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Let ABCD be a parallelogram. Extend BC past  B to F, and let E be the intersection of AB and DF. If the areas of triangles BEF and ADE are 4 and 9, respectively, find the area of parallelogram .

 

 Dec 1, 2020
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Note  that triangle EAD is similar to  triangle  EBF

The scale factor  from   EBF  to  EAD = sqrt  (4/9)   = 2/3

 

Therefore

EB / EA  =  2/3

EB = (2/3)EA    and

EA = (3/2)EB

 

Then AB =  EA + EB = (3/2)EB + EB =  (5/2)EB  = DC

 

And   the height of   EBF  is  (2/3)  that of the height of  EAD...so  height of EAD = (3/2) height of EBF

 

Therefore  EBCD is a trapezoid  with bases  EB and DC    and  height = height of EAD

 

So  the  area  of  EBCD  =  (1/2) (height of EAD)  (EB + DC)  =

 

(1/2)[ (3/2)height of EBF ]   ( EB + (5/2)EB)  =

 

(1/2) (3/2) (height of EBF)   [( 7/2)EB ]  =

 

(3/2) (7/2)  *  (1/2) height of EBF *  EB

 

But (1/2) height of EBF  * EB =  Area of  EBF

 

So area of  trapezoid  EBCD =  (3/2)(7/2) (4)  =  21

 

So...the area of ABCD  = area of EAD  +  area of  EBCD  =   9 + 21    = 30

 

cool cool cool

 Dec 1, 2020
edited by CPhill  Dec 1, 2020

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