Suppose there is a cone of height 3h,slant height 3l & radius r.
Dividing the cone into two equal volumes.ie,cutting at height h from the base & parallel to the base.
So resulting area of top object(which is a cone) is
pi*r1*2l(LSA) + Area of base
r1^2 = (2l)^2 - (2h)^2 = 4(l^2 - h^2)
r / r1 = 3/2 -> r1 = 2r/3
so, Total surface area of top cone = pi*2r*2l/3(LSA) + Area of base.
Now,area of bottom object = area of top base + area of bottom base + LSA
LSA = LSA of original cone - LSA of top cone
= pi*r*3l - pi*2r*3l/3 = pi*r*l
Area of bottom base = pi*r*r
So, comparing both objects' areas,
pi*4*rl/3 + area of base = pi*r*l + area of top base + pi*r*r
area of base of top object = area of top base of bottom object,so both will be cancelled.
4rl/3= rl + r*r
rl/3 = r*r
ie l = 3r
so, the areas will be equal only when l = 3r. For all other cases, areas will be unequal.
Conclusion: For any other object too, there will be a special case where the areas of the resulting objects with equal volume, will be equal, but for all other cases,it'll be unequal.