The area (in sq. units) of the region consist | vedclass

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(B) The given inequalities are $|x| \le 1 + |y|$ and $|y| \le 1$.Since the region is symmetric about both the $X$-axis and $Y$-axis,we can calculate t

Vedclass · Accepted Answer

$6$

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(B) The given inequalities are $|x| \le 1 + |y|$ and $|y| \le 1$.Since the region is symmetric about both the $X$-axis and $Y$-axis,we can calculate the area in the first quadrant and multiply it by $4$.In the first quadrant,$x \ge 0$ and $y \ge 0$,so the inequalities become $x \le 1 + y$ and $y \le 1$,which means $x - y \le 1$ and $y \le 1$.The region in the first quadrant is a trapezoid bounded by $x=0, y=0, y=1$,and $x=1+y$.At $y=0, x=1$. At $y=1, x=2$.The area of this trapezoid is $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height} = \frac{1}{2} 	imes (1 + 2) 	imes 1 = \frac{3}{2}$.Total area $= 4 	imes \frac{3}{2} = 6$ sq. units.

The area (in sq. units) of the region consisting of points $(x,y)$ on the $X-Y$ plane which satisfy $|x| \le 1 + |y|$ and $|y| \le 1$ is:

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