Question

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete
revolutions to move once over to level a playground. Find the area of the playground
in m2
.

Answer 1

$\huge \underline \mathbb {SOLUTION:-}$

A roller is shaped like a cylinder.

Let h be the height of the roller and r be the radius.

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2 πrh

= 2 x 22/7 x 42 x 120

= 31680 cm²

Area of field = 500 × CSA of roller

= (500 × 31680) cm²

= 15840000 cm²

= 1584 m²

Therefore, area of playground is 1584 m².

Answer 2

$\huge\underline\bold{Question}$

1. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in ${m}^{2}$

$\huge\underline\bold{Solution}$

The roller is in the form of a cylinder

$\huge\underline\bold{given}$

• diameter = 84 cm
• Radius of the roller(r) = $\dfrac{84}{2}$cm = 42 cm
• Length of the roller (h) = 120 cm

Curved surface area of the roller

= 2πrh

= 2 x $\dfrac{22}{7}$ x 42 x 120 ${cm}^{2}$

= 2 x 22 x 6 x 120 ${cm}^{2}$

= 31680 ${cm}^{2}$

Now, area of the playground levelled in one revolution of the roller = 31680 ${cm}^{2}$

= $\dfrac{31680}{10000}$${m}^{2}$

∴ Area of the playground levelled in 500

revolutions

= 500 x $\dfrac{31680}{10000}$${m}^{2}$

= 1584$\bold{ {m}^{2}}$