Respuesta :
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, [tex]\frac{AP.BQ}{AP.BC} =\frac{AP.BQ}{AB.PC} =\frac{x.4y}{7y.3x} =\frac{4xy}{21xy} =\frac{4}{21}[/tex]
And, [tex]\frac{AA.QP}{AA.BC} =\frac{AQ.AP}{AB.AC} =\frac{3y.1x}{7y.4x} =\frac{3xy}{28xy} =\frac{3}{28}[/tex]
Answer:
The required ratios are:
[tex]\frac{AP.BQ}{AP.BC}=\frac{4}{21}[/tex] and [tex]\frac{AA.QP}{AA.BC} =\frac{3}{28}[/tex]
Step-by-step explanation:
Given information:
In [tex]\triangle ABC[/tex] the point [tex]p[/tex] ∈ [tex]AC[/tex] with [tex]AP[/tex]
Where:
[tex]PC=1:3[/tex]
Point [tex]q[/tex] ∈ [tex]AB[/tex]
Suppose:
[tex]AP=x[/tex]
[tex]PC=3x[/tex]
Thus:
[tex]AC=AP+AC\\AC=4x[/tex]
Similarly:
Suppose :
[tex]AQ=3y\\QB=4y[/tex]
So:
[tex]AB=AQ+QB\\AB=3y+4y\\AB=7y[/tex]
Now, on putting the values for required values:
One can calculate:
[tex]\frac{AP.BQ}{AP.BC} =\frac{4xy}{21xy}\\\\ \frac{AP.BQ}{AP.BC}=\frac{4}{21}[/tex]
And:
[tex]\frac{AA.QP}{AA.BC} =\frac{3xy}{28xy} \\\\\frac{AA.QP}{AA.BC} =\frac{3}{28} \\[/tex]
Hence, the required ratios are (4/21) and (3/28) respectively.
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