### $ABCD$ is a trapezium in which $AB || DC$ and $AB = 2DC$. If the diagonals of trapezium intersect each other at a point $O$, find the ratio of the areas of $\Delta AOB$ and $\Delta COD$. A C D O B

$4:1$
1. Given: A trapezium $ABCD$ in which $AB || DC$ and $AB = 2 DC$. Its diagonals intersect each other at the point $O$.
2. Here, we have to find the ratio of \begin{aligned} \dfrac { ar( \Delta AOB ) } { ar( \Delta COD ) } = ? \end{aligned}
3. In $\Delta AOB$ and $\Delta COD$ , we have \begin{aligned} &\angle AOB = \angle COD &&[ \text{ Vertically opposite angles } ] \\ &\angle OAB = \angle OCD &&[ \text{ Alternate interior angles } ] \\ \therefore \space \space& \Delta AOB \sim \Delta COD &&[ \text{ By AA-similarity } ] \\ \end{aligned}
4. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides. \begin{aligned} \therefore \space\space \dfrac { ar(\Delta AOB) } { ar(\Delta COD) } = \dfrac { AB^2 } { DC^2 } =& \dfrac { (2 \times DC)^2 } { DC^2 } &&[\because \text{ AB = 2DC }] \\ =& \dfrac { 4 \times DC^2 } { DC^2 } = \dfrac { 4 } { 1 } \end{aligned} Hence, $ar(\Delta AOB) : ar(\Delta COD) = 4:1$. 