Question

Sa se determine perimetrul triunghiului ABC ale carui varfuri sunt punctele A(-1,3) B(-2,0) si C(0,3) Multumesc!

Answer 1

$AB= \sqrt{(XB-XA)^{2}+ (YB-YA)^{2}} = \sqrt{(-2+1)^{2}+ (3-0)^{2}}$
$AB=\sqrt{(-1)^{2}+ (3)^{2}}=\sqrt{1+ 9}=\sqrt{10}$
$BC= \sqrt{(XC-XB)^{2}+ (YC-YB)^{2}} = \sqrt{(0+2)^{2}+ (3-0)^{2}}$
$BC=\sqrt{(2)^{2}+ (3)^{2}}=\sqrt{4+ 9}=\sqrt{13}$
$AC= \sqrt{(XC-XA)^{2}+ (YC-YA)^{2}} = \sqrt{(0+1)^{2}+ (3-3)^{2}}$
$AC=\sqrt{(1)^{2}+ (0)^{2}}=\sqrt{1+ 0}=\sqrt{1}$=1

Deci Perimetru ABC=AB+BC+AC= √10+√13 +1