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

Mar 2, 2016

$P = \sqrt{5} + \sqrt{17} + \sqrt{26} \approx 11.4582$

#### Explanation:

We will use the distance formula, which states that the distance between the points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

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 $\left(3 , 6\right)$ and $\left(1 , 5\right)$:

$\sqrt{{\left(1 - 3\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$

Side length between $\left(1 , 5\right)$ and $\left(2 , 1\right)$:

$\sqrt{{\left(2 - 1\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{{1}^{2} + {4}^{2}} = \sqrt{17}$

Side length between $\left(2 , 1\right)$ and $\left(3 , 6\right)$:

$\sqrt{{\left(3 - 2\right)}^{2} + {\left(6 - 1\right)}^{2}} = \sqrt{{1}^{2} + {5}^{2}} = \sqrt{26}$

Thus the perimeter of the triangle is

$P = \sqrt{5} + \sqrt{17} + \sqrt{26} \approx 11.4582$