A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is $ 24 \mathrm{~m} $. The height of the cylindrical portion is $ 11 \mathrm{~m} $ while the vertex of the cone is $ 16 \mathrm{~m} $ above the ground. Find the area of canvas required for the tent.
Given:
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is \( 24 \mathrm{~m} \).
The height of the cylindrical portion is \( 11 \mathrm{~m} \) while the vertex of the cone is \( 16 \mathrm{~m} \) above the ground.
To do:
We have to find the area of canvas required for the tent.
Solution:
Diameter of the base of the tent $= 24\ m$
This implies,
Radius of the tent $r=\frac{24}{2}$
$= 12\ m$
Total height of the tent $H= 16\ m$
Height of the cylindrical part $h_1 = 11\ m$
Height of the conical part $h_2 =16-11$
$= 5\ m$
Slant height of the cone $l=\sqrt{r^2+h_2^2}$
$=\sqrt{(12)^{2}+(5)^{2}}$
$=\sqrt{144+25}$
$=\sqrt{169}$
$=13 \mathrm{~m}$
Area of the canvas required for the tent $=$ Curved surface area of the cone part $+$ Curved surface area of the cylindrical part
$=\pi r l+2 \pi r h_{1}$
$=\pi r(l+2 h_{1})$
$=\frac{22}{7} \times 12(13+2 \times 11)$
$=\frac{264}{7} \times(13+22)$
$=\frac{264}{7} \times 35$
$=1320 \mathrm{~m}^{2}$
The area of canvas required for the tent is $1320\ m^2$.
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