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 π = 3.14).

Given,

Height of tent (h) = 8 m

Radius of tent (r) = 6 m

By formula,

Slant height of tent (l) = $\sqrt{r^2 + h^2}$

Substituting values we get :

$l = \sqrt{(6)^2 + (8)^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} = 10 m.$

By formula,

Curved surface area of cone = πrl

= 3.14 × 6 m × 10 m

= 188.4 m².

Area of tarpaulin used to make tent will be equal to curved surface area of tent.

So,

Area of the tarpaulin = Width of the tarpaulin × Length of the tarpaulin

⇒ 188.4 = 3 × length of the tarpaulin

⇒ Length of the tarpaulin = $\dfrac{188.4}{3}$ = 62.8 m

Extra length of the material = 20 cm = $\dfrac{20}{100}$ m = 0.2 m

Total length required = 62.8 m + 0.2 m = 63 m.

Hence, the required length of the tarpaulin is 63 m.