Question

Find the perimeter of the triangle with vertices (0,8),(6,0) and origin ; (9,3),(1,-3) and origin .

Answer 1

Perimeter is the sum of all sides of the triangle.
Side of a triangle with the ends being  (x1,y1) and (x2,y2) is
$\sqrt{(x2-x1)^2 + (y2-y1)^2}$

The first triangle: (0,8) , (6,0) (0,0) 
perimeter = $\sqrt{6^2+8^2} + \sqrt{6^2+0^2} + \sqrt{0^2+8^2} = 10 + 6 + 8 = 24$

The second triangle: (9,3) (1,-3) (0,0)
perimeter : $\sqrt{6^2+8^2} + \sqrt{9^2+3^2} + \sqrt{1^2+(-3)^2} = 10 + \sqrt{90} + \sqrt{10} = 10 + 4*\sqrt{10}$