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Question-205880




Question Number 205880 by cortano12 last updated on 01/Apr/24
Commented by cortano12 last updated on 01/Apr/24
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Answered by mr W last updated on 02/Apr/24
Commented by mr W last updated on 02/Apr/24
Δ_(BCE) =((square)/4)=((12^2 )/4)=36  ((EF)/(FB))=((EC)/(AB))=(1/2) ⇒((EF)/(EB))=(1/3)  A_1 =(1/3)Δ_(BCE)   ΔBGH∼ΔBCE  ((BH)/(BE))=(1/( (√(1^2 +2^2 ))))=(1/( (√5)))  A_2 =((1/( (√5))))^2 Δ_(BCE) =(1/5)Δ_(BCE)   shaded A_3 =(1−(1/3)−(1/5))Δ_(BCE) =(7/(15))Δ_(BCE)                         =(7/(15))×36=((84)/5)=16.8 ✓
$$\Delta_{{BCE}} =\frac{{square}}{\mathrm{4}}=\frac{\mathrm{12}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{36} \\ $$$$\frac{{EF}}{{FB}}=\frac{{EC}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\frac{{EF}}{{EB}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\Delta_{{BCE}} \\ $$$$\Delta{BGH}\sim\Delta{BCE} \\ $$$$\frac{{BH}}{{BE}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${A}_{\mathrm{2}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \Delta_{{BCE}} =\frac{\mathrm{1}}{\mathrm{5}}\Delta_{{BCE}} \\ $$$${shaded}\:{A}_{\mathrm{3}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)\Delta_{{BCE}} =\frac{\mathrm{7}}{\mathrm{15}}\Delta_{{BCE}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{7}}{\mathrm{15}}×\mathrm{36}=\frac{\mathrm{84}}{\mathrm{5}}=\mathrm{16}.\mathrm{8}\:\checkmark \\ $$
Answered by A5T last updated on 01/Apr/24
Commented by A5T last updated on 01/Apr/24
((sin(90−θ))/(y=FH))=((sin(90−β))/(BF)); ((sinβ)/(BF))=((sinθ)/x)  x=((BFsinθ)/(sinβ));(1/y)=((sin(90−β))/(BFsin(90−θ)))⇒(x/y)=((tanθ)/(tanβ))=((2/1)/(1/2))=4  (((1/2)×4y×2z)/((1/2)×5y×3z))=(8/(15))=(([AGF])/([ACH]))⇒(8/(15))×36=[AGF]=19.2  ⇒[CGFH]=[ACH]−[AGF]=36−19.2=16.8
$$\frac{{sin}\left(\mathrm{90}−\theta\right)}{{y}={FH}}=\frac{{sin}\left(\mathrm{90}−\beta\right)}{{BF}};\:\frac{{sin}\beta}{{BF}}=\frac{{sin}\theta}{{x}} \\ $$$${x}=\frac{{BFsin}\theta}{{sin}\beta};\frac{\mathrm{1}}{{y}}=\frac{{sin}\left(\mathrm{90}−\beta\right)}{{BFsin}\left(\mathrm{90}−\theta\right)}\Rightarrow\frac{{x}}{{y}}=\frac{{tan}\theta}{{tan}\beta}=\frac{\frac{\mathrm{2}}{\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{4} \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}{y}×\mathrm{2}{z}}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}{y}×\mathrm{3}{z}}=\frac{\mathrm{8}}{\mathrm{15}}=\frac{\left[{AGF}\right]}{\left[{ACH}\right]}\Rightarrow\frac{\mathrm{8}}{\mathrm{15}}×\mathrm{36}=\left[{AGF}\right]=\mathrm{19}.\mathrm{2} \\ $$$$\Rightarrow\left[{CGFH}\right]=\left[{ACH}\right]−\left[{AGF}\right]=\mathrm{36}−\mathrm{19}.\mathrm{2}=\mathrm{16}.\mathrm{8} \\ $$

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