Quote by HallsofIvy
No, it's not. The radius of the inscribed circle is the distance from the center of the hexagon to the center of one side. The length of one side of the regular hexagon is equal to the distance from the center of the hexagon to a vertex of the hexagon. If we take "x" as the length of a side of the hexagon, then drawing the line segment from the center of the hexagon to the center of a side gives a right triangle with one leg of length x/2 and hypotenuse of length x. The length of the other leg is [itex]\sqrt{x^2 x^2/4}= x\sqrt{3}/2[/itex].
That is, the radius of a circle inscribed in a hexagon of side length x is
[tex]\frac{x\sqrt{3}}{2}[/tex]
NOT equal to the side length. It is the radius on the circle circumscribed about a hexagon that is equal to a side of the hexagon.

Draw it out measure it, whatever you need to do. You're saying the same things I've heard before. It's is not almost equal, nearly equal, it is equal. Debate it all you want. you're overlooking the fact that when the hexagon is inscribed, the radius is right there, already equal, just as a hexagon is. The radius of the circle is just that, just because it happens to be one side of a triangle, it's still the radius, ya know the point from the center of a circle to the perimeter? what are you saying? Ty for your time.
Will anyone take me seriously please? If not disprove me. But you can't, there's the problem. I know I'm right. I just want help going further. This theorem is old news.