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Inside a regular hexagon ABCDEF is drawn triangle ACE. What is the ratio of the area of triangle ACE to that of the hexagon?

 May 5, 2020
 #1
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Let   the side of the  hexagon  =  S

 

Its  area  =   6(1/2)S^2 sin60“    =    3S^2sin 60°

 

We can find  the  side  of  the  triangle  using the Law of Cosines

 

This is

 

sqrt  [ S^2  + S^2  - 2S^2 cos (120°)]  =  sqrt [  2S^2 + S^2]  =  S√3

 

The area of  this equilateral triangle  is  (1/2)(S√3)^2 sin60° = (3/2)S^2sin60°

 

So  the ratio  is

 

(3/2)S^2 sin 60°

_____________   =    (3/2)  / 3   =    1  /  2

     3S^2 sin 60°

 

 

cool cool cool

 May 5, 2020

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