P is a variable point in the square formed by the lines x = $\pm 1$ and y = $\pm 1$, P moves such that its distance from the origin is less than its distance from any side of the square. The area traced by the point P is __ sq.units

$\begin{array}{r}\sqrt{x^2+y^2}<|1-x|,|1+x|,|1-y|,|1+y|\\ \Rightarrow x^2+y^2<(1-x{)}^2,(1+x{)}^2,(1-y{)}^2,(1+y{)}^2\\ \Rightarrow x^2+y^2<x^2-2x+1,x^2+2x+1\\ y^2-2y+1,y^2+2y+1\\ \Rightarrow y^2<1-2x,y^2<1+2x,x^2<1-2y\text{ and }{x}^2<2y+1\end{array}$

A₁ = area of triangle $=\frac{(\sqrt{2}-1{)}^2}{2}$

$\text{ Area of the region OEFO = area ofDOEH }+\text{ area of HEFH }$

$\begin{array}{r}=\frac{1}{2}(\sqrt{2}-1{)}^2+{\int }_{\sqrt{2}-1}^{1/2} \sqrt{1-2x}dx=\\ \frac{1}{2}(\sqrt{2}-1{)}^2+{\left[(1-2x{)}^{3/2}\frac{2}{3}\cdot \frac{1}{2}(-1)\right]}_{\sqrt{2}-1}^{1/2}\\ =\frac{1}{2}(3-2\sqrt{2})+\frac{1}{3}(3-2\sqrt{2}{)}^{3/2}\\ =1/2(3-2\sqrt{2})+\frac{1}{3}{\left[(\sqrt{2}-1{)}^2\right]}^{3/2}\\ =\frac{1}{2}(3-2\sqrt{2})+\frac{1}{3}(\sqrt{2}-1{)}^3\\ \left.\left.=\frac{1}{2}(3-\sqrt{2})+\frac{1}{3}[2\sqrt{2}-1-3\sqrt{2})\sqrt{2}-1\right)\right]\\ =\frac{1}{2}(3-2\sqrt{2})+\frac{1}{3}[5\sqrt{2}-7]\\ =\frac{1}{6}[9-6\sqrt{2}+10\sqrt{2}-14]=\frac{1}{6}[4\sqrt{2}-5]\end{array}$

$$\begin{gathered} \text{ Similarly, area OEGO }=\frac{1}{6}(4\sqrt{2}-5) \\ \text{ Therefore, area of S lying in first quadrant } \\ =\frac{2}{6}(4\sqrt{2}-5)=\frac{1}{3}(4\sqrt{2}-5) \end{gathered}$$