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Fynjy0 [20]
3 years ago
12

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

Mathematics
1 answer:
finlep [7]3 years ago
8 0

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

</span>d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2

Data:

x_B = 3 &#10;&#10;x_A = -5 &#10;&#10;y_B = -5 &#10;&#10;y_A = 5 &#10;&#10;d_ {AB} =?

Solving: distance from A to B

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

<span>Solving:
 
distance from A to C</span>

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

Data:

x_C = -5&#10; &#10;x_A = -5&#10; &#10;y_C = 1 &#10;&#10;y_A = 5 &#10;&#10;d_ {AC} =?


d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 &#10;&#10;d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 &#10;&#10;d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2  &#10;&#10;d^ 2_ {AC} = 0 + 16  &#10;&#10;d^ 2_ {AC} = 16 &#10;&#10;d_ {AC} = \sqrt {16} &#10;&#10;\boxed {d_ {AC} = 4}&#10;<span>
</span>

Solving:

distance from B to C

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

Data:

x_C = -5 &#10;&#10;x_B = 3 &#10;&#10;y_C = 1 &#10;&#10;y_B = -5 &#10;&#10;d_ {BC} =?


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

<span>Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:
</span>
p = d_ {AB} + d_ {AC} + d_ {BC}  &#10;&#10;

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

&#10;\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark
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