A) 2 sq units
B) 3 sq units
C) 4 sq units
D) 6 sq units
Correct Answer: C
Solution :
\[y=\left| x-1 \right|=\left\{ \begin{matrix} x-1, & x>0 \\ -x+1, & x0 \\ \end{matrix} \right.\] Solving \[y=x-1\] and \[y=3-x\] \[\Rightarrow \] \[x-1=3-x\] \[\Rightarrow \] \[x=2\] and \[y=3-2=1\] Now, \[A{{B}^{2}}={{(2-1)}^{2}}+(1-{{0}^{2}})=1+1=2\] \[\Rightarrow \] \[AB=\sqrt{2}\] and \[B{{C}^{2}}={{(0-2)}^{2}}+{{(3-1)}^{2}}=4+4=8\] \[BC=2\sqrt{2}\] Area of rectangle ABCD = AB \[\times \] BC \[\Rightarrow \] \[=\sqrt{2}\times 2\,\sqrt{2}=4\] sq unitsYou need to login to perform this action.
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