Question 852725
If {{{x}}}= side of the square
{{{x*sqrt(2)}}}= diagonal of the square, and
{{{x+x*sqrt(2)=(1+sqrt(2))*x=1}}}= perimeter of triangle.
So {{{x=1/(1+sqrt(2))=(1-sqrt(2))/(1-2)=sqrt(2)-1}}} , and
{{{highlight(4(sqrt(2)-1)=4sqrt(2)-4)}}}= perimeter of square.