Anyone here get their Higher School Certificate (or non-Australian
equivalent?) Draw a line the scaled length of the longest side. Using a
compass, draw arcs from the endpoints for the 2nd longest and shortest
lengths. With a ruler or pair of dividers, find two points on those arcs
that are the scaled distance of the remaining sides apart. This gives a
diagram of one of an infinite number of solutions. Construct a sqaure with
sides the length of the shortest side and place it flush against that
shortest side. Extend the opposite side of the box. You have now converted
your shape - ALL of the possible shapes - into 3 triangles and 1 box, all of
which have the length of at least one side defined, some with two, and a
second side in common with something else. Solve these as simultanious trig
equations until you have the lengths of 2 sides defined for all 3 triangles.
Area of a triangle is half length times height, so the area of the resulting
shape, regardless of which unique solution has been found for the overall
shape, is the sum of the areas of the square and the three triangles.
Mike Bourke