Question

The vertices of a triangle are located at A (-5, -5), B (3, 4), C (3, -5). What is the perimeter of the triangle?

Answer 1

Answer:

34.6 units

Step-by-step explanation:

Given data

A (-5, -5), B (3, 4)

distance is

$d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((3 +5)^2 + (4 +5)^2)\\\\d= \sqrt((8)^2 + (9)^2)\\\\d= \sqrt(64 +81)\\\\d= \sqrt(145)\\\\d= 12.04$

B (3, 4), C (3, -5)

$d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((3 -3)^2 + (-5-4)^2)\\\\d= \sqrt((0)^2 + (-9)^2)\\\\d= \sqrt(0 +81)\\\\d= \sqrt(81)\\\\d= 9$

C (3, -5), A (-5, -5)

$d= \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)\\\\d= \sqrt((-5 -3)^2 + (-5-5)^2)\\\\d= \sqrt((-8)^2 + (-10)^2)\\\\d= \sqrt(64 +100)\\\\d= \sqrt(164)\\\\d= 12.80$

Hence the perimeter

= 12.80+9+12.80

=34.6 units