Find the area, in square units, of the quadrilateral whose vertices taken in order are (0, 0), (6, 0), (8, 5) and (0, 5).

The side from (0, 0) to (6, 0) lies on $y=0$ and the side from (0, 5) to (8, 5) lies on $y=5$, so the figure is a trapezium with parallel sides 6 and 8 units and height 5 units. Area = ½ × $(6+8)\times 5=35$ square units. (The shoelace sum 0 + 30 + 40 + 0 = 70, halved, gives the same result.)