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# 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.