Question

Find the perimeter of Triangle ABC with the verticies A(-5,5) B(3,-5) and C(-5,1

Answer 1

Hello!
 
Let's calculate a distance between two points using the Pythagorean theorem:

$d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2$

Data:

$x_B = 3 x_A = -5 y_B = -5 y_A = 5 d_ {AB} =?$

Solving: distance from A to B

$d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2 d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2 d^ 2_ {AB} = 64 + 100 d^ 2_ {AB} = 164 d_{AB} = \sqrt {164} d_{AB} = \sqrt {2 ^ 2 * 41} \boxed {d_ {AB} = 2 \sqrt {41}}$

Solving:
 
distance from A to C

$d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2$

Data:

$x_C = -5 x_A = -5 y_C = 1 y_A = 5 d_ {AC} =?$

$d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2 d^ 2_ {AC} = 0 + 16 d^ 2_ {AC} = 16 d_ {AC} = \sqrt {16} \boxed {d_ {AC} = 4}$

Solving:

distance from B to C

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2$

Data:

$x_C = -5 x_B = 3 y_C = 1 y_B = -5 d_ {BC} =?$

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2 d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2 d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2 d^ 2_ {BC} = 64 + 36 d^ 2_ {BC} = 100 d_ {BC} = \sqrt {100} \boxed {d_ {BC} = 10}$

Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:

$p = d_ {AB} + d_ {AC} + d_ {BC}$

$p = 2\sqrt {41} + 4 + 10$

$\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark$