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ABCDEF is a regular hexagon and angle AOF= 90 degree.

FO is parallel to ED.

enter image description here

What is the ratio of the triangle to the hexagon?

Give a hint so that i can get to the solution?

Thanks in advance.

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Do you mean the ratio of area? –  Lazar Ljubenović Apr 1 '12 at 8:06
    
@Lazar yes.edited the question. –  vikiiii Apr 1 '12 at 8:07
    
are you familiar with the so-called 30-60-90 triangle (which has sides $1,\sqrt3,2$ up to similarity)? –  bgins Apr 1 '12 at 8:11
    
@bgins no.i will search this on google and try again. –  vikiiii Apr 1 '12 at 8:11
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It says that $AF=2\cdot FO$ and $AO=\sqrt3 \cdot FO$. You also know that $FO \perp AO$, right? Also, if you reflect triangle $AFO$ about the vertical line $AO$, the two together give you an equilateral triangle which is one sixth the area of the total hexagon. –  bgins Apr 1 '12 at 8:13
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2 Answers

up vote 2 down vote accepted

With regards to the first answer, as exceptionally well put as it was, I'd like to add a "caption answer" to it.


I assume you have some idea about a regular hexagon as being comprised of 6 congruent equilateral triangles. This implies that the Area of equilateral triangle $$A_\Delta=\frac16A_{\text{hex}}$$ Where $A_{\text{hex}}$ is the Area of your hexagon.

Now notice that $\Delta AOF$ is a right triangle and hence it follows, from the fact that $AO$ bisects one of the small triangles, that: $$A_\Delta=2A_{AOF}$$ Together, you get $$A_{AOF}=\frac1{12}A_{\text{hex}}$$ OR better yet, $$\frac{A_{AOF}}{A_{\text{hex}}}=\frac1{12}$$
Hope it helps!

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@downvoter I am sorry, but why the downvote?, did I make a mistake in my answer? I am no whiz kid so sorry about whatever bothered you so much but it would have been better if you'd pointed it out –  Bidit Acharya Apr 1 '12 at 8:31
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$\hskip 1.7in$ hexagon

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cool visual explanation –  Bakudan Apr 1 '12 at 11:35
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