Let side = a and apothem = h. Note this is the height of a triangle.

Area of a triangle = half base times height = 0.5 ah.

So area of hexagon = 6 times 0.5 ah = 3ah. But perimenter = P = 6a so area = 0.5 times 6ah = 0.5 hP.

Using Pythag on a half triangle h = √ [a^2 - (0.5a)^2] = √3 a/2

so area = 3ah = 3a .√3.a/2 = 3√3 a^2 /2.

Bob

]]>You're talking about a regular hexagon I think; so the answer is yes. You can divide the hexagon into 6 equilateral triangles each with side = the same as for the hexagon itself. Let's say side = a. Then perimeter = 6a and diagonal = 2a.

Bob

Thanks, Bob, but someone elsewhere has pointed out to me an error in my thinking.

I've been thinking about this in relation to my sqi ar ^2 idea.

But with sqi ar ^2 I was using the centre of the square to the centre of a side for the 'radius'. To be consistent I should be using that for the 'radius' of the regular hexagon

I found out that's called the apothem (which means I can dispense with the 'ar' for 'radius' in my formula and now use 'a' for apothem.

So,

sqi(a^2)

And,

hxi(a^2)

respectively, are the formulae

But I got 2 different answers for the value of hxi

Because I got 2 different answers for the area of the regular hexagon (sides 6cm)

I got 2 different formulae for the area online;

(3√ 3/(2))*a^2 where a=length of sides

and

0.5(a)(P) where a=apothem and P= perimeter

apothem seems to equal 5.5cm (?)

]]>Bob

]]>pi for a circle = 3.14

pi for a square (calling that sqi) = 4

pi for a hexagon (calling that hxi) = 3?

]]>