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

## 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?

## 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.