# How do you find the perimeter of a triangle whose coordinates are X(0,2), Y(4,-1), Z(-2,-1)?

Mar 13, 2018

$11 + \sqrt{13}$

#### Explanation:

You can use Pythagorus' Theorem (${a}^{2} + {b}^{2} = {c}^{2}$)to find the lengths of the sides.

Therefore the length of side XY is $\sqrt{{\left(4 - 0\right)}^{2} + {\left(\left(- 1\right) - 2\right)}^{2}} = \sqrt{{4}^{2} + {\left(- 3\right)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

Side YZ: $\sqrt{{\left(\left(- 2\right) - 4\right)}^{2} + {\left(\left(- 1\right) - \left(- 1\right)\right)}^{2}} = \sqrt{{\left(- 6\right)}^{2} + {0}^{2}} = \sqrt{36} = 6$

Side XZ: $\sqrt{{\left(\left(- 2\right) - 0\right)}^{2} + {\left(\left(- 1\right) - 2\right)}^{2}} = \sqrt{{\left(- 2\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$

Perimeter = XY + YZ + XZ = $5 + 6 + \sqrt{13}$ = $11 + \sqrt{13}$