# Find the ratio of the area of the triangles and hexagon?

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

FO is parallel to ED.

What is the ratio of the triangle to the hexagon?

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

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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
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

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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