Question

Me ajudeem ,
Qual a hipotenusa de BC ?

Answer 1

Pelo Teorema de Pitágoras,

$(x+3)^2=(x+1)^2+(x-1)^2$

$x^2+6x+9=x^2+2x+1+x^2-2x+1$

$x^2-6x-7=0$

$\Delta=(-6)^2-4\cdot1\cdot(-7)=36+28=64$

$x>0$,

$x=\dfrac{6+8}{2}=7$

$BC=7+3=10$