A delivery truck travels 180 km from a nearby town to Matatiele at an average speed of 90 km/h. If it departs at 05:30, determine whether it will arrive before the event starts. Show all calculations.

The delivery truck travels a distance of 180 km at an average speed of 90 km/h.
The formula for time is $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

$$\text{Time taken} = \frac{180 \text{ km}}{90 \text{ km/h}} = 2 \text{ hours}$$

The truck departs at 05:30 and travels for 2 hours.

$$\text{Arrival time} = 05:30 + 2 \text{ hours} = 07:30$$

The question asks to determine whether it will arrive before the event starts. The event start time is not provided in the given information. Assuming a common event start time for such scenarios, for example, 08:30, the truck would arrive before the event starts. However, without the explicit event start time, a definitive answer cannot be given.
If the event starts at 08:30, then 07:30 is before 08:30.

$$\boxed{\text{Arrival time: 07:30. (Cannot confirm if before event starts without event start time.)}}$$

The original table has a diameter of 1.6 m. Madlamini adds an extra 20 cm all around the edge for folding and stitching. This means 20 cm is added to the radius on each side, effectively adding 20 cm to the diameter twice.
First, convert the extra length from centimetres to metres:

$$20 \text{ cm} = \frac{20}{100} \text{ m} = 0.2 \text{ m}$$

The total increase in diameter will be $0.2 \text{ m} + 0.2 \text{ m} = 0.4 \text{ m}$.
Now, calculate the new diameter:

$$\text{New diameter} = \text{Original diameter} + \text{Total extra length}$$ $$\text{New diameter} = 1.6 \text{ m} + 0.4 \text{ m} = 2.0 \text{ m}$$ $$\boxed{\text{2.0 m}}$$

The new diameter of the tablecloth is 2.0 m.
The radius ($r$) is half of the diameter:

$$r = \frac{2.0 \text{ m}}{2} = 1.0 \text{ m}$$

The formula for the area of a circle is $\text{Area} = \pi r^2$.

$$\text{Area} = \pi (1.0 \text{ m})^2$$ $$\text{Area} = \pi \times 1.0 \text{ m}^2$$ $$\text{Area} = \pi \text{ m}^2 \approx 3.14159 \text{ m}^2$$

Rounding to two decimal places, which is common for such problems:

$$\text{Area} \approx 3.14 \text{ m}^2$$ $$\boxed{\text{3.14 m}^2}$$