Let Q be the convex hull of a nonconvex bounded polyhedron P

Let Q be the convex hull of a non-convex bounded polyhedron P. (Polyhedron is a three-dimensional solid whose surface is constructed of two-dimensional polygons connected via their edges.) Is it true that the surface area of Q is not greater than the surface area of P ?

Solution

this in false in 3-dimensions , now for the reason

let us connect the convex hull with a nonconvex polyhedron with a very small surface area by thickening a spanning tree, for example. The convex hull Q will have surface area at least as great as the surface area of the convex hull of the points on the polyhedron P . So the

when K = 2, it is true. K=2 represents 2- dimensional spce, so in 2 dimension the convex hull operation decreases the number of times that a line intersects the boundary, so according to the Cauchy surface area formula, it decreases the perimeter.

but for 3 dimensional non convex polyhedrons the statement is false.