Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD

Since AB || CD,


∴∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)


 



 


In ΔAOB and ΔCOD,


 


∠AOB = ∠COD (Vertically opposite angles)


 


∠OAB = ∠OCD (Alternate interior angles)


 


∠OBA = ∠ODC (Alternate interior angles)


 


ΔAOB ~ΔCOD (By AAA similarity)



Since, AB = 2 CD


Therefore,




Therefore, the ratio of the areas of triangles AOB and COD is 4:1 


 

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