A pencil whose ends are regular hexagons was cut from a cylindrical piece of wood with the least waste possible If the original piece was 8in long and 12in in diameter What is the volume of the pencil?

Answer 1

I'm assuming that you actually mean that the diameter of the original piece of wood 1/2 (0.5) inches rather than 12 inches.

First, you have to find the area of the regular hexagon that forms the cross section of the pencil. To do this, divide the hexagonal cross section into 6 equal equilateral triangles. The lines dividing up these triangles will each by half of the diameter in length...0.25 inches.

Then, divide (bisect) each triangle from the center of the hexagon to the outermost edge. This bisecting line will be the altitude of the triangle. Since each of the 6 triangles is an equilateral triangle, the outermost edge is the same length as the other two sides, making the outermost edge 0.5 inches. So, when you bisect this, the base of the right triangle formed is 0.125 inches.

The height of a 30-60-90 triangle is the square root of 3 times the short side.

So, calculate the height of the triange, multiply it by 1/2 x the base to get the area. Double that to get the area of one of the equilateral triangles, and then multiply by 6 to get the area of the hexagon.

Mutiply that by the length of the unsharpened pencil, and you have about 10.39 cubic inches.