Here, Diameter of cone, `d = 10.5m`
So, Radius of cone, `= 10.5/2m`
Height of cone, `= 3m`
Volume of the given heap that is in the form of a cone, `V = 1/3pir^2h`
`V = 1/3**22/7**10.5/2**10.5/2**3=(33*10.5)/4 = 86.625m^3`
Area of canvas required to cover the heap will be its curved surface area.
Curved Surface area of a cone, `A = pirl=pirsqrt(h^2+r^2)`
`A = 22/7**10.5/2sqrt((10.5/2)^2+3^2) = 99.77m^2`