A football pitch measures 105 m by 68 m. A security fence is to be placed 4 m away from the edge of the pitch on all sides. Calculate the perimeter of the fence.

The pitch dimensions are $105 \text{ m}$ by $68 \text{ m}$.
The fence is placed $4 \text{ metres}$ away from the edge on all sides. This means the length and width of the fenced area will each increase by $4 \text{ m}$ on both sides (total $8 \text{ m}$).

$$\text{Fenced length} = 105 \text{ m} + (2 \times 4 \text{ m}) = 105 \text{ m} + 8 \text{ m} = 113 \text{ m}$$ $$\text{Fenced width} = 68 \text{ m} + (2 \times 4 \text{ m}) = 68 \text{ m} + 8 \text{ m} = 76 \text{ m}$$

The perimeter of a rectangle is given by $P = 2 \times (\text{length} + \text{width})$.

$$P = 2 \times (113 \text{ m} + 76 \text{ m})$$ $$P = 2 \times (189 \text{ m})$$ $$P = 378 \text{ m}$$

The total length (perimeter) of the security fencing required is $\boxed{\text{378 m}}$.

Question 2.2.5

The measured length on Annexure B is $42 \text{ mm}$.
The map scale is $1 : 1250$. This means $1 \text{ mm}$ on the map represents $1250 \text{ mm}$ in reality.

$$\text{Actual length} = \text{Map length} \times \text{Scale factor}$$ $$\text{Actual length} = 42 \text{ mm} \times 1250 = 52500 \text{ mm}$$

Convert the actual length to metres: $1 \text{ m} = 1000 \text{ mm}$.

$$\text{Actual length} = \frac{52500 \text{ mm}}{1000 \text{ mm/m}} = 52.5 \text{ m}$$

The boundary line is $0.1 \text{ m}$ wide.

$$\text{Area of boundary line} = \text{Actual length} \times \text{Width}$$ $$\text{Area of boundary line} = 52.5 \text{ m} \times 0.1 \text{ m} = 5.25 \text{ m}^2$$

The specialized safety paint covers $5 \text{ m}^2$ per litre.

$$\text{Litres of paint needed} = \frac{\text{Area of boundary line}}{\text{Paint coverage per litre}}$$ $$\text{Litres of paint needed} = \frac{5.25 \text{ m}^2}{5 \text{ m}^2/\text{litre}} = 1.05 \text{ litres}$$

The paint is sold in 5-litre tins. Since you cannot buy a fraction of a tin, you must round up to the next whole number.

$$\text{Number of tins} = \lceil \frac{\text{Litres of paint needed}}{\text{Litres per tin}} \rceil$$ $$\text{Number of tins} = \lceil \frac{1.05 \text{ litres}}{5 \text{ litres/tin}} \rceil = \lceil 0.21 \rceil = 1 \text{ tin}$$

Each tin costs $\text{R}849.50$.

$$\text{Total cost} = \text{Number of tins} \times \text{Cost per tin}$$ $$\text{Total cost} = 1 \times \text{R}849.50 = \text{R}849.50$$

The total cost of the paint tins required to paint this boundary line is $\boxed{\text{R}849.50}$.

Question 2.3.1

This question requires "Annexure C (BMI Chart for Adults)" which is not provided in the image. Therefore, I cannot identify the specific BMI value for the spectator.