A) \[\frac{1}{2}\]
B) 4
C) 3
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
We know that diagonals of a parallelogram divides it into two triangle of equal areas. In the given figure. ABCD is a parallelogram and AC is the diagonal of ABCD. \[\therefore \]\[ar(\Delta \Alpha \Beta C)=\frac{1}{2}ar\](parallelogram ABCD) ?(i) Now, In \[\Delta \Alpha {\mathrm O}Q\]and\[\Delta COP\] \[\angle QAO=\angle PCO\](Alternate angles) OA = OC (diagonals of a parallelogram bisect each other) \[\angle AOQ=\angle COP\](vertically opposite angles) \[\Rightarrow \]\[\Delta AOQ\cong \Delta COP\](by ASA congruency) \[\Rightarrow \]\[ar(\Delta \Alpha {\mathrm O}Q)=ar(\Delta COP)\] \[\Rightarrow \]\[ar(\Delta \Alpha {\mathrm O}Q)+ar\](quadrilateral AOPB) \[=ar(\Delta COP)+ar\](quadrilateral AOPB) \[\Rightarrow \]ar(quadrilateral ABPQ)=\[ar(\Delta \Alpha \Beta C)\] \[=\frac{1}{2}ar\](parallelogram ABCD) [Using (i)] \[\therefore \]\[k=\frac{1}{2}.\]
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