Question

What length of tarpaulin3 m wide will be required to make conical tent of height 8 m
und 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 (Usea= 3,14),​

Answer 1

Answer:

Tarpaulin is the curved surface area of tent

The height of tent h=8m

And radius of tent r=6m

Then slant height of tent l=

r

2

+h

2

=

(6)

2

+(8)

2

=

36+64

=

100

=10

Curved surface area of tent =3.14×6×10=188.4m

2

Area of tarpaulin material = area of tent

length×breadth=Curved surface area of tent

lenght×3=188.4

⇒Length=

3

188.4

=62.8m

The wastes in tarpaulin =20cm=0.2m

Then length of tarpaulin =62.8+0.2=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, $\sf{l^2=r^2+h^2}$

__________________________

$\sf{Here,}$

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

$\large\sf{→l^2=(36+64)}$

$\large\sf{→l^2=100}$

$\large\sf{→l=√100}$

$\large\sf{→l=10}$

__________________________

$\sf{Again,}$

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.