A rectangle is constructed in the complex pla | vedclass

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(D) The vertex of the rectangle in the first quadrant is given by the complex number $z = a + ib\sqrt{3}$,which corresponds to the point $(a, b\sqrt{3

Vedclass · Accepted Answer

$4ab\sqrt{3}$

Vedclass · Answer

(D) The vertex of the rectangle in the first quadrant is given by the complex number $z = a + ib\sqrt{3}$,which corresponds to the point $(a, b\sqrt{3})$ in the Cartesian plane.Since the rectangle is centered at the origin $(0, 0)$ and its sides are parallel to the axes,the vertices of the rectangle are $(a, b\sqrt{3})$,$(-a, b\sqrt{3})$,$(-a, -b\sqrt{3})$,and $(a, -b\sqrt{3})$.The length of the rectangle along the $X$-axis is $2|a| = 2a$ (assuming $a, b > 0$).The height of the rectangle along the $Y$-axis is $2|b\sqrt{3}| = 2b\sqrt{3}$.Therefore,the area of the rectangle is $	ext{Length} 	imes 	ext{Height} = (2a) 	imes (2b\sqrt{3}) = 4ab\sqrt{3}$.

$A$ rectangle is constructed in the complex plane with its sides parallel to the axes and its centre situated at the origin. If one of the vertices of the rectangle is $a + ib\sqrt{3}$,then the area of the rectangle is

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