Question

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

Answer 1

Tarpaulin is the curved surface area of tent

The height of tent $h=8m$

And radius of tent $r=6m$

Then slant height of tent $l=\sqrt{r^2+h^2}=\sqrt{(6{)}^2+(8{)}^2}=\sqrt{36+64}=\sqrt{100}=10$

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

Area of tarpaulin material $=$ area of tent

$length\times breadth=Curvedsurfaceareaoftent$

$lenght\times 3=188.4$

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

The wastes in tarpaulin $=20cm=0.2m$

Then length of tarpaulin $=62.8+0.2=63m$