Question

The radius of a conical tent is $7m$ and the height is $24m$. Calculate the length of the rectangular canvas used to make the tent. If the width of the rectangular canvas is $4m$.

Answer 1

Hint: Here, in this question we first need to calculate the area of canvas in terms of an unknown length. Then, next we calculate the curved surface of the cone and further equate them in order to get the length of the canvas used. We also take the value of $\pi $ as $\pi = \dfrac{{22}}{7}$.

Complete step by step answer:
Given is, the canvas is rectangular in shape. Let us assume that its dimensions are length and width.
We are given that width of canvas is $4m$i.e., $w = 4m$.
Area of a rectangle=$length \times width$

Then, accordingly the area of canvas =$length \times 4{m^2}$-----(1)
We are given that the tent is in conical shape, so let us assume that its dimensions are radius $\left( {r = 7m} \right)$, height$\left( {h = 24m} \right)$ and slant height be $l$.

As we already know that the curved surface area of the cone =$\pi rl$
The relation between radius, height and slant height is given by:
$ \Rightarrow l = \sqrt {{h^2} + {r^2}} $
By using the above equation, we find the value of slant height of the tent i.e., $l$.
$
   \Rightarrow l = \sqrt {{{\left( {24} \right)}^2} + {{\left( 7 \right)}^2}} \\
   \Rightarrow l = \sqrt {576 + 49} \\
   \Rightarrow l = \sqrt {625} \\
 $
$ \Rightarrow l = 25m$
Further proceeding,
The curved surface area of tent = The curved surface area of cone
$
   = \pi rl \\
   = \dfrac{{22}}{7} \times 7 \times 25 \\
   = 22 \times 25 \\
 $
$ = 550\;{m^2}$----(2)
We are given that conical tents are made out of rectangular canvas. So,
Area of rectangular canvas = Curved surface area of conical tent
Using equation (1) and (2) we get,
$
   \Rightarrow length \times 4 = 550 \\
   \Rightarrow length = \dfrac{{550}}{4} \\
   \Rightarrow length = 137.5\;m \\
 $
Therefore, the length of the rectangular canvas used is $137.5\;m$.

Note:
While solving such types of questions, the key component is to learn the formulas of different shapes. Like in this given case, we were in need of formulas of area of rectangle, which was $length \times width$ and the curved surface area of cone, which was $\pi rl$.