Question

In △ABC, point P∈AC with AP:PC=1:3, point Q∈AB so that AQ:QB=3:4, Find the ratios A(PBQ) : A(PBC) and A(AQP) : A(ABC)

Answer 1

Answer:

The ratio APBQ : APBC is 4:21, AAQP : AABC is 3:28

Step-by-step explanation:

Data provided in the question

AP : PC = 1 : 3

So let us assume AP = x and PC = 3x.

And, If AQ : QB = 3 : 4

So, let us assume AQ = 3y and QB = 4y.

Now we have to find the area of ΔAQP and ΔABC

$A_{AQP}=\dfrac{1}{2}\cdot AP\cdot AQ\cdot \sin\angle A=\dfrac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A;$

$A_{ABC}=\dfrac{1}{2}\cdot AC\cdot AB\cdot \sin\angle A=\dfrac{1}{2}\cdot (x+3x)\cdot (3y+4y)\cdot \sin\angle A=\dfrac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA$

Therefore

$\dfrac{A_{APQ}}{A_{ABC}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA}=\dfrac{3}{28}.$

Now

$\dfrac{A_{APQ}}{A_{ABP}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot x\cdot (3y+4y)\cdot \sin\angle A}=\dfrac{3}{7}.$

Now after solving these two ratios we can find

$A_{ABP}=\dfrac{7}{3}A_{APQ}\Rightarrow A_{PBQ}=A_{APB}-A_{APQ}=\dfrac{7}{3}A_{APQ}-A_{APQ}=\dfrac{4}{3}A_{APQ}$

$A_{ABC}=\dfrac{28}{3}A_{APQ}\Rightarrow A_{PBC}=A_{ABC}-A_{APB}=\dfrac{28}{3}A_{APQ}-\dfrac{7}{3}A_{APQ}=7A_{APQ}.$

Therefore

$\dfrac{A_{PBQ}}{A_{PBC}}=\dfrac{\frac{4}{3}A_{APQ}}{7A_{APQ}}=\dfrac{4}{21}.$

Hence, we applied the above equation so that we can get to know the ratios