Question

The area enclosed by the curve $|x+y|+|x-y|-11=0$ is

• A. $121$
• B. $144$
• C. $100$
• D. $169$

Answer 1

The correct option is A $121$

Here, $|x+y|+|x-y|=11$

When $(x+y)>0,(x-y)>0$

$(x+y)+(x-y)=11$

$x=\frac{11}{2}$

$(x+y)>0,(x-y)<0$

$(x+y)+(y-x)=11$

$y=\frac{11}{2}$

$(x+y)<0,(x-y)>0$

$-(x+y)+(x-y)=11$

$y=\frac{-11}{2}$

$(x+y)+(y-x)=11$

$x=\frac{-11}{2}$

The enclosed surface is a square of side $11$ unit.

Thus, area enclosed $=121sq$. units.

$\therefore \left(A\right)$