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daca triunghiul isoscel ABC are baza BC=12 si aria=48, atunci perimetrul triunghiului=?

Răspuns :

A=BC*AD/2
48=12*AD/2
AD=8
In ΔADB, mD=90, AD=8, BD=BC/2=6
AB²=AD²+BD²
AB²=64+36
AB²=100
AB=10
P=2*10+12=20+12=32