Fie AB, BC si AC catetele si ipotenuza.
Conform T. Pitagora=>AC^2=AB^2+BC^2=>AC^2=80=>AC=4√5 (cm)
Fie BD mediana coresp. ipotenuzei (conform teoremei medianei)=>BD=1/2×AC=>AD=2√2 (cm)
Fie AE mediana coresp. catetei BC=>BE=EC=BC/2=4/2=2( cm)
In ▲ABE, m(B)=90°=>AE^2=BE^2+AB^2=>AE^2=4+64=>AE=√68=2√17
Fie CF mediana coresp. catetei AB=>AF=FB=AB/2=8/2=4 (cm)
In ▲FBC, m(B)=90°=> FC^2=BC^2+BF^2=> FC^2=16+16=>FC=4√2
Medianele triunghiului au lungimile egale cu: 4√5, 2√17, 4√2.
Sper c-am fost de ajutor! :)
