Estude! Com Prof. Caju

Na figura se AO = OB = x, AQ = 3 e BP = 4.
Calcular "x", sendo MN² = AM²+BN²
A) 1
B) 2
C) 3
D) 4
E) 5

[tex3]
\mathsf{
ON=x−BN\\
OM=x−AM\\
MN^2=AM^2+BN^2\\
MN^2=ON^2+OM^2=(x−BN)^2+(x−AM)^2=2x^2−2BNx−2AMx+\underbrace{AM^2+BN^2}_{MN^2}\\
\therefore MN^2=2x^2−2BNx−2AMx+MN^2\\
x^2−BNx−AMx=x(x−BN−AM)=0
\therefore \cancel{x=0} \vee x=AM+BN\\
Potência(B):BP^2=x.BN=4^2=16(I)\\
Potência(A):AQ^2=x.AM =3^2=9(II)\\
(I)+(II)9+16=25=x.AM+x.BN=x.(AM+BN)=x^2\\
\therefore x=-5 \vee \boxed{x=5}
}[/tex3]
(Solução:LionHeart)