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sesenic [268]
4 years ago
6

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.

Mathematics
1 answer:
weqwewe [10]4 years ago
5 0

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

Explanation:

Here, ABC is a triangle where, P and Q are the midpoints of the edges AC and AB respectively,

Now, according to the question ,AP:PC=1:3 let AP=x and PC=3x where x is any real number. Thus, AC=AP+AC= x+3x= 4x

Similarly,  AQ:QB=3:4 let AQ=3y and QB=4y where y is also any real number. Thus, AB=AQ+QB=3y+4y=7y

Since, \frac{AP.BQ}{AP.BC} =\frac{AP.BQ}{AB.PC} =\frac{x.4y}{7y.3x} =\frac{4xy}{21xy} =\frac{4}{21}

And, \frac{AA.QP}{AA.BC} =\frac{AQ.AP}{AB.AC} =\frac{3y.1x}{7y.4x} =\frac{3xy}{28xy} =\frac{3}{28}

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