Question

what length of 3 m wide will be required to make conical tent of height 8 m and base radius 6 m ? assume that the extra lenth of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use π=3.14).​

Answer 1

Answer:

Given a conical tent with

Height ( h ) = 8m

Radius ( r ) = 6m

So, its length = l2 = h2 + r2

= l2 = ( 8 )2m + ( 6 )m2

= l2 = 64m + 36m

= l2 = 100m

= l = 10m

Curved surface area of the cone = TTrl

= 3.14 x 6m x 10m

= 188.4 cm2

The extra length of material that will be required for stiching margins and wastages in cutting is approx 20 cm = 0.2m

So, its length = l - 0.2m

and its width = 3m

area of sheet = CSA of tent

so the area of the trapaulln = l x b

= ( l - 0.2m ) x 3m = 188.4m2

= l - 0.2m = 62.8m2

= l = 63m

Therefore, length of the trapaulin sheet required is 63m.

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Step-by-step explanation:

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, $\sf{l^2=r^2+h^2}$

__________________________

• $\sf{Here,}$

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

$\large\sf{→l^2=(36+64)}$

$\large\sf{→l^2=100}$

$\large\sf{→l=√100}$

$\large\sf{→l=10}$

__________________________

• $\sf{Again,}$

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)

_________________________

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

$\large\sf{[(L-0.2)×3]=188.4}$

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

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

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

$\large\sf{L=63}$

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