Question

Prove that the surface area of the outside pyramid is greater then the surface area of the inside pyramid and prove that the perimeter of the outside pyramid is greater then the perimeter of the inside pyramid.

Answer 1

Let inner tetrahedron have edge lengths $(a,b,c,d)$ and outer cone $(A,B,C,D)$.

We have to place the smaller tetrahedron comfortably inside the bigger one so that one vertex of the outer coincide with an inner tetrahedron vertex and other vertices do not touch any of outer tetrahedron faces. By comfortably we mean that avoid touching of outer solid by rotation of inner solid is an enclosing arrangement it will be possible to find corresponding sides.

$(a<A,b<B,c<C,d<D)$ and so $(a+b+c+d)<(A+B+C+D)$

Next Surface Area

Since each face area $=\sqrt{s(s-a)(s-b)(s-c)}$

where $2s=(a+b+c)$

Smaller corresponding tetrahedron faces areas are less in area than bigger tetrahedron face areas so sum is less.