07 Oct '08 13:06>
Originally posted by joe shmoActually, I was wrong, and my mistake was somewhere in the second triangle...
that looks to technical to be false, so you da person geepamoogle.
I did find the easiest method and it says 108 as the answer.
First I find the smallest rectangular window that fits all 4 points, such that each boundary has at least 1 point on it. That rectangle in this case will be 9x16 rectangle with an area of 144.
X from -4 to +5, Y from -9 to +7
Now I can chop off rectangles and right triangles until all the remaining area is within the quadrilateral indicated.
Now (-3,5) is the only point not on the edge, so I take off the 1x2 rectangle from it to the outside corner, leaving an area of 142 left.
That leaves 4 right triangles to carve away. The 1x11 and 7x2 both border the rectangle we took out. There is also a 1x16 and a 9x3.
Total area of these four are (11+14+16+27)/2 = 34, which leaves an Area of 108 in the quadrilateral.
I did relook at my numbers from before, and I still get 37.5 for my first Heron's triangle, but the second should be 70.5, so I must have messed up somewhere in it.