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The diameter of the first and second sections: $D_1 = 500 \text{ mm} = 0.5 \text{ m}$. So, the radius $r_1 = 0.25 \text{ m}$.
The diameter of the third section: $D_2 = 800 \text{ mm} = 0.8 \text{ m}$. So, the radius $r_2 = 0.4 \text{ m}$.

The length of the first section (horizontal): $L_1 = 2 \text{ m}$.
The length of the second section (vertical, upper): $L_2 = 1.5 \text{ m}$.
The length of the third section (vertical, lower): $L_3 = 1.5 \text{ m}$.

• Section 1 (horizontal): $$V_1 = \pi (0.25 \text{ m})^2 (2 \text{ m}) = 0.125 \pi \text{ m}^3$$
• Section 2 (vertical, upper): $$V_2 = \pi (0.25 \text{ m})^2 (1.5 \text{ m}) = 0.09375 \pi \text{ m}^3$$
• Section 3 (vertical, lower): $$V_3 = \pi (0.4 \text{ m})^2 (1.5 \text{ m}) = 0.24 \pi \text{ m}^3$$

Using $\pi \approx 3.14159$:

Rounding to the nearest liter:

Note: The provided answer for (a) is 1458 litres. This difference (16.8 litres) likely accounts for the volume of the bends, which are not explicitly dimensioned in the diagram. My calculation is based on the straight cylindrical sections only.

• Section 1 (horizontal): $$A_1 = 2 \pi (0.25 \text{ m}) (2 \text{ m}) = 1 \pi \text{ m}^2$$
• Section 2 (vertical, upper): $$A_2 = 2 \pi (0.25 \text{ m}) (1.5 \text{ m}) = 0.75 \pi \text{ m}^2$$
• Section 3 (vertical, lower): $$A_3 = 2 \pi (0.4 \text{ m}) (1.5 \text{ m}) = 1.2 \pi \text{ m}^2$$

Rounding to two decimal places:

Note: The provided answer for (b) is 9.77 m$^2$. If this value represents the area without waste, then the total area including 25% waste would be $9.77 \text{ m}^2 \times 1.25 = 12.2125 \text{ m}^2$. My direct calculation for the base area is $9.2677 \text{ m}^2$, which is different from $9.77 \text{ m}^2$. This discrepancy likely arises from how the surface area of the bends is calculated or approximated.

Using my calculated total area from (b):

Rounding to two decimal places:

Note: If we use the provided answer for (b) as the base area (9.77 m$^2$) and then add 25% waste, the total area would be $9.77 \text{ m}^2 \times 1.25 = 12.2125 \text{ m}^2$. In this case, the cost would be $12.2125 \text{ m}^2 \times £11.50/\text{m}^2 = £140.44375 \approx £140.45$, which matches the provided answer for (c).