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) \[\Rightarrow \,\,\,k=\]You need to login to perform this action.
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