Construim inaltimea in triunghiul isoscel:
AN⊥BC; N∈BC
⇒ΔANB este dreptunghic in ∡N.
ΔABC isoscel; AN-h ⇒ BN=NC=BC/2=24/2=12cm
sin∡B=4/5
sin∡B=cateta opusa/ipotenuza=AN/AB
⇒AN/AB=4/5 ⇒AB=4AN/5
Aplic T.P. in ΔANB: [tex]AN^{2}[/tex]+[tex]NB^{2}[/tex]=[tex]AB^{2}[/tex]
⇒16[tex]AN^{2}[/tex]/25=[tex]AN^{2} +BN^{2}[/tex]
[tex]BN^{2}[/tex]=16[tex]AN^{2}[/tex]/25 -[tex]AN^{2}[/tex]
[tex]BN^{2}[/tex]=9[tex]AN^{2}[/tex]/25
BN=3AN/5
3AN/5=12
AN=12*5/3
AN=20cm
AN/AB=4/5 ⇒ 20/AB=4/5 ⇒ 20*5=AB*4
⇒AB= 25cm=AC
P=AB+AC+BC=25+25+24=74cm
A=b*h/2
AN-inaltimea
BC-baza
A=AN*BC/2=20*24/2=240[tex]cm^{2}[/tex]
