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