Let a vector $\mathrm{\alpha }\hat{\mathrm{i}}+\mathrm{\beta }\hat{\mathrm{j}}$ obtained by rotating the vector $\sqrt{3}\hat{\mathrm{i}}+\hat{\mathrm{j}}$ by an angle ${45}^0$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $\left(\mathrm{\alpha },\mathrm{\beta }\right),\left(0,\mathrm{\beta }\right)$ and (0, 0) is equal to

$$\begin{gathered} \left|\mathrm{\alpha i}+\mathrm{\beta j}\right|=\left|\sqrt{3}\mathrm{i}+\mathrm{j}\right| \\ {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=3+1=4 \\ \cos {45}^0=\frac{\sqrt{3}\mathrm{\alpha }+\mathrm{\beta }}{4} \Rightarrow \sqrt{3}\mathrm{\alpha }+\mathrm{\beta }=2\sqrt{2} \\ \Rightarrow \mathrm{\beta }=2\sqrt{2}-\sqrt{3}\mathrm{\alpha } \\ {\mathrm{\alpha }}^2+8+3{\mathrm{\alpha }}^2-4\sqrt{6}\mathrm{\alpha }=4 \\ {\mathrm{\alpha }}^2-\sqrt{6}\mathrm{\alpha }+1=0 \Rightarrow \mathrm{\alpha }=\frac{\sqrt{6}\pm \sqrt{6-4}}{2} \\ =\frac{\sqrt{6}\pm \sqrt{2}}{2} =\frac{\sqrt{3}\pm 1}{\sqrt{2}} \\ \mathrm{\alpha }=\frac{\sqrt{3}+1}{\sqrt{2}} \mathrm{\beta }=2\sqrt{2}-\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right) \\ =\frac{4-3-\sqrt{3}}{\sqrt{2}}=\frac{1-\sqrt{3}}{\sqrt{2}} \\ \mathrm{area} \mathrm{of} {∆}^{\mathrm{le}}=\frac{1}{2}\left|\mathrm{\alpha \beta }-0\right|=\left|\frac{\mathrm{\alpha \beta }}{2}\right| \\ =\left|\frac{3-1}{2}\times \frac{1}{2}\right| =\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$