Answer :

We have,


r adius = 9/2 = 4.5 m


height = 3.5 m


slant height l == = 5.70 m


Volume of the heap =


= ×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


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