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 wastage in cutting is approximately 20 cm (use te=3.14)
​

Answer 1

Answer:

I think this is a question from our NCERT books

Step-by-step explanation:

first find the CSA of conical tent that will be equal to 3m×8m +the area of cone base (include 20m extra(

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.