Too Little Info. Right?

Let the radius of the big circle be $r$ . By Power of a Point, we have $r(r-6)=(r-4)^2$ Solving this equation gives $r=8$ . Thus, the area of the square is $(2r)^2=\boxed{256}$ .

6 is 3/4 of the half, 4 is 1/2, so 1/4 of half is 2. 6+2=8(2)= 16. area of square is equal to side times side so (16)(16)=256

diameter = 6 + 6 + 4 = 16 = 2r = side of the square

Notice how an inscribed triangle can be drawn in the smaller circle by connecting the three bold dots. From our knowledge of circles, we can see that this is a right triangle. Let the radius of the larger circle be $r$ . From the Altitude theorem, we can construct:

$\sqrt{(r-6)(r)}=r-4$

Squaring and ignoring the extraneous solution produces

$r^2-6r=r^2-8r+16$

$2r=16$

$r=8$

We see that the side length of the square is $2r$ , or $16$ , and thus the area is $16^2=\boxed{256}$ . Great problem @Mardokay Mosazghi ! :D