What's on the side?

$1]$ First lay down the inner segments as shown

$2]$ Now, extend the smaller segment of length $3$ till the end as shown

$3]$ Now, this forms a Right Angled Triangle with side lengths equal to $3$ and $12 + 9 = 21$ .

$4]$ Now, by Pythagoras Theorem calculate the Hypotenuse $(H)$

$H^{2} = 3^{2} + 21^{2}$ i.e, $H = 15\sqrt{2}$

$5]$ Thus rearranging , back we get,

Thus, we get the Hypotenuse of the Square equals $15\sqrt{2}$ .Hence, we get,

$\sqrt{x^{2} + x^{2}} = \sqrt{ (15\sqrt{2})^{2} }$

Therefore, $x^{2} = 15^{2}$

$ANSWER : x = \boxed{15}$

Since the extended segments from the vertices form two right angles (thus their sum is $180^{\circ}$ ) as given, we can extend the segment of length $9$ to the segment of length $12$ , which creates the hidden square in the center. In this case, the largest square is the combination of four right triangles and a mini square.

Thus, the answer to this problem follows the special Ppythagorean primitive triple, which gives $\boxed{x=15}$