Let a vector $\alpha \widehat i + \beta \widehat j$ be obtained by rotating the vector $\sqrt 3 \widehat i + \widehat j$ 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 :