Given that, $|x| – y \leq 1 ; y \geq 0$ and $y \leq 1.$
We know that, $|x| = \left\{\begin{matrix} x ; & x \geq 0 & \\ -x ; & x < 0 & \end{matrix}\right.$
Let us assume $|x| – y = 1 ; y = 0 ; y = 1$
$ \Rightarrow y = |x| – 1 \quad \longrightarrow (1)$
We know that, $y = |x|$ graph.
Then, $y = |x| – 1$ graph will be
So, final graph for equation $(1).$
We know that, $\text{area of a trapezium} = \frac{1}{2} \text{(sum of parallel sides)} \times \text{Distance between them}$
So, the required area $ = \frac{1}{2} (4+2) \times 1 = \frac{6}{2} = 3 \; \text{unit square.}$
$$\text{(OR)}$$
$\text{The area of ABCD = Area of PQCD – Area of PAD – Area of QBC}$
$\qquad = (4 \times 1) – \frac{1}{2} \times 1 \times 1 – \frac{1}{2} \times 1 \times 1$
$\qquad = 4 – \frac{1}{2} – \frac{1}{2} $
$\qquad = 4 – \left( \frac{2}{2} \right)$
$\qquad = 4 – 1 $
$\qquad = 3 \; \text{unit square}.$
Correct Answer$: 3$