Question

Find the perimeter of triangle ABC, with coordinates A(-3, 0), B(0, 4), and C(3, 0)

Answer 1

$Perimeter\ = AB+BC+AC\\\\ AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(0+3)^2+(4-0)^2}=\\\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\\\\ BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}=\sqrt{(3-0)^2+(0-4)^2}=\\\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\\\\ AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}=\sqrt{(3+3)^2+(0-0)^2}=\\\sqrt{6^2+0^2}=\sqrt{36+0}=\sqrt{36}=6\\\\Perimeter=5+5+6=16$