The area bounded by circles $x^2+y^2=r^2, r=1,$1 and rays given by $2x^2-3xy-2y^2=0, \left(y>0\right)$ is 

$$\begin{gathered} Given pair of lines \\ 2{\mathrm{x}}^2-3\mathrm{xy}-2{\mathrm{y}}^2=0 \\ \Rightarrow 2x^2-4xy+xy-2y^2=0 \\ \Rightarrow (2\mathrm{x}+\mathrm{y}) (\mathrm{x}-2\mathrm{y})=0 \\ \Rightarrow y=-2x,y=\frac{x}{2} \\ Here slopes of two lines are -2 and \frac{1}{2} and their product is 1 \\ Hence two lines are perpendicular to each other \\ So,required \mathrm{Area}=\frac{\pi R^2}{4}-\frac{\pi r^2}{4} \\ \Rightarrow \mathrm{A}=\frac{\mathrm{\pi }{\left(2\right)}^2}{4}-\frac{\mathrm{\pi }{\left(1\right)}^2}{4} \\ \Rightarrow \overline{)\mathrm{A}=\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \end{gathered}$$