How are the values $3\sqrt{2}$ and $\sqrt{2}$ determined? Those values $\sqrt{2}$ and $3\sqrt{2}$

How do they suppose to match with $MB=BN=2$?


Draw the bottom square $ABCD$!

I've added the point $T$: the centre of the square $ABCD$.
The computation is essentially not different from the one given by mathlove in the earlier answer: the square $MBNT$ has sides of length $2$, its diagonal $TB$ has length $2 \sqrt{2}$, and $S$ is in the middle, so $SB = \sqrt{2}$.
And, of course, analoguously, $ABCD$ has sides of length $4$, its diagonal $DB$ has length $4 \sqrt{2}$, so $DS = 3\sqrt{2}$.
