question_answer

If diagonals of a parallelogram ABCD intersect at point O in such a way that Area of \[\Delta OAB\] = k x Area of parallelogram ABCD. Then the value of k is:

A) 1                                 

B) \[\frac{1}{2}\]

C) \[\frac{1}{3}\]                          

D) \[\frac{1}{4}\]

Correct Answer: D

Solution :

               \[\therefore \]     Diagonals of a parallelogram bisect each other. \[\Rightarrow \]    AO is the median of\[\Delta ADB\]. \[\Rightarrow \]    \[ar(\Delta AOB)=\frac{1}{2}~ar(\Delta ADB)\]    ... (i) [A median of a A divides it into two triangles of equal areas] Also, \[~ar(\Delta ABD)\]=     \[\frac{1}{2}~ar\](parallelogram ABCD) ... (ii) [A diagonal of a parallelogram divides it into two As of equal areas] \[\therefore \]  From (i) & (ii) \[ar(\Delta AOB)=\frac{1}{2}~\times \frac{1}{2}~ar\] (parallogram ABCD) = \[\frac{1}{4}\] ar(parellogram ABCD) \[\Rightarrow \,\,\,k=\]