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### Calculating the Area of a Hexagon

Date: 2/10/96 at 13:50:55
From: Anonymous
Subject: Area of a hexagon

I've been trying to find a simple way to calculate the area of a
hexagon given only its width between two parallel sides.  I've been
told that this isn't possible..that you need to know more about the
shape first,  but I don't agree.

If you draw a line through the center of a hexagon to connect two
opposing sides like a 'T' crossed on the top and bottom, then do the
same thing for the other two directions, the lines form a kind of pie
shape with six slices.  The outer lines will naturally meet each other,
forming the hexagon's perimeter perfectly, so long as the three cross
lines' angles were correct and equal.  This is not an oblong or oddly
sized hexagon...just a regular symmetrical one.

So, given all of this, I think there should be a simple formula to
calculate the hex's area, similar to the formula used to get the area of
a circle.  So if you've got a hexagon that's 1 meter across, how do
you  find its area?  If you can give me the answer to this, I would
really  appreciate it.

Date: 6/13/96 at 21:5:26
From: Doctor Jim
Subject: Re: Area of a hexagon

Assuming the hexagon is regular, that is, all sides and angles are
equal, then you are correct that there is a fairly easy way to find the
area: find the area of one of the six triangles and multiply by 6.

Since the hexagon is one meter across, then the height of the
triangle is 1/2 meter.  Using the Pythagorean theorem gives us that
half of the base of the triangle is (1/2)/squareroot3 or 1/(2sqrt3),
which means the base of the triangle is 1/sqrt3. The area of the
triangle is then 1/2*1/2*1/sqrt3 = 1/(4sqrt3) and the area of the
hexagon is 6*1/(4sqrt3) = 3/(2sqrt3)~.87squares.

-Doctor Jim,  The Math Forum

Date: 12/26/2008 at 14:51:56
From: Chickenhawk
Subject: Area of a hexagon

Your solution works only for the special case where the measurement across the flats equals one. One meter, millimeter, inch or whatever.

For example, as a fitter and turner I was recently required to find the cross sectional area of a piece of rusty old 2 1/2" hexagonal steel bar. It measured 63.4mm across the flats. (Steel bar always seems to be provided a bit under size and it WAS rusty.)

I invented the formula

The square of (1/2 times the across-the-flats-measurement), times 6,
times the tangent of 30 equals the crossectional area.

for myself and solved the problem. (I couldn't find anything in the engineering books we had that could do the job.)

The solution you offered can still be used to solve general problems of this type if you multiply the result, 3/(2sqrt3), by the square of the across-the-flats measurement.  (This still works when that measurement is 1.)

As a matter of interest, 3/(2sqrt3) is the cosine of 30 degrees.

So! Now we are left with a general formula for finding the area of regular hexagons:

(x)^2  * cos 30 = A

where

x is the across-the-flats measurement

A is the area in square units of that measurement
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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