Question : Diagonals of a trapezium $ABCD$ with $AB \parallel CD$ intersect each other at the point $O$. If $AB = 2CD$, then the ratio of the areas of $\triangle AOB$ and $\triangle COD$ is:
Option 1: $4:1$
Option 2: $1:16$
Option 3: $1:4$
Option 4: $16:1$
Correct Answer: $4:1$
 
  Solution :
 
 
 Given: $ABCD$ is a given trapezium in which $AB \parallel CD$ and $AB = 2CD$.
 
  In $\triangle AOB$ and $\triangle COD$
  
  $\angle AOB=\angle COD$  (Vertically opposite triangle)
  
  $\angle 1=\angle 2$  (Alternate interior angle)
  
  $\angle AOB \sim \angle COD$  ( by AA similarity)
  
  The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
  
  $\frac{ar(\triangle AOB)}{ar(\triangle COD)}=(\frac{AB}{CD})^2=(\frac{2CD}{CD})^2=(\frac{4}{1})$
  
  Hence, the correct answer is $4:1$.
 
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