Curved Surface Area Cylinder | Exercise 13.2 | Q4

Question 4: 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.

Video Explanation

Explanatory Answer | Exercise 13.2 Question 4

Area of the playground = area covered by the roller in 500 revolutions.
In each revolution, the roller covers an area that is equal to its curved surface area.
The roller is in the shape of a cylinder. So, we have to compute the curved surface area of cylinder.
Therefore, the area of the playground = 500 × curved surface area of cylinder.

Given Data:
Height of the cylinder = length of the roller = 120 cm = \\frac{120}{100}) = \\frac{12}{10}) m.
Diameter of the cylinder = 84 cm. So, its radius = 42 cm = \\frac{42}{100}) cm = \\frac{21}{50}) cm

Curved Surface Area of Cylinder = 2πrh
Take π = \\frac{22}{7})

Area of the playground = 500 × curved surface area of cylinder (cylindrical roller)
= 500 × 2 × \\frac{22}{7}) × \\frac{21}{50}) × \\frac{12}{10})
= 500 × \\frac{2×22×3×12}{50×10})
= 44 × 36
Area of the playground = 1584 m²

Exercise 13.1 | Surface Area of Cubes & Cuboids

1. Class 9 Exercise 13.1 Question 3 | Height of hall ▶
2. Class 9 Exercise 13.1 Question 6 | TSA of Herbarium ▶
3. Class 9 Exercise 13.1 Question 7 | Shanti Sweets ▶

Exercise 13.2 | Surface Area of Cylinders

1. Class 9 Exercise 13.2 Question 3 | CSA TSA of Metal Pipe ▶
2. Class 9 Exercise 13.2 Question 4 | Roller & Playground ▶
3. Class 9 Exercise 13.2 Question 9 | Cylindrical Petrol Tank ▶
4. Class 9 Exercise 13.2 Question 10 | CSA of Decorative Lampshade ▶

Exercise 13.3 | Surface Area of Cones

1. Class 9 Exercise 13.3 Question 5 | Length of Tarpaulin 3 m wide ▶
2. Class 9 Exercise 13.3 Question 8 | Cost of painting conical barricades ▶

Exercise 13.4 | Surface Area of Spheres & Hemispheres

1. Class 9 Exercise 13.4 Question 8 | Outer CSA of hemispherical bowl ▶
2. Class 9 Exercise 13.4 Question 9 | Surface area of sphere & Cylinder ▶