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

Question

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?

Thanks in advance.

Answer 1

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

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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!

Answer 2

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