What length of tarpaulin 3m wide will be required to make a conical tent of height 8m and base radius 6m? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm `(U s e\ pi=3. 14)`

Explanation

Let r, h and l be the radius, height and slant height of the tent, respectively.
Given, r = 6 m, h = 8 m
`because l = sqrt(r^(2)+h^(2))`
`therefore l = sqrt(6^(2)+8^(2)=sqrt(36+64)=sqrt(100)=10m`
Area of the canvas used for the tent = Curved surface area of the cone
`= pi rl=3.14xx6xx10=188.4 m^(2)`
`therefore` Length of tarpaulin required `=("Area of tarpaulin required")/("Width of tarpaulin")`
`=(188.4)/(3) " " (because "width of tarpaulin = 3, given")`
= 62.8
The extra material required for stitching margins and cutting = 20 cm = 0.2 m
Hence, the total length of tarpaulin required = 62.8+0.2 = 63 m