The area enclosed by the graphs of |x + y| = | vedclass

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(D) The given equations are $|x + y| = 2$ and $|x| = 1$.From $|x + y| = 2$,we have $x + y = 2$ or $x + y = -2$.From $|x| = 1$,we have $x = 1$ or $x =

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$8$

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(D) The given equations are $|x + y| = 2$ and $|x| = 1$.From $|x + y| = 2$,we have $x + y = 2$ or $x + y = -2$.From $|x| = 1$,we have $x = 1$ or $x = -1$.Substituting $x = 1$ into $x + y = 2$ gives $y = 1$. Point: $(1, 1)$.Substituting $x = 1$ into $x + y = -2$ gives $y = -3$. Point: $(1, -3)$.Substituting $x = -1$ into $x + y = 2$ gives $y = 3$. Point: $(-1, 3)$.Substituting $x = -1$ into $x + y = -2$ gives $y = -1$. Point: $(-1, -1)$.The vertices of the enclosed region are $(1, 1), (1, -3), (-1, -1),$ and $(-1, 3)$.This region is a parallelogram.The base length (vertical distance between $(1, 1)$ and $(1, -3)$) is $|1 - (-3)| = 4$ units.The height (horizontal distance between $x = 1$ and $x = -1$) is $|1 - (-1)| = 2$ units.Area of the parallelogram = $	ext{base} 	imes 	ext{height} = 4 	imes 2 = 8$ square units.Hence,the correct answer is $(D) 8$.

The area enclosed by the graphs of $|x + y| = 2$ and $|x| = 1$ is

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