question_answer

In a trapezium \[\Delta \,BCD\], AB is parallel to DC and AB = 2 DC. If AC and BD meet at 0, then area of \[\Delta \,AOB\] is equal to

A)  the area of \[\Delta \,COD\]

B)  twice the area of \[\Delta \,COD\]

C)  thrice the area of \[\Delta \,COD\]

D)  four times the area of \[\Delta \,COD\]

Correct Answer: D

Solution :

 Given,       \[AB=2CD\]             .....(i) Let PM be perpendicular to AB and CD through O.                  Since \[\Delta OCD\] and \[\Delta OAB\] are similar, therefore \[\therefore \] \[\frac{Area\,\,\Delta \,AOB}{Area\,\,\Delta \,DOC}=\frac{{{(AB)}^{2}}}{{{(DC)}^{2}}}=\frac{4}{1}\] \[\therefore \]Area \[\Delta \,AOB=4\]Area \[\Delta \,DOC\]