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Answer :
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.