12. In the given figure, AD is a median of ∆ABC and P
is a point on AC such that : ar(∆ADP): ar∆ABD) =2:3
Find : (i) AP: PC (ii) ar(∆PDC): ar(∆ABC).
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AP/PC = 2 , ar(∆PDC) / ar(∆ABC) = 1/3
Step-by-step explanation:
AD is a median of ∆ABC
=> Area of ΔABD = Area of ΔACD
Area of ΔADP / Area of ΔACD = AP/AC
=> Area of ΔADP / Area of ΔABD = AP/AC
=> 2/3 = AP/AC
=> AC = 3AP/2
PC = AC - AP = 3AP/2 - AP = AP/2
PC = AP/2
=> AP/PC = 2
ar(∆PDC) = Area of ΔACD - Area of ΔADP
=> ar(∆PDC) = Area of ΔABC - 2 * Area of ΔABC/3
=> ar(∆PDC) = Area of ΔABC/3
=> ar(∆PDC) / ar(∆ABC) = 1/3
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