ so the total area added to all three edges is
$$3\times {9A\over 5}$$ which is ${3\over 5}$th of the area of the
original triangle. With side $1$ unit the area of the Stage $0$
triangle is ${\sqrt{3} \over 4}$ so the area inside the Von Koch
curve is $${2\sqrt 3\over 5}.$$ The magnification factor is $3$ and
the generator contains $4$ copies of the line segment it replaces
so the dimension is given by $$3^d = 4$$ which gives $d={\log 4
\over \log 3} \approx 1.262$.