What length of tarpaulin 3, m wide will be re | vedclass

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(B) Given: Base radius $(r) = 6\, m$,Height $(h) = 8\, m$.Slant height $(l) = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,

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$63.0$

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(B) Given: Base radius $(r) = 6\, m$,Height $(h) = 8\, m$.Slant height $(l) = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\, m$.Curved surface area of the tent $= \pi rl = 3.14 	imes 6 	imes 10 = 188.4\, m^2$.Let the length of the tarpaulin be $L$. Since the width is $3\, m$,the area is $L 	imes 3 = 188.4\, m^2$.$L = \frac{188.4}{3} = 62.8\, m$.Extra length for margins and wastage $= 20\, cm = 0.2\, m$.Total length required $= 62.8\, m + 0.2\, m = 63.0\, m$.

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

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