Question 31172: Hello! I have tried this problem over and over again but I can't seem to get it. Please help.
A regular hexagon has sides four feet long. Find the area of the triangle formed by connecting alternate vertices.
I know your supposed to show your previous work, but honestly, I don't even know where to start! I really need your help. Thank you so very much.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! not sure exactly which is the triangle you are asking for. Label the vertices as A, B, C, D, E, F then draw a line between A and C...you mean the triangle ABC?
If so, we have Area=(1/2)absinC --> the notation here is nothing to do with the labelling of the vertices.
a=AB
b=BC
and a=b=4
angle C is angle ABC. This is an interior angle of a hexagon.
To find the interior angle of an hexagon:
A hexagon is 6sided. This can be divided into 4 triangles.
4 triangles have 4*180 angles between them --> 720degrees.
So each interior angle = 720/6 --> 120degrees
so area= (1/2)*4*4*sin120
8sin120
8*sqrt(3)/2
4sqrt(3) is the area of triangle ABC.
Hope this helps
jon.
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