A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is $24\ m$. The height of the cylindrical portion is $11\ m$ while the vertex of the cone is $16\ m$ above the ground. Find the area of the 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 $24\ m$. The height of the cylindrical portion is $11\ m$ while the vertex of the cone is $16\ m$ above the ground.

To do:

We have to the area of the canvas required for the tent.

Solution:

Diameter of the cylindrical part $= 24\ m$

This implies,

Radius of the cylindrical part $(r) = \frac{24}{2}$

$=12\ m$

Height of the cylindrical part $= 11\ m$

Total height of the tent $= 16\ m$

This implies,

Height of the conical portion $= 16 - 11$

$= 5\ m$

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

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

$=\sqrt{144+25}$

$=\sqrt{169}$

$=13 \mathrm{~m}$

Area of the canvas used $=2 \pi r h+\pi r l$

$=\pi r(2 h+l)$

$=\frac{22}{7} \times 12(2 \times 11+13)$

$=\frac{264}{7}(22+13)$

$=\frac{264}{7} \times 35$

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

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Updated on: 10-Oct-2022

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