Question

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

Answer 1

Step 1: Find the volume of the conical heap

Diameter of the conical heap$=10.5m$

The radius of the conical heap$=\frac{10.5}{2}=5.25m$ (Radius is half of the diameter)

Height of the conical heap$=3m$

The volume of the conical heap $=$Volume of cone$=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

$\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\times \frac{22}{7}\times 5.25\times 5.25\times 3=86.625{\mathrm{m}}^3$

Thus, the volume of the conical wheat heap$=86.625m^3$

Step 2: Find the area of the canvas

Area of the canvas required $=$Curved surface area of the cone$=\mathrm{\pi rl}$(where $l$ is the slant height)

$$\begin{gathered} \Rightarrow l=\sqrt{r^2+h^2} \\ \Rightarrow l=\sqrt{5.{25}^2+3^2} \\ \Rightarrow l=\sqrt{27.625+9} \\ \Rightarrow l=\sqrt{36.5625} \\ \Rightarrow l=6.05m \end{gathered}$$

$\mathrm{\pi rl}=\frac{22}{7}\times 5.25\times 6.05=99.82{\mathrm{m}}^2$

Thus, the area of the canvas required is $99.82m^2$.

Hence, the volume of the conical wheat heap $=86.625m^3$ and the area of the canvas required is $99.82m^2$.