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

Go back to  'Ex.13.7'

## Question

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.

## Text Solution

Reasoning:

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.

Steps:

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}