Question:
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is 24 m. The height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas required for making the tent.
(Use π = 22/7)
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 $1320 \mathrm{~m}^{2}$