What length of tarpaulin 3m 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).

Given :
Height (h) of conical tent = 8 cm
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)
\(\Rightarrow \) \(l^2= {8}^2 + {6}^2 {m}^2\)
\(\Rightarrow \) \(l^2= 100 {m}^2\)
\(\Rightarrow \) \({l}^2 = {10}^2 m\)
\(\Rightarrow \) l = 10 m

Therefore, the slant height Of the tent is 10 m.

We know, CSA of tent
= \({\pi}rl\)
= [\(3.14 × 6 × 10 m\)]
= \(188.4 {m}^2\)

Now, let the length of tarpaulin sheet required be l.

As, 20 cm will be wasted, therefore, the effective length will be (l- 0.2) m.

Breadth of tarpaulin = 3m ...(Given)

We have, Area of sheet = CSA of tent
\(\Rightarrow \) [(l - 0.2) × 3]m = \(188.4 {m}^2\)
\(\Rightarrow \) (l - 0.2) m = 62.8 \({m}^2\)
\(\Rightarrow \) l = 73 m.

\(\therefore \) the length of the required tarpaulin sheet will be 63 m.