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I am kinda curious in something. When looking at a hexagon (6-sided entity) and wanting to calculate the area, could you in theory treat the hexagon as 4 triangles and one square in the middle like this?

hexagon

Here is my calculations, if this can be done
$\left(\frac{C×A/2}{1/2}×4\right) + (B×A)$

An Alien
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Nemanja Vuksanovic
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  • You should be dividing C x A/2 by 2 not by 1/2 to get area of one of the four triangles. Another note is your hexagon is not a "regular" hexagon without some conditions on A,B,C. – coffeemath Nov 01 '21 at 14:33
  • I see that, sorry, I am not very familiar with the LaTex formatting. Second of, what are you referring to when you say "conditions"? – Nemanja Vuksanovic Nov 01 '21 at 14:35
  • @coffeemath Furthermore, just a sidenote. You can just "rotate" the hexagon to get the "regular" hexagon. Hope I understood you correctly. – Nemanja Vuksanovic Nov 01 '21 at 14:40
  • In your drawing, if you keep A and B like they are, and make C shrink toward zero, you can easily see it won't be a regular hexagon. Een rotating it won't change that. By the way "regular" hexagon means all six sides have same length. So that would impose some limits on what A,B,C can be to make it regular. – coffeemath Nov 01 '21 at 14:47
  • In the latex for your formula with the longer fraction line, there is a \frac followed by two expressions in curly brackets, and in the second of these you now have 1/2. Just edit and change that to a 2 to fix it. – coffeemath Nov 01 '21 at 15:05
  • @coffeemath I see, I misunderstood what you meant with "regular" hexagon, pardon me for that. Could you also elaborate what you mean with "conditions"? – Nemanja Vuksanovic Nov 01 '21 at 15:11
  • Let us continue this discussion in chat. – coffeemath Nov 01 '21 at 15:15

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