Question

The diameter of a roller is 84cm and its length is 120cm. It takes 500 complete

revolutions to move once over to level a play ground. Find the area of the play ground.


explain step by step clearly ​

Answer 1

Answer:

the roller takes 500 complete revolutions to roll once over that playground. Since a roller is in cylindrical shape. Lateral surface area = 2*Pie*r*h = 2*22/7*42cm*120cm = 31680cm2.

hope this helps!

Answer 2

$\bf{\underline{Question:-}}$

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

$\bf{\underline{Given :-}}$

• Diameter of roller = 84cm so radius = 84/2 = 42 cm
• length(h) of roller is 120cm
• it take 500 complete revolution to move once over to level the play ground .

$\bf{\underline{TO\:FIND:-}}$

• Area = ?

$\bf{\underline{SOLUTION:-}}$

We know,

• Curved surface area of cylinder = 2πrh

→ $\sf 2 × \frac{22}{7}×42 × 120$

→ $\sf 2 × 22 × 6 × 120$

→ $\sf → 44 × 720$

→ $\sf → 31680 cm^2$

• CURVED SURFACE AREA 31680 cm^$\bf ^2$ = 1 REVOLUTION

$\bf{\underline{ACCORDING\:TO\: QUESTION:-}}$

• The roller takes 500 complete revolutions to move once over to level a play playground.

Area of playground = 31680 × 500

Area of playground = 15840000 cm$\sf ^2$

• Changing $\bf cm^2$ in $\bf Metre^2$

So

→ 15840000 ÷ 10000

→ 1584$\bf m^2$

$\bf{\underline{HENCE:-}}$

• AREA OF PLAYGROUND = $\bf 1584m^2$