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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 $ABCD$.

 Mar 9, 2024
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A     3    E    4    B

 

                F             

 

C                      D

 

[EAD] = (1/2) (3/7)AB * AC

9 = (3/14)AB * 10

9 = (30/14)AB

9 = (15/7)AB

9 ( 7/15)  = AB

(63/15) = AB  =   21/5  

 

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

 

cool cool cool

 Mar 9, 2024

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