# What is the perimeter of a triangle with corners at (2 ,5 ), (9 ,2 ), and (3 ,8 )?

Apr 12, 2018

$\sqrt{58} + \sqrt{72} + \sqrt{10}$

#### Explanation:

This requires you to apply the distance formula three times. Recall that the distance between two 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 distance between $\left(2 , 5\right)$ and $\left(9 , 2\right)$ is ${D}_{1} = \sqrt{{\left(9 - 2\right)}^{2} + {\left(2 - 5\right)}^{2}} = \sqrt{49 + 9} = \sqrt{58}$.

The distance between $\left(9 , 2\right)$ and $\left(3 , 8\right)$ is ${D}_{2} = \sqrt{{\left(3 - 9\right)}^{2} + {\left(8 - 2\right)}^{2}} = \sqrt{36 + 36} = \sqrt{72}$.

The distance between $\left(3 , 8\right)$ and $\left(2 , 5\right)$ is ${D}_{3} = \sqrt{{\left(2 - 3\right)}^{2} + {\left(5 - 8\right)}^{2}} = \sqrt{1 + 9} = \sqrt{10}$.

The perimeter is thus $P = {D}_{1} + {D}_{2} + {D}_{3} = \sqrt{58} + \sqrt{72} + \sqrt{10}$.