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


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

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

 Video Solution
Surface Areas And Volumes
Ex 13.2 | Question 7

Text Solution



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?

  1.  The inner diameter and the depth of the well.
  2.  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.


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.


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

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

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