Question

What length if tarpaulin 3m wide will be required to make a conical tent of height 8m and base radius 6? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm ( use pie = 3.14)

Answer 1

Breadth of tarpaulin= 3m.

length of tarpaulin=?

so area of tarpaulin = CSA of conical tent

l×b= 3.14×radius×slant height

slant height= √h^2+r^2

= √8^2+6^2

= √64+36

= √100

= 10m

therefore

l×b= 3.14× 6 ×10

l×3 = 188.4

l= 62.8m

waste cloth= 20 cm = 0.2 m

therefore total length of tarpaulin is 62.8+0.2m

=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.