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Area bounded by lines $y=2+x, y=2-x$ and $x=2$ is
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$4$
Obviously, triangle $A C B$ is right angled at $C$.
$\therefore$ Required area $=\frac{1}{2} \times A C \times B C$
$=\frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2}=4$sq. unit.
$\therefore$ Required area $=\frac{1}{2} \times A C \times B C$
$=\frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2}=4$sq. unit.
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