What is the perimeter of a triangle with the vertices located at (1,3), (2,6), and (0,4), rounded to the nearest hundredth? A. 7.40 unitsB. 9.07 units C. 8.49 unitsD. 7.07 units

Accepted Solution

Answer: OPTION AStep-by-step explanation: You can plot the points given in the problem, as you can see in the figure attached. Apply the formula for calculate the distance between two points, which is: [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex] Distance AB: [tex]d_{AB}=\sqrt{(2-1)^2+(6-3)^2}=3.162units[/tex]Distance AC: [tex]d_{AC}=\sqrt{(0-1)^2+(4-3)^2}=1.414units[/tex] Distance BC: [tex]d_{BC}=\sqrt{(2-0)^2+(6-4)^2}=2.828units[/tex] The perimeter is: [tex]P=3.162units+1.414units+2.828units\\P=7.40units[/tex]