Math, asked by sidhant014, 7 months ago

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 ​

Answers

Answered by Anonymous
4

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!

Answered by Anonymous
3

\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
Similar questions