Surface Areas & Volumes | Exercise 13.2 | Q11

NCERT Solutions for Class 9 Maths | Chapter 13 | Curved Surface Area of Cylinder

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 ?


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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

 


NCERT Solutions for Class 9 Math | Chapter 13 Video Solutions



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