Răspuns
5(1+2√3)cm.
Explicație pas cu pas:
Daca Δ ABC-isoscel => ∡B=∡C= (180°-120°):2 =30°
Fie AD⊥BC, BE⊥AC si CF⊥AB -cele trei inaltimi in ΔABC.
Daca AD⊥BC => Δ ADC-dreptunghic in D iar BD=BC/2 (inaltimea in Δ isoscel cade pe mijlocul bazei)
In Δ ABD, sin ∡B=AD/AB <=> sin 30°=AD/10 <=> 1/2=AD/2 => AD=5cm.
cos 30°= BD/BA <=> √3/2=BD/10 => BD=5√3 cm
=>BC=2·BD=2·5√3=10√3.
Aria ΔABC=AD·BC/2= (5·10√3)/2 =25√3
Dar Aria ABC=AC·BE/2=AB·CF/2
=> 25√3=(10·BE)/2=(10·CF)/2 => BE=CF=25√3/5=5√3
=> suma lungimilor inaltimilor in Δ ABC este:
AD+BE+CF= 5+5√3+5√3=5+10√3=5(1+2√3)
