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I really just dont know where to start on this one to be honest. I thought about making a triangle BEA or CBG but not sure how that would work. Thansk for the help.

 Feb 6, 2020
 #1
avatar+2646 
+2

\

Aha! CBD ~ CGA!

 

Both are 30-60-90.

 

1. Set CG = \(x\), and set CB = \(y\)

 

2. CD is \(x+1\)

 

3. Based on 30-60-90 properties, \(\frac{x+1}{2}=y\)

 

4. Write a similarity proportion

 

\(\frac{y}{x+1}=\frac{x}{y+13}\)

 

 

So now we have two equations...

 

Have fun solving!

 Feb 6, 2020
 #3
avatar
+2

I actually got this far after 2 hours but I can't figure out EA. Here's what I have thus far... GA = 6.5 + root3/3, CD = 5.08611, the intersection of the auxilliary line you drew and line FGA I called H. HA is 6.5. 

Guest Feb 6, 2020
 #2
avatar+17 
+3

Here is my solution :

https://imgur.com/a/hhC4nVl

 

I just used some basic geometry theories like Equilateral Triangle, Pythago, etc.

 

Hope it can help :)

 Feb 6, 2020
 #4
avatar+319 
+3

Triangles  BCD, ACG, and GDN are similar;  all of them are  30-60-90

Let's denote the intersection point of AF and BD with a leter N

BN = 13 * tan(30°) = 7.505

DN = 1 / cos (30°)  = 1.155

BD = BN + DN = 8.66

CD = BD / cos (30°) = 10

Circle radius r = CD / 2 = 5

EG = sqrt [ r² - ( r - 1 )² ] = 3

CG = CD - GD = 9

AG = CG / tan (30°)  = 15.588

EA = AG - EG = 12.588    indecision

 Feb 6, 2020
edited by Dragan  Feb 6, 2020
edited by Dragan  Feb 6, 2020
edited by Dragan  Feb 6, 2020
 #6
avatar+107414 
+3

Thanks guys,

That is really impressive.

I followed your logic and then drew the pic.

Here it is.

 

I have coloured the 3 similar triangles that are used in the solution.  

 

 Feb 6, 2020

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