Question Number 208344 by mr W last updated on 13/Jun/24
Commented by mr W last updated on 13/Jun/24
$${find}\:{green}\:{area}=? \\ $$
Commented by naka3546 last updated on 13/Jun/24
$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}? \\ $$
Commented by mr W last updated on 13/Jun/24
$${yes} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{Good}\:\mathrm{one}\:\mathrm{sir} \\ $$
Answered by A5T last updated on 13/Jun/24
Commented by A5T last updated on 13/Jun/24
![((sin45)/y)=((sin90)/4)⇒y=2(√2)⇒BC=3y=6(√2)⇒BH=2(√(10)) ((sinHBC)/4)=((sin45)/(2(√(10))))=(((√2)/2)/((2(√(10)))/1))=((√5)/(20))⇒sinHBC=((√5)/5) ⇒cosHBC=((2(√5))/5); (((2(√5))/5)/(6(√2)))=(1/(BF))⇒BF=3(√(10)) ((sinGHF)/(6(√2)))=(((√5)/5)/4)⇒sinGHF=(((6(√(10)))/5)/4)=((3(√(10)))/(10)) ((AE)/(EB))×((BF)/(FH))×((HG)/(GA))=1⇒((AE)/(EB=6(√2)−AE))=((FH)/(BF))=(1/3) ⇒4AE=6(√2)⇒AE=((3(√2))/2) [green]=(1/2)[((3(√2))/2)×2(√2)+4×(√(10))×((3(√(10)))/(10))+6(√2)×2(√2)] =(1/2)[6+12+24]=21](https://www.tinkutara.com/question/Q208348.png)
$$\frac{{sin}\mathrm{45}}{{y}}=\frac{{sin}\mathrm{90}}{\mathrm{4}}\Rightarrow{y}=\mathrm{2}\sqrt{\mathrm{2}}\Rightarrow{BC}=\mathrm{3}{y}=\mathrm{6}\sqrt{\mathrm{2}}\Rightarrow{BH}=\mathrm{2}\sqrt{\mathrm{10}} \\ $$$$\frac{{sinHBC}}{\mathrm{4}}=\frac{{sin}\mathrm{45}}{\mathrm{2}\sqrt{\mathrm{10}}}=\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\mathrm{2}\sqrt{\mathrm{10}}}{\mathrm{1}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{20}}\Rightarrow{sinHBC}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\Rightarrow{cosHBC}=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}};\:\:\:\frac{\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}}{\mathrm{6}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{{BF}}\Rightarrow{BF}=\mathrm{3}\sqrt{\mathrm{10}} \\ $$$$\frac{{sinGHF}}{\mathrm{6}\sqrt{\mathrm{2}}}=\frac{\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}}{\mathrm{4}}\Rightarrow{sinGHF}=\frac{\frac{\mathrm{6}\sqrt{\mathrm{10}}}{\mathrm{5}}}{\mathrm{4}}=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$$\frac{{AE}}{{EB}}×\frac{{BF}}{{FH}}×\frac{{HG}}{{GA}}=\mathrm{1}\Rightarrow\frac{{AE}}{{EB}=\mathrm{6}\sqrt{\mathrm{2}}−{AE}}=\frac{{FH}}{{BF}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{4}{AE}=\mathrm{6}\sqrt{\mathrm{2}}\Rightarrow{AE}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left[{green}\right]=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4}×\sqrt{\mathrm{10}}×\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}}+\mathrm{6}\sqrt{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{6}+\mathrm{12}+\mathrm{24}\right]=\mathrm{21} \\ $$
Answered by MM42 last updated on 13/Jun/24
$${B}=\frac{\mathrm{1}}{\mathrm{2}}{S}_{{FGC}} \\ $$$$\Delta{AGE}\sim\Delta{FGC}\Rightarrow\frac{{A}}{{B}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${C}=\mathrm{2}{S}_{{FCH}} =\mathrm{2}{B} \\ $$$${C}=\mathrm{4}{A}\:\:\&\:\:{B}=\mathrm{2}{A} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{144}\Rightarrow{a}=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}\sqrt{\mathrm{2}}×\mathrm{4}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{12} \\ $$$$\Rightarrow{S}={A}+{B}+{C}=\mathrm{21}\:\:\checkmark\: \\ $$$$ \\ $$
Commented by MM42 last updated on 13/Jun/24
Answered by mr W last updated on 13/Jun/24
Commented by mr W last updated on 13/Jun/24
$${a}={side}\:{length}\:{of}\:{square} \\ $$$${S}={area}\:{of}\:{square} \\ $$$$\sqrt{\mathrm{2}}{a}=\mathrm{3}×\mathrm{4}\:\Rightarrow{a}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{2}}} \\ $$$${S}={a}^{\mathrm{2}} =\left(\frac{\mathrm{12}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\mathrm{72} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{S}}{\mathrm{2}}=\frac{{S}}{\mathrm{6}} \\ $$$${GC}=\mathrm{2}×{AG} \\ $$$$\frac{{B}+{B}}{{C}}=\mathrm{2}^{\mathrm{2}} \:\Rightarrow{B}=\mathrm{2}{C} \\ $$$${A}+{B}={B}+{B}+\mathrm{2}{C}\:\Rightarrow{A}={B}+\mathrm{2}{C}=\mathrm{2}{B} \\ $$$$\Rightarrow{B}=\frac{{A}}{\mathrm{2}}=\frac{{S}}{\mathrm{12}}\:\Rightarrow{C}=\frac{{B}}{\mathrm{2}}=\frac{{S}}{\mathrm{24}} \\ $$$${green}\:{area}\:={A}+{B}+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{24}}\right){S}=\frac{\mathrm{7}}{\mathrm{24}}{S}=\mathrm{21}\:\checkmark \\ $$
Answered by cherokeesay last updated on 13/Jun/24
