Given the coordinates of the vertices of the triangle ABC A (-5; 0) B (7; 9) C (5; -5)

Given the coordinates of the vertices of the triangle ABC A (-5; 0) B (7; 9) C (5; -5) Find: The lengths of the sides AB, BC, CA and the perimeter.

We have the coordinates of the vertices of the triangle:

A (-5; 0), B (7; 9), C (5; -5).

Find the lengths of the sides of the triangle:

| AB | = ((7 + 5) ^ 2 + (9 – 0) ^ 2) ^ (1/2) = (144 + 81) ^ (1/2) = 15;

| BC | = ((5 – 7) ^ 2 + (-5 – 9) ^ 2) ^ (1/2) = (4 + 196) ^ (1/2) = 200 ^ (1/2);

| AC | = ((5 + 5) ^ 2 + (-5 – 0) ^ 2) ^ (1/2) = (100 + 25) ^ (1/2) = 125 ^ (1/2).

Find the perimeter of the triangle:

P (ABC) = 15 + 200 ^ (1/2) + 125 ^ (1/2).