# What is the perimeter of a triangle with corners at (7 ,2 ), (8 ,3 ), and (4 ,4 )?

Jul 29, 2017

$9.14$ $\text{units}$

#### Explanation:

To find the perimeter of a triangle, add up the lengths of all of its sides. Use the distance formula, $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$.

Find the distance between $\left(7 , 2\right)$ and $\left(8 , 3\right)$:

$d = \sqrt{{\left(8 - 7\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

Find the distance between $\left(8 , 3\right)$ and $\left(4 , 4\right)$:

$d = \sqrt{{\left(4 - 8\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{{\left(- 4\right)}^{2} + {1}^{2}} = \sqrt{17}$

Find the distance between $\left(7 , 2\right)$ and $\left(4 , 4\right)$:

$d = \sqrt{{\left(4 - 7\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {2}^{2}} = \sqrt{13}$

The perimeter is $\sqrt{2} + \sqrt{17} + \sqrt{13}$, or about $9.14$ $\text{units}$.