Question

Find the perimeter of triangle XYZ with vertices X(3,7), Y(-1,-5) & Z (6,-4).

Answer 1

$YZ = \sqrt{(-5+4)^{2}}+(-1 - 6)^{2} \sqrt{1+49} \sqrt{50} --- XY= \sqrt{(-5-7)^{2}}+(-1 - 3)^{2} \sqrt{144+16} \sqrt{160} --- XZ= \sqrt{(-4-7)^{2}}+(6 - 3)^{2} \sqrt{121+9} \sqrt{130} to get perimeter you add all the sides together XY + XZ + YZ \sqrt{130} + \sqrt{50} + \sqrt{160} = 31.123 (3 d.p.)$