Radius (r) of the heap =(10.52)m=5.25 m
Height (h) of heap = 3 m
Volume of the heap =13πr2h
=(13×227×(5.25)2×3)m3
=86.625 m3
Therefore, the volume of the heap of wheat is 86.625 m3.
Area of canvas required = CSA of cone
=πrl=πr√r2+h2
=[227×5.25×√(5.25)2+(3)2]m2
=(227×5.25×6.05)m2
=99.825 m2
Therefore 99.825 m2canvas will be required to protect the heap from the rain.