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Given regular pentagon ABCDE a circle can be drawn that is tangent to DE at D and to AB at A. What is the ratio of the radius to the circle to side AB?

 

 Apr 20, 2021
 #1
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Let  O  be  the  center  of the  circle

Then, by symmetry, EO  ies on  the  diameter  of  the  circle

 

Form triangle EOD

 

Angle CDO  = 90°

Angle EDC  =108°

So angle EDO  =  108 - 90   =   18 °

And angle  DEO  =  (1/2) 108  = 54° 

And angle EOD =  180  - 54 - 18  =  108°

 

Let  the side of  the pentagon =  S =  AB

Let  the radius of the  circle = R

 

The  by  the Law of Sines

 

AB / sin EOD  =  R / sin DEO

 

AB / sin 108  =  R /  sin 54

 

R / AB  =   (sin 54 / sin 108)    =   ( sqrt   [ 5 + sqrt 5 ]  / sqrt 10)    ≈   .85  

 

 

cool cool cool

 Apr 20, 2021

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