A road roller is 2.1 m long and has a diameter of 1.4 m. How many complete revolutions must it make to level a playground measuring 110 m by 63 m? (Use π = 22/7.)

One revolution flattens a strip equal to the roller's curved surface, 2πrh = $2\times (22/7)\times 0.7\times 2.1=9.24$ sq m. The playground is $110\times 63=6930$ sq m, so the count = 6930/9.24 = 750. The trap 1500 drops the 2 and uses πrh; 375 uses the diameter 1.4 m as the radius; 4500 uses the circular end face π$r^2=1.54$ sq m instead of the curved surface, but it is the curved surface that touches the ground.