Question 1116483
<pre>{{{drawing(800,800,-9,9,-9,9,circle(0,0,8),
line(0,0,8cos(pi/6),8sin(pi/6)),
line(0,0,-8cos(pi/6),8sin(pi/6)),
blue(line(-8,0,8,0)),
line(2.893780615,1.670725017,2.893780615,7.458286248),
line(-2.893780615,1.670725017,-2.893780615,7.458286248),
line(-2.893780615,7.458286248,2.893780615,7.458286248),
line(-2.893780615,1.670725017,2.893780615,1.670725017),
locate(0,0,O),locate(3,1.63,B),locate(-3.1,1.63,A),
locate(.7,.4,"30°"), locate(-1.2,.4,"30°"),locate(-.3,.5,"120°"),
locate(2.8,7.85,C), locate(-2.9,7.85,D),
red(line(0,0,2.893780615,7.458286248)),

locate(1.6,0,x*sqrt(3)),locate(3,1,x),
green(line(2.893780615,1.670725017,2.893780615,0),line(-2.893780615,1.670725017,-2.893780615,0)),locate(-2.893780615,0,P), locate(2.893780615,0,Q),
locate(3,4.5,2sqrt(3)x)



 )}}}


Let the square be ABCD.

Side of the square = AB = PQ = 2OQ.

Since triangle OBQ is a 30°,60°,90° triangle, its longer
leg is {{{sqrt(3)}}} times its shorter leg.

We let its shorter leg BQ be x, then its longer leg OQ
is {{{x*sqrt(3)}}}.

Since {{{x*sqrt(3)}}} = OQ = PQ/2 = AB/2,

BC = 2&#8729;OQ = {{{2x*sqrt(3)}}}.

OC = radius = 8

Using Pythagorean theorem on right triangle OQC

OQ² + QC² = OC²

{{{(x*sqrt(3))^2+(x+2*x*sqrt(3))^2=8^2}}}

Solve that and get 

{{{x = 4/sqrt(4 + sqrt(3))}}}

So one side of the square is twice that, or 2x, which is

{{{8/sqrt(4 + sqrt(3))}}} = side of square.

Therefore the area of the square is

{{{(8/sqrt(4 + sqrt(3)))^2}}}

or 

{{{64/(4+sqrt(3))}}}

Edwin</pre>