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.
Section | Number of girls per thousand boys |
|---|---|
Scheduled Caste (SC) | 940 |
Scheduled Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward districts | 950 |
Non-backward districts | 920 |
Rural | 930 |
Urban | 910 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
