The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and the x-axis is:

The correct option is 16 sq units.

To find the area of the region bounded by the given lines and the x-axis, we first express the equation of the boundary line in terms of $y$ as a function of $x$.

The equation of the line is given by:
$x+2y=12$
Subtracting $x$ from both sides, we get:
$2y=12-x$
Dividing by 2, we obtain:
$y=\frac{12-x}{2}$

The region is bounded by the vertical lines $x=2$ and $x=6$, and the x-axis (where $y=0$). The area $A$ under the curve $y=f\left(x\right)$ from $x=a$ to $x=b$ is given by the definite integral:

$A={\int }_a^{b}ydx$

Substituting the given boundaries $a=2$, $b=6$, and the function for $y$, we have:

$A={\int }_2^6\frac{12-x}{2}dx$

We can factor out the constant $\frac{1}{2}$ and integrate term by term:

$A=\frac{1}{2}{[}_{12}^x$

Now, we evaluate the antiderivative at the upper limit ($x=6$) and the lower limit ($x=2$):

At $x=6$:
$12\left(6\right)-\frac{6^2}{2}=72-\frac{36}{2}=72-18=54$

At $x=2$:
$12\left(2\right)-\frac{2^2}{2}=24-\frac{4}{2}=24-2=22$

Subtracting the value at the lower limit from the value at the upper limit:

$A=\frac{1}{2}(54-22)$
$A=\frac{1}{2}\left(32\right)=16$

Therefore, the area of the bounded region is 16 sq units.