A tent of height $77 \mathrm{dm}$ is in the form a right circular cylinder of diameter $36 \mathrm{~m}$ and height $44 \mathrm{dm}$ surmounted by a right circular cone. Find the cost of the canvas at Rs $3.50$ per $\mathrm{m}^{2}$. [Use $\pi=22 / 7$ ]
The height of the tent is 77dm = 7.7m. The height of the upper portion of the tent is
$44 \mathrm{dm}=4.4 \mathrm{~m}$. Therefore, the height of the cylindrical part is $77-44=33 \mathrm{dm}=3.3 \mathrm{~m}$. The radius of the cylindrical part is $\frac{36}{2}=18 \mathrm{~m}$.
Let the slant height of the cone part is l m. Then, we have
$l^{2}=(18)^{2}+(3.3)^{2}$
$\Rightarrow l^{2}=324+10.89=334.89$
$\Rightarrow \quad l=18.3$
Therefore, the slant height of the cone part is 18.3 m.
The curved surface area of the cylindrical part is
$S=2 \pi \times 18 \times 4.4 \mathrm{~m}^{2}$
The curved surface area of the cone part is
$S_{1}=\pi \times 18 \times 18.3 \mathrm{~m}^{2}$
Therefore, the total curved surface area of the tent is
$S+S_{1}=2 \pi \times 18 \times 4.4+\pi \times 18 \times 18.3$
$=18 \pi(8.8+18.3)$
$=18 \pi \times 27.10$
The cost of canvas per $\mathrm{m}^{2}$ is Rs $3.50$. Hence, the total cost for canvas in Rs is
$=18 \times \frac{22}{7} \times 27.10 \times 3.50$
$=5365.80$
Hence total cost is Rs.5365.80