Question

Calculate the perimeter of
the triangle with vertices
A(-2, -1), B (10, -1), and
R(4,4). State your answer in
simplest radical form.
please help !

Answer 1

Answer:

Perimeter of ∆ABR = $12 + 2\sqrt{61}$

Step-by-step explanation:

Given, A(-2, -1), B(10, -1), and R(4, 4).

Perimeter of ∆ABR = $\overline{AB} + \overline{BR} + \overline{AR}$

Distance between A(-2, -1) and B(10, -1) using distance formula:

$\overline{AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(10 -(-2))^2 + (-1 -(-1))^2}$

$\overline{AB} = \sqrt{(12)^2 + (0)^2}$

$\overline{AB} = \sqrt{(144 + 0)}$

$\overline{AB} = \sqrt{144} = 12$

Distance between B(10, -1) and R(4, 4):

$\overline{BR} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 10)^2 + (4 -(-1))^2}$

$\overline{BR} = \sqrt{(-6)^2 + (5)^2}$

$\overline{BR} = \sqrt{(36 + 25)}$

$\overline{BR} = \sqrt{61}$

Distance between A(-2, -1) and R(4, 4):

$\overline{AR} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 -(-2))^2 + (4 -(-1))^2}$

$\overline{AR} = \sqrt{(6)^2 + (5)^2}$

$\overline{AR} = \sqrt{(36 + 25)}$

$\overline{AR} = \sqrt{61}$

Perimeter of ∆ABR = $\overline{AB} + \overline{BR} + \overline{AR}$

Perimeter of ∆ABR = $12 + \sqrt{61} + \sqrt{61}$

Perimeter of ∆ABR = $12 + 2\sqrt{61}$