The region represented by |x - y| leq 2 and | | vedclass

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(D) The given inequalities are $|x - y| \leq 2$ and $|x + y| \leq 2$.These can be written as $-2 \leq x - y \leq 2$ and $-2 \leq x + y \leq 2$.This re

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square of side length $2\sqrt{2}$ units

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(D) The given inequalities are $|x - y| \leq 2$ and $|x + y| \leq 2$.These can be written as $-2 \leq x - y \leq 2$ and $-2 \leq x + y \leq 2$.This represents the region bounded by the lines $x - y = 2$,$x - y = -2$,$x + y = 2$,and $x + y = -2$.The vertices of this region are the intersection points of these lines:$1$) $x - y = 2$ and $x + y = 2$ gives $(2, 0)$.$2$) $x + y = 2$ and $x - y = -2$ gives $(0, 2)$.$3$) $x - y = -2$ and $x + y = -2$ gives $(-2, 0)$.$4$) $x + y = -2$ and $x - y = 2$ gives $(0, -2)$.The distance between consecutive vertices (e.g.,$(2, 0)$ and $(0, 2)$) is $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.Since all sides are equal to $2\sqrt{2}$ and the diagonals are perpendicular (slopes are $1$ and $-1$),the region is a square.The area of the square is $(	ext{side})^2 = (2\sqrt{2})^2 = 8$ sq. units.Thus,the region is a square of side length $2\sqrt{2}$ units.

The region represented by $|x - y| \leq 2$ and $|x + y| \leq 2$ is bounded by a

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