Question

What length of tarpaulin $3m$ wide will be required to make conical tent of height $8m$ and base radius $6m$? 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

Tarpaulin will be on the curved surface area of tent.

Curved Surface area of tent$=\pi rl$

$r=6$m, $h=8$m

Let slant height be $=l$

We know that $l^2=h^2+r^2$

$l^2=8^2+6^2=64+36=100$

$\Rightarrow l=10$m

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

Now, Area of tarpaulin material$=$Area of tent

$\Rightarrow l\times b=$Area of tent

$\Rightarrow l\times 3=$Area of tent

$\Rightarrow l=\frac{1}{3}\times$Area of tent

$\Rightarrow l=\frac{188.4}{3}=62.8$m

Now, given that margin is $20$cm

Total length$=$Length calculated$+$Margin

$=62.8$m$+20$cm$=62.8+\frac{20}{100}=62.8+0.2$m$=63$m