$$ABCD$$ 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 $$8$$. Since $$B,E,G$$ are collinear, $$BG=BE+EG=6+8=14$$. A diagonal of a square bisects a right angle, so $$\angle BGC=45^\circ$$. By the cosine rule, $$BC^2=14^2+(4\sqrt{2})^2-2 \times 14 \times 4\sqrt{2}\cos45^\circ=116$$, which is the area of square $$ABCD$$.