In the diagram, WXYZ is a square of length 7 cm. Calculate, correct to three significant figures, the area of the shaded region. [Take pi = 22/7]

Given: Speed $= 850 \text{ km/h}$, Distance $= 3,400 \text{ km}$.

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$ $$\text{Time} = \frac{3,400 \text{ km}}{850 \text{ km/h}}$$ $$\text{Time} = 4 \text{ hours}$$

Given: Fuel consumption rate $= 5,000 \text{ litres/hour}$.

$$\text{Fuel consumed} = \text{Fuel consumption rate} \times \text{Time}$$ $$\text{Fuel consumed} = 5,000 \text{ litres/hour} \times 4 \text{ hours}$$ $$\text{Fuel consumed} = 20,000 \text{ litres}$$

The quantity of fuel consumed is $\boxed{20,000 \text{ litres}}$.

1. b) Find the average speed required to reduce the fuel consumption by 10%.

Original fuel consumption rate $= 5,000 \text{ litres/hour}$.
Reduction by 10% means the new consumption is $100\% - 10\% = 90\%$ of the original.

$$\text{New fuel consumption rate} = 0.90 \times 5,000 \text{ litres/hour}$$ $$\text{New fuel consumption rate} = 4,500 \text{ litres/hour}$$

Let $F_1$ be the original fuel consumption rate and $S_1$ be the original speed.
Let $F_2$ be the new fuel consumption rate and $S_2$ be the new speed.
Since fuel consumption is proportional to speed:

$$\frac{F_1}{S_1} = \frac{F_2}{S_2}$$ $$\frac{5,000 \text{ litres/hour}}{850 \text{ km/h}} = \frac{4,500 \text{ litres/hour}}{S_2}$$

$$S_2 = \frac{4,500 \text{ litres/hour} \times 850 \text{ km/h}}{5,000 \text{ litres/hour}}$$ $$S_2 = \frac{4,500 \times 850}{5,000} \text{ km/h}$$ $$S_2 = \frac{9 \times 850}{10} \text{ km/h}$$ $$S_2 = 9 \times 85 \text{ km/h}$$ $$S_2 = 765 \text{ km/h}$$

The average speed required is $\boxed{765 \text{ km/h}}$.