A model pool has a length of 30 cm. Its cross-section is a trapezium with parallel sides 4.5 cm and 1.5 cm, and the distance between them is 45 cm. The actual pool has a length of 27 m. Calculate the

The cross-section is a trapezium.
The lengths of the parallel sides are $h_1 = 4.5 \text{ cm}$ and $h_2 = 1.5 \text{ cm}$.
The perpendicular distance between the parallel sides (height of the trapezium) is $b = 45 \text{ cm}$.
The area of a trapezium is given by $A = \frac{1}{2}(h_1 + h_2)b$.

$$A_m = \frac{1}{2}(4.5 \text{ cm} + 1.5 \text{ cm}) \times 45 \text{ cm}$$ $$A_m = \frac{1}{2}(6 \text{ cm}) \times 45 \text{ cm}$$ $$A_m = 3 \text{ cm} \times 45 \text{ cm}$$ $$A_m = 135 \text{ cm}^2$$

The length of the model pool is $L_m = 30 \text{ cm}$.
The volume of the model pool is the area of the cross-section multiplied by its length.

$$V_m = A_m \times L_m$$ $$V_m = 135 \text{ cm}^2 \times 30 \text{ cm}$$ $$V_m = 4050 \text{ cm}^3$$

The length of the model pool is $L_m = 30 \text{ cm}$.
The length of the actual pool is $L_a = 27 \text{ m}$. Convert this to centimeters:

$$L_a = 27 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 2700 \text{ cm}$$

The linear scale factor $k$ is the ratio of the actual length to the model length.

$$k = \frac{L_a}{L_m} = \frac{2700 \text{ cm}}{30 \text{ cm}}$$ $$k = 90$$

The volume scale factor is $k^3$.

$$k^3 = 90^3 = 729000$$

The volume of the actual pool $V_a$ is the volume of the model pool multiplied by the volume scale factor.

$$V_a = V_m \times k^3$$ $$V_a = 4050 \text{ cm}^3 \times 729000$$ $$V_a = 2952450000 \text{ cm}^3$$

We know that $1000 \text{ cm}^3 = 1 \text{ litre}$.

$$V_a = \frac{2952450000 \text{ cm}^3}{1000 \frac{\text{cm}^3}{\text{litre}}}$$ $$V_a = 2952450 \text{ litres}$$

The capacity of the actual pool is $\text{2952450 litres}$.