Math CentralQuandaries & Queries


Question from Vicki, a student:

I need to find out the area of a hexagon (please explain thoroughly; I'm not the best at math) and why/how that particular formula works.
Also, it would be nice if there was an example question with a series of steps (as in not just finding the area and being given the information to do so) that involved both the perimeter and area of a hexagon.
Thank you.

There isn't,as far as I know, any elegant formula for the area of a hexagon (or other polygon with several sides). Instead, unless it has some very special properties, you break it up into triangles and add their area.

Dividing up: Draw your hexagon, and add a set of non-crossing diagonals that break it up into triangles. If you can make one of the triangles right, or with one vertex an easily-computed distance from the opposite side, this is good.

How do you find the area? That depends on how the data for the polygon are given. It's better to think of these as a toolkit that you decide which bits of to use as you go, rather than as a single algorithm to apply the same way each time.

Data given as edge lengths and angles: You _may_ have enough data to use

1/2 bh

(half of base times height) but height may not be easily available.

You will probably need lengths of diagonals. These will be found using the sine law

sin(A)/a = sin(B)/b = sin(C)/c

where A,B,C are the measures of the angles opposite sides of length a,b,c; and the cosine law (which is like Pythagoras' theorem for non-right triangles)

a2 = b2 + c2 - 2bc cos(A)
b2 = c2 + a2 - 2ca cos(B)
c2 = a2 + b2 - 2ab cos(C)

Areas are found using Heron's formula in terms of edge lenghts a,b,c and the semiperimeter (a+b+c)/2

Area2 = s(s-a)(s-b)(s-c)

or from two edges and the included angle using (eg)

Area = (1/2) ab sin(C).

Data given as coordinates of points: Usually less "preprocessing" is needed here. The area of the triangle with vertices


is given by the absolute value of

[(y1-x1)(z2-x2) - (y2-x2)(z1-x1)]/2

Good Hunting!



If your hexagon is regular you can use Stephen's approach.


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