Question

Figure shows a regular hexagonal prism and a solid of 12 congruent triangular faces. The vertices of this polyhedron are at the center of the faces of the prism. What is the ratio between the volume of the solid and the volume of the prism?

Answer 1

http://www.revista.vestibular.uerj.br/lib/spaw2/uploads/images/2013/ED/Matematica/Questao_10.png

Answer 2

Assume side of the hexagon is a and the height of the hexagon prism is h.
Volume of prism = area of base * height.

Answer 3

ok..

Answer 4

The internal angles of a hexagon are 120 degrees. SO half the angle is 60 degrees.
tan(60) = x / (a/2). x = a/2*tan(60) = sqrt(3) * a/2
Area of the triangle = 1/2 * base * height = 1/2 * a * sqrt(3) * a/2
= sqrt(3)/4 * a^2
Area of hexagon = 6 * area of triangle = 6 * sqrt(3)/4 * a^2 = 3sqrt(3)/2 * a^2

Volume of hexagon prism = Area of hexagon base * height
= 3sqrt(3)/2 * a^2 * h

Answer 5

wait, why did you use the tan(60)?

Answer 6

I get it, right

Answer 7

The polyhedron vertices are located at the center of the faces of the prism.
Find the area of the triangle OAB. I will leave this to you.
Note: AC = CB = a/2.