We have,
radius = 9/2 = 4.5 m
height = 3.5 m
∴ slant height l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(4.5)^2+(3.5)^2}\)
= 5.70 m
∴ Volume of the heap = \(\frac{1}{3}\)πr2h
= \(\frac{1}{3}\)×3.14×(4.5)2×3.5
= 74.1825 m3
Area of canvas required = πrl
= 3.14×4.5×5.7 m2
= 80.54 m2