# What is the perimeter of a triangle with corners at (2 ,6 ), (4 ,5 ), and (3 ,1 )?

Jan 26, 2018

See a solution process below:

#### Explanation:

The perimeter of a triangle is:

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

Where: ${s}_{1}$, ${s}_{2}$ and ${s}_{3}$ are the lengths of the sides of the triangle.

We need to find the distance between each of the points in the problem. 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 (2, 6) and (4, 5) is:

${s}_{1} = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{5} - \textcolor{b l u e}{6}\right)}^{2}}$

${s}_{1} = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}}$

${s}_{1} = \sqrt{4 + 1}$

${s}_{1} = \sqrt{5}$

Distance Between (2, 6) and (3, 1) is:

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

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

${s}_{2} = \sqrt{1 + 25}$

${s}_{2} = \sqrt{26}$

Distance Between (4, 5) and (3, 1) is:

${s}_{3} = \sqrt{{\left(\textcolor{red}{3} - \textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{5}\right)}^{2}}$

${s}_{3} = \sqrt{{\left(- 1\right)}^{2} + {\left(- 4\right)}^{2}}$

${s}_{3} = \sqrt{1 + 16}$

${s}_{3} = \sqrt{17}$

The Perimeter of the Triangle is:

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

$p = \sqrt{5} + \sqrt{26} + \sqrt{17}$

If you need a single number answer:

$p = 11.458$ rounded to the nearest thousandth.