2
$\begingroup$

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.

  • 1
    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.2012-04-01

2 Answers 2

3

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!

8

$\hskip 1.7in$ hexagon