Question

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

• A. 1 sq unit
• B. $2\sqrt{2}$ sq unit
• C. $\sqrt{2}$ sq unit
• D. 2 sq unit

Answer 1

Answer: (D)

Key Idea: $|x|=\{\begin{array}{ccc}x& if& x\ge 0\\ -x& if& x<0\end{array}$ Given curve $|x|+|y|=1$ $\therefore$ Respective lines are $x+y=1$ ...(i) $x-y=1$ ...(ii) $-x+y=1$ ...(iii) $-x-y=1$ ...(iv) Points of intersections of Eqs. of (i), (ii), (iii) and (iv) are $(-1,0),$ (0,1), (1, 0) and $(0,-1)$ .
$\therefore$ $AC=\sqrt{0^2+{(1+1)}^2}=2$ $BD=\sqrt{{(1+1)}^2+0^2}=2$ $\therefore$ $Area=\frac{1}{2}\times 2\times 2squnit$ $=2squnit$ Note: Area can also be determined using the side length of the square, $AB=\sqrt{2}$ $\therefore$ Area $={(\sqrt{2})}^2$ sq unit = 2 sq unit.