# 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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Updated on: 10-Oct-2022

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