Question

what length of tarpaulin 3 m 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 wastages in cutting is approximately 20 cm

Answer 1

For conical tent.....given..height= 8m, radius=6m....therefore length = root of r square + h square = root of 6 square + 8 square= root of 36+64= root of 100= 10m...width of the tarpaulin = 3m....therfore length of the tarpaulin = 188.4÷3= 62.8m...extra length of material required= 20cm = 0.2m(convertng cm to m)....Actual length of tarpaulin required= 62.8m + 0.2m = 63m..........

Answer 2

$\large{\mathcal{\underline{\:\:\:\:\:\:Solution\:\:\:\:\:\:\:\:}}}$

Hight of conical tent,h=8 m

Radius of base of tent,r=6 m

Stant height of tent, $\sf{l^2=r^2+h^2}$

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$\sf{Here,}$

$\large\sf{l^2=(6^2+8^2)}$

$\large\sf{→l^2=(36+64)}$

$\large\sf{→l^2=100}$

$\large\sf{→l=√100}$

$\large\sf{→l=10}$

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$\sf{Again,}$

CSA of conical tent,

$\large\sf{πrl}$

$\large\sf{=3.14×6×10}$

$\large\sf{=188.4\:m^2}$

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Let the length of tarpaulin sheet required be L

As, 20 cm will be wasted,therefore,

Effective length will be (L - 0.2 m)

Effective length will be (L - 0.2 m)Breadth of tarpaulin = 3 m (given)

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$\sf{Area\:of\:sheet\:=CSA\:of\:tent}$

$\large\sf{[(L-0.2)×3]=188.4}$

$\large\sf{L-0.2=\dfrac{188.4}{3}}$

$\large\sf{L-0.2=62.8}$

$\large\sf{L=62.8+0.2}$

$\large\sf{L=63}$

Therefore, the length of the tarpaulin required is 63 m.