Hello!
Let's calculate a distance between two points using the Pythagorean theorem:
[tex]d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 [/tex]
Data:
[tex]x_B = 3
x_A = -5
y_B = -5
y_A = 5
d_ {AB} =? [/tex]
Solving:
distance from A to B
[tex]d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2
d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2
d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2
d^ 2_ {AB} = 64 + 100
d^ 2_ {AB} = 164
d_{AB} = \sqrt {164}
d_{AB} = \sqrt {2 ^ 2 * 41}
\boxed {d_ {AB} = 2 \sqrt {41}} [/tex]
Solving:
distance from A to C[tex]d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2[/tex]
Data:
[tex]x_C = -5
x_A = -5
y_C = 1
y_A = 5
d_ {AC} =? [/tex]
[tex]d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2
d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2
d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2
d^ 2_ {AC} = 0 + 16
d^ 2_ {AC} = 16
d_ {AC} = \sqrt {16}
\boxed {d_ {AC} = 4}
[/tex]
Solving:
distance from B to C
[tex]d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2 [/tex]
Data:
[tex]x_C = -5
x_B = 3
y_C = 1
y_B = -5
d_ {BC} =?[/tex]
[tex]d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2
d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2
d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2
d^ 2_ {BC} = 64 + 36
d^ 2_ {BC} = 100
d_ {BC} = \sqrt {100}
\boxed {d_ {BC} = 10} [/tex]
Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:
[tex]p = d_ {AB} + d_ {AC} + d_ {BC}
[/tex]
[tex]p = 2\sqrt {41} + 4 + 10 [/tex]
[tex]
\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark[/tex]
