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Junior Divisional 2013/1

Posted: Wed Dec 10, 2014 5:41 pm
by barnik
$O$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$.The ratio $AD:AB=2:5$ and the ratio $AF:AC=2:5$. The area of triangle $ABC$ is $50\text m^2$. What is the difference between the area of $ADPF$ and triangle $PON$?

Re: Junior Divisional 2013/1

Posted: Thu Dec 11, 2014 2:58 pm
by tanmoy
$(ABC)=50m^{2}.\therefore (AON)=\frac{50}{4}=\frac{25}{2}m^{2}.(ADF)=50\times \frac{4}{25}=8$
$(DPF)=\frac{16}{25}(PON).(DPO)=(FPN)=\frac{4}{5}(PON).$
$(DONF)=\frac{25}{2}-8=\frac{9}{2}$
$\therefore \frac{16}{25}(PON)+\frac{4}{5}(PON)+\frac{4}{5}(PON)+(PON)=\frac{9}{2}$
$\therefore (PON)=\frac{25}{18}.\therefore (DPF)=\frac{8}{9}$
$\therefore (ADPF)=8+\frac{8}{9}=\frac{80}{9}m^{2}.
\therefore (ADPF)-(PON)=\frac{80}{9}-\frac{25}{18}=\frac{15}{2}m^{2}$ :)