Question

What length of tarpaulin $3$ 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. ($\pi =3.14$)

Answer 1

Tarpaulin will be on the curved surface of the tent.

Curved surface area of the tent$=\pi rl$

$r=6m,h=8m$ and let $l$ be the slant height.

We know that,

$l^2=r^2+h^2$

$=6^2+8^2={10}^2$

$\therefore l=10m$

$\therefore$ Curved surface area of the tent$=\pi rl=3.14\times 6\times 10=188.4m^2$

Now,

Area of tarpaulin material$=$ Area of tent

$\Rightarrow Length\times Breadth=Areaoftent$

$\Rightarrow Length\times 3=Areaoftent$

$\Rightarrow Length=\frac{1}{3}\times Areaoftent=\frac{188.4}{3}=62.8m$

$L=62.8m$

Margin$=20cm$

$\therefore$Total length$=62.8+0.2=63m$