Question

If AD is median of ΔABC and P is a point on AC such that ar(ΔADP) : ar(ΔABD) = 2 : 3, then ar(ΔPDC) : ar(ΔABC) is ______ .

• A. 3 : 5
• B. 2 : 5
• C. 1 : 5
• D. 1 : 6

Answer 1

The correct option is D 1 : 6

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 ADC have equal base BD 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