# What is the perimeter of a triangle with corners at (6 ,4 ), (9 ,2 ), and (5 ,7 )?

Jan 8, 2017

$P = \sqrt{13} + \sqrt{41} + \sqrt{10}$

#### Explanation:

Each side of the triangle can be measured by the distance formula:

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

Let's use labels for corners:

A(6;4);B(9;2);C(5;7)

Then

${d}_{A B} = \sqrt{{\left(9 - 6\right)}^{2} + {\left(2 - 4\right)}^{2}} = \sqrt{13}$

${d}_{B C} = \sqrt{{\left(5 - 9\right)}^{2} + {\left(7 - 2\right)}^{2}} = \sqrt{41}$

${d}_{A C} = \sqrt{{\left(5 - 6\right)}^{2} + {\left(7 - 4\right)}^{2}} = \sqrt{10}$

Then the perimeter of the given triangle is:

$P = \sqrt{13} + \sqrt{41} + \sqrt{10}$