How many candy bars fit in such a box?

Question

A chocolate company has a new candy bar in the shape of a prism whose base is a
1-inch equilateral triangle and whose sides are rectangles that measure 1 inch by 2 inches.
These prisms will be packed in a box that has a regular hexagonal base with 2-inch edges,
and rectangular sides that are 6 inches tall. How many candy bars fit in such a box?

Answer 1

BASE AREA OF THE BOX:

 

  Regular hexagon with 2 inch edges contains 180*(6-2) = 180*4 = 720 degrees

  That's 120 degrees in each of the corner pockets.

  Drawing line segments from the center to each vertex creates 6 congruent isosceles triangles.

  Selecting any one of these six triangles and dropping the height from the center to the edge

  creates a 30-60-90 right triangle with height equal to the square root of 3 (rad3) and base measure 1.

  So the area of each of these triangles is the square root of 3 , or in slang terms "rad 3", which means

   the base area is 6*rad3.

 

The volume of the box is therefore 36*rad3

 

 BASE AREA OF THE CANDY:

 

    Dropping the height from one vertex to the opposite side of this equilateral triangle

    creates 2 right triangles with base measure 1/2 and hypotneuse 1. So the height

    must be rad3/2.  The area of the entire equilateral triangle is rad3/4, which is the base area.

 

   The volume of the candy is rad3/2

 

 

36*rad3  divided by rad3/2 =

 

36 rad3 times 2/rad3  <--- KFC : keep , change, flip

 

 36 * 2    <--    rad3 cancels out

 

72 candies will fill the box.

 

We are ignoring optimization issues.

That is, some pieces of candy may have to be broken in order to fit.