Question

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 20 cm (Use pi = 3.14).

Answer 1

Given is a conical tent with :

Height( h )  =  8 meter

Radius ( r ) = 6 meter

$So, its length = l^{2} = h^{2} + r^{2}$

$= l^{2}=8^{2} + 6^{2}$

$= l^{2} =[tex]64 m^{2}  + 36 m^{2}$

$= l^{2}=100m$

$\therefore\:l\:=10m$

Curved surface area of the cone = $\pi r l$

= 3.14 x 6m x 10m

$= 188.4 cm^{2}$

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

So, its length = l - 0.2 m

And its width = 3 meter

Area of sheet = CSA of tent

So the area of the tarpaulin = l x b

$= ( l - 0.2m ) x 3m = 188.4m^{2}$

= l - 0.2m = 62.8m

$\therefore\:length\:=63m$

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

Thanks

Answer 2

63 m long trampoline is required.



STEP-BY-STEP EXPLANATION


Height of tent = 8 m

Radius of tent = 6 m

The shape of the tent is conical.

So,
The slant height of the conical tent is equal to :-

$l = \sqrt{ {(h)}^{2} + {(r)}^{2} } \\ \\ l = \sqrt{64 + 36} = \sqrt{100} = 10 \: m$


Now,

We have
R= 6 m
L= 10 m

And also it is given that value of pie is 3.14


Finding Curved surface area of tent :-
(Area of trampoline required)

$\pi \: r \: l = 3.14 \times 6 \times 10 \\ \\ \pi \: r \: l = 188.4$


So,
The curved surface area of cone is 188.40 m²


Now,

Length of the trampoline = x m
Breadth is given that is 3 m

So,

Area of trampoline = Curved surface area of tent

So,

$3x = 188.4 \\ \\ x = \frac{188.4}{3} = 62.80 \: m$


Now

Extra length required is 20 cm

So,

$20 \: cm = \frac{20}{100} = 0.2m$


So,

Actual length of trampoline required is equal to :-


$62.8 + 0.2 = 63 \: m$


Hence,

Total length of trampoline required is 63 m