how do you calclulate the area of a reguleat hexagon?

[dragonstar57]

side length of 6

[math] \frac{6*6*6\sqrt{3}}{2} [/math].

what now?

how do i do that?

ps the answer is [math]54 \sqrt{3}[/math] I just don't know how to get there

Edited June 8, 2012 by dragonstar57

[swansont]

It might help you to recognize that a regular hexagon is comprised of 6 equilateral triangles.

[dragonstar57]

It might help you to recognize that a regular hexagon is comprised of 6 equilateral triangles.

that is how i figured out the a was[math] 6\sqrt{3} [/math].

Edited June 8, 2012 by dragonstar57

[Joatmon]

You have a formula and you have an answer. If the formula produces a different value to the answer then either the formula or the answer is wrong. If you have been given one of them then surely the other must be wrong. Are you trying to derive a formula or were you given both pieces of information?

[dragonstar57]

You have a formula and you have an answer. If the formula produces a different value to the answer then either the formula or the answer is wrong. If you have been given one of them then surely the other must be wrong. Are you trying to derive a formula or were you given both pieces of information?

given both

[math] \frac{asn}{2} [/math].

n=6 b/c it is a hexagon

s=6: given

a=6 √3 because it is a long leg of a 30,60,90 special triangle which means it is s√3

Edited June 8, 2012 by dragonstar57

[Joatmon]

You have 6 equilateral triangles. I expect you know the formula for the area of any triangle. I expect you can calculate the height of each equilateral triangle. Perhaps you have been given some incorrect information?

 

edit - its UK bedtime now, but hope this is some help.

Edited June 8, 2012 by Joatmon

[dragonstar57]

You have 6 equilateral triangles. I expect you know the formula for the area of any triangle. I expect you can calculate the height of each equilateral triangle. Perhaps you have been given some incorrect information?

 

edit - its UK bedtime now, but hope this is some help.

yeah i tried that i get 106 root 3

[Joatmon]

I have a couple of problems. I'm not too sure what you are trying to prove. Also, if I were asked to calculate the area of a regular hexagon I think I would do it differently to the way you seem to have been taught.

Anyway let's see if I can help a bit.

If I understand the formula you give and work it out I get approx 187.

I make the answer you give approx 93.5 (which is half what I got for the formula).

Assuming I understand the formula, one of these must be wrong.

As you probably know your hexagon fits neatly into a circle of radius 6 so it's area will be a bit less than such a circle.

I make the area of such a circle approx 113.1 which rules out the formula you have been given and points you toward the answer given.(assuming I understand the formula).

In your formula it looks to me as if one of the 6's should be a 3 - alternatively perhaps the 2 should be a 4.

Now how would I solve the problem myself?

As pointed out by Swansont your hexagon can be divided into 6 equilateral triangles.

I make the height of any one of the triangles Approx 5.196

I therefore make the area of any one of the triangles Approx 15.588

And so I make the area of the whole hexagon Approx 93.5. (Which agrees with the answer given)

Note: I used Pythagoras to calculate triangle height. This may be where your method differs from mine?

Because this is homework I have not shown any formulae used.

 

Also because I'm quite rusty I would appreciate it if anyone else spots a mistake

Edited June 9, 2012 by Joatmon

[swansont]

that is how i figured out the a was[math] 6\sqrt{3} [/math].

 

Do you find it curious that your answer is twice what it should be?

 

You have a base of 6.

 

[math] 6\sqrt{3} > 6 [/math] . Does that make sense?

[Joatmon]

Do you find it curious that your answer is twice what it should be?

 

You have a base of 6.

 

[math] 6\sqrt{3} > 6 [/math] . Does that make sense?

 

This is why I wondered about my interpretation of the formula - The 6 being the same size as the 3 left me in doubt about whether I was dealing with a root or a multiplication. Thanks for clearing that up.

 

Later edit: did a bit of scraping away the rust and I think the formula given in #1 is incorrect.

It seems to me that it should have been (6*6*6*(3^0.5))/4.

Edited June 9, 2012 by Joatmon

[Auto Engineer]

side length of 6

[math] \frac{6*6*6\sqrt{3}}{2} [/math].

what now?

how do i do that?

ps the answer is [math]54 \sqrt{3}[/math] I just don't know how to get there

 

Although we could show you how it is done, this would not help you to fully understand how to solve problems, which is what you are up against.

 

It would be better if you put some effort into trying to solve the problem first and ensure you have the right information from the question, then we can if necessary guide you where you might make slight errors along the way.

 

To help you along the correct path have a look and read of this link;

 

http://en.wikipedia.org/wiki/Hexagon

 

Auto Engineer

[swansont]

Later edit: did a bit of scraping away the rust and I think the formula given in #1 is incorrect.

It seems to me that it should have been (6*6*6*(3^0.5))/4.

 

Yes. There is a missing factor of two in the denominator, either from the formula for area of a triangle (1/2 b*h) or in the trig function (cos30 = sin60 = ([math]\sqrt{3}/2[/math])

[Joatmon]

Yes. There is a missing factor of two in the denominator, either from the formula for area of a triangle (1/2 b*h) or in the trig function (cos30 = sin60 = ([math]\sqrt{3}/2[/math])

Thanks - relieved waking up the rather old grey matter didn't lead to a cock-up!

[doG]

[math] \frac{6*6*6\sqrt{3}}{2} [/math].

This is wrong. Back up and start over.

 

ps the answer is [math]54 \sqrt{3}[/math] I just don't know how to get there

This is right.

 

There is another approach you can try that might help you understand the area of a polygon in general. In the case of your hexagon it can be divided into 6 equilateral triangles which is unique to hexagons among polygons in general. It can also be divided into right triangles as can any other polygon. This fact can be used to derive a general equation for the area of any polygon with n sides of length s.

Edited June 9, 2012 by doG

[dragonstar57]

I used the side length as the base when locating the apothem

so instead of [math] \frac{6*6*6\sqrt{3}}{2} [/math].

it should be

[math] \frac{6*6*3\sqrt{3}}{2} [/math].

[doG]

I used the side length as the base when locating the apothem

so instead of [math] \frac{6*6*6\sqrt{3}}{2} [/math].

it should be

[math] \frac{6*6*3\sqrt{3}}{2} [/math].

Yay! You can see the equation for this at MathWorld.