Experts please answer it fast...

Dear Student,


2020-09-16.jpg (3000×2059)
Given : BP :  PC  =  1 : 2 , Let ratio coefficient  =  x , So BP  =  x and PC  =  2 x , Then , BC  =  BP + PC  =  x + 2 x  = 3

Let height =  h1 with corresponding base is ' BC ' .

We know area of triangle  = 12×Base × Height , So

Area of ABC  = 12×BC × h1 = 12×3 x × h1= 3 x h12

And

Area of APC  = 12×PC × h1 = 12×2 x × h1= = x h1

Then ,

Area of  ABCArea of  APC = 3 x h12 x h1= 3 x h1 2 x h1 = 32Area of  ABC = 32×Area of  APC                              --- ( 1 )

Also given : PQ :  QA  =  2 : 3 , Let ratio coefficient  =  y , So PQ  = 2 y and QA  =  3 y , Then , AP  =  PQ + QA  =  2 y + 3 y = 5

Let height =  h2 with corresponding base is ' AP ' .

Then

Area of APC  = 12×AP × h2 = 12×5 y × h2= 5 y h22

And

Area of AQC  = 12×QA × h2 = 12×3 y × h2= 3 y h22

Then ,

Area of  APCArea of  AQC = 5 y h223 y h22= 5 y h23 y h2 = 53Area of  APC = 53×Area of  AQC
Substitute that value in equation 1 we get

Area of  ABC  = 32 × 53×Area of  AQCArea of  ABC  = 52 ×Area of  AQCArea of  AQC Area of  ABC  = 25Area of  AQC  : Area of  ABC  =2 : 5                    ( Hence proved )




Regards

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