# Ex.13.1 Q2 Surface Areas and Volumes - NCERT Maths Class 9

## Question

The length, breadth and height of a room are \(5 \,\rm{m}, 4 \,\rm{m},\) and \(3\,\rm{ m}\) respectively. Find the cost of white washing the walls of the room and ceiling at the rate of \(\begin{align} \text{Rs 7.50 per }\rm{m^2}? \end{align}\)

## Text Solution

**What is known:**

Medium

**What is unknown:**

The lenght, breadth and height of a room are \(5m\), \(4m\), and \(3m\) respectively.

**Reasoning:**

Since the four walls and ceiling are to be whitewashed. So, it has \(5\) faces only, excluding the base.

Hence, area of the room to be whitewashed can be obtained by adding area of the ceiling to the lateral surface area of the cuboidal room.

Lateral surface area of cuboid \(=2(l+b)h\)

The cost of white washing the walls of the room and ceiling will be equal to area of the room to be whitewashed multiplied by rate of the whitewashing.

**Steps:**

\[\begin{align} \\\text{length} &= 5\,\rm{m} \\\text{breadth} &= \rm{4\,m} \\\text{height} &= \rm{3 \,\,m} \end{align}\]

Surface are of \(5\) faces \(=\) Area of the \(4\) walls and ceiling \(=\) \(lb+2(l+b)h\)

\[\begin{align}&=(5m\!\times\!4m)\!+\!2\!\times\!(5m\!+\!4m)\!\times\!3m\\ &= 20m^2+2\times9m\times3m \\&=20m^2+54m^2 \\&=74m^2\end{align}\]

The cost of white washing the walls of the room and ceiling = \(Rate\times area\)

\[\begin{align}&=₹ 7.50/m^2\times 74m^2\\ &= ₹\,555\end{align}\]

**Answer:**

Cost of white washing the walls of the room and the ceiling \(= Rs. 555\)