Find the radius r

Consider my diagram, $d= r\sqrt{2}$ . Then by pythagorean theorem, we have

$(r\sqrt{2}+r+5)^2=5^2+5^2$

$r\sqrt{2}+r+5=\sqrt{50}$

$r(\sqrt{2}+1)=\sqrt{50}-5$

$r=\dfrac{\sqrt{50}-5}{\sqrt{2}+1} \approx 0.85786$

The diagonal of the square of side 5 cm = 5 $\sqrt2$ cm

And as diagonal lengths are equal, we can form the following linear equation to solve for r:

r ( $\sqrt2$ + 1) + 5 = 5 $\sqrt2$

r = 0.857 cm