Q. The area enclosed within the curve $ |x|+|y|=1 $ is
JamiaJamia 2006
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.
