interesting geometry challenge

[Thomas Klemm]

(09-18-2014 02:34 PM)Peter Murphy Wrote:  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.

There are programs in the General Software Library:
HP-48
HP-42S

Or then you can use Pick's theorem:


i = 6
b = 8
A = i + b/2 − 1 = 6 + 4 - 1 = 9

Cheers
Thomas