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.You need to login to perform this action.
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