Question

A heap of rice in the form of a cone of base diameter $24m$ and height $3.5m$. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Answer 1

Hint: The base radius is half of the base diameter. Also, the formula used for calculating the slant height of a cone is $l = \sqrt {{r^2} + {h^2}} $, where $r$ is the radius of cone and $h$ is the height of cone. $l = \sqrt {{r^2} + {h^2}} $

Complete step-by-step answer:
Given, base diameter of heap of rice$ = 24m$
Therefore, base radius of heap of rice, $r = \dfrac{{24}}{2} = 12m$
Also given, height of heap of rice, $h$=$3.5m$
It is given to the heap of rice in the form of a cone.
So, volume of rice=volume of cone $ = \dfrac{1}{3}\pi {r^2}h$
$ \Rightarrow $ Volume of rice=$\dfrac{1}{3} \times \dfrac{{22}}{7} \times {\left( {12} \right)^2} \times 3.5$
$ \Rightarrow $ Volume of rice=$\dfrac{{22 \times 12 \times 12 \times 3.5}}{{3 \times 7}}$
$ \Rightarrow $ Volume of rice=$22 \times 12 \times 4 \times 0.5$
$ \Rightarrow $ Volume of rice=$528{m^3}$
Canvas required to just cover the heap =Curved surface area of cone = $\pi rl$, where $l$ is the slant height of cone.
The slant height of a cone is calculated by
$l = \sqrt {{r^2} + {h^2}} $
$ \Rightarrow l = \sqrt {{{\left( {12} \right)}^2} + {{\left( {3.5} \right)}^2}} $
$ \Rightarrow l = \sqrt {144 + 12.25} $
\[ \Rightarrow l = \sqrt {156.25} \]
\[ \Rightarrow l = 12.5m\]
Thus, canvas required to just cover the heap=$\pi rl$
\[ \Rightarrow \]Canvas required to just cover the heap= $\dfrac{{22}}{7} \times 12 \times 12.5$
\[ \Rightarrow \] Canvas required to just cover the heap= $\dfrac{{22 \times 150}}{7}$
\[ \Rightarrow \] Canvas required to just cover the heap=$471.428{m^2}$

Canvas required to cover the heap is $471.428{m^2}$

Note: Here we calculate the curved surface area of the cone , because we have calculated the canvas required to just cover the heap. If we have to calculate the total canvas required to cover the heap including its base, then we must calculate the total surface area of the cone.