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