$$ABCD$$ and $$CEFG$$ are two squares such that the extension of $$GE$$, a diagonal of $$CEFG$$, passes through $$B$$. Given $$BE=6\text{ cm}$$ and $$CG=4\sqrt{2}\text{ cm}$$, then the area of square $$ABCD$$ in $$\text{cm}^2$$ is

The side of square $$CEFG$$ is $$CG=4\sqrt{2}$$, so its diagonal $$GE$$ is $$\sqrt{(4\sqrt{2})^2+(4\sqrt{2})^2 }= 8$$cm

In a square, the diagonals bisect each other perpendicularly. 

Thus, $$ CH = EH = \dfrac{8}{2} = 4$$cm

In $$\triangle BHC$$ we can apply Pythagoras theorem to find the length of BC

$$BH = 6+4 = 10, CH = 4 \implies BC = \sqrt{10^2 + 4^2} = \sqrt{116}$$cm

Thus, Area of the square is

$$BC^2 = 116\text{cm}^2$$