Question

find the perimeter of triangle formed by the points A(5,2) B(4,7) and C(7,-4)

Answer 1

Perimeter$=AB+BC+CA$

$AB=\sqrt{{(5-4)}^2+{(2-7)}^2}=\sqrt{1+25}=\sqrt{26}BC=\sqrt{{(7-4)}^2+{(-4-7)}^2}=\sqrt{3^2+{11}^2}=\sqrt{9+121}=\sqrt{130}CA=\sqrt{{(7-5)}^2+{(-4-2)}^2}=\sqrt{4+36}=\sqrt{40}$

$\therefore$ Perimeter $=(\sqrt{26}+\sqrt{130}+\sqrt{40})$ sq.units