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Question
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution
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 criterion)
`:.(ar(ΔAOB))/(ar(COD)) = ((AB)/(CD))^2`
Since AB = 2 CD
`:. (ar(triangleAOB))/(ar(triangleCOD)) =((2CD)/(CD))^2 =4/1=4:1`
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