A tent is of the shape shown in the given figure. The area of the floor under the tent is $64\mathrm{\pi }$ m².

If 1 m² of canvas costs Rs. 5, then what is the cost of the canvas required for making the tent?

The area of the floor under the tent is $64\mathrm{\pi }$ m².

$\therefore \mathrm{\pi }{r}^2=64\mathrm{\pi }$

$\Rightarrow r=8 \mathrm{m}$

Height ($h$) of the conical part = 6 m

Slant height ($l$) of the conical part $=\sqrt{r^2+h^2}=\sqrt{8^2+6^2}=10 \mathrm{m}$ 

Height ($h_1$) of the cylindrical part = (18 − 6) m = 12 m

Therefore, the total surface area of the tent

= CSA of cone + CSA of cylinder

$=\mathrm{\pi }rl+2\mathrm{\pi }r{h}_1$

$=\mathrm{\pi }r[l+2h_1]$

$=\frac{22}{7}\times 8\left[10+2\times 12\right]$

$=\frac{22}{7}\times 8\times 34$

$=854.8 {\mathrm{m}}^2$

Cost of 1 m² of canvas = Rs. 5

∴ Cost of 854.8 m² of canvas = 5 × 854.8 = Rs. 4274

Thus, the cost of the canvas required for making the tent is Rs. 4274.