Ex.13.3 Q8 Surface Areas and Volumes Solution - NCERT Maths Class 9

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Question

A bus stop is barricaded from the remaining part of the road, by using \(50\) hollow cones made of recycled cardboard. Each cone has a base diameter of \(40\rm\, cm\) and height \(1\rm\, m.\) If the outer side of each of the cones is to be painted and the cost of painting is \(\rm Rs\, 12\) per \(\rm m^{2}\) . What will be the cost of painting all these cones?

(Use \(\pi = 3.14\) and take \(\sqrt{1.04}= 1.02\))

Text Solution

Reasoning:

Curved surface area of a right circular cone of base radius \(r\) and slant height l is \( = \pi rl \) and \(l = \sqrt {{r^2} + {h^2}} \)

What is known?

Base diameter and height of the cone number of cones and cost per \(\rm\,{m^2}. \)

What is unknown?

Cost of painting the \(50\) cones.

Steps:

\(\begin{align} \text{Base diameter} &=40 {\rm{cm}}\\2r & =40 {\rm{cm}} \\ r &= \frac{{40}}{2} \\ &= 20\,\,{\rm{cm}}\\ & = 0.2 \, {\rm{m}}\; \begin{bmatrix} \because 100\,{\rm{cm}} \\ = 1\,{\rm{m}}\end{bmatrix} \end{align}\)

Height \( h = 1\rm\, m\)

\[\begin{align}l &= \sqrt {{r^2} + {h^2}} \\ &= \sqrt {{{(0.2)}^2} + {{(1)}^2}} \\ &= \sqrt {1.04} \\ &= 1.02\rm \,m \end{align}\]

Curved surface are \(=\) \(\begin{align}\pi rl \end{align}\)

\[\begin{align} &= \pi \times 0.2 \times 1.02\\ &= 3.14 \times 0.2 \times 1.02\\ &= 0.64056\,\,\rm\,{m^2} \end{align}\]

Curved surface area of \(50\) cones

\[\begin{align} &= 0.64056 \times 50\\ &= 32.028\,\,\rm{m^2} \end{align}\]

Cost of painting per \(\begin{align}\rm{m^2} \end{align}\) \(=\rm Rs\, 12\)

Cost of painting \(32.028\rm\,{m^2}\)

\[\begin{align} &= \rm Rs\,\,32.028 \times 12 \\ &= \rm Rs\,\,\,\,384.34\,\,\rm\,{m^2} \end{align}\]

Answer:

Cost of painting all the cones

\(=\rm Rs\, 384.34\)

  
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