Question

What length of tarpaulin $3m$ wide will be required to make a conical tent of height $8m$ and base radius $6m$? Assume that extra length of material that will be required for stitching margins and wastage in cutting is approximately $20cm$ (use $\pi = 3.14$)

Answer 1

Hint: Use the formula $l = \sqrt {{r^2} + {h^2}} $, for finding the slant height of the conical tent and then find the curved surface area of the cone by using the standard formula, $CSA = \pi rl$, where $CSA = Curved Surface Area$.

Complete step-by-step answer:
Given, h= height of the conical tent= $8m$, r=Base radius= $6m$


Using the standard formula for finding the slant height of the conical tent i.e., $l = \sqrt {{r^2} + {h^2}} - (1)$, where l is the slant height, r is the base radius and h is the height of the conical tent.


 Substituting the values of $r,h$in equation 1, we get


$l = \sqrt {{6^2} + {8^2}} = 10m$


Curved Surface Area of the cone, $CSA = \pi rl = 3.14 \times 6 \times 10{m^2}$


According to the question, the tarpaulin is $3m$.


Therefore, length of $3m$ wide tarpaulin required $ = \dfrac{{CSA of Conical Tent}}{{Width Of Tarpaulin}} = \dfrac{{3.14 \times 6 \times 10}}{3} = 62.8m$


Given in the question that extra length of material required for stitching margins and cutting is $20cm = 0.2m$.


Therefore, total length required will be $ = 62.8 + 0.2 = 63m$.

Note: Whenever such type of questions appear, the always write down the things given in question and then use the standard formula of finding the slant height of the conical tent, as mentioned in the solution which is $l = \sqrt {{r^2} + {h^2}} $and then use this slant height to find the curved surface area of the conical tent. To get the length of tarpaulin required divide the curved surface area of the cone by the width of the tarpaulin material and then add the extra length of material which is required for stitching and cutting.