Question

5. 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 20cm (Use Tt= 3.14).
meaning
​

Answer 1

Answer:

h = Height of tent = 8 m

r = Base radius = 6 m

Therefore,

l=

r

2

+h

2

=10 m

Curved surface area of the cone = πrl=3.14×6×10 m

2

Therefore, length of 3 m wide tarpaulin required =

3

3.14×6×10

=62.8 m

Extra length required = 0.2 m

Therefore, total length required = 63 m

Step-by-step explanation:

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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.