Explanation

(c)
$Given,\ \ \ y = |x - 1| = \left\{ \begin{matrix}
{x - 1,x > 1} \\
{- x + 1,x \leq 1}
\end{matrix} \right.\$
$and\ \ \ y = 3 - |x| = \left\{ \begin{array}{r}
3 + x,x \leq 0 \\
3 - x,x > 0
\end{array} \right.\$
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On solving $y = x - 1$ and $y = 3 - x,\$we get $x = 2,y = 1$
Now, ${AB}^{2} = {(2 - 1)}^{2} + {(1 - 0)}^{2} = 2$
$\Longrightarrow \ \ AB = \sqrt{2}$
$and\ {BC}^{2} = (0 - 2)^{2} + (3 - 1)^{2} = 8$
$\Longrightarrow \ \ BC = 2\sqrt{2}$
$\therefore\ \$Area of rectangle $ABCD = AB \times BC$
$= \sqrt{2} \times 2\sqrt{2}$=4 sq units