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Question Number 72579 by TawaTawa last updated on 30/Oct/19

Answered by mind is power last updated on 30/Oct/19

CF.AD=DC.AE=AB.AE ⇒AD=((16.8)/(10))=((64)/5)=12.8cm 2)ar(ABCD)=DC.AD.sin(∠CDA) ar(EFGH)=ar(ABCD)−ar(EBF)−ar(FCG)−ar(GDH)−ar(DHE) =ar(ABCD)−4ar(EBF) ar(EBF)=(1/2).((AB)/2).((BF)/2).sin(∠EBF)=((DC.AC.sin(∠CDA))/8)=((ar(ABCD))/8) ⇒ar(EFGH)=ar(ABCD)−(4/8)ar(ABCD)=((ar(ABCD))/2) let a=∠ADC 3)ar(APB)=ar(ABCD)−{ar(ADP)+ar(PCD)} =((AD.DC)/)sin(a)−{((AD.DP)/2).sin(a)+((PC.AD)/2)sin(a)}=AD.DCsin(a)−((AD.DCsin(a))/2) =(1/2).Ar(ABCD) =ar(BQC)

$$\mathrm{CF}.\mathrm{AD}=\mathrm{DC}.\mathrm{AE}=\mathrm{AB}.\mathrm{AE} \\ $$$$\Rightarrow\mathrm{AD}=\frac{\mathrm{16}.\mathrm{8}}{\mathrm{10}}=\frac{\mathrm{64}}{\mathrm{5}}=\mathrm{12}.\mathrm{8cm} \\ $$$$\left.\mathrm{2}\right)\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{DC}.\mathrm{AD}.\mathrm{sin}\left(\angle\mathrm{CDA}\right) \\ $$$$\mathrm{ar}\left(\mathrm{EFGH}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)−\mathrm{ar}\left(\mathrm{EBF}\right)−\mathrm{ar}\left(\mathrm{FCG}\right)−\mathrm{ar}\left(\mathrm{GDH}\right)−\mathrm{ar}\left(\mathrm{DHE}\right) \\ $$$$=\mathrm{ar}\left(\mathrm{ABCD}\right)−\mathrm{4ar}\left(\mathrm{EBF}\right) \\ $$$$\mathrm{ar}\left(\mathrm{EBF}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{AB}}{\mathrm{2}}.\frac{\mathrm{BF}}{\mathrm{2}}.\mathrm{sin}\left(\angle\mathrm{EBF}\right)=\frac{\mathrm{DC}.\mathrm{AC}.\mathrm{sin}\left(\angle\mathrm{CDA}\right)}{\mathrm{8}}=\frac{\mathrm{ar}\left(\mathrm{ABCD}\right)}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{ar}\left(\mathrm{EFGH}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)−\frac{\mathrm{4}}{\mathrm{8}}\mathrm{ar}\left(\mathrm{ABCD}\right)=\frac{\mathrm{ar}\left(\mathrm{ABCD}\right)}{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{a}=\angle\mathrm{ADC} \\ $$$$\left.\mathrm{3}\right)\mathrm{ar}\left(\mathrm{APB}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)−\left\{\mathrm{ar}\left(\mathrm{ADP}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)\right\} \\ $$$$=\frac{\mathrm{AD}.\mathrm{DC}}{}\mathrm{sin}\left(\mathrm{a}\right)−\left\{\frac{\mathrm{AD}.\mathrm{DP}}{\mathrm{2}}.\mathrm{sin}\left(\mathrm{a}\right)+\frac{\mathrm{PC}.\mathrm{AD}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{a}\right)\right\}=\mathrm{AD}.\mathrm{DCsin}\left(\mathrm{a}\right)−\frac{\mathrm{AD}.\mathrm{DCsin}\left(\mathrm{a}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{Ar}\left(\mathrm{ABCD}\right)\:=\mathrm{ar}\left(\mathrm{BQC}\right) \\ $$$$ \\ $$

Commented by TawaTawa last updated on 30/Oct/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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