# What is the perimeter of a triangle with corners at (9 ,2 ), (2 ,3 ), and (4 ,1 )?

Oct 6, 2016

$\sqrt{50} + \sqrt{8} + \sqrt{26}$

#### Explanation:

We know the distance between two points P(x1,y1) and Q(x2,y2) is given by PQ= $\sqrt{{\left(x 2 - x 1\right)}^{2} + {\left(y 2 - y 1\right)}^{2}}$
First we have to calculate the distance between (9,2)(2,3) ; (2,3)(4,1) and (4,1)(9,2) to get the lengths of the sides of triangles.
Hence lengths will be $\sqrt{{\left(2 - 9\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{{\left(- 7\right)}^{2} + {1}^{2}} = \sqrt{49 + 1} = \sqrt{50}$
$\sqrt{{\left(4 - 2\right)}^{2} + {\left(1 - 3\right)}^{2}} = \sqrt{{\left(2\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8}$
and
$\sqrt{{\left(9 - 4\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{{5}^{2} + {1}^{2}} = \sqrt{26}$
Now the perimeter of the triangle is $\sqrt{50} + \sqrt{8} + \sqrt{26}$