Question 1045516
we know AB=CD
AC=BD={{{sqrt(5)}}}
perimeter={{{2AC+2AB}}}
let {{{AB}}}={{{x}}} then {{{AD}}}={{{2x}}}
you have right angle triangle
{{{sqrt(5)^2}}}+{{{x^2}}}={{{(2x)^2}}}
{{{x}}}={{{sqrt(5/3)}}}

{{{p}}}={{{2*sqrt(5)+2*(sqrt(5/3))}}}