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Question-84892

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Question Number 84892 by Power last updated on 17/Mar/20
Commented by Power last updated on 17/Mar/20
please see
$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{see}} \\ $$
Commented by Power last updated on 17/Mar/20
sir mr W please
$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mr}}\:\boldsymbol{\mathrm{W}}\:\:\boldsymbol{\mathrm{please}} \\ $$
Commented by Power last updated on 17/Mar/20
sir mind is power pls
$$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:\mathrm{pls} \\ $$
Commented by Power last updated on 18/Mar/20
solution sir plese
$$\mathrm{solution}\:\mathrm{sir}\:\mathrm{plese} \\ $$
Answered by mr W last updated on 18/Mar/20
Commented by Power last updated on 18/Mar/20
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$$$ \\ $$
Commented by mr W last updated on 18/Mar/20
make NG//AP PR//AB, ((DP)/(AD))=((PR)/(AB))=(2/3) ⇒((DP)/(AP))=(2/(3+2))=(2/5), ((AD)/(AP))=(3/(3+2))=(3/5) (Δ_(DPR) /Δ_(DAB) )=(((PR)/(AB)))^2 =((2/3))^2 =(4/9) (Δ_(DAB) /Δ_(PAB) )=((AD)/(AP))=(3/5) (Δ_(PAB) /Δ_(ABC) )=((BP)/(BC))=(1/3) (Δ_(DPR) /Δ_(ABC) )=(Δ_(DPR) /Δ_(DAB) )×(Δ_(DAB) /Δ_(PAB) )×(Δ_(PAB) /Δ_(ABC) ) ⇒(Δ_(DPR) /Δ_(ABC) )=(4/9)×(3/5)×(1/3)=(4/(45)) ((NG)/(AP))=((BG)/(BP))=((BN)/(AB))=(1/3) ⇒NG=((AP)/3) ⇒BG=((BP)/3)=(1/3)×((BC)/3)=((BC)/9) ⇒GC=BC−BG=(8/9)×BC ((EP)/(NG))=((PC)/(GC))=(2/3)×((BC)/(GC))=(2/3)×(9/8)=(3/4) ((EP)/(DP))=((EP)/(NG))×((NG)/(AP))×((AP)/(DP)) ⇒((EP)/(DP))=(3/4)×(1/3)×(5/2)=(5/8) ((DE)/(DP))=((8−5)/8)=(3/8) similarly ((DF)/(DR))=(3/8) ⇒ΔDEF∼ΔDPR (Δ_(DEF) /Δ_(DPR) )=(((DF)/(DR)))^2 =((3/8))^2 =(9/(64)) (Δ_(DEF) /Δ_(ABC) )=(Δ_(DEF) /Δ_(DPR) )×(Δ_(DPR) /Δ_(ABC) ) ⇒(Δ_(DEF) /Δ_(ABC) )=(9/(64))×(4/(45))=(1/(80)) i.e. the area of each shaded small triangle is (1/(80)) of the area of the big triangle ABC, therefore the sum of areas of all shaded regions is (3/(80)) of the area of the triangle ABC. A_(shaded) =(3/(80))Δ_(ABC) =(3/(80))×35=((21)/(16))
$${make}\:{NG}//{AP} \\ $$$${PR}//{AB},\:\frac{{DP}}{{AD}}=\frac{{PR}}{{AB}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{DP}}{{AP}}=\frac{\mathrm{2}}{\mathrm{3}+\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{5}},\:\frac{{AD}}{{AP}}=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\Delta_{{DPR}} }{\Delta_{{DAB}} }=\left(\frac{{PR}}{{AB}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\frac{\Delta_{{DAB}} }{\Delta_{{PAB}} }=\frac{{AD}}{{AP}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\Delta_{{PAB}} }{\Delta_{{ABC}} }=\frac{{BP}}{{BC}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\Delta_{{DPR}} }{\Delta_{{ABC}} }=\frac{\Delta_{{DPR}} }{\Delta_{{DAB}} }×\frac{\Delta_{{DAB}} }{\Delta_{{PAB}} }×\frac{\Delta_{{PAB}} }{\Delta_{{ABC}} } \\ $$$$\Rightarrow\frac{\Delta_{{DPR}} }{\Delta_{{ABC}} }=\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{45}} \\ $$$$ \\ $$$$\frac{{NG}}{{AP}}=\frac{{BG}}{{BP}}=\frac{{BN}}{{AB}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{NG}=\frac{{AP}}{\mathrm{3}} \\ $$$$\Rightarrow{BG}=\frac{{BP}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{BC}}{\mathrm{3}}=\frac{{BC}}{\mathrm{9}} \\ $$$$\Rightarrow{GC}={BC}−{BG}=\frac{\mathrm{8}}{\mathrm{9}}×{BC} \\ $$$$\frac{{EP}}{{NG}}=\frac{{PC}}{{GC}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{{BC}}{{GC}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{9}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{EP}}{{DP}}=\frac{{EP}}{{NG}}×\frac{{NG}}{{AP}}×\frac{{AP}}{{DP}} \\ $$$$\Rightarrow\frac{{EP}}{{DP}}=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$\frac{{DE}}{{DP}}=\frac{\mathrm{8}−\mathrm{5}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${similarly} \\ $$$$\frac{{DF}}{{DR}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow\Delta{DEF}\sim\Delta{DPR} \\ $$$$\frac{\Delta_{{DEF}} }{\Delta_{{DPR}} }=\left(\frac{{DF}}{{DR}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{64}} \\ $$$$\frac{\Delta_{{DEF}} }{\Delta_{{ABC}} }=\frac{\Delta_{{DEF}} }{\Delta_{{DPR}} }×\frac{\Delta_{{DPR}} }{\Delta_{{ABC}} } \\ $$$$\Rightarrow\frac{\Delta_{{DEF}} }{\Delta_{{ABC}} }=\frac{\mathrm{9}}{\mathrm{64}}×\frac{\mathrm{4}}{\mathrm{45}}=\frac{\mathrm{1}}{\mathrm{80}} \\ $$$${i}.{e}.\:{the}\:{area}\:{of}\:{each}\:{shaded}\:{small} \\ $$$${triangle}\:{is}\:\frac{\mathrm{1}}{\mathrm{80}}\:{of}\:{the}\:{area}\:{of}\:{the}\:{big} \\ $$$${triangle}\:{ABC},\:{therefore}\:{the}\:{sum}\:{of} \\ $$$${areas}\:{of}\:{all}\:{shaded}\:{regions}\:{is}\:\frac{\mathrm{3}}{\mathrm{80}}\:{of} \\ $$$${the}\:{area}\:{of}\:{the}\:{triangle}\:{ABC}. \\ $$$${A}_{{shaded}} =\frac{\mathrm{3}}{\mathrm{80}}\Delta_{{ABC}} =\frac{\mathrm{3}}{\mathrm{80}}×\mathrm{35}=\frac{\mathrm{21}}{\mathrm{16}} \\ $$

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