Solve this

Solution:

Radius of the conical tent, r = 5 m

Area of the base of the conical tent $=\pi r^{2}=\frac{22}{7} \times 5^{2}=\frac{550}{7} \mathrm{~m}^{2}$

Average area occupied by a student on the ground $=\frac{5}{7} \mathrm{~m}^{2}$

∴ Number of students who can sit in the tent

$=\frac{\text { Area of the base of the conical tent }}{\text { Average area occupied by a student on the ground }}$

$=\frac{\frac{550}{7}}{\frac{5}{7}}$

$=110$

Thus, the number of students who can sit in the tent is 110.

Let the slant height of the tent be l m.

Curved surface area of the tent = 165 m2

$\therefore \pi r l=165$

$\Rightarrow \frac{22}{7} \times 5 \times l=165$

$\Rightarrow l=\frac{165 \times 7}{22 \times 5}$

$\Rightarrow l=10.5 \mathrm{~m}$

Let the height of the tent be h m.

$h=\sqrt{l^{2}-r^{2}}=\sqrt{(10.5)^{2}-5^{2}}=\sqrt{110.25-25}=\sqrt{85.25} \approx 9.23 \mathrm{~m}$

$\therefore$ Volume of the tent $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times(5)^{2} \times 9.23 \approx 241.74 \mathrm{~m}^{3}$

$\therefore$ Volume of the tent $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times(5)^{2} \times 9.23 \approx 241.74 \mathrm{~m}^{3}$