# What is the perimeter of a triangle with corners at (9 ,2 ), (6 ,3 ), and (4 ,1 )?

Mar 10, 2018

See a solution process below:

#### Explanation:

To find the perimeter of the triangle formed by these three points we need to find the distance between each pair of points.

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$

Distance Between (9, 2) and (6, 3)

$d = \sqrt{{\left(\textcolor{red}{6} - \textcolor{b l u e}{9}\right)}^{2} + {\left(\textcolor{red}{3} - \textcolor{b l u e}{2}\right)}^{2}}$

$d = \sqrt{- {3}^{2} + {1}^{2}}$

$d = \sqrt{9 + 1}$

$d = \sqrt{2}$

Distance Between (9, 2) and (4, 1)

$d = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{9}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{2}\right)}^{2}}$

$d = \sqrt{- {5}^{2} + {\left(- 1\right)}^{2}}$

$d = \sqrt{25 + 1}$

$d = \sqrt{26}$

Distance Between (6, 3) and (4, 1)

$d = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{6}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{3}\right)}^{2}}$

$d = \sqrt{- {2}^{2} + {\left(- 2\right)}^{2}}$

$d = \sqrt{4 + 4}$

$d = \sqrt{8}$

Therefore the perimeter of the triangle is:

$p = \sqrt{2} + \sqrt{26} + \sqrt{8}$

We can factor out a $\sqrt{2}$ if desired:

$p = \sqrt{2} + \sqrt{2 \cdot 13} + \sqrt{2 \cdot 4}$

$p = \sqrt{2} + \sqrt{2} \sqrt{13} + \sqrt{2} \sqrt{4}$

$p = 1 \sqrt{2} + \sqrt{2} \sqrt{13} + \sqrt{2} \sqrt{4}$

$p = \left(1 + \sqrt{13} + \sqrt{4}\right) \sqrt{2}$

$p = \left(1 + \sqrt{13} + 2\right) \sqrt{2}$

$p = \left(3 + \sqrt{13}\right) \sqrt{2}$

Or

$p = 3 \sqrt{2} + \sqrt{13} \sqrt{2}$

$p = 3 \sqrt{2} + \sqrt{13 \cdot 2}$

$p = 3 \sqrt{2} + \sqrt{26}$