Question 842936
Wanted is a circle enclosing {{{320*ft^2}}} of AREA.  Width of a circle would best be understood as its diameter.


{{{pi*(d/2)^2=320}}} and you can substitute 3.14 for pi whenever you are ready.
{{{(d/2)=sqrt(320/pi)}}}
{{{d=2*sqrt(320/pi)}}}
{{{d=2*sqrt(2^5*2*5/pi)}}}
{{{highlight(d=16*sqrt(5/pi))}}}------the diameter