question_answer

ABCD is a trapezium in which \[\mathbf{AB}\parallel \mathbf{CD}\] and \[AB=3CD\]Its diagonals intersect other at O then the ratio of the areas of the triangles AOB and COD is:

A)  1 : 2                            

B)  2 : 1             

C)  9 : 1

D)  1 : 4

Correct Answer: C

Solution :

(c): let PQ be the perpendicular distance between the two parallel sides ab and cd. So, \[\frac{Area\,of\,\Delta AOB}{Area\,of\,\Delta COD}=\frac{\frac{1}{2}AB\times OX}{\frac{1}{2}CD\times OY}\] =\[\frac{3CD\times 3OY}{CD\times OY}=\frac{9}{1}\]