Hong Kong
^@ 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^@.
Answer:
^@ 4:1 ^@
- Given: A trapezium ^@ ABCD ^@ in which ^@AB || DC^@ and ^@AB = 2 DC^@. Its diagonals intersect each other at the point ^@O^@.
- Here, we have to find the ratio of @^ \begin{aligned} \dfrac { ar( \Delta AOB ) } { ar( \Delta COD ) } = ? \end{aligned} @^
- 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} @^
- 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^@.
