A tent is in the form of a right circular cylinder surmounted by a cone.

Solution:

Surface area of cylindrical part

$=2 \pi r h$

$=2 \times \frac{22}{7} \times \frac{24}{2} \times 11$

$=829.71 \mathrm{~m}^{2}$

Height of cone

$=16-11$

 

$=5 \mathrm{~m} .$

Surface area of conical part

$=\pi r$

$I=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(5)^{2}+(12)^{2}}$

 

$=13 \mathrm{~m}$

S.A. of conical part

$=\frac{22}{7} \times 12 \times 13$

$=490.29 \mathrm{~m}^{2}$

Total area

$=490.29+829.71$

$=1319.99$

 

$=1320 \mathrm{~m}^{2}$

So canvas required will be $1320 \mathrm{~m}^{2}$