Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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!

share|cite|improve this answer

$\hskip 1.7in$ hexagon

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.