Question

Find what length of canvas $2{\text{ m}}$ in width in required to make a conical tent $20{\text{ m}}$ in diameter and $42{\text{ m}}$ in slant height allowing $10\% $ for folds and the stitching. Also find the cost of the canvass at the rate of ${\text{Rs}}{\text{. 80}}$ per meter.

Answer 1

Hint: In this question we have to find the cost of the canvas for a conical tent. Since the cost of the canvas is given per meter of the length of the canvas so we have to find the length of the canvas first. Now we know that the tent is in a conical shape so using the formula for the surface area of the cone we calculate the surface area of the canvas. Then using the total area of the canvas which is the sum of the surface area of the canvas and the area allowed for folds and the stitching we can find the length of the canvas.

Complete step-by-step answer:
Given:

The width of the canvas of the tent $w = 2{\text{ m}}$
The diameter of the conical tent $d = 20{\text{ m}}$
So, the radius of the conical tent
$
r = \dfrac{d}{2}\\
r = \dfrac{{20}}{2}\\
r = 10{\text{ m}}
$
The slant height of the conical tent $l = 42{\text{ m}}$
The cost rate of the canvas $ = \;{\text{Rs}}{\text{. 80 per m}}$
Since the tent is in conical shape so the area of the tent is equal to the curved surface area for the cone.
So, the surface area of the conical tent,
$\Rightarrow A = \pi r\left( {r + l} \right)$
Substituting the values in the formula we get,
$
\Rightarrow A = \dfrac{{22}}{7} \times 10 \times \left( {10 + 42} \right)\\
A = 1634.3{\text{ }}{{\text{m}}^2}
$
Also, it is given that only $10\% $ of this area is allowed for folds and the stitching.
So, according to the question,
The area of the canvas allowed for folds and the stitching
$
{A_{allow}} = 10\% {\text{ of 1634}}{\text{.3}}\\
{A_{allow}} = \dfrac{{10}}{{100}} \times 1634.3\\
{A_{allow}} = 163.43{\text{ }}{{\text{m}}^2}
$
So, the total canvas required is the sum of the surface area of the conical tent and the area of the canvas allowed for folds and the stitching.
$
\Rightarrow {A_{total}} = A + {A_{allow}}\\
\Rightarrow {A_{total}} = 1634.3 + 163.43\\
\Rightarrow {A_{total}} = 1797.73{\text{ }}{{\text{m}}^2}
$
Now to find the length of the canvas $\left( l \right)$, we know that the total area of the canvas is the product of the length and width of the canvas.
So,
${A_{total}} = l \times w$
Solving for the length of the canvas we get,
$
l = \dfrac{{{A_{total}}}}{w}\\
l = \dfrac{{1797.73}}{2}\\
l = 898.865{\text{ m}}
$
So, the cost of canvas for the total length $\left( l \right)$ of the canvas is given by –
$
\Rightarrow {\text{Cost = Rs}}{\text{. 80 per m }} \times {\text{ 898}}{\text{.865 m}}\\
\Rightarrow {\text{Cost = Rs}}{\text{. }}71909.2
$
Therefore, the cost of the canvas is ${\text{Rs}}{\text{. }}71909.2$.

Note: It should be noted that all units of the dimensions of the conical shape should be the same, this should always be done before any calculation to avoid any complications. For example, in the given question all the units are already the same (in meters), so we do not need to convert them.