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 :
Radius (r) of heap = \(\frac{10.5}{2} m = 5.25 m\)
Height (h) of heap = 3 m

As we know that,

Volume of heap
= \(\frac{1}{3} {\pi} {r}^2 h\)
= \(\frac{1}{3} × \frac{22}{7} × {5.25}^2 × 3\) \({m}^3\)
= 86.625 \({m}^3\)

Therefore, the volume of the heap of wheat is 86.625 \({m}^3\).

But, Area of canvas required = CSA of cone

So, we get,
= \({\pi} r l\)
= \({\pi} × r × (\sqrt{(r)^2 + (h)^2})\)
= \(\frac{22}{7} × 5.25 × (\sqrt{(5.25)^2 + (3)^2})\) \({m}^2\)
= \(\frac{22}{7} × × 5.25 × 6.05\) \({m}^2\)
= 99.825 \({m}^2\)

Therefore, 99.825 \({m}^2\) canvas will be required to protect the heap from rain.