# How do you find the perimeter of a triangle with vertices X(3, 0), Y(7, 4), and Z(10, 0)?

Jan 20, 2017

$4 \left(3 + \sqrt{2}\right) = 17.657$, nearly.

#### Explanation:

$\vec{X Y} = < 7 , 4 > - < 3 , 0 > = < 4 , 4 >$.

So,$X Y = \sqrt{{4}^{2} + {4}^{2}} = 4 \sqrt{2}$.

$\vec{Y Z} = < 10 , 0 > - < 7 , 4 > = < 3 , - 4 >$

So, $Y Z = \sqrt{{3}^{2} + {\left(- 4\right)}^{2}} = 5$

$\vec{Z X} = < 3 , 0 > - < 10 , 0 > = < - 47 - 4 >$

So, $Z X = \sqrt{{7}^{2} + 0} = 7.$

The perimeter = XY +YZ + ZX = 4sqrt2+5+7#

$= 4 \left(3 + \sqrt{2}\right) = 17.657$, nearly.