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avatar+1213 

 

Given regular pentagon ABCDE a circle can be drawn that is tangent to DE at D and to AB at A. In degrees, what is the measure of minor arc AD?

 Jan 13, 2019
 #1
avatar+4594 
0

180(n-2), 180*3=540. So, 540/5=108 and the answer is 108.

 Jan 13, 2019
 #3
avatar+111437 
0

That would be true if E were the center of the circle.....but clearly....that's not the case....angle DEA will subtend an arc > 108°

 

 

cool cool cool

CPhill  Jan 13, 2019
 #2
avatar+111437 
+1

Let us  let the center of the circle be at F

 

Drawing radii from F to both D and A  will result in right angles CDF  and  BAF

 

And ABCDF will form an irregular pentagon.....the sum of the interior angles of this pentagon = 540°

 

Angles DCB and ABC =  108°

 

So....we can find angle AFD as

 

540 - 108 - 108 - 90 - 90  = 144°

 

But AFD is a central angle in the circle....so....its measure = minor arc AD   =  144°

 

 

cool cool cool

 Jan 13, 2019

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