Raymond B. answered • 07/08/22
Math, microeconomics or criminal justice
three possible quadrilaterals with those vertices
QRPNQ has area 6
QNRPQ has area = 6
QRNPQ has area = 3
just looking at the graph visually suggests this is the smallest of the three possible quadrilateral areas.
plot the points, connect the points with straight line segments (3 different ways to do it), then
chop up the irregular quadrilateral(s) up into triangles or rectangles, add up their areas
hb/2 for triangle areas, hb for rectangle areas. h=height, b=base
no guarantee the above arithmetic calculators are correct, but it seems to be in that general range: 3 to 6 . Look for answers from others. Someone may have figured this out more.
