Circle inscribe a hexagon

shahar

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Circle inscribes a hexagon.
1668170300943.png
What is the circumference of the hexagon? (Hint: draw the triangle AOB and calculate the length side of the hexagon).
My solution -
The hexagon is regular hexagon (*1) because it inscribe by circle.
The angle AOB is part of a isosceles triangle and the size AO equals to the size OB.
O.K, from here I'm stuck.
Is *1 is explained right?
How I find the side of AB?
I knows that when I find AB I can calculate the circumference by multiplying it by 6.
 
Circle inscribes a hexagon.
View attachment 34497
What is the circumference of the hexagon? (Hint: draw the triangle AOB and calculate the length side of the hexagon).
My solution -
The hexagon is regular hexagon (*1) because it inscribe by circle.
The angle AOB is part of a isosceles triangle and the size AO equals to the size OB.
O.K, from here I'm stuck.
Is *1 is explained right?
How I find the side of AB?
I knows that when I find AB I can calculate the circumference by multiplying it by 6.
First, you are misusing the word "inscribe"; the hexagon is inscribed in the circle.

But the key to this problem is that triangle AOB is not merely isosceles; it is equilateral. Can you see why?
 
OK. I saw it but how does I explain that the trangle is equilateral? What is the reason? Which axioms can I assure that fact?
 
I see that it is 60 degree. But Why?
If you watch carefully the construction using a compos, you see that each side has length equal to the radius:
i.e. AB=r|\overline{AB}|=r
 
If you watch carefully the construction using a compos, you see that each side has length equal to the radius:
i.e. AB=r|\overline{AB}|=r
I know that question can be explained by Euclid Axioms, Isn't it?
 
I see that it is 60 degree. But Why?
I know that question can be explained by Euclid Axioms, Isn't it?

Have you thought about how to answer your own question? That's what I tried to lead you into by asking a question. Please make some attempt to think, or it is useless to try to teach you anything.

Do you see that there are six congruent angles meeting at the center? If they add up to 360 degrees, what is the measure of each of them?

Then, do you see how that implies that each of the triangles is equilateral?
 
Have you thought about how to answer your own question? That's what I tried to lead you into by asking a question. Please make some attempt to think, or it is useless to try to teach you anything.

Do you see that there are six congruent angles meeting at the center? If they add up to 360 degrees, what is the measure of each of them?

Then, do you see how that implies that each of the triangles is equilateral?
Thanks. Now I see. All the six angles create 360 degrees angle so because they are equal every angle is 60 degrees.
 
Circle inscribes a hexagon.
What is the circumference of the hexagon?
How I find the side of AB?
I knows that when I find AB I can calculate the circumference by multiplying it by 6.
I suspect English may not be your first language, @shahar, yes?

There’s a number of (linguistic) “errors” in your OP.

1. As referred to already, the circle circumscribes the hexagon and the hexagon is inscribed in(side) the circle.
2, A hexagon does not have a circumference, rather it has a perimeter.

The problem does not require reference to Euclidean "Axioms" but just the well-known properties of the simple geometric shapes: hexagons & triangles (though these were described by Euclid).

You (now) seem to have grasped the idea that a regular hexagon comprises six equilateral triangles since all it's internal angles are equal (to 60°; ie: 360° ÷ 6) (See diagram, below.)
Then, considering (your) triangle, AOB, which you correctly identified as isosceles (since OA is the same length as OB) but, since the apex (at O) is 60°, then the other two angles must add to 120° (180° - 60°) and since it is
isosceles they must also be the same (ie: 120° ÷ 2 = 60°), therefore, the triangle is not just isosceles but actually equilateral!
330px-Regular-polygon-6-annotated.jpg
And, since we now know that the length of AB is the same as OA & OB (since ΔAOB is equilateral)....
I knows that when I find AB I can calculate the circumference by multiplying it by 6.
No, you may calculate it's perimeter by multiplying AB by 6, however, if it's actually the circumference of the circle that you need to calculate then, since you know it's diameter (2 × OA) you can easily calculate (using π\displaystyle \pi) the circumference of the circle to any stated level of accuracy (eg: any given number of decimal places) if that is what is actually required. ?
 
Yes, which you'd posted after the concept had been addressed in posts #9,#10. ;)
"posts #9,#10" were both submitted whilst I was in the process of composing my post (#11) but, having read them (both) before I submitted my post I, at least, acknowledged the fact that "reference" had been made to the measure of the internal angles (though not the explicit 'calculation': 360/6=60 which I (had already) included in my composition and saw no need to remove it as it simply stated explicitly what the OP had already grasped).
Of course, had @Steven G bothered to read my post, then s/he would have realised that their post (#12) was entirely redundant and unneccesary but, of course, none of the Elite bother to pay any attention to anything I post (regardless of how helpful to OP's it might be) since I became persona non grata in here (mainly due to your 'attentions').
Doubtless the remainder (if not all) of this post will (also) be edited (or deleted) as "unnecessary diatribe" as you are the 'exception', combing through everything I write to nitpick over any small thing you can find to 'accuse' me of.
Even, as here, springing to the defence (of the indefensible) of a fellow Elite who hasn't bothered to read what was previously posted (by little old me) before adding their own tautologically worthless comment (and then having the effrontery to express surprise/shock(?) at my pointing out their unwarranted contribution!
Please stop picking on me, @Otis, it's really beneath your dignity; please just do as your fellow Elite do and ignore me. ?
(Unless there is an error in my Maths, then I will happily welcome any corrections offered. ?)
 
The Thin-Skinned Highlander – There can be only one!! :p
  \;
 
The Thin-Skinned Highlander – There can be only one!! :p
  \;
Those kilts. The skin gets chafed raw by the cold, the damp, and the heather. Let’s avoid thick-skinnedism.
 
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