question_answer

ABCD is a trapezium in which AB | | CD and AB = 2 CD. Its diagonals intersect each other at O, then the ratio of the areas of the triangles AOB and COD is :

A) 1: 2                              

B) 2: 1

C) 4: 1                              

D) 1: 4

Correct Answer: C

Solution :

Let PQ be the perpendicular distance between the two parallel sides AB and CD. So, \[\frac{area\,\,of\,\Delta \,AOB}{area\,\,of\,\Delta \,COD}\] \[=\,\,\,\frac{\frac{1}{2}AB\times OP}{\frac{1}{2}CD\times OQ}[\because \,OP=\,2OQ]\] \[=\frac{2CD\times 2OQ}{CD\times OQ}=\frac{4}{1}\]