Area fractal pentagrams I When I saw this image I was a little curious.
How can I find the area of this fractal?

 A: First,
$ Area\ of\ Star\ with\ 5\ petals\ = Area\ of\ pentagram\ ->\  step\ 1 $  
please refer http://mathworld.wolfram.com/Pentagram.html for area formula of pentagram. because the tricky part is identifying the GP.
Idea is that from each of the 5 triangles (assuming the new triangle has side one third the length of base triangle on which it forms ,as fractals are regular figures)  there are 2 triangles coming up .
after iteration 1 added area =area of 10 equilaterla triangles of side $ \frac{a}3 $
after iteration 2 added area =area of 20 equilaterla triangles of side $ \frac{a}{3^2} $
after iteration 3 added area =area of 40 equilaterla triangles of side $ \frac{a}{3^3} $
so as this is infinite as per your problem this goes on and on 
Area = Area of STAR (as found in step 1) + area added after infinite iteration (let this be k)
$k\ =  10\times(\sqrt(3)/4)(\frac{a}{3})^2\ +\ 20\times(\sqrt(3)/4)(\frac{a}{3^2})^2\ +\ 40\times(\sqrt(3)/4)(\frac{a}{3^3})^2\ ...$
$k\ = 10\times(\sqrt(3)/4)a^2 [ \frac{1}{3^2} + \frac{2}{3^4} + \frac{2^2}{3^6}  +\ ..... ]$
this within square brackets is a infinite GP with common ratio of $\frac{2}{3^2}\ $and first term is  $\frac{1}{3^2} $
as summation of GP with infinite series is $\frac{firstterm}{1-commonratio} $
$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{\frac{1}{3^2}}{1-\frac{2}{3^2}}]$
$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$
so 
$Area = Area\ of\ PENTAGRAM\ +  10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$
A: Each segment of the pentagram is the initiator of the fractal. Take its length to be 1. Now the generator consists of 2 line segments each of length $\frac{1}{3}$.
Hence on each iteration $n$ the area can be expressed as follows:
$$A_n=10\sum_{k=0}^{n}2^kS_k +S_{p}$$
Where $S_p$ is the area of the regular pentagon and:
$$S_k=\frac{1}{2}\frac{1}{3^{2k}}\sin\frac{\pi}{5}$$
is the area of the "k-th generation" petal.
$$10\sum_{k=0}^{n}2^kS_k=5\pi\sin\frac{\pi}{5}\sum_{k=0}^{\infty}\left(\frac{2}{9}\right)^k=\frac{45\pi}{7}\sin\frac{\pi}{5}$$
The area of the pentagon is:
$$S_p=\frac{t^2\sqrt{25+10\sqrt{5}}}{4}$$
where $t=2\sin\frac{\pi}{10}$
Finally,
$$A=\frac{45\pi}{7}\sin\frac{\pi}{5}+\sqrt{25+10\sqrt{5}}\left(\sin\frac{\pi}{10}\right)^2$$
Or equivalently
$$A=\frac{45\pi\sqrt{2(5-\sqrt{5})}}{28}+\frac{\sqrt{25+10\sqrt{5}}}{4\phi^2}$$
or any other way you wish to think of it.
