The area enclosed within the curve |x| + |y| | vedclass

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(C) The equation $|x| + |y| = 1$ represents a square with vertices at $(1, 0)$,$(0, 1)$,$(-1, 0)$,and $(0, -1)$.The distance between consecutive verti

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The equation $|x| + |y| = 1$ represents a square with vertices at $(1, 0)$,$(0, 1)$,$(-1, 0)$,and $(0, -1)$.The distance between consecutive vertices (e.g.,$(1, 0)$ and $(0, 1)$) is the side length $s = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.The area of a square is given by $s^2$.Therefore,the area $= (\sqrt{2})^2 = 2$ square units.Alternatively,the area of the square formed by $|x| + |y| = a$ is $2a^2$. Here $a = 1$,so the area is $2(1)^2 = 2$.

The area enclosed within the curve $|x| + |y| = 1$ is

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