question_answer

ABCD is a parallelogram. Two lines\[l\]and m are parallel to AD. Line\[l\]meets AB and CD at P and S respectively. Line m meets AB and CD at Q and R respectively. X is any point on CD between R and S. If\[ar(\Delta DPX)+ar(\Delta CQX)=kar(ABCD),\]find k.

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) \[=\frac{1}{2}(\Delta \Alpha \Beta CD)\] \[\therefore \]      \[k=\frac{1}{2}\]