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Quadrilateral $ABCD$ is a parallelogram.  Let $E$ be a point on $\overline{AB},$ and let $F$ be the intersection of lines $DE$ and $BC.$  The area of triangle $EAD$ is $9.$  If $AE:EB = 3:4$ and the height of parallelogram $ABCD$ is $10$, then find the area of parallelogram $ABCD$.

 May 24, 2024

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avatar+129567 
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                             F                     

 

 

    A    3    E     4    B

 

                                     10

 

    D                       C

 

[EAD]  =  9

Therefore  

9 = (1/2) AE * BC

9 = (1/2) AE * 10

9 / 5 =  AE =  (3/7)AB

 

So

AB = (7/3)AE = (7/3)(9/5)  = 63/15 = 21/5 

So

[ABCD ] = AB * BC  =  (21 / 5) (10)    =  42

 

 

cool cool cool

 May 24, 2024

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