Question 733933
<pre>
We can't tell whether the square is inscribed in or circumscribed
about the circle.

That is, we can't tell which of these you mean:

{{{drawing(300,300,-1.5,1.5,-1.5,1.5, circle(0,0,1), rectangle(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2))}}}   {{{drawing(300,300,-1.5,1.5,-1.5,1.5, circle(0,0,1), rectangle(-1,-1,1,1))}}}

In either case we draw a diameter of the circle (in green)

{{{drawing(300,300,-1.5,1.5,-1.5,1.5, circle(0,0,1), green(line(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2)),rectangle(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2))}}}   {{{drawing(300,300,-1.5,1.5,-1.5,1.5, green(line(-1,0,1,0)),circle(0,0,1), rectangle(-1,-1,1,1))}}}

In either case we find the diameter using the formula

C = <font face="symbol">p</font>d

20 = <font face="symbol">p</font>d

{{{20/pi}}} = d

So the diameter of the circle in either case is {{{20/pi}}} cm

{{{drawing(300,300,-1.5,1.5,-1.5,1.5, circle(0,0,1), green(line(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2)),

locate(0,0,20/pi),

rectangle(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2))}}}   {{{drawing(300,300,-1.5,1.5,-1.5,1.5, green(line(-1,0,1,0)),circle(0,0,1),
locate(0,0,20/pi),
 rectangle(-1,-1,1,1))}}}

In the first case, the green line is a diagonal of the square and
forms two isosceles right traingles.  So the two side are equal,
say x,

{{{drawing(300,300,-1.5,1.5,-1.5,1.5, circle(0,0,1), green(line(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2)),

locate(0,0,20/pi),locate(0,-.71,x), locate(.73,0,x),

rectangle(-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2,sqrt(2)/2))}}}

By the Pythagorean theorem,

x² + x² = {{{(20/pi)^2}}}

    2x² = {{{(20/pi)^2}}}

Divide both sides by 2

     x² = {{{(20/pi)^2/2}}}

Take positive square roots of poth sides

      x = {{{(20/pi)/sqrt(2)}}}

      x = {{{(20/pi)*(1/sqrt(2))}}}

      x = {{{(20/pi)*(sqrt(2)/2)}}}

      x = {{{20sqrt(2)/(2pi)}}}

Divide top and bottom by 2

      x = {{{10sqrt(2)/pi}}}

Since it has 4 sides, the perimeter

is {{{4*expr(10sqrt(2)/pi)}}} cm

or {{{40sqrt(2)/pi}}} cm

In the second case each of the four sides of the
square equals the diameter of the circle. So the 
primeter is {{{4*expr(20/pi)}}} or {{{80/pi}}} cm.  

Edwin</pre>