Given M(k; 0), P(0; 2), and N(3; 4) are collinear points. Find the value of k. Then, determine the size of heta, the obtuse angle between the lines AR (y = -1/2x + 4) and MN.

Given that M, P, and N are collinear points, the gradient of MP must be equal to the gradient of PN.
The coordinates are $M(k; 0)$, $P(0; 2)$, and $N(3; 4)$.

Calculate the gradient of MP:

$$m_{MP} = \frac{y_P - y_M}{x_P - x_M} = \frac{2 - 0}{0 - k} = \frac{2}{-k}$$

Calculate the gradient of PN:

$$m_{PN} = \frac{y_N - y_P}{x_N - x_P} = \frac{4 - 2}{3 - 0} = \frac{2}{3}$$

Since $m_{MP} = m_{PN}$:

$$\frac{2}{-k} = \frac{2}{3}$$ $$-k = 3$$ $$k = -3$$

The value of $k$ is $\boxed{-3}$.

First, find the gradients of the two lines.
For line AR, the equation is $y = -\frac{1}{2}x + 4$. The gradient is $m_{AR} = -\frac{1}{2}$.
For line MN, using points $P(0; 2)$ and $N(3; 4)$, the gradient is $m_{MN} = \frac{4-2}{3-0} = \frac{2}{3}$.

Let $\alpha$ be the angle of inclination of line MN and $\beta$ be the angle of inclination of line AR.
For line MN:

$$\tan \alpha = m_{MN} = \frac{2}{3}$$ $$\alpha = \arctan\left(\frac{2}{3}\right) \approx 33.690^\circ$$

For line AR:

$$\tan \beta = m_{AR} = -\frac{1}{2}$$

Since the gradient is negative, $\beta$ is an obtuse angle.

$$\beta = 180^\circ + \arctan\left(-\frac{1}{2}\right)$$ $$\beta = 180^\circ - 26.565^\circ \approx 153.435^\circ$$

From the diagram, $\theta$ is the angle between the lines. We can find $\theta$ using the relationship between the angles of inclination. In the triangle formed by the intersection of the two lines and the x-axis, $\beta$ is an exterior angle.
Therefore, $\theta = \beta - \alpha$.

$$\theta = 153.435^\circ - 33.690^\circ$$ $$\theta = 119.745^\circ$$

The size of $\theta$ is $\boxed{119.75^\circ}$ (rounded to two decimal places).

From Step 1, $M(k; 0) = (-3; 0)$.
R is the x-intercept of line AR, $y = -\frac{1}{2}x + 4$.
To find the x-intercept, set $y=0$:

$$0 = -\frac{1}{2}x + 4$$ $$\frac{1}{2}x = 4$$ $$x = 8$$

So, $R(8; 0)$.

The length of MR is the distance between $M(-3; 0)$ and $R(8; 0)$.

$$MR = \sqrt{(x_R - x_M)^2 + (y_R - y_M)^2}$$ $$MR = \sqrt{(8 - (-3))^2 + (0 - 0)^2}$$ $$MR = \sqrt{(11)^2 + 0^2}$$ $$MR = \sqrt{121}$$ $$MR = 11$$

The length of MR is $\boxed{11 \text{ units}}$.

The vertices of the triangle are $M(-3; 0)$, $N(3; 4)$, and $R(8; 0)$.
The base of the triangle can be MR, which lies on the x-axis.
The length of the base $MR = 11$ units (from Step 3).
The height of the triangle is the perpendicular distance from point N to the x-axis, which is the absolute value of the y-coordinate of N.
Height $h = |y_N| = |4| = 4$ units.

The area of a triangle is given by the formula:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area}_{\triangle MNR} = \frac{1}{2} \times MR \times h$$ $$\text{Area}_{\triangle MNR} = \frac{1}{2} \times 11 \times 4$$ $$\text{Area}_{\triangle MNR} = 11 \times 2$$ $$\text{Area}_{\triangle MNR} = 22$$

The area of $\triangle MNR$ is $\boxed{22 \text{ square units}}$.