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

Given,

Diameter of the conical heap (d) = 10.5 m

Radius of the conical heap (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 m

Height of the conical heap (h) = 3 m

By formula,

Volume of the conical heap (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 27.5625 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 82.6875 \\[1em] = \dfrac{1819.125}{21} \\[1em] = \text{86.625 m}^3$

By formula,

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

Substituting values we get :

$l = \sqrt{(5.25)^2 + (3)^2} \\[1em] = \sqrt{27.5625 + 9} \\[1em] = \sqrt{36.5625} \\[1em] = 6.047 \text{ m}$

The area of the canvas required to cover the heap of wheat (A) = Curved surface area of conical heap = πrl

Substituting values we get :

$A = \dfrac{22}{7} \times 5.25 \times 6.047 \\[1em] = \dfrac{698.4285}{7} \\[1em] = 99.775 \text{ m}^2.$

Hence, the volume of the conical heap is 86.625 m³ and the area of the canvas required is 99.775 m².