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.
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?