Question

In $\Delta ABC$, AP : PB = 2 : 3, PO is parallel to BC and is extended to Q so that CQ is parallel to BA. [3 MARKS]

Find:
i) area $\Delta APO$ : area $\Delta ABC$
ii) area $\Delta APO$ : area $\Delta CQO$

Answer 1

Part (i): Application of Formula: 1 Mark
Calculation: 0.5 Mark

Part (ii): Application of Formula: 1 Mark
Calculation: 0.5 Mark


Given: AP : PB = 2 : 3 and $PO\parallel BC$
i) As $PO\parallel BC$
$\therefore \angle AOP=\angle ACB$ [Corresponding angles]
$\angle APO=\angle ABC$ [Corresponding angles]
$\therefore \Delta APO\sim \Delta ABC$ [By AA similarity]
Now $\frac{AP}{PB}=\frac{2}{3}$ [Given]
$\Rightarrow \frac{PB}{AP}=\frac{3}{2}\Rightarrow \frac{PB}{AP}+1=\frac{3}{2}+1$
$\Rightarrow \frac{PB+AP}{AP}=\frac{3+2}{2}=\frac{5}{2}$
$\Rightarrow \frac{AB}{AP}=\frac{5}{2}\Rightarrow \frac{AP}{AB}=\frac{2}{5}$
$\therefore \frac{Areaof\Delta APO}{Areaof\Delta ABC}=\frac{A{P}^2}{A{B}^2}$
$={\left(\frac{2}{5}\right)}^2=\frac{4}{25}$
$\Rightarrow$ Area of $\Delta APO$ : Area of $\Delta ABC$ = 4 : 25

ii) In $\Delta APO$ and $\Delta CQO$,
$\angle AOP=\angle QOC$ [Vertically opposite angles]
$\angle OAP=\angle OCQ$ [Interior alternate angles]
$\therefore \Delta APO\sim \Delta CQO$ [By AA similarity]
$\Rightarrow \frac{AP}{CQ}=\frac{AO}{OC}$
$\frac{AP}{CQ}=\frac{AP}{PB}=\frac{2}{3}$
$\therefore \frac{Areaof\Delta APO}{Areaof\Delta CQO}=\frac{A{P}^2}{C{Q}^2}$
$=\frac{2^2}{3^2}=\frac{4}{9}$
$\Rightarrow$ Area of $\Delta APO$ : Area of $\Delta CQO=4:9$