What is the perimeter of a triangle with corners at #(3 ,6 )#, #(1 ,5 )#, and #(2 ,1 )#?

Question

What is the perimeter of a triangle with corners at #(3 ,6  )#, #(1 ,5  )#, and #(2 ,1  )#?

Sophia Alfred · Accepted Answer

#P=sqrt5+sqrt17+sqrt26approx11.4582#

We will use the distance formula, which states that the distance between the points #(x_1,y_1)# and #(x_2,y_2)# is

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

The triangle is made up of three line segments. We can determine the length of each side through the distance formula and then add them for the entire perimeter of the triangle.

Side length between #(3,6)# and #(1,5)#:

#sqrt((1-3)^2+(5-6)^2)=sqrt(2^2+1^2)=sqrt5#

Side length between #(1,5)# and #(2,1)#:

#sqrt((2-1)^2+(1-5)^2)=sqrt(1^2+4^2)=sqrt17#

Side length between #(2,1)# and #(3,6)#:

#sqrt((3-2)^2+(6-1)^2)=sqrt(1^2+5^2)=sqrt26#

Thus the perimeter of the triangle is

#P=sqrt5+sqrt17+sqrt26approx11.4582#

Sadie Adams · Answer

undefined To find the perimeter of a triangle with corners at (3, 6), (1, 5), and (2, 1), we use the distance formula to calculate the lengths of each side of the triangle:

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

1. Distance between (3, 6) and (1, 5):
$$ d_1 = \sqrt{(1 - 3)^2 + (5 - 6)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} $$

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

3. Distance between (2, 1) and (3, 6):
$$ d_3 = \sqrt{(3 - 2)^2 + (6 - 1)^2} = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} $$

Finally, we sum up the lengths of the sides to find the perimeter:
$$ \text{Perimeter} = d_1 + d_2 + d_3 = \sqrt{5} + \sqrt{17} + \sqrt{26} $$