A garden roller has a diameter of 84 cm and a length of 120 cm. How much ground does it level in 500 complete revolutions? Give the answer in square metres. Use π = $\frac{22}{7}$.

Radius = 84 ÷ 2 = 42 cm. One revolution levels the curved surface: 2πrh $=2\times \frac{22}{7}\times 42\times 120=31,680{\text{cm}}^2$. In 500 revolutions: 15,840,$000{\text{cm}}^2$. Since $1{\text{m}}^2=10,000{\text{cm}}^2$, that is $1584{\text{m}}^2$. Trap $3168{\text{m}}^2$: putting the diameter 84 cm into the radius slot, which doubles the answer. Trap 15,$840{\text{m}}^2$: dividing by 1000 instead of 10,000. Trap $158.4{\text{m}}^2$: dividing by 100,000.