ABCD is a trapezium in which AB is parallel to DC and 2AB=3DC. The diagonals AC and BD intersect at O. What is the ratio of the area of \Delta AOB to that of \Delta DOC?

Explanation

Since AB \parallel DC, \Delta AOB \sim \Delta COD. The ratio of their areas is equal to the square of the ratio of their corresponding sides. Hence, \frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta DOC)} = \left(\frac{AB}{DC}\right)^2. Given 2AB = 3DC, we have \frac{AB}{DC} = \frac{3}{2}. The area ratio is \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 9:4.

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