Ex.13.7 Q9 Surface Areas and Volumes Solution - NCERT Maths Class 9

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A heap of wheat is in the form of a cone whose diameter is \(10.5\rm\,m\) and height is \(3\rm\, m.\) Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

 Video Solution
Ex exercise-13-7 | Question 9

Text Solution


Volume of the right circular cone is \(\begin{align}\frac{1}{3} \end{align}\)times of the volume of a cylinder. Curved surface area of the cone =\(\begin{align}\pi rl \end{align}\)

What is  known?

Diameter and height of the cone.

What is  unknown?

Volume of the cone and area of the canvas to cover the heap.


Diameter \(= 2r = 10.5 \rm\,m\)

\(\begin{align}r = \frac{{10.5}}{2} = 5.25m \end{align}\)

Height \((h) = 3 \rm\,m\)

Volume of cone 

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \frac{{22}}{7} \times {(\frac{{10.5}}{2})^2} \times 3 \\ &= 86.625\,\,\,\rm\,{m^3} \end{align}\]

Slant height 

\[\begin{align} &= \sqrt {{r^2} + {h^2}} \\&= \sqrt {{{(5.25)}^2} + {{(3)}^2}} \\ &= \sqrt {27.5625 + 9} \\ &= \sqrt {36.5625}\\ &= 6.05\,\,\rm\,m \end{align}\]

The area of the canvas to cover the heap of wheat \(=\) surface area of the cone \(=\) \(\begin{align}\pi rl \end{align}\)

\[\begin{align}& = \frac{{22}}{7} \times \frac{{10.5}}{2} \times 6.05 \\ &= 99.825\,\,\rm\,{m^2} \end{align}\]


Volume of the cone \(\begin{align} = 86.625\,\,\rm\,{m^3} \end{align}\)

Area of the canvas required \(\begin{align} = 99.825\,\,\rm\,{m^2} \end{align}\)