Math, asked by anonymousblah, 3 months ago

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m
and base radius 6 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).​

Answers

Answered by Anonymous
13

\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}

___________________________

  • \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}

____________________________

  • \sf{Again,}

CSA of conical tent,

\large\sf{πrl}

\large\sf{=3.14×6×10}

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

____________________________

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)

_________________________

\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.

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