Question

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 $\pi$ = 3.14).

Answer 1

Radius, r = 6 m
Height, h = 8 m
Slant height, l = $\sqrt{r² + h²}$
$= \sqrt{(6)² + (8)²}$
$= \sqrt{36 + 64}$
$= \sqrt{100}$
= 10 m

The curved surface area =$\pi rl$
= 3.14 × 6m × 10m
= 188.4 m2

Width of the tarpaulin = 3m
Area of the tarpaulin = 188.4 m2

∴ Area of the tarpaulin = width of the tarpaulin × length of the tarpaulin

188.4 m2 = 3 × length of the tarpaulin

$⇒$ Length of the tarpaulin = $\frac{188.4 m^2}{3}$
= 62.8 m

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

Actual length required = 62.8m + 0.2m = 63m

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