question_answer

In the given trapezium ABCD, \[AB||CD\] and \[AB=2CD.\] If area of \[\Delta AOB=84c{{m}^{2}},\] then the area of \[\Delta COD\] is  

A)  \[22\,sq.\,cm\]

B)         \[25\,sq.\,cm\]

C)         \[21\,sq.\,cm\]

D)         \[24\,sq.\,cm\]

Correct Answer: C

Solution :

In \[\Delta AOB\]and \[\Delta COD,\] we have \[\angle AOB=\angle COD\] [Vertically opposite angles]             \[\angle OAB=\angle OCD\]  [Alternate interior angles]: \[\therefore \]  \[\Delta \,AOB\tilde{\ }\Delta COD\] [By AA similarity] \[\Rightarrow \]\[\frac{ar(\Delta AOB)}{ar(\Delta COD)}=\frac{A{{B}^{2}}}{C{{D}^{2}}}=\frac{{{(2CD)}^{2}}}{C{{D}^{2}}}\] \[\left[ \because \,AB=2CD \right]\] \[\Rightarrow \]\[\frac{84}{ar\,(\Delta COD)}=\frac{4}{1}\]\[\Rightarrow \]\[ar(\Delta COD)=21\,sq.\,cm\]