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(D) Given: Diameter of the conical heap $= 10.5\, m$,so radius $(r) = \frac{10.5}{2} = 5.25\, m$. Height $(h) = 3\, m$.$1$. Volume of the heap:Volume

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$99.825$

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(D) Given: Diameter of the conical heap $= 10.5\, m$,so radius $(r) = \frac{10.5}{2} = 5.25\, m$. Height $(h) = 3\, m$.$1$. Volume of the heap:Volume $= \frac{1}{3} \pi r^2 h = \frac{1}{3} 	imes \frac{22}{7} 	imes (5.25)^2 	imes 3 = \frac{22}{7} 	imes 27.5625 = 86.625\, m^3$.$2$. Area of the canvas:The canvas required is equal to the curved surface area of the cone,which is $\pi rl$.First,find the slant height $(l) = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} = 6.0467\, m$ (approx $6.05\, m$).Curved Surface Area $= \pi rl = \frac{22}{7} 	imes 5.25 	imes 6.0467 = 99.825\, m^2$.Thus,the required area of the canvas is $99.825\, m^2$.

$A$ 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. (in $, m^{2}$)

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