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

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.

Ibelieve 4mn because it’s like you’re combining them