Surface Areas & Volumes | Exercise 13.2 | Q11

Question 11: The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitiors, how much cardboard was required to be bought for the competition ?

Explanatory Answer

The penholder is a cylinder with a circular base. It is open at the top.
Area of cardboard required for 1 penholder = curved surface area of cylinder + area of the circular base

Radius of penholder = 3 cm
Height of penholder = 10.5 cm = \\frac{21}{2}) cm

Curved surface area = 2πrh
Area of circular base = πr²
Take π = \\frac{22}{7})

Area of cardboard required for 35 competitiors = 35 × Area required for 1 penholder
= 35 × (curved surface area of cylinder + area of circular base)
= 35 × (2πrh + πr²) = 35 π r (2h + r)
= 35 × \\frac{22}{7}) × 3 × (2 × \\frac{21}{2}) + 3)
= 35 × \\frac{22}{7}) × 3 × 24
= 110 × 72
= 7920 cm²

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 ▶