question_answer

    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

Correct Answer: D

Solution :

                Key Idea: \[|x|=\left\{ \begin{matrix}    x & if & x\ge 0  \\    -x & if & x<0  \\ \end{matrix} \right.\] 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\[(0,-1)\]. \[\therefore \] \[AC=\sqrt{{{0}^{2}}+{{(1+1)}^{2}}}=2\]                 \[BD=\sqrt{{{(1+1)}^{2}}+{{0}^{2}}}=2\] \[\therefore \]  \[Area=\frac{1}{2}\times 2\times 2\,sq\,unit\] \[=2\,sq\,unit\] 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.