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Se considera dreptunghiul DEFG,in care m(

Răspuns :

a)m(GDF)=2*m(EDF)⇒x+2x=90⇒3x=90⇒x=30°⇒m(GDF)=30° si m(EDF)=60°
Deci m(EFD)=30°
Dupa th unghiului de 30° aflam DF:
DE=1/2*DF⇒DF=2*DE⇒DF=2*5[tex] \sqrt{6} [/tex] =10[tex] \sqrt{6} [/tex]
Acum folosim teorema lui Pitagora
EF²=DF²-DE²
EF²=(10[tex] \sqrt{6} [/tex])²-(5[tex] \sqrt{6} [/tex])²
EF²=100*6-25*6
EF²=600-150
EF²=450
EF=15[tex] \sqrt{2} [/tex]

Adr=L*l⇒Adr=5[tex] \sqrt{6} [/tex] *15[tex] \sqrt{2} [/tex] =150[tex] \sqrt{3} [/tex]