Question

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).​

Answer 1

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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