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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
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B
2 : 5
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C
1 : 5
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D
1 : 6
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Solution

The correct option is D 1 : 6

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


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