Question

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

Answer 1

Answer:

In △AOB and △COD,

∠AOB=∠COD [Vertically opposite angles]

∠OAB=∠OCD [Since AB||CD with AC as traversal alternate angle are equal]

Hence, △AOB∼△COD [AA similarity]

We know that if two triangle re similar, then

Ration of areas is equal to square of ratio of its corresponding sides

Hence,

Area △COD / Area △AOB

=( AB / CD )²

= ( 2CD / CD ) ²

= ( 2 / 1 ) ²

= 4 / 1

∴Area △AOB:Area △COD=4:1

Hence ratio of areas is 4:1.