Showing posts from November, 2010

More estimates

Last time, I talked about the height of a hexagon; the strategy I used works with any shape with a pair of parallel sides.  The parallel sides allow you to stack the polygons and add up the heights without having to worry about angles. For instance, it will work with a regular octagon.  The picture to the left shows 5 stacked octagons that are slightly more than 12 square (otherwise in TileLand the top purple square would be wiped out).  That means that 5 octagons > 12 squares or that  1 octagon > 12/5 =2.4 (since they are unit squares). 
For those who like to exact lengths, you can break the height of an octagon up into three pieces.  This diagram below demonstrates the breakdown: two half squares of length 1/ root 2  (since the diagonal is 1) and a rectangle with the long side of length 1.  Root 2 plus 1 is 1.41421+1 = 2.41421.   

Look forward to when I tackle odd sided polygons that don't have parallel sides.

Root 3

Who said approximating can't be fun?  This is an activity that I designed to get geometry students to come up with estimates for the height of a hexagon.  The tool I asked them to use was Tileland.  Tileland is a world of polygon paths (well known to those who follow this blog regularly) that doesn't allow for polygons to overlap (the top green squares replaced vertical squares that were there previously).  The following polygon path demonstrates an estimate for the height of a hexagon with unit length sides that is pretty good.  Here the height of four hexagons is slightly less than seven squares; this means that the height of one hexagon is slight less than 7/4 = 1.75.  Using the diagram of the hexagon composed of equilateral triangles and the Pythagorean Theorem, it is easy to show that the hexagon's height is actually root 3. Thus root 3 must be slightly under 1.75.  It is pretty close since root 3 is about 1.732--so within 0.02.

The reason I prefer to use types of appr…