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Let a vector $\alpha \hat{i} + \beta \hat{j}$ be obtained by rotating the vector $\sqrt{3}\hat{i} + \hat{j}$ by an angle $45°$ 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
(1) $\frac{1}{2}$
(2) $1$
(3) $\frac{1}{\sqrt{2}}$
(4) $2\sqrt{2}$
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