How do you find the perimeter of #triangleABC# with vertices A (-5, -2), B(-2,-2), and C(-5, 2)?

Question

Claire Akin · Accepted Answer

#12" units. "# Using the Distance Formula, #AB^2={(-5)-(-2)}^2+{(-2)-(-2)}^2=9# #:. AB=3# Similarly, #BC=5, and, AC=4# Hence, #"the Perimeter="AB+BC+AC=12" units."#

Aaliyah Anderson · Answer

undefined To find the perimeter of triangle ABC with vertices A(-5, -2), B(-2, -2), and C(-5, 2), you use the distance formula to find the length of each side of the triangle, and then add up the lengths of all three sides.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Using this formula, you calculate the distance between points A and B, B and C, and C and A. Then, you add these distances together to find the perimeter of triangle ABC.

Let's calculate the distances:

1. Distance between A and B:
$$ d_{AB} = \sqrt{(-2 - (-5))^2 + (-2 - (-2))^2} = \sqrt{(3)^2 + (0)^2} = \sqrt{9} = 3 $$

2. Distance between B and C:
$$ d_{BC} = \sqrt{(-5 - (-2))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

3. Distance between C and A:
$$ d_{CA} = \sqrt{(-5 - (-5))^2 + (2 - (-2))^2} = \sqrt{(0)^2 + (4)^2} = \sqrt{16} = 4 $$

Now, add the distances together to find the perimeter:
$$ \text{Perimeter} = d_{AB} + d_{BC} + d_{CA} = 3 + 5 + 4 = 12 $$

Therefore, the perimeter of triangle ABC is 12 units.