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(A) $(i)$ The area of the ground covered by the tent is the area of the circular base: $	ext{Area} = \pi r^{2} = \frac{22}{7} 	imes 5^{2} = \frac{55

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(A) $(i)$ The area of the ground covered by the tent is the area of the circular base: $	ext{Area} = \pi r^{2} = \frac{22}{7} 	imes 5^{2} = \frac{550}{7} \, m^{2}$. Number of students $= \frac{	ext{Total ground area}}{	ext{Area per student}} = \frac{550/7}{5/7} = \frac{550}{5} = 110$. $(ii)$ The curved surface area of the cone is $165 \, m^{2}$. $\pi r l = 165 \Rightarrow \frac{22}{7} 	imes 5 	imes l = 165$. $l = \frac{165 	imes 7}{110} = 10.5 \, m$. Using $l^{2} = r^{2} + h^{2}$,we find the height $h$: $h = \sqrt{l^{2} - r^{2}} = \sqrt{(10.5)^{2} - 5^{2}} = \sqrt{110.25 - 25} = \sqrt{85.25} \approx 9.233 \, m$. Volume of the cone $= \frac{1}{3} \pi r^{2} h = \frac{1}{3} 	imes \frac{22}{7} 	imes 25 	imes 9.233 \approx 242.1 \, m^{3}$.

$A$ conical tent is made from a piece of cloth with an area of $165 \, m^{2}$. The radius of the base of the tent is $5 \, m$.
$(i)$ If one student occupies an average of $\frac{5}{7} \, m^{2}$ of ground area,how many students can sit in the tent?
$(ii)$ Find the volume of the cone.

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$A$ conical tent is made from a piece of cloth with an area of $165 \, m^{2}$. The radius of the base of the tent is $5 \, m$. $(i)$ If one student occupies an average of $\frac{5}{7} \, m^{2}$ of ground area,how many students can sit in the tent? $(ii)$ Find the volume of the cone.