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

Perimeter of ∆ABR =
`12 + 2√(61)`

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} = √((x_2 - x_1)^2 + (y_2 - y_1)^2) = √((10 -(-2))^2 + (-1 -(-1))^2)`

`\overline{AB} = √((12)^2 + (0)^2)`

`\overline{AB} = √((144 + 0))`

`\overline{AB} = √(144) = 12`

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

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

`\overline{BR} = √((-6)^2 + (5)^2)`

`\overline{BR} = √((36 + 25))`

`\overline{BR} = √(61)`

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

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

`\overline{AR} = √((6)^2 + (5)^2)`

`\overline{AR} = √((36 + 25))`

`\overline{AR} = √(61)`

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

Perimeter of ∆ABR =
`12 + √(61) + √(61)`

Perimeter of ∆ABR =
`12 + 2√(61)`