The volume of a truncated cone would be the volume of an untruncated cone minus the volume of the bit chopped off.

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 ...

A truncated cone is basically a normal cone with the top cut off. That means that to find the volume, we need the radius of the top circle, the radius of the bottom circle and the height. If we only know one the height and the volume, then we cannot find the other two because it is two variables but only one equation.