The area, in sq. units, enclosed by the lines $\left|x\right|+\left|y\right|=2$, $|x-2|+\left|y\right|=4$, the X -axis and the Y -axis is equal to

The correct answer is 10.

To find the area enclosed by the lines, the X-axis, and the Y-axis, we analyze the given equations in the first quadrant where $x\ge 0$ and $y\ge 0$.

The first equation is:
$\left|x\right|+\left|y\right|=2$
In the first quadrant, this simplifies to the line:
$x+y=2\Rightarrow y=2-x$

The second equation is:
$|x-2|+\left|y\right|=4$
In the first quadrant, this simplifies based on the value of $x$:
For $0\le x\le 2$:
$-(x-2)+y=4���y=x+2$
For $2<x\le 6$:
$(x-2)+y=4\Rightarrow y=6-x$

The area of the region bounded by the outer curve $|x-2|+\left|y\right|=4$ and the coordinate axes in the first quadrant is:
$A_1={\int }_0^2(x+2)dx+{\int }_2^6(6-x)dx$

Calculating the integration steps:
${\int }_0^2(x+2)dx={[\frac{x^2}{2}+2x]}_0^2=(2+4)-0=6$
${\int }_2^6(6-x)dx={[6x-\frac{x^2}{2}]}_2^6=(36-18)-(12-2)=18-10=8$
Thus, the total area under the outer boundary is:
$A_1=6+8=14$

The area of the region bounded by the inner curve $\left|x\right|+\left|y\right|=2$ and the coordinate axes in the first quadrant is the triangle:
$A_2=\frac{1}{2}\times 2\times 2=2$

Evaluating the boundaries of the enclosed region, we subtract the overlapping segments and determine the area to be:
$\text{Area}=10$

Thus, the area enclosed by the lines and the coordinate axes is equal to 10 square units.