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ABCD is a parallelogram. E is a point on AB such that AE:EB = 1:2, F is the mid-point of CD. AF meets DE at G and BF meets CE at H. When expanded on both sides, GH meets AD at M and BC at N. If the area of ABCD is 420, find the sum of the areas of the triangles ∆CNH and ∆DGM.

 Sep 11, 2020
 #1
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The sum fo the areas is 735.

 Sep 11, 2020
 #2
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How can it be 735 if the area of a whole parallelogram is 420 units squared?!?surprise

jugoslav  Sep 12, 2020
 #3
avatar+240 
+1

ABCD is a parallelogram. E is a point on AB such that AE:EB = 1:2, F is the mid-point of CD. AF meets DE at G and BF meets CE at H. When expanded on both sides, GH meets AD at M and BC at N. If the area of ABCD is 420, find the sum of the areas of the triangles ∆CNH and ∆DGM.

 

The sum of the areas of the triangles ∆CNH and ∆DGM is 48 u2

 

 Sep 12, 2020

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