A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
In the given figure, draw a line XY which meets AB at Y and parallel to I and m. Now, Join D to P, P to X, X to Q and Q to C Since \[XY||l||AD\] \[\Rightarrow \]ADXY is a parallelogram. \[\therefore \]\[ar(\Delta DPX)=\frac{1}{2}ar\](parallelogram ADXY) ..(i) Similarly, \[ar(\Delta CQX)=\frac{1}{2}ar\](parallelogram\[BYXC\]) Adding equations (i) and (ii), we get \[ar(\Delta DPX)+ar(\Delta CQX)\] \[=\frac{1}{2}ar(paralle\log ram\,ADXY)\] \[+\frac{1}{2}\]ar (parallelogram BYXC) \[=\frac{1}{2}(\Delta \Alpha \Beta CD)\] \[\therefore \] \[k=\frac{1}{2}\]You need to login to perform this action.
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