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How can I calculate the area of a hexagon which all of it's angles are equal given 3 of it's sides?
Edit:
I forgot the constraint that opposite sides have same length, e.g. for hexagon $ABCDEF$
$AB = DE$
$BC = EF$
$CD = FA$

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If you take an arbitrary (large enough) equilateral triangle and cut off its corners you can fit the sides which cut off the corners to the numbers you are given, but different equilateral triangles will give different areas. Note also that if you have a horizontal (bottom) side and the two sides adjacent to it, you can create a pentagon with two sides from the ends of the bowl you have, and then a hexagon by cutting off the top corner with a horizontal line, but this is not fixed and different choices give different areas. So the three sides are insufficient to determine the area. –  Mark Bennet Aug 31 '11 at 6:11
    
@Mark Bennet: you are right, I forgot a constraint, see the edit. –  Dani Aug 31 '11 at 14:26
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2 Answers 2

As Mark already explained, you can not find the area of a hexagon with only the length of three of its sides and all the angles known.

Formulae for the area, perimeter, and other interesting facts about hexagons can be found on the Wikipedia page.

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@anon: It doesn't mean the hexagon is regular; the sides could differ. Mark's comment shows that we don't know the other sides. –  joriki Aug 31 '11 at 9:13
    
@joriki: Oh yes, you're quite right. Doh. –  anon Aug 31 '11 at 9:22
    
@Glen Wheeler: you are right, I forgot a constraint, see the edit. –  Dani Aug 31 '11 at 14:27
    
See the first computation here. It provides a nice direct way to calculate these areas. –  pgadey Feb 27 at 15:03
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Given your constraints, opposite sides of the hexagon are parallel. You can draw diagonals $AF$ and $BD$, which are also parallel and equal and $AD$. The length of $AF$ and $BD$ can be found from the law of cosines and the area of the triangles from Heron's formula (among other ways.) Then the angles can be assessed to determine $AD$ and the area of the other two triangles.

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