Actually, to four decimal places, it's 3.1416, but that's only a minor quibble.
No, you're not. You're constructing *chords* using a length that is one-half of the diameter.
No, you aren't. Think about it some more. Draw if if necessary.
There is no error, regardless of the size of the hexagon, and no mythical seventh side is ever needed.
[snip]Helloooooooo!!! If you've just used the compass to draw that circle, it's
*already* set at exactly the distance between the center and the circumference. No "adjustment" is necessary, and any change in the setting of the compass is guaranteed to introduce error.This should be interesting... a "proof" of a false proposition...
Hold it right there, cowboy. The sides of a hexagon are _straight_lines_, not circular arcs.
The remainder of the "proof" is ignored, since it rests on a demonstrably false proposition.
[snip] [snip *another* clumsy construction]Holy smokes! Talk about making things more complicated than they need to be! Yes, there is a geometric method for drafting a perfect hexagon. Several, in fact. The simplest one is the method using points around the circumference of a circle. Here's another:
Mark a point, A. Set the compass to any arbitrary radius. Draw an arc. Mark any point, B, on the arc. Connect A and B. Draw another arc, centered at B, that intersects the first arc. Mark the intersection C. ABC is an equilateral triangle.
Draw a third arc, centered at C. Draw fourth arc, centered at A, that intersects the third arc. Mark the intersection D. ACD is also an equilateral triangle.
Draw a fifth arc, centered at D, and a sixth, centered at A, that intersects the fifth arc. Mark the intersection E. ADE is also an equilateral triangle... and BCDE is half a hexagon.
Continue in a similar manner to construct points F and G, which define the equilateral triangles AEF, AFG, *and* AGB -- *and* the hexagon BCDEFG.
Now put your compass point at A and draw a circle.
Where are points B, C, D, E, F, and G?
Seems that you could use a refresher on the fundamentals yourself.