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(B) Let the square have vertices at $(\pm 1, \pm 1)$. The area of the square is $2 	imes 2 = 4$ square units.The intersection of the diagonals is the

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$2$

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(B) Let the square have vertices at $(\pm 1, \pm 1)$. The area of the square is $2 	imes 2 = 4$ square units.The intersection of the diagonals is the origin $O(0, 0)$.$A$ point $P(x, y)$ is nearer to $O$ than to a vertex $V(1, 1)$ if $PO $x^2 + y^2 $x^2 + y^2 $0 Similarly,considering all four vertices,the region is defined by the inequalities $|x| + |y| This region is a square with vertices at $(1, 0), (0, 1), (-1, 0), (0, -1)$.The side length of this inner square is $\sqrt{1^2 + 1^2} = \sqrt{2}$.The area of this region is $(\sqrt{2})^2 = 2$ square units.

$A$ point $P$ moves inside a square of area $4$ square units such that it is nearer to the point of intersection of its diagonals than any vertex. The area of the region traced by $P$ is

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