Help me with this!
- Thread starter ssolaming
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I would observe that, if we call \(r\) the radius of the circle in which the regular hexagon can be inscribed, then:
[MATH]\overline{HI}=\frac{1}{2}r[/MATH]
And:
[MATH]2\overline{FI}+\overline{HI}=2r[/MATH]
See what you can do with that...
[MATH]\overline{HI}=\frac{1}{2}r[/MATH]
And:
[MATH]2\overline{FI}+\overline{HI}=2r[/MATH]
See what you can do with that...
- Joined
- Nov 24, 2012
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To add a bit of clarity to my previous post, consider the following diagram of a regular hexagon inscribed within a circle of radius \(r\), and decomposed into 6 equilateral triangles having side lengths \(r\):
Now, in your diagram, we can see that \(\triangle GHI\) and \(\triangle GDE\) are similar, with the latter having twice the altitude of the former. And so the base of the latter, which is \(r\) will be twice the base of the former.
The rest of what I previously posted now follows.
Now, in your diagram, we can see that \(\triangle GHI\) and \(\triangle GDE\) are similar, with the latter having twice the altitude of the former. And so the base of the latter, which is \(r\) will be twice the base of the former.
The rest of what I previously posted now follows.