Answer: Option D
:
D

Given : AD is median of $\Delta$ ABC
$\therefore BD=DC$
ar(ΔADP) : ar(ΔABD) = 2 : 3
To find:ar(ΔPDC) : ar(ΔABC)
Construction: Draw XY $\parallel$ BC
Median divides the triangle into two equal areas and
Triangle ABD and ADChave equal baseBD and CD and are within the same parallels XY and BC.
$\therefore$ area $\Delta$ ABD = area $\Delta$ ADC...(i)
area $\Delta$ ABD: area $\Delta$ ABC = 1 : 2 ...(ii)
area $\Delta$ ADP : area $\Delta$ ABD = 2 : 3 …(iii)
area $\Delta$ ADC = area $\Delta$ ADP +area $\Delta$ PDC
area $\Delta$ ABD= area $\Delta$ ADP +area $\Delta$ PDC
area $\Delta$ PDC
=area $\Delta$ ABD -area $\Delta$ ADP
=area $\Delta$ ABD - $\frac{2}{3}$area $\Delta$ ABD
=$\frac{1}{3}$area $\Delta$ ABD
$\therefore$area $\Delta$ PDC : area $\Delta$ ABD = 1 : 3...(iv)
$\frac{area\Delta PDC}{area\Delta ABC}=\frac{1}{3}\times \frac{1}{2}$ ….. (from equations (i) and (iv)
area $\Delta$ PDC : area $\Delta$ ABC = 1 : 6