A regular hexagon A has the midpoints of its edges joined to form a smaller hexagon B. This process is repeated by joining the midpoints of the edges of the hexagon B to get a third hexagon C. What is the ratio of the area of hexagon C to the area of hexagon A?

Follow up question: What is the ratio of the are of hexagon B to the area of hexagon A.

Question

A regular hexagon A has the midpoints of its edges joined to form a smaller hexagon B. This process is repeated by joining the midpoints of the edges of the hexagon B to get a third hexagon C. What is the ratio of the area of hexagon C to the area of hexagon A?

Follow up question: What is the ratio of the are of hexagon B to the area of hexagon A.

anonymous · Answer

I drew the figure. I suspect there's something possible computing with the triangles formed between the hexagons. I'm stuck there. Please help me. Thanks

anonymous · Answer

@mathmale  @Mathbreaker  @skullpatrol

anonymous · Answer

dude finally did it...
its 16:9

anonymous · Answer

its 9:16 , but i don't know how my classmates solved it. please elaborate. :)

anonymous · Answer

and please do answer the follow up question., thanks.

anonymous · Answer

@Mertsj please help me with this. Thanks.

mertsj · Answer

http://ph.answers.yahoo.com/question/index?qid=20100605044206AAorb8W

mertsj · Answer

That seems to have the explanation.  We are limited here because the draw function is not working.

anonymous · Answer

@Mertsj can you elaborate how to algebraically solve this

y^2 = [(1/2)x]^2 + [(1/2)x]^2 - 2*(1/2)x*(1/2)x*cos120 deg.

and end up here: (y/x)^2 = 3/4

mertsj · Answer

y^2=1/4x^2+1/4x^2-1/2x^2(-1/2)

mertsj · Answer

y^2=1/4x^2+1/4x^2+1/4x^2

mertsj · Answer

y^2=3/4x^2
y^2/x^2=3/4
(y/x)^2=3/4