Let the equilateral triangle be ABC, and the
inscribed square be DEFG:



We draw in the green median CH from vertex C to the bottom
side AB, which is also the perpendicular bisector of AB,
and also the bisector of the angle C.



We find the length of CH by the Pythagorean theorem

, 










Triangle ADG is similar to triangle AHC,

so 

, , so

 

, 

Substitute in





Multiply tops and bottoms of the fractions on both
sides by 2.

   

Cancel t's on the right sides:



Multiply both sides by 







factor out s on the left:



Divide both sides by 



Punching the left side out on a calculator



To the nearest thousandth, 



Edwin