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

Identifying the region by finding the intersection points:

• The $x$-axis corresponds to $y=0$
  • When $y=0$ in $x+2y=12$, we get $x=12$
  • Our region is bounded by $x=2$ and $x=6$, which are vertical lines

---

Expressing $y$ in terms of $x$ for the sloped boundary:

From $x+2y=12$, I can solve for $y$:

$2y = 12-x$

$y = \dfrac{12-x}{2} = 6-\dfrac{x}{2}$

---

Finding intersection points to define the region completely:

At $x=2$: $y = 6-\dfrac{2}{2} = 5$

At $x=6$: $y = 6-\dfrac{6}{2} = 3$

---

Calculating the area using integration:

$\text{Area} = \int_{2}^{6} (6-\dfrac{x}{2}) \, dx$

$\text{Area} = [6x - \dfrac{x^2}{4}]_{2}^{6}$

$\text{Area} = \left(6(6) - \dfrac{6^2}{4}\right) - \left(6(2) - \dfrac{2^2}{4}\right)$

$\text{Area} = \left(36 - \dfrac{36}{4}\right) - \left(12 - \dfrac{4}{4}\right)$

$\text{Area} = \left(36 - 9\right) - \left(12 - 1\right)$

$\text{Area} = 27 - 11 = 16$

---

Alternatively, using the trapezoid formula:

$\text{Area} = \dfrac{1}{2} \times \text{width} \times (\text{height}_1 + \text{height}_2) = \dfrac{1}{2} \times 4 \times (5 + 3) = 16$

Therefore, the area of the region is 16 square units.