ABCD is a trapezium in which AB∥DC and AB=2DC.O is the point of intersection of the diagonals. The ratio of the areas of △AOB and △COD is:
A
1:2
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B
2:1
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C
4:1
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D
1:4
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Solution
The correct option is C4:1
Given, AAB=2CD ....(1)
In △s AOB and COD ∠AOB=∠COD (vert. opp. ∠s) ∠OAB=∠DCO (DC∥AB, alt. ∠s are equal) ∴△AOB∼△COD (AA similarity) ⇒ar(△AOB)ar(△COD)=AB2CD2 {Ratio of areas of two similar △s is equal to the ratio of the squares of the corresponding sides} =(2CD)2CD2=4CD2CD2=41. ....From (1)