+0  
 
0
11
1
avatar+662 

I get 10, but that's not the right answer.

 

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 $EBF$ is $5,$ and the area of triangle $EAD$ is $5.$  Find the area of parallelogram $ABCD$.

 Dec 14, 2023
 #1
avatar+126978 
+1

See this

 

 

 

AE = EB  =  2.5

Height of EAD = Height of EBF =  4

Area of each triangle =  (1/2) ( 4) (2.5)  =  5

 

Triangle FDC  has a base and  height twice that of EAD.....so....its area = 2^2 * 5   = 20

 

[ FDC ]  - [ EBF]   =  20  -  5 =   15   = area of EBCD

 

[ EAD ] +  [ EBCD ]  =  5 + 15  =    20    =   [ ABCD  ]

 

 

cool cool cool

 Dec 14, 2023

4 Online Users

avatar