Ex.13.3 Q4 Surface Areas and Volumes Solution - NCERT Maths Class 9

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Question

A conical tent is \(10\,\rm  m\) high and the radius of its base is \(24\,\rm  m.\) Find

i. Slant height of the tent.

ii. Cost of the canvas required to make the tent, if the cost of \( 1 \, \rm {m^2} \) canvas is \(\rm Rs. \,70.\)

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-3 | Question 4

Text Solution

Reasoning:

Curved surface area of the cone of base radius \(r\) and slant height \(l\) is\(\begin{align}\pi rl \end{align}\). Where, \(\begin{align}l = \sqrt {{r^2} + {h^2}} \end{align}\) using the Pythagoras Theorem. And cost of canvas required will be the product of area and cost per meter square of canvas.

What is known?

Height of the cone and its base radius.

What is unknown?

i. Slant height of the tent.

Steps:

Slant height \(l = \sqrt {{r^2} + {h^2}} \)

Radius \((r) = 24\rm\, m\)

Height \((h) = 10\rm \, m\)

\[\begin{align}l &= \sqrt {{r^2} + {h^2}} \\ l  &= \sqrt {{{(24)}^2} + {{(10)}^2}} \\ & = \sqrt {576 + 100} \\ &= \sqrt {676} \end{align}\]

Slant height of the conical tent  \(= 26\, \rm m\)

ii. Cost of canvas required to make if \(1 \rm{m^2} \) canvas is \(\rm Rs\, 70.\)

Canvas required to make the tent is equal to the curved surface area of the cone.

Curved surface area of the cone \( = \pi rl \)

Radius \((r) = 24\rm\, m\)

Slant height \(\begin{align}(l) = 26\,\, \rm m \end{align}\)

\[\begin{align}CSA = \frac{{22}}{7} \times 24 \times 26\,\, \rm {m^2} \end{align}\]

\(\begin{align}\therefore \end{align}\) Cost of \(\begin{align}\frac{{22}}{7} \times 24 \times 26\,\rm {m^2} \end{align}\) canvas \(\begin{align} = \frac{{22}}{7} \times 24 \times 26 \times 70 = \rm Rs\,\,137280 \end{align}\)

Cost of the canvas required to make the tent \(= \rm Rs. 137280\)

Answer:

(i) Slant height of the tent is \(26\,\rm m.\)

(ii) The cost of the canvas is \(\rm Rs. 137280\)

  
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