Question

The diameter of a garden roller is 1.4 m and it is 2 m long. If it needs 100 revolutions to
level the garden once . Then What will be the area of the garden ?​

Answer 1

Answer:

∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2

Answer 2

Answer:

880 $m^{2}$

Step-by-step explanation-

The garden roller is obviously a cylinder.

The diameter of the circle side of the garden roller is 1.4m

Therefore,

The radius of the circle side of the garden roller is 0.7m

The height of the garden roller is 2m

The garden roller will move laterally, (only the cylindrical surface is to be considered),so we need to find out only the lateral surface area of the cylinder.

The lateral surface area-

Formula-$2$π$rh$

= 2 ×22/7×7/10×2 $m^{2}$

=8.8 $m^{2}$(Area of garden covered in one revolution by the roller)

Therefore the area of garden covered in 100 revolutions-

=8.8×100 $m^{2}$

=880 $m^{2}$(FINAL ANSWER)