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

Commented by naka3546 last updated on 16/Feb/18

Luas is Area in Indonesia .

$${Luas}\:\:{is}\:\:{Area}\:\:{in}\:\:{Indonesia}\:. \\ $$

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

Let us draw PN⊥AD ar(APD)=(1/2)×AD×PN −−(1) ar(FCD)=(1/2)×FC×2PN ar(FCD)=(1/2)×((1/2)×AD)×2PN =(1/2)×AD×PN −−−(2) ⇒ar(APD)=ar(FCD) ⇒ar(APD)−ar(EPD)=ar(FCD)−ar(EPD) ⇒ar(FCPE)=ar(ADE)

$${Let}\:{us}\:{draw}\:{PN}\bot{AD} \\ $$$${ar}\left({APD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{PN}\:−−\left(\mathrm{1}\right) \\ $$$${ar}\left({FCD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{FC}×\mathrm{2}{PN}\: \\ $$$${ar}\left({FCD}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{1}}{\mathrm{2}}×{AD}\right)×\mathrm{2}{PN}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{PN}\:−−−\left(\mathrm{2}\right) \\ $$$$\Rightarrow{ar}\left({APD}\right)={ar}\left({FCD}\right)\: \\ $$$$\Rightarrow{ar}\left({APD}\right)−{ar}\left({EPD}\right)={ar}\left({FCD}\right)−{ar}\left({EPD}\right) \\ $$$$\Rightarrow{ar}\left({FCPE}\right)={ar}\left({ADE}\right) \\ $$

Answered by ajfour last updated on 16/Feb/18

(1/(11)) .

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

Commented by naka3546 last updated on 16/Feb/18

how to get (1/(11)) ?

$${how}\:{to}\:{get}\:\:\:\frac{\mathrm{1}}{\mathrm{11}}\:\:? \\ $$

Answered by ajfour last updated on 16/Feb/18

Commented by ajfour last updated on 16/Feb/18

let entire area(of //gm) = 4A^� 4A^� = a^� ×b^� since ar(EDA)=ar(FCPE) ar(EDA)+ar(EPD)= ar(FCPA)+ar(EPD) ⇒ ar(DAP)= ar(DFC) (1/2)(a^� ×(b^� /2))= (1/2)(λa^� ×b^� ) ⇒ ((a^� ×b^� )/4) = A^� =(1/2)( λa^� ×b^� ) ⇒ 𝛌 =(1/2) .....(i) as DE^(−) =DP^(−) +PE^(−) ⇒ ρ(b^� +λa^� )=(b^� /2)+μ(a^� −(b^� /2)) ⇒ ρ(b^� +(a^� /2))=(b^� /2)+μ(a^� −(b^� /2)) so comparing coefficients of a^� and b^� we get 𝛒=(1/2)−(𝛍/2) and (𝛒/2)=𝛍 ⇒ 2ρ =1−μ and ρ=2μ ⇒ 4μ = 1−μ or 𝛍=(1/5) ; 𝛒=(2/5) .....(ii) ((ar(DPE))/(ar(ABFE))) = ((μ×ar(DAP))/(ar(ABCD)−2×ar(DAP)+ar(DEP))) =((μA^� )/(4A^� −2A^� +μA^� )) and with μ=(1/5) ((ar(DPE))/(ar(ABFE)))=((1/5)/(2+(1/5))) = (1/(11)) .

$${let}\:{entire}\:{area}\left({of}\://{gm}\right)\:=\:\mathrm{4}\bar {{A}} \\ $$$$\mathrm{4}\bar {{A}}\:=\:\bar {{a}}×\bar {{b}} \\ $$$${since}\:{ar}\left({EDA}\right)={ar}\left({FCPE}\right) \\ $$$${ar}\left({EDA}\right)+{ar}\left({EPD}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ar}\left({FCPA}\right)+{ar}\left({EPD}\right) \\ $$$$\Rightarrow\:{ar}\left({DAP}\right)=\:{ar}\left({DFC}\right) \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\bar {{a}}×\frac{\bar {{b}}}{\mathrm{2}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\lambda\bar {{a}}×\bar {{b}}\right) \\ $$$$\Rightarrow\:\:\:\frac{\bar {{a}}×\bar {{b}}}{\mathrm{4}}\:=\:\bar {{A}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\lambda\bar {{a}}×\bar {{b}}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{\lambda}\:=\frac{\mathrm{1}}{\mathrm{2}}\:.....\left({i}\right) \\ $$$$\:{as}\:\:\:\overline {{DE}}\:=\overline {{DP}}+\overline {{PE}} \\ $$$$\Rightarrow\:\:\:\rho\left(\bar {{b}}+\lambda\bar {{a}}\right)=\frac{\bar {{b}}}{\mathrm{2}}+\mu\left(\bar {{a}}−\frac{\bar {{b}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\rho\left(\bar {{b}}+\frac{\bar {{a}}}{\mathrm{2}}\right)=\frac{\bar {{b}}}{\mathrm{2}}+\mu\left(\bar {{a}}−\frac{\bar {{b}}}{\mathrm{2}}\right) \\ $$$${so}\:{comparing}\:{coefficients}\:{of} \\ $$$$\bar {{a}}\:{and}\:\bar {{b}}\:\:{we}\:{get} \\ $$$$\:\:\:\:\:\:\boldsymbol{\rho}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\boldsymbol{\mu}}{\mathrm{2}}\:\:\:\:{and}\:\:\:\frac{\boldsymbol{\rho}}{\mathrm{2}}=\boldsymbol{\mu} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}\rho\:=\mathrm{1}−\mu\:\:\:{and}\:\:\rho=\mathrm{2}\mu \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{4}\mu\:=\:\mathrm{1}−\mu\:\:\: \\ $$$$\:\:\:\:{or}\:\:\:\boldsymbol{\mu}=\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:\:;\:\:\boldsymbol{\rho}=\frac{\mathrm{2}}{\mathrm{5}}\:.....\left({ii}\right) \\ $$$$\:\:\frac{{ar}\left({DPE}\right)}{{ar}\left({ABFE}\right)}\:=\:\frac{\mu×{ar}\left({DAP}\right)}{{ar}\left({ABCD}\right)−\mathrm{2}×{ar}\left({DAP}\right)+{ar}\left({DEP}\right)} \\ $$$$\:\:\:=\frac{\mu\bar {{A}}}{\mathrm{4}\bar {{A}}−\mathrm{2}\bar {{A}}+\mu\bar {{A}}}\:\:\:\:{and}\:{with}\:\mu=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\frac{{ar}\left({DPE}\right)}{{ar}\left({ABFE}\right)}=\frac{\mathrm{1}/\mathrm{5}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{5}}}\:=\:\frac{\mathrm{1}}{\mathrm{11}}\:\:. \\ $$

Answered by mrW2 last updated on 16/Feb/18

Commented by mrW2 last updated on 17/Feb/18

let S=Area of ABCD ΔADP=(S/4)=S_1 +S_3 ΔCDF=S_1 +S_3 =(S/4) ⇒FC=((BC)/2) CG=AD=BC ((FG)/(CG))=(3/2) ((Area of ΔEFG)/(Area of ΔHCG))=(((FG)/(CG)))^2 ⇒((S_3 +S_1 +S_3 )/S_3 )=((3/2))^2 =(9/4) 8S_3 +4S_1 =9S_3 ⇒S_3 =4S_1 S_3 +S_1 =5S_1 =(S/4) ⇒S_1 =(S/(20)) ⇒S_3 =4S_1 =(1/5)S S_2 +S_3 =(3/4)S ⇒S_2 =(3/4)S−S_3 =(3/4)S−(1/5)S=((11)/(20))S=11S_1 ⇒(S_1 /S_2 )=(1/(11))

$${let}\:{S}={Area}\:{of}\:{ABCD} \\ $$$$\Delta{ADP}=\frac{{S}}{\mathrm{4}}={S}_{\mathrm{1}} +{S}_{\mathrm{3}} \\ $$$$\Delta{CDF}={S}_{\mathrm{1}} +{S}_{\mathrm{3}} =\frac{{S}}{\mathrm{4}} \\ $$$$\Rightarrow{FC}=\frac{{BC}}{\mathrm{2}} \\ $$$${CG}={AD}={BC} \\ $$$$\frac{{FG}}{{CG}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{Area}\:{of}\:\Delta{EFG}}{{Area}\:{of}\:\Delta{HCG}}=\left(\frac{{FG}}{{CG}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{S}_{\mathrm{3}} +{S}_{\mathrm{1}} +{S}_{\mathrm{3}} }{{S}_{\mathrm{3}} }=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{8}{S}_{\mathrm{3}} +\mathrm{4}{S}_{\mathrm{1}} =\mathrm{9}{S}_{\mathrm{3}} \\ $$$$\Rightarrow{S}_{\mathrm{3}} =\mathrm{4}{S}_{\mathrm{1}} \\ $$$${S}_{\mathrm{3}} +{S}_{\mathrm{1}} =\mathrm{5}{S}_{\mathrm{1}} =\frac{{S}}{\mathrm{4}} \\ $$$$\Rightarrow{S}_{\mathrm{1}} =\frac{{S}}{\mathrm{20}} \\ $$$$\Rightarrow{S}_{\mathrm{3}} =\mathrm{4}{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}}{S} \\ $$$${S}_{\mathrm{2}} +{S}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{4}}{S} \\ $$$$\Rightarrow{S}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}{S}−{S}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{4}}{S}−\frac{\mathrm{1}}{\mathrm{5}}{S}=\frac{\mathrm{11}}{\mathrm{20}}{S}=\mathrm{11}{S}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{11}} \\ $$

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