In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5PQ, find area of ∆AQC : area of ∆ABC.

Given, PC = 2BP

From figure,

BC = BP + PC = BP + 2BP = 3BP.

$\dfrac{PC}{BC} = \dfrac{2BP}{3BP} = \dfrac{2}{3}$

PC = $\dfrac{2}{3}$ BC

Let AD be the altitude.

Area of △ABC = $\dfrac{1}{2}$ × BC × AD ……(i)

Area of △APC = $\dfrac{1}{2}$ × PC × AD

= $\dfrac{1}{2} \times \dfrac{2}{3} BC \times AD$ ……..(ii)

Dividing (ii) by (i) we get,

$\Rightarrow \dfrac{\text{Area of △APC}}{\text{Area of △ABC}} = \dfrac{\dfrac{1}{2} × \dfrac{2}{3}BC × AD}{\dfrac{1}{2} \times BC \times AD} \\[1em] \Rightarrow \dfrac{\text{Area of △APC}}{\text{Area of △ABC}} = \dfrac{2}{3} \\[1em] \Rightarrow \text{Area of △APC} = \dfrac{2}{3}\text{Area of △ABC} ……….(iii)$

Given,

QA = 5PQ

From figure,

AP = AQ + QP = 5PQ + PQ = 6PQ.

$\dfrac{AQ}{AP} = \dfrac{5PQ}{6PQ} = \dfrac{5}{6}$.

AQ = $\dfrac{5}{6}AP$

Let CE be the altitude.

Area of △AQC = $\dfrac{1}{2}$ × AQ × CE ……(iv)

Area of △APC = $\dfrac{1}{2}$ × AP × CE …….(v)

Dividing (iv) by (v) we get,

$\Rightarrow \dfrac{\text{Area of △AQC}}{\text{Area of △APC}} = \dfrac{\dfrac{1}{2} \times AQ × CE}{\dfrac{1}{2} \times AP \times CE} \\[1em] \Rightarrow \dfrac{\text{Area of △AQC}}{\text{Area of △APC}} = \dfrac{\dfrac{1}{2} \times \dfrac{5}{6}AP × CE}{\dfrac{1}{2} \times AP \times CE} \\[1em] \Rightarrow \dfrac{\text{Area of △AQC}}{\text{Area of △APC}} = \dfrac{5}{6} \\[1em] \Rightarrow \text{Area of △AQC} = \dfrac{5}{6}\text{Area of △APC} \\[1em] \Rightarrow \text{Area of △AQC} = \dfrac{5}{6} \times \dfrac{2}{3} \text{ Area of △ABC (From eq iii)} \\[1em] \Rightarrow \text{Area of △AQC} = \dfrac{5}{9} \text{ area of △ABC } \\[1em] \Rightarrow \dfrac{\text{Area of △AQC}}{\text{Area of △ABC}} = \dfrac{5}{9}$

∴ area of △AQC : area of △ABC = 5 : 9.

Hence, area of △AQC : area of △ABC = 5 : 9.