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# 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.