Question

Find the perimeter of the triangle whose vertices are $(3,2),(7,2)$ and $(7,5)$

Answer 1

Let $A(x_1,y_1)=(3,2),B(x_2,y_2)=(7,2)$ and $C(x_3,y_3)=(7,5)$

$\therefore$ Distance between two points $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{{(7-3)}^2+{(2-2)}^2}=\sqrt{16+0}=4$units

$BC=\sqrt{{(5-7)}^2+{(5-2)}^2}=\sqrt{4+9}=\sqrt{13}$units

$CA=\sqrt{{(7-7)}^2+{(2-5)}^2}=\sqrt{0+9}=3$units

Perimeter of $△ABC=AB+BC+CA=4+\sqrt{13}+3=7+\sqrt{13}$ units.