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 ______ .
Given : AD is median of Δ ABC
∴BD=DC
ar(ΔADP) : ar(ΔABD) = 2 : 3
To find: ar(ΔPDC) : ar(ΔABC)
Construction: Draw XY ∥ 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.
∴ area Δ ABD = area Δ ADC...(i)
area Δ ABD : area Δ ABC = 1 : 2 ...(ii)
area Δ ADP : area Δ ABD = 2 : 3 … (iii)
area Δ ADC = area Δ ADP + area Δ PDC
area Δ ABD = area Δ ADP + area Δ PDC
area Δ PDC
= area Δ ABD - area Δ ADP
= area Δ ABD - 23area Δ ABD
=13 area Δ ABD
∴area Δ PDC : area Δ ABD = 1 : 3...(iv)
areaΔPDCareaΔABC=13×12 ….. (from equations (i) and (iv)
area Δ PDC : area Δ ABC = 1 : 6