What is the simplest way for solving the question of class 9th of exercise 13.2 of math of question no.11 give me the best and simple way for solving this question in simple way of Surface areas and Volumes chapter The students of Vidyalaya were asked to participate in a competition for making and decorating penholders 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 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)
AnilSinghBoraGuru
The students of Vidyalaya were asked to participate in a competition for making and decorating penholders 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 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7) Q.11
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Radius of the circular end of cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh+πr2
= 2×(22/7)×3×10.5+(22/7)×32= 1584/7
Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm2
So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be needed for the competition.