# What is the perimeter of a triangle with corners at (1 ,5 ), (8 ,3 ), and (4 ,1 )?

Mar 2, 2016

Perimeter is $7.28 + 4.472 + 5 = 11.752$

#### Explanation:

To find the perimeter of a triangle with corners at $\left(1 , 5\right) , \left(8 , 3\right)$, and $\left(4 , 1\right)$, we need the length of all sides of the triangle i.e. distance between each pair of points. Let the points be $A , B , C$ respectively.

So $A B$ is $\sqrt{{\left(8 - 1\right)}^{2} + {\left(3 - 5\right)}^{2}} = \sqrt{{7}^{2} + {\left(- 2\right)}^{2}} = \sqrt{49 + 4} = \sqrt{53} = 7.28$

$B C$ is $\sqrt{{\left(8 - 4\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 4.472$

$C A$ is $\sqrt{{\left(1 - 4\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {4}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Hence perimeter is $7.28 + 4.472 + 5 = 11.752$