Determine the number of rectangles and hexagons in the given figure.

In a compound figure built from several overlapping elementary regions, classify every candidate rectangle by how many elementary regions it is made of (1-region, 2-region, 3-region, and so on up to the largest combination), count each class separately, then add the classes together.

A hexagon in such a figure is any closed six-vertex boundary you can trace using the figure's labelled points; find every one by systematically tracing candidate six-sided boundaries around the diagram rather than guessing.

Label the figure as shown, with the outer points A-P and the inner points Q-W marking every intersection.

Total rectangles = 4 + 4 + 4 + 5 + 4 + 2 + 4 + 2 + 1 = 30.

Tracing every six-vertex boundary the same way gives these hexagons:

• CDEKLM

• CEUKMQ

• CFHJMQ

• BEUKNP

• BFHJNP

That gives 5 distinct hexagons.

Cross-checking against the other pairings shows why each is off:

• 32 rectangles over-counts by double-reading some of the larger multi-component rectangles (an eight- or ten-component rectangle counted from two overlapping directions), and 3 hexagons under-counts by missing two of the six-vertex boundaries that share edge points with the outer border.

• 28 rectangles under-counts by dropping an entire four-component class (shapes such as CEKM or VGHS, each built from two adjacent quadrant pieces).

• 3 hexagons under-counts for the same reason as above - two boundary hexagons sharing edge points with the outer border get missed when the trace isn't carried all the way around the figure.

So the figure contains 30 rectangles and 5 hexagons.