Here are some ways of finding the area of non-regular hexagons

1. Tessellating Hexagons - method 1

With a tessellating hexagon that has all opposite sides parallel and of equal length:

Choose any three alternate vertices. Join two of them with a line (length A); drop a perpendicular from the third vertex to this line (length B)

The hexagon area is simply A x B

This works because parallellograms can be laid over the tessellation grid so that the vertices coincide with those we just chose. The area of the parallellogram is therefore the same as that of the hexagon.

This diagram shows the hexagon dissected to fit into the parallellogram.

2. Tessellating Hexagons - method 2

With a tessellating hexagon that has reflective symmetry about a line through two opposite vertices:

Take the measurements shown:

The area of the centre rectangle is A x B. The top and bottom triangle can be combined into a rectangle (by horizontal shearing and then translation) with area A x (C-B)/2

Adding these two elements and simplifying, we get the area of the hexagon as: {A x (B + C)}/2

3. Three squares joined at the corners

The 3 squares touch each other, and we wish to find the area of the whole hexagon. That will be the area of the 3 squares, plus the area of the 4 triangles. Let's label some points:

Now we concentrate on the 2 triangles ABC and CDE:

Rotate CDE counterclockwise through 90 degrees, so points B and E coincide:

The area of the triangles is one half x base length x height; both triangles have the base length and height, and hence the same area.

The same can be said of each of the 3 outer triangles - all 4 triangles in the hexagon have the same area. We can use Hero's formula to find that area for the middle triangle with sides x, y and z.

(Hero was an ancient Greek bloke, sometimes called Heron. Derivations of his formula can be found here and here)

4. Equilateral Hexagon with two rightangles

In this diagram, all sides are of equal length s:

Cut off the 2 blue triangles, which conveniently form a square together, and add the square's area s x s to the calculated area of the remaining green rectangle s x s x sqrt(2) (using Pythagoras on a blue triangle to find the length of its hypoteneuse). The final area is therefore (s x s).(1 + sqrt(2))

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