Question

$\begin{array}{c}\text{The area enclosed by the curve ,}\\ |x+y|+|x-y|-11=0\text{is}\end{array}$:

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

Answer 1

The correct option is A $121$

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

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

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

$2x=11$

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

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

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

$2y=11$

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

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

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

$-2y=11$

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

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

$-2x=11$

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

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

$\therefore$ Area enclosed $=121$ $uni{t}^2$