interesting geometry challenge

[Peter Murphy]

The surveyor's formula works here.

Simplify by doubling all linear dimensions (which quadruples all areas). Take the center of the small square as the origin of coordinates, and the coordinates of the vertices in question are then (0,0), (3,1), (3,-3), and (1,-3).

Then the shoelace version of the surveyor's formula yields 9 as the area of the figure.

See here:

<http://en.wikipedia.org/wiki/Shoelace_formula>