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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