Question 1209520
<pre>

{{{drawing(400,400,-.5,2.5,-2.5,.5,

line(0,0,2,0),line(0,0,0,-2),line(0,-2,2,-2),line(2,-2,2,0),

line(0,0,1,-sqrt(3)),line(2,0,1,-sqrt(3)), line(0,0,2,-2),

locate(0,-2,R), locate(2,-2,U), locate(.95,-1.75,E),
locate(-.01,.15,S), locate(1.25,-1.3,A), locate(1.98,.15,Q),
locate(1.7,0,60^o), locate(1.8,-1.6,45^o), locate(1.81,-.27,30^o),
locate(2.05,-1,x),locate(1.34,-1.15,105^o)
 )}}}

I think it's pretty obvious how I got the angles I have marked the
degree measures of above.

We can use the ASA area formula, 

{{{Area = (S^2*sin(ANGLE[1])sin(ANGLE[2]))/(2*sin(ANGLE[3])))}}}

although we can't get the exact answer that way, since the third angle is
{{{180^o-30^o-45^o=105^o}}}, which is not a special angle.
[Of course, we could break up 105<sup>o</sup> as 45<sup>o</sup>+60<sup>o</sup>.] 

{{{Area =(x^2*sin(30^o)sin(45^o))/(2*sin(105^o))}}}{{{""=""}}}{{{0.1830127019x^2}}}

That's the approximate answer of the exact solutions Ikleyn and greenestamps
got.

Edwin</pre>