```Question 117289
I found only this for you:

Iteration {{{n}}} in the construction adds {{{2^n}}} squares of size {{{((1/2)sqrt(2))^2n}}} for a total area of {{{1}}}.

Thus the area of the tree might seem to grow without bound in the limit
{{{n ->infinity}}}
However, some of the squares overlap starting at the order {{{5}}} iteration, and the tree actually has a {{{finite}}}{{{ area}}} because it fits inside a {{{6*4}}} box.
It can be shown easily that the area
{{{A }}}of the {{{Pythagoras }}}{{{tree}}} must be in the range

{{{5 < A <18}}}, which can be narrowed down further with extra effort.

Little seems to be known about the actual value of {{{ A}}}.

Let {{{n=10}}}
{{{2^10(1/2*sqrt(2))^20=1}}}
{{{2^10(.5*1.4142))^20=1}}}
{{{2^10(.707))^20=1}}}

{{{1024(.707)^20=1}}}
{{{1024(.000976)=1}}}
{{{1=1}}}

Let {{{n=100}}}
{{{2^100(1/2*sqrt(2))^200=1}}}
{{{2^100(.5*1.4142))^200=1}}}
{{{2^100(.707))^200=1}}}

{{{1.2676*10^30*1.046*10^-31=1}}}
{{{1.3*10^30*1/(1.046*10^31)=1}}}

{{{1.3(1/(1.046*10))=1}}}

{{{1.3*.9 =1}}}

{{{1.17=1}}} ....{{{1.17}}} could be rounded to {{{1}}}

I hope it will help.```