The volume of a normal Cone = π × r² × (h/3)

If the cone were chopped off at "y" then we would remove a smaller cone, whose height would be (h-y) and whose base radius would be r × (h-y)/h

Hence the "removed cone"s volume would = π × (r × (h-y)/h)² × ((h-y)/3)

So the Volume of the truncated cone = π × r² × (h/3) - π × (r × (h-y)/h)² × ((h-y)/3)

The areas of a normal cone are:

Base = π × r²

Surface Area of Side = π × r × s (where s = side length)

The areas of a truncated cone are (h=height of normal cone, y=height of truncated cone):

Base = π × r²

Top = π × (r × (h-y)/h)²

Surface Area of Side = π × r × s × (y/h)

That's a start, but I gotta go ...

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