question_answer

If AD is median of \[\Delta ABC\]and P is a point on AC such that \[ar(\Delta ADP):ar(\Delta ABD)=2:3,\]then\[ar(\Delta PDC):\] \[ar(\Delta ABC)\]is                  

A)  1 : 5                

B)         5 : 1                

C)         1 : 6                            

D)         3 : 5                

Correct Answer: C

Solution :

Let \[ar(\Delta \Alpha DP)=2x\] and \[ar\,\Delta ABD=ar\Delta ADC=3x\] [AD is the median of\[\Delta \Alpha \Beta C\]]\[ar\Delta \Alpha DC=ar\Delta ADP+ar\Delta DPC\] \[3x+2x+ar\Delta DPC\Rightarrow ar\,\Delta DPC=x\] a\[ar\Delta \Alpha \Beta C=ar\Delta ABD+ar\Delta ADC=3x+3x=6x\] \[\therefore \]      \[\frac{ar(\Delta PDC)}{ar(\Delta ABC)}=\frac{x}{6x}-1L6.\]