Thanks!
Hello, Jammix!
Is there any more information?
I suspect that I know the solution . . . but it's only a guess.
Circumscribe a square about a semicircle.
Square $\displaystyle ABCD$ is circumscribed about the (tilted) semicircle.Code:R B * - -.-.*.o.*.-.- * C | .*:::::|:::::*.| | *:::::::|:::::::o S |*::::::::|:::::* | |:::::::::|:::* r | *:::::::::|:* * Q o:-:-:-:-:O - - - o *:::::::* : * |:::::*r : | | ::* : | A * o - - - * - - - * D P E
The circle has center $\displaystyle O$ and radius $\displaystyle r\!:\;OP = OQ = OR = OS = r.$
$\displaystyle \Delta OEA$ is an isosceles right triangle.
Hypotenuse $\displaystyle PO = r \quad\Rightarrow\quad OE = \tfrac{\sqrt{2}}{2}r$
Let $\displaystyle x = AB$, the side of the square.
Then: .$\displaystyle x \:=\:RO + OE \:=\:r + \tfrac{\sqrt{2}}{2}r $
Therefore: .$\displaystyle x \:=\:\left(\tfrac{2 + \sqrt{2}}{2}\right)r$