a hard pentagon questionABCDE

Oct 2012
680
4
Istanbul
ABCDE is a regular pentagon.CF is perpendicular to AB. KL is perpendicular to AC. |FB|=|KL|.|DE|=6 units. Find area of triangle AKC?
https://postimg.org/image/8yw954k9z/

Question and image are very clear. I don't want any solution with trigonometry and golden ratio. I don't want to hear any comment about my drawing. If you know the answer you are wellcome.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey kastamonu.

We ask that people put in their own attempts before giving further guidance.
 
May 2009
612
334
You've already asked this question about six months ago here. I'm wondering why is it necessary to solve this without trig or use of the golden ratio.


01
 
Oct 2012
680
4
Istanbul
I solved it without trig. but if there is another way I would like to learn.
 
Oct 2012
680
4
Istanbul
This is my solution with golden:
In triangle BCA AC=6phi(GOLDEN RATIO)
In triangle FBC LET FB be X
sin18=1/2phi
x/6=1/2phi
X=6/2phi
area is AC.KL/2
AC=6phi
KL=6/2phi
6phi.6/2phi divided by 2=9
If you have a solution ignoring trig or golden you are wellcome.
 
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