Question 1037390
{{{s = r*theta}}} and s = 2r ==> {{{theta = 2}}}.

The area of a circular sector is {{{A = (r^2*theta)/2}}}.

==> {{{A = (r^2*theta)/2 = r^2}}}

Since r = s/2, it follows that {{{A = s^2/4}}}, or {{{s^2 = 4A}}}, or 

{{{s = 2sqrt(A)}}}.