What will be area of $|x|+|y|\leq 1$?. It is required to obtain the area of $|x|+|y|\leq 1$. Since, the required region will just be a square with the side one. So area should be $1$. Am I right? Thanks. The answer is in the range $1.90$ to $2.10$.
 A: The area is the square with the corners $(1,0),(0,1),(-1,0),(0,-1)$. The sides has length $ \sqrt{2}$.
A: Here's the graph of $|x|+|y|\le 1$

As Gono has pointed out in the comments, it's a square of side $\sqrt 2$ units and not $1$ unit. 
Thus the area is $2$ units squared.
A: All has been said by Gimusi, Abcd, and Fred.
1) Length of sides of the quadrilateral.
a) First quadrant boundary: $y= -x +1$, $x,y$-intercepts $=1$.
Hypotenuse of the right  triangle formed by the intercepts: $\sqrt{1+1} =√2.$
Repeat the argument for the other quadrants.
Hence we have a quadrilateral with side length $√2$ (a rhombus).
2) Angle between the sides of the quadrilateral.
a) First quadrants: $y= -x +1.$
b) Second quadrant: $y=x+1.$
Note : The $2$ lines are perpendicular since $m_1m_2=-1.$
Repeat the argument for the other quadrants.
Hence : Area of interest  is bounded by  a square of side length $√2.$
A: $$\int_{x=-1}^1\int_{y=|x|-1}^{1-|x|}dy\,dx=\int_{x=-1}^12(1-|x|)\,dx=\left.\left(2x-x|x|\right)\right|_{-1}^1=2.$$
A: Considering the 4 cases, by definition of absolute value, we have that $|x|+|y|\leq 1$ corresponds to


*

*first quadrant: $x+y\le 1 \implies y\le -x+1$

*second quadrant: $-x+y\le 1 \implies y\le x+1$

*third quadrant: $-x-y\le 1 \implies y\ge -x-1$

*fourth quadrant: $x-y\le 1 \implies y\ge x-1$

