a circle of radius 3cm crosses the centre of a square of side 2cm.The positive difference between the area of non overlapping portions is ?

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$$\begin{gathered} \mathrm{Question} \mathrm{is} \mathrm{fairly} \mathrm{simple}. \\ \mathrm{Let} \mathrm{area} \mathrm{of} \mathrm{overlapped} \mathrm{part} \mathrm{be} \mathrm{x}. \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{non}-\mathrm{overlapped} \mathrm{part} \mathrm{of} \mathrm{cirlce}={\mathrm{\pi r}}^2-\mathrm{x}=\mathrm{\pi }{\left(3\right)}^2-\mathrm{x}=\left(9\mathrm{\pi }-\mathrm{x}\right) {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{non}-\mathrm{overlapped} \mathrm{part} \mathrm{of} \mathrm{square}={\mathrm{side}}^2-\mathrm{x}=2^2-\mathrm{x}=\left(4-\mathrm{x}\right) {\mathrm{cm}}^2 \\ \Rightarrow \mathrm{Required} \mathrm{difference}=\left(9\mathrm{\pi }-\mathrm{x}\right)-\left(4-\mathrm{x}\right)=9\mathrm{\pi }-\mathrm{x}-4+\mathrm{x} \\ \mathbf{\Rightarrow }\mathbf{Required}\mathbf{ }\mathbf{difference}\mathbf{=}\left(\mathbf{9}\mathbf{\pi }\mathbf{-}\mathbf{4}\right)\mathbf{ }{\mathbf{cm}}^{\mathbf{2}} \end{gathered}$$

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