Question 927882
<pre>
{{{drawing(400,400,-2.3,2.3,-2.3,2.3,


line(1.73205081,0,0,1.73205081),
line(0,1.73205081,-1.73205081,0),
line(-1.73205081,0,0,-1.73205081),
line(0,-1.73205081,1.73205081,0),
locate(0,0,O), locate(-.05,1.9,B),locate(-.97,1.05,A),
line(2,0,1.0,1.73205081),
line(1.0,1.73205081,-1.0,1.73205081),
line(-1.0,1.73205081,-2,0),
line(-2,0,-1.0,-1.73205081),
line(-1.0,-1.73205081,1.0,-1.73205081),
line(1.0,-1.73205081,2,0),
circle(0,0,sqrt(3)),
green(line(0,0,1.0,1.73205081)),
locate(1,1.9,D),
green(line(0,0,0,1.73205081),line(-sqrt(3)/2,sqrt(3)/2,0,0)))
  )}}} 

The area of the square is 48 m² 
So a side of the square is &#8730;<span style="text-decoration: overline">48</span> = &#8730;<span style="text-decoration: overline">16×3</span> = 4&#8730;<span style="text-decoration: overline">3</span> 

AB is half a side of the square, so it's 2&#8730;<span style="text-decoration: overline">3</span>

OA is also 2&#8730;<span style="text-decoration: overline">3</span>

So by the Pythagorean theorem hypotenuse OB of right triangle OAB is

{{{sqrt((2sqrt(3))^2+(2sqrt(3))^2)=sqrt(4*3+4*3)=sqrt(12+12)=sqrt(24)=sqrt(4*6)=2sqrt(6)}}}

OB is the raius of the circle so by that formula for a circle's area:

{{{A=pi*r^2=pi*(2sqrt(6))=2pi*sqrt(6)}}}

Angle BDO is half of an interior angle of a regular hexagon. An
interior angle of a regular hexagon is given by:
{{{((n-2)*"180°")/n=((6-2)*"180°")/6="120°"}}}

So angle BDO is 60°, so right triangle BDO is a 30°-60°-90°,
so its hypotenuse OD is twice BD.  So by the Pythagorean theorem
{{{OB^2+BD^2=OD^2}}}
{{{24+BD^2=(2*BD)^2}}}
{{{24+BD^2=4*BD^2}}}
{{{24=3*BD^2}}}
{{{8=BD^2}}}
{{{sqrt(8)=BD}}}
{{{2sqrt(2)=BD}}}

BD is half a side of the hexagon, so one side of the
hgecagon is {{{4sqrt(2)}}}, so the perimeter of the
hexagon is 6 times that, or {{{24sqrt(2)}}}

Edwin</pre>