Algebra in Geometry!

Let the side of the square is $S$ .

First, note that $\triangle ADK$ is congruent with triangle of $\triangle KDN$ (side, angle, angle).

So, $AK = KN = 6 \space \text{cm}$ , then $KB = (S-6) \space \text{cm}$ .

Second, note that $\triangle NDL$ is congruent with $\triangle CDL$ (side, side, angle).

So, $LC=LN=4 \space \text{cm}$ , then $BL=(S-4) \space \text{cm}$ .

Third, we move to triangle of $KBL$ . Put it into Pythagoras Formula.

$\begin{aligned} KB^2 + BL^2 & = KL^2 \\ (S-6)^2 + (S-4)^2 & = 10^2 \\ S^2 - 12S + 36 + S^2 - 8S + 16 & = 100\\ 2S^2 - 20S - 48 & = 0 \\S^2 - 10S - 24 & =0 \\ (S-12)(S+2) & =0 \end{aligned}$

Then, we have $S=12 \space \text{or}\space-2$

The side of a square is always positive, so $S=12cm$