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Question Number 30079 by naka3546 last updated on 16/Feb/18

Commented by naka3546 last updated on 16/Feb/18

$$LuasisAreainIndonesia.$$

Answered by $@ty@m last updated on 16/Feb/18

$$LetusdrawPN\mathrm{\perp }AD$$$$ar\left(APD\right)=\frac{1}{2}\times AD\times PN--\left(1\right)$$$$ar\left(FCD\right)=\frac{1}{2}\times FC\times 2PN$$$$ar\left(FCD\right)=\frac{1}{2}\times \left(\frac{1}{2}\times AD\right)\times 2PN$$$$=\frac{1}{2}\times AD\times PN---\left(2\right)$$$$\Rightarrow ar\left(APD\right)=ar\left(FCD\right)$$$$\Rightarrow ar\left(APD\right)-ar\left(EPD\right)=ar\left(FCD\right)-ar\left(EPD\right)$$$$\Rightarrow ar\left(FCPE\right)=ar\left(ADE\right)$$

Answered by ajfour last updated on 16/Feb/18

$$\frac{1}{11}.$$

Commented by naka3546 last updated on 16/Feb/18

$$howtoget\frac{1}{11}?$$

Answered by ajfour last updated on 16/Feb/18

Commented by ajfour last updated on 16/Feb/18

$$letentirearea\left(of//gm\right)=4\overline{A}$$$$4\overline{A}=\overline{a}\times \overline{b}$$$$sincear\left(EDA\right)=ar\left(FCPE\right)$$$$ar\left(EDA\right)+ar\left(EPD\right)=$$$$ar\left(FCPA\right)+ar\left(EPD\right)$$$$\Rightarrow ar\left(DAP\right)=ar\left(DFC\right)$$$$\frac{1}{2}\left(\overline{a}\times \frac{\overline{b}}{2}\right)=\frac{1}{2}\left(\lambda \overline{a}\times \overline{b}\right)$$$$\Rightarrow \frac{\overline{a}\times \overline{b}}{4}=\overline{A}=\frac{1}{2}\left(\lambda \overline{a}\times \overline{b}\right)$$$$\Rightarrow \mathit{\lambda }=\frac{1}{2}.....\left(i\right)$$$$as\overline{DE}=\overline{DP}+\overline{PE}$$$$\Rightarrow \rho (\overline{b}+\lambda \overline{a})=\frac{\overline{b}}{2}+\mu (\overline{a}-\frac{\overline{b}}{2})$$$$\Rightarrow \rho (\overline{b}+\frac{\overline{a}}{2})=\frac{\overline{b}}{2}+\mu (\overline{a}-\frac{\overline{b}}{2})$$$$socomparingcoefficientsof$$$$\overline{a}and\overline{b}weget$$$$\mathit{\rho }=\frac{1}{2}-\frac{\mathit{\mu }}{2}and\frac{\mathit{\rho }}{2}=\mathit{\mu }$$$$\Rightarrow 2\rho =1-\mu and\rho =2\mu$$$$\Rightarrow 4\mu =1-\mu$$$$or\mathit{\mu }=\frac{1}{5};\mathit{\rho }=\frac{2}{5}.....\left(ii\right)$$$$\frac{ar\left(DPE\right)}{ar\left(ABFE\right)}=\frac{\mu \times ar\left(DAP\right)}{ar\left(ABCD\right)-2\times ar\left(DAP\right)+ar\left(DEP\right)}$$$$=\frac{\mu \overline{A}}{4\overline{A}-2\overline{A}+\mu \overline{A}}andwith\mu =\frac{1}{5}$$$$\frac{ar\left(DPE\right)}{ar\left(ABFE\right)}=\frac{1/5}{2+\frac{1}{5}}=\frac{1}{11}.$$

Answered by mrW2 last updated on 16/Feb/18

Commented by mrW2 last updated on 17/Feb/18

$$letS=AreaofABCD$$$$\mathrm{\Delta }ADP=\frac{S}{4}=S_1+S_3$$$$\mathrm{\Delta }CDF=S_1+S_3=\frac{S}{4}$$$$\Rightarrow FC=\frac{BC}{2}$$$$CG=AD=BC$$$$\frac{FG}{CG}=\frac{3}{2}$$$$\frac{Areaof\mathrm{\Delta }EFG}{Areaof\mathrm{\Delta }HCG}={\left(\frac{FG}{CG}\right)}^2$$$$\Rightarrow \frac{S_3+S_1+S_3}{S_3}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$$$$8S_3+4S_1=9S_3$$$$\Rightarrow S_3=4S_1$$$$S_3+S_1=5S_1=\frac{S}{4}$$$$\Rightarrow S_1=\frac{S}{20}$$$$\Rightarrow S_3=4S_1=\frac{1}{5}S$$$$S_2+S_3=\frac{3}{4}S$$$$\Rightarrow S_2=\frac{3}{4}S-S_3=\frac{3}{4}S-\frac{1}{5}S=\frac{11}{20}S=11S_1$$$$\Rightarrow \frac{S_1}{S_2}=\frac{1}{11}$$

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