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# A tent is cylindrical to a height of $4.8{\rm{m}}$and conical above it. The radius of the base is $4.5{\rm{m}}$ and the total height of the tent is $10.8{\rm{m}}$. Find the approximate area of the canvas required for the tent in square meters.

Last updated date: 13th Jun 2024
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Hint:
Here we need to find the approximate area of the canvas used for the tent. For that, we will find the curved surface area of the conical part of the tent using the formula and substituting the given values in the formula, we will get the curved surface area of the conical part. Then we will find the curved surface area of the cylindrical part of the tenet. We will use the formula of the curved surface area of the cylinder and then substitute the given values in the formula to get the value. Then we will add the curved surface area of both the conical part and the cylindrical part to get the required area.

Formula used:
We will use the following formulas:
1) Curved surface area of cone $= \pi rl$, where $r$ is the radius and $l$ is the slant height of the cone.
2) Curved surface area of cylinder $= 2\pi rh$, where $r$ is the radius and $h$ is the slant height of the cone.

Complete step by step solution:
Here we need to find the approximate area of the canvas used for the tent. It is given that the tent has conical as well as the cylindrical part.
The radius of the conical part and the cylindrical part of the tent are the same.
The radius of the conical part and radius of cylindrical part $= 4.5{\rm{m}}$
Now, we will draw the figure of the tent first.

The height of conical above $= 10.8 - 4.8 = 6{\rm{m}}$
Now, we will calculate the slant height of the conical part.
$l = \sqrt {{{\left( {4.5} \right)}^2} + {{\left( 6 \right)}^2}}$
On squaring and then adding terms, we get
$\Rightarrow l = \sqrt {20.25 + 36} = \sqrt {56.25}$
On further simplification, we get
$\Rightarrow l = 7.5$
Now, we will calculate the curved surface area of the conical part.
Substituting $l = 7.5$ and $r = 4.5$ in the formula Curved surface area of cone $= \pi rl$, we get
Curved surface area of cone $= \pi \times 4.5 \times 7.5$
On multiplying the terms, we get
$\Rightarrow$Curved surface area of cone $= 33.75\pi {{\rm{m}}^2}$
Now, we will calculate the curved surface area of the cylindrical part.
Substituting $h = 4.8$ and $r = 4.5$ in the formula Curved surface area of cylinder $= 2\pi rh$,
we get
Curved surface area of cylinder $= 2 \times \pi \times 4.5 \times 4.8$
On multiplying the terms, we get
$\Rightarrow$ Curved surface area of cone $= 43.2\pi {{\rm{m}}^2}$
Now, we will add the curved surface area of the conical part and curved surface area of the cylindrical part to get the approximate area of the canvas required.
Area of canvas $= 33.75\pi + 43.2\pi$
$\Rightarrow$ Area of canvas $= 76.95\pi$
$\Rightarrow$Area of canvas $= 76.95 \times \dfrac{{22}}{7} = 241.85{{\rm{m}}^2}$
Hence, the approximate area of the canvas required is equal to $241.85{{\rm{m}}^2}$.