What is the area of the square?

Let the radius of the larger circle be $R$ . Then the segments inside the smaller circle are $R$ , $R - 6$ , $R - 4$ , and $R - 4$ , as follows:

Then by the intersecting chords theorem , $R(R - 6) = (R - 4)^2$ , which solves to $R = 8$ , which means the area of the square is $(2 \cdot R)^2 = (2 \cdot 8)^2 = \boxed{256}$ .

$2R-6=2r$

$r=R-3$

$3^2$ + $(R-4)^2$ = $(R-3)^2$ $\rightarrow 1$

Since $R>4$ , Therefore solving $1$ leads to $R=8$

$s=2R$

$A= s^2$

$A= 256$