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Let ABCDEF be a regular hexagon.  Let P be the midpoint of BC, and let Q be the midpoint of EF.  Find the ratio of the area of APDQ to the area of ABCDEF.

 May 13, 2020
 #1
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Answer is E, because you have to: b + b then its e. Quick Math!

 May 13, 2020
 #2
avatar+111456 
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See the following  image

 

 

 

 

Let the side of the hexagon  =  S

 

Then the  area  of the hexagon  =  6(1/2)S^2 sqrt (3) / 2   =   3sqrt(3)S^2 / 2  =  (3/2) sqrt (3)S^2

 

Note that  the  area  of  triangle  ABP  =

 

(1/2)(AB)(BP) sin ABP  =

 

(1/2)(S)(1/2)S * sin (120)  =

 

(1/2) (S) (1/2)S * sqrt (3) / 2 =

 

(1/8)sqrt (3) S^2

 

Using symmetry  we  have  three other triangles  congruent to triangle  ABP

 

So....the total area  of these  4  triangles =  (1/2)sqrt(3) S^2 

 

So...the area of  APDQ =  (3/2 - 1/2)sqrt(3)S^2  = sqrt (3)S^2

 

So.....the  ratio of the area of APDQ  to  the area of the hexagon  =

 

 sqrt (3)S^2                  2

_____________   =   ___   

(3/2) sqrt (3) S^2         3

 

 

cool cool cool

 May 13, 2020

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