# Hexagon area question.

Tags: hexagon
 P: 19 I've found a formula for the area of a regular hexagon,but it seems to falter when i try to finds its area using the apothem sometimes,i know the formula is not wrong because i derived and verified it's authenticity so that can't be it. I heard that by not utilizing the apothem formula, you may be off by a small decimal amount,which is what usually happens to me. Does anyone know what i am talking about? Im willing to provide an example and my formula if anyone has trouble understanding what im talking about. Thank you :)
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,534 Unless you provide more information, it is not clear what you are talking about. A given hexagon has only one area; your formula is correct or it isn't. BTW, you can determine the area of a hexagon (or any polygon) without necessarily using the apothem.
P: 19
 Quote by SteamKing Unless you provide more information, it is not clear what you are talking about. A given hexagon has only one area; your formula is correct or it isn't. BTW, you can determine the area of a hexagon (or any polygon) without necessarily using the apothem.
Thank you mister:)

The formula is ##6x^2√3##, the only difference is that ##x## is half of a side length. (ex:6 is the side length, 3 is ##x##)

For example for a regular hexagon of side length 8 and apothem 7,I get 166.2768...,however in a entirely different youtube video this very same problem gets an area of EXACTLY 168 (using the apothem formula, that is).

That is the confusion that is making me question whether it's right or not.

P: 957
Hexagon area question.

 Quote by shadowboy13 For example for a regular hexagon of side length 8 and apothem 7 [...]
There is no such regular hexagon. With a side length 8, the apothem would be √48 ~= 6.928 and with apothem 7 the side length would be √(196/3) ~= 8.08
P: 19
 Quote by jbriggs444 There is no such regular hexagon. With a side length 8, the apothem would be √48 ~= 6.928 and with apothem 7 the side length would be √(196/3) ~= 8.08
Thank you :)

Don't know why several videos on youtube have been mentioning that those sort of hexagons are "regular" but whatever.

Thanks again :)
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,534 Without URLs for these videos, it is hard to check them out.
P: 19
 Quote by SteamKing Without URLs for these videos, it is hard to check them out.

That's on the very first page of list results mind you, so i wasn't scavenging for results, the fact that the person in that video said that it was a regular hexagon is what got me confused, because like i said, my formula clearly works.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,553 That's almost amusing. As jbriggs444 said, a hexagon with side length 8 would have apothegm of slightly greater than 6.9. Clearly, that has been rounded to "7" but that should have been said. And the area, as given in that video is only approximate although, again, that is not said. One can calculate, more generally, that if a polygon has n sides, then drawing lines from the center to each vertex divides it into n triangles, each with central angle $\frac{2\pi}{n}$ the altitude of each triangle, an apothegm of the polygon, gives a right triangle with one leg the apothegm, the other leg half a side, and angle $\frac{\pi}{n}$. So taking "a" to be the length of the apothegm and "s" the length of a side, $\frac{s/2}{a}= tan(\pi/n)$ so that $a= s/(2 tan(\pi/n))$. In particular, a hexagon with side length 8 has $a= 4/tan(\pi/6)= 4/\sqrt{3}= 6.9282...$ The "area of the octagon" calculation is, similarly, an approximation.
P: 957
 Quote by HallsofIvy In particular, a hexagon with side length 8 has $a= 4/tan(\pi/6)= 4/\sqrt{3}= 6.9282...$
Nitpick on error in intermediate step...

$tan(\pi/6)=1/\sqrt{3}$

so

$4/tan(\pi/6)=4\sqrt{3}=6.9282...$

 Related Discussions Calculus & Beyond Homework 1 Precalculus Mathematics Homework 7 Precalculus Mathematics Homework 4 Calculus & Beyond Homework 6 General Math 4