# A circus tent is cylindrical to a height of $3m$ and conical above it. If its diameter is $105m$ and the slant height of the conical portion is $53m$, calculate the length of the canvas $5m$ wide to make the required tent.

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Hint: Area of the canvas required will be equal to the total surface area of the tent,using this information we can visualise the problem easily.

We know that the total surface area of the cone is $\pi rl,$ where $r$ is the radius of its base and $l$ is its slant height.

On the other hand, the total surface area of the cylinder is $2\pi rh,$ where $r$ is the radius of its base and $h$ is its height.

In this case, the cone is surmounted on the cylinder. So, the radius will be the same for cone and cylinder.

Thus, the total surface area of the tent will be $\pi rl + 2\pi rh$ and this will be equal to the area of canvas required. So,

Area of tent $ = \pi rl + 2\pi rh.$

Area of canvas is length times width i.e., $L \times B.$

$

\Rightarrow L \times B = \pi rl + 2\pi rh \\

\Rightarrow L \times B = \pi r(l + 2h) \\

$

Given, from the question, $B = 5m,h = 3m,r = \dfrac{{105}}{2}m$ and $l = 53m.$Putting all these values in above equation:

$

\Rightarrow L \times 5 = \dfrac{{22}}{7} \times \dfrac{{105}}{2} \times (53 + 6), \\

\Rightarrow L \times 5 = 11 \times 15 \times 59 \\

\Rightarrow L = 11 \times 3 \times 59 \\

\Rightarrow L = 1947. \\

$

Thus, the length of the required canvas is $1947m.$

Note: In this case, the canvas will have to cover the outer surface of the tent that’s why we are considering the total surface area and not volume. If we have to calculate the capacity or strength of the tent, then we will have to consider its volume.

We know that the total surface area of the cone is $\pi rl,$ where $r$ is the radius of its base and $l$ is its slant height.

On the other hand, the total surface area of the cylinder is $2\pi rh,$ where $r$ is the radius of its base and $h$ is its height.

In this case, the cone is surmounted on the cylinder. So, the radius will be the same for cone and cylinder.

Thus, the total surface area of the tent will be $\pi rl + 2\pi rh$ and this will be equal to the area of canvas required. So,

Area of tent $ = \pi rl + 2\pi rh.$

Area of canvas is length times width i.e., $L \times B.$

$

\Rightarrow L \times B = \pi rl + 2\pi rh \\

\Rightarrow L \times B = \pi r(l + 2h) \\

$

Given, from the question, $B = 5m,h = 3m,r = \dfrac{{105}}{2}m$ and $l = 53m.$Putting all these values in above equation:

$

\Rightarrow L \times 5 = \dfrac{{22}}{7} \times \dfrac{{105}}{2} \times (53 + 6), \\

\Rightarrow L \times 5 = 11 \times 15 \times 59 \\

\Rightarrow L = 11 \times 3 \times 59 \\

\Rightarrow L = 1947. \\

$

Thus, the length of the required canvas is $1947m.$

Note: In this case, the canvas will have to cover the outer surface of the tent that’s why we are considering the total surface area and not volume. If we have to calculate the capacity or strength of the tent, then we will have to consider its volume.

Last updated date: 21st Sep 2023

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