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

Jul 17, 2017

See a solution process below:

#### Explanation:

To find the perimeter of the triangle we must find the length of each segment with the endpoints in the problem. Then we can add these lengths to find the perimeter of the triangle.

We need to find the length of segments:

${d}_{1}$ is the distance between $\left(1 , 2\right) \text{ and } \left(2 , 3\right)$

${d}_{2}$ is the distance between $\left(2 , 3\right) \text{ and } \left(4 , 1\right)$

${d}_{3}$ is the distance between $\left(4 , 1\right) \text{ and } \left(1 , 2\right)$

The formula for calculating the distance between two points is:

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

Length ${d}_{1}$:

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

Length ${d}_{2}$:

${d}_{2} = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{3}\right)}^{2}} = \sqrt{{2}^{2} + {\left(- 2\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8}$

$= \sqrt{4 \cdot 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2}$

Length ${d}_{3}$:

${d}_{3} = \sqrt{{\left(\textcolor{red}{1} - \textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{1}\right)}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

$= \sqrt{5 \cdot 2} = \sqrt{5} \sqrt{2}$

The perimeter $p$ is:

$p = {d}_{1} + {d}_{2} + {d}_{3}$

Substituting gives:

$p = \sqrt{2} + 2 \sqrt{2} + \sqrt{5} \sqrt{2}$

$p = 1 \sqrt{2} + 2 \sqrt{2} + \sqrt{5} \sqrt{2}$

$p = \left(1 + 2 + \sqrt{5}\right) \sqrt{2}$

$p = \left(3 + \sqrt{5}\right) \sqrt{2}$

Or

$p = 7.405$ rounded to the nearest thousandth.