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Question-190269




Question Number 190269 by normans last updated on 30/Mar/23
Commented by a.lgnaoui last updated on 02/Apr/23
Answered by a.lgnaoui last updated on 02/Apr/23
The Area of  △AEB  ABC and △AFC  are equal to  yellow △ABC=30 (AB=AC=BE=CF=a)  sum of Area=30×3=90  Area of BCI  ?  ABC(triangle isocele and rectangle  )⇒  Area=(a^2 /2)=30⇒a=2(√(15))   BC=AB(√2)   BC=2(√(30)) ⇒ AD=BD=(√(30))   DI=AI−AD=10−(√(30))   Area(△BCI)=DI×BD=BC×DI=(10−(√(30)) )(√(30))  =10(√(30)) −30  ⇒Area of great triangle(EFI ) is  90+10(√(30)) −30=60+10(√(30))   triangles(EMI)  (FNI) are equal  their area=Area trapeze(EFMN)−(60+10(√(30)) )  Area(trapeze)=(((EF+MN)/2))×10  =(((2BC+MN))/2)×10=100 +10(√3)   Area(equal triangles)=(100+10(√3) )−(60+10(√(30)) )=40    ⇒the Area of  forms colored is  40+area(ABCI)=40+10(√(30))     ⇒Total Area=94,77
$${The}\:{Area}\:{of}\:\:\bigtriangleup{AEB}\:\:{ABC}\:{and}\:\bigtriangleup{AFC} \\ $$$${are}\:{equal}\:{to}\:\:{yellow}\:\bigtriangleup{ABC}=\mathrm{30}\:\left({AB}={AC}={BE}={CF}={a}\right) \\ $$$${sum}\:{of}\:{Area}=\mathrm{30}×\mathrm{3}=\mathrm{90} \\ $$$${Area}\:{of}\:{BCI}\:\:? \\ $$$${ABC}\left({triangle}\:{isocele}\:{and}\:{rectangle}\:\:\right)\Rightarrow \\ $$$${Area}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{30}\Rightarrow{a}=\mathrm{2}\sqrt{\mathrm{15}}\:\:\:{BC}={AB}\sqrt{\mathrm{2}}\: \\ $$$${BC}=\mathrm{2}\sqrt{\mathrm{30}}\:\Rightarrow\:{AD}={BD}=\sqrt{\mathrm{30}}\: \\ $$$${DI}={AI}−{AD}=\mathrm{10}−\sqrt{\mathrm{30}}\: \\ $$$$\boldsymbol{{Area}}\left(\bigtriangleup\boldsymbol{{BCI}}\right)={DI}×{BD}={BC}×{DI}=\left(\mathrm{10}−\sqrt{\mathrm{30}}\:\right)\sqrt{\mathrm{30}} \\ $$$$=\mathrm{10}\sqrt{\mathrm{30}}\:−\mathrm{30} \\ $$$$\Rightarrow{Area}\:{of}\:{great}\:{triangle}\left({EFI}\:\right)\:{is} \\ $$$$\mathrm{90}+\mathrm{10}\sqrt{\mathrm{30}}\:−\mathrm{30}=\mathrm{60}+\mathrm{10}\sqrt{\mathrm{30}}\: \\ $$$${triangles}\left({EMI}\right)\:\:\left({FNI}\right)\:{are}\:{equal} \\ $$$${their}\:{area}={Area}\:{trapeze}\left({EFMN}\right)−\left(\mathrm{60}+\mathrm{10}\sqrt{\mathrm{30}}\:\right) \\ $$$$\boldsymbol{{Area}}\left(\boldsymbol{{trapeze}}\right)=\left(\frac{{EF}+{MN}}{\mathrm{2}}\right)×\mathrm{10} \\ $$$$=\frac{\left(\mathrm{2}{BC}+{MN}\right)}{\mathrm{2}}×\mathrm{10}=\mathrm{100}\:+\mathrm{10}\sqrt{\mathrm{3}}\: \\ $$$$\boldsymbol{{Area}}\left(\boldsymbol{{equal}}\:\boldsymbol{{triangles}}\right)=\left(\mathrm{100}+\mathrm{10}\sqrt{\mathrm{3}}\:\right)−\left(\mathrm{60}+\mathrm{10}\sqrt{\mathrm{30}}\:\right)=\mathrm{40} \\ $$$$ \\ $$$$\Rightarrow{the}\:{Area}\:{of}\:\:{forms}\:{colored}\:{is} \\ $$$$\mathrm{40}+{area}\left({ABCI}\right)=\mathrm{40}+\mathrm{10}\sqrt{\mathrm{30}}\: \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{{Total}}\:\boldsymbol{{Area}}=\mathrm{94},\mathrm{77} \\ $$$$ \\ $$

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