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 ????=3.14).

Answer 1

Answer:

63 m

Step-by-step explanation:

Assuming Tent without cloth at Base

=> Lateral surface area = πRL

R = radius = 6m

h = 8 m

L = √R² + h² = √6² + 8² = 10 m

Lateral  Surface Area = 3.14 * 6 * 10 = 188.4 m²

Wastage = 3 * 20/100  = 0.6 m²

Total tarpaulin required = 188.4 + 0.6 = 189 m²

Length of tarpaulin = 189/3 = 63 m

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)

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

$\large\sf{[(L-0.2)×3]=188.4}$

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

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

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

$\large\sf{L=63}$

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