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Ex.13.2 Q11 Surface Areas and Volumes - NCERT Maths Class 9

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Question

The students of a 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 \(3 \,\rm{}cm\) and height \(10.5 \,\rm{}cm.\) 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?

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-2 | Question 11

Text Solution

 

Reasoning:

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh. \end{align}\)

And total surface area =\(\begin{align}2\pi r(r + h) \end{align}\). So, total amount of cardboard required will be the product of surface area of pen holders and total number of participating students.

What is the known?

Radius and height of the card board.

What is the unknown?

Card board required to be bought for \(35\) competitors.

Steps:

radius(r) \(\begin{align}= 3\,\,\rm{}cm \end{align}\)

height(h) \(\begin{align} = 10.5\,\,\rm{}cm \end{align}\)

Card board required for \(1\) competitor = Total surface area of the cylinder

Since the pen holder is open at the top.

One area of \(\begin{align}\pi {r^2} \end{align}\) should be subtracted.

Cardboard required for \(1\) competitor

\[\begin{align} &= 2\pi rh + \pi {r^2}\\&= 2 \times \frac{{22}}{7} \times 3 \times 10.5 + \frac{{22}}{7} \times 3 \times 3\\ &= 198 + \frac{{198}}{7} = 198(1 + \frac{1}{7})\\ &= \frac{{198 \times 8}}{7}\,c{m^2} \end{align}\]

Cardboard required for \(35\) competitions:

\[\begin{align} = \frac{{198 \times 8}}{7} \times 35 = 7920\,\,\rm{}c{m^2} \end{align}\]

\(\begin{align}7920\,\,\rm{}c{m^2} \end{align}\) of cardboard was required to be bought for the competition.