Believe it or not the solution to this problem did not come as easy as it should have to our group. For several days we sat together throwing our ideas at one another. Many fell just short of the mark, (or so we thought). It was not until yesterday that we found a formula that could be applied to all types of polygons, albeit loosely. However, after realizing that this would work, with a lot of extra figuring, we were given a hint to help make a previous method work that is much easier.In trying to develop a formula we simply decided to brainstorm. Something that we immediately thought of was dividing the figure into triangles and then adding area. However, the number of triangles chosen did not have to be equal and could be chosen arbitrarily, so we tried to refine it. Eventually we decided to place the figure in a parallelogram, so that as many sides as possible were flush with the parallelogram. The following drawings, done artfully by Kari Mankey, illustrate exactly what we mean by this. Our First Solution
We quickly decided that by placing the figures in the parallelogram a basic equation to find the area of all of the figures follows.
a(parallelogram) - a(triangles that lie outside original figure) = a(original figure)
A few problems arise from this however, that are easily worked out. The square, rectangle and parallelogram can be bounded by a parallelogram that lies directly over the original figure; i.e., there are no triangles to be subtracted from the original area.
Mindy Mueller, our resident trapezoid wizard, threw out an idea early on that we thought was excellent, except for one hitch. The formula to find the area of a trapezoid can be used to find the area of all the figures except the triangle (a triangle only has one base). But upon realizing we could substitute zero in for the negative area triangles in our original solution,why couldn't we substitute zero in for the other base of the triangle? Eureka!
When applied to the problem at hand, this formula works well. However, it will not work for any polygon, just the ones listed. Our Second Solution
This solution works for all of the figures listed if b2 for the triangle is taken to be zero. If you don't believe us try it yourself. A=1/2h(b1+b2)
This page is maintained by Drew Schaub, Head Clown at Bozo & Co.
