The figure above shows an infinite series of triangles ,in which $r_1 >r_2>r_3>..............$ I am stuck on the following problem that says: 


The figure above shows an infinite series of triangles ,in which $r_1 >r_2>r_3>..............$ What is the total length 
  of the solid line segments in the figure? The options are: 
  
  
*
  
*$\frac{r_1}{r_2}+\frac{r_2}{r_3}+.......$ 
  
*$\frac{r_1^2}{r_1-r_2}$  
  
*$\frac{r_2^2}{r_1+r_2}$  
  
*$\frac{r_1-r_2}{r_1^2}$ 

I have no clue how to tackle it. Can someone explain in details ? Thanks in advance for your time.

Let $T$ be the clockwise rotation of angle $\theta$ followed by a dilation of ratio $k=r_2/r_1$. The second triangle $\triangle_{2}$ is the image under $T$ of the first triangle $\triangle_{1}$. More generally,
the $n+1^\mathrm{st}$ triangle $\triangle_{n+1}$is the image of $\triangle_{n}$ under $T$.
So the $n^\mathrm{th}$ segmant is the image under $T^{n-1}$ of the first one. Consequently,
$
r_n=k^{n-1}r_1
$. It follows that
$$\sum_{n=1}^\infty r_n=r_1\sum_{n=1}^\infty k^{n-1}=\frac{r_1}{1-k}=\frac{r_1}{1-r_2/r_1}
=\frac{r_1^2}{r_1-r_2}.$$
