Semicircle in a square

Consider a semicircle whose diameter endpoints touch two adjacent sides of the square. It is intuitively obvious that such a semicircle of maximal area will be tangent to both of the other sides of the square.

Since the figure is symmetrical in the diagonal BD, angle QPB = 45°.

Consider the point X on AD at which the semicircle is tangent to AD. A line extended from X that is perpendicular to the tangent will be parallel to AB, and will also pass through the middle of the semicircle diameter. Let the line meet BC at Y.

OY = $\frac { r }{ \sqrt { 2 } }$

Hence 1 = AB = $r$ + $\frac { r }{ \sqrt { 2 } }$ = $r(1+\frac { 1 }{ \sqrt { 2 } } )$

Thus r = $\frac { 1 }{ 1+\frac { 1 }{ \sqrt { 2 } } }$ = $2-\sqrt { 2 }$ = $\boxed { 0.586 }$