Question

The diameter of a roller is 84 cm and its length is 120 cm. It takes 750 complete revolutions to move once over to level a playground. Find the area of the playground.
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Answer 1

Answer:

$\bigstar{\bold{Area\:of\:playground=2376\:m^{2} }}$

Step-by-step explanation:

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

• Diameter of the roller = 84 cm
• Length of the roller = 120 cm
• Number of revolutions = 750

$\Large{\underline{\bf{To\:Find:}}}$

• Area of the playground

$\Large{\underline{\bf{Concept:}}}$

First find the curved surface area of the roller by using the formula. Next multiply the CSA of the roller with number of revolutions to get the area of the playground.

$\Large{\underline{\bf{Solution:}}}$

➡ Given that the it takes 750 revolutions for the roller to level a playground.

➡ First we have to find the curved surface area of the roller.

➡ Here the roller is in the shape of a cylinder.

➡ CSA of a cylinder is given by,

   CSA of a cylinder = 2 π r h

    where r is the radius of the cylinder

     h is the height of the cylinder

➡ Hence CSA of the roller is given by,

    CSA of roller = 2 × 22/7 × 84/2 × 120

    CSA of roller = 443520/14

   CSA of roller = 31680 cm²

➡ Hence the curved surface area of the roller is 31680 cm²

➡ Now we have to find the area of the playground.

➡ Area of playground is given by,

  Area of playground = CSA of roller × Number of revolutions

   Area of playground = 31680 × 750

   Area of playground = 23760000 cm²

   Area of playground = 2376 m²

➡ Therefore area of playground is 2376 m².

    $\boxed{\bold{Area\:of\:playground=2376\:m^{2} }}$

Answer 2

Answer:

★Areaofplayground=2376m

2

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

• Diameter of the roller = 84 cm
• Length of the roller = 120 cm
• Number of revolutions = 750

$\Large{\underline{\bf{To\:Find:}}}$

• Area of the playground

$\Large{\underline{\bf{Concept:}}}$

• First find the curved surface area of the roller by using the formula. Next multiply the CSA of the roller with number of revolutions to get the area of the playground.

$\Large{\underline{\bf{Solution:}}}$

• ➡ Given that the it takes 750 revolutions for the roller to level a playground.

• ➡ First we have to find the curved surface area of the roller.

• ➡ Here the roller is in the shape of a cylinder.

• ➡ CSA of a cylinder is given by,

CSA of a cylinder = 2 π r h

• where r is the radius of the cylinder

• h is the height of the cylinder

• ➡ Hence CSA of the roller is given by,

• CSA of roller = 2 × 22/7 × 84/2 × 120

• CSA of roller = 443520/14

CSA of roller = 31680 cm²

• ➡ Hence the curved surface area of the roller is 31680 cm²

• ➡ Now we have to find the area of the playground.

• ➡ Area of playground is given by,

Area of playground = CSA of roller × Number of revolutions

• Area of playground = 31680 × 750

• Area of playground = 23760000 cm²

• Area of playground = 2376 m²

• ➡ Therefore area of playground is 2376 m².

$\boxed{\bold{Area\:of\:playground=2376\:m^{2} }}$