# Ex.13.2 Q7 Surface Areas and Volumes - NCERT Maths Class 9

## Question

The inner diameter of a circular well is \(3.5\, \rm{m}\,.\) It is \(10 \,\rm{m}\) deep. Find

- Its inner curved surface area
- The cost of plastering this curved surface at the rate of \(40\) per\(\begin{align}\rm{{m^2}} \end{align}\)

## Text Solution

**Reasoning:**

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

**What is the known?**

- The inner diameter and the depth of the well.
- Cost of plastering the curved surface area per\(\begin{align}\rm{{m^2}} \end{align}\)

1.Its inner curved surface area

**What is the unknown?**

The inner curved surface area.

**Steps:**

Diameter \(= 2r = 3.5 \rm{m}\)

\(\begin{align} r &= \frac{{3.5}}{2}\,\,\rm{m}\\height = h &= 10\,\,\rm{m}\end{align}\)

The Curved surface area

\[\begin{align}&= 2\pi rh\\& = 2 \times \frac{{22}}{7} \times \frac{{3.5}}{2} \times 10\\& = 110\,\,{m^2} \end{align}\]

2. Cost of plastering the curved surface area per \(\begin{align}\rm{{m^2}} \end{align}\).

**What is the unknown?**

Cost of plastering the curved surface area.

**Steps:**

Cost of plastering \(\begin{align}\,1\,\rm{{m^2}} = Rs\,\,40 \end{align}\)

Cost of plastering

\(\begin{align}110\,\,\rm{{m^2}} = 110 \times 40 = Rs\,\,4400 \end{align}\)

Its inner curved surface \(\begin{align} = 110\,\,\rm{{m^2}} \end{align}\)

Cost of plastering the curved surface

\(= \rm{Rs}\, 4400\)