What is the perimeter of a triangle with corners at #(6 ,4 )#, #(9 ,2 )#, and #(5 ,1 )#?

Question

What is the perimeter of a triangle with corners at #(6 ,4  )#, #(9   ,2  )#, and #(5 ,1 )#?

Violet Aldrich · Accepted Answer

Perimeter of triangle ABC #= color (green)(10.891)#

Using distance formula, let us calculate the lengths of the triangle.

#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

Given A (6,4), B (9,2), C(5,1)

#vec(AB) = sqrt((9-6)^2 + (2-4)^2) = 3.6056#

#vec(BC) = sqrt((9-5)^2+(2-1)^2) = 4.1231#

#vec(CA) = sqrt((6-5)^2 + (4-1)^2) = 3.1623#

Perimeter of #Delta ABC = vec(AB) + vec(BC) + vec(CA)#

#=> (3.6056 + 4.1231 + 3.1623) = color(green)(10.891)#

Violet Aldrich · Answer

undefined To find the perimeter of a triangle with corners at the given coordinates, you can calculate the distance between each pair of points and then add up those distances.

Using the distance formula $ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $, where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of two points:

1. Distance between (6, 4) and (9, 2):
$$ d_1 = \sqrt{(9 - 6)^2 + (2 - 4)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} $$

2. Distance between (9, 2) and (5, 1):
$$ d_2 = \sqrt{(5 - 9)^2 + (1 - 2)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} $$

3. Distance between (5, 1) and (6, 4):
$$ d_3 = \sqrt{(6 - 5)^2 + (4 - 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} $$

Now, add up the distances to find the perimeter:
$$ \text{Perimeter} = d_1 + d_2 + d_3 = \sqrt{13} + \sqrt{17} + \sqrt{10} $$
$$ \text{Perimeter} \approx 3.61 + 4.12 + 3.16 $$
$$ \text{Perimeter} \approx 10.89 $$

So, the perimeter of the triangle is approximately 10.9 units.