Question

In △ABC, point P∈ AC with AP:PC=1:3, point Q∈ AB so that AQ:QB=3:4, Find the ratios APBQ : APBC and AAQP : AABC.

Answer 1

Answer: APBQ : APBC=4:21 and AAQP : AABC= 3:28.

Explanation: given : Here ABC is a triangle where AP:PC=1:3

let AP=1x and PC=3x, where x is the common value of AP and PC.

And AQ:QB = 3:4, similarly, let AQ=3y and QB= 4y, where y is the common value of AQ and QB.

Thus, AC= AP+PC= 1x+3x=4x and AB= AQ+QB= 3y+4y=7y

since, we have to find out APBQ : APBC= AP×BQ: AB×PC= 1x×4y: 7y×3x=4xy:21xy=4:21

So, APBQ : APBC=4:21

Now, AAQP : AABC= AQ×AP:AB×AC= 3y×1x:7y×4x=3xy:28xy=3:28

Thus, AAQP : AABC= 3:28.