# What is the perimeter of a triangle with corners at (1 ,5 ), (6 , 2 ), and (5 ,7 )?

Jan 5, 2018

See a solution process below:

#### Explanation:

The perimeter of an object is the length of the outer edge of the object. To solve this problem we need to determine the distance between:

• $\left(1 , 5\right) \mathmr{and}$(6, 2)
• $\left(6 , 2\right) \mathmr{and}$(5, 7)
• $\left(5 , 7\right)$ and $\left(1 , 5\right)$

The formula for calculating the distance between two points is:

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

Distance Between $\left(1 , 5\right)$ and $\left(6 , 2\right)$:

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

${d}_{1} = \sqrt{{5}^{2} + {\left(- 3\right)}^{2}}$

${d}_{1} = \sqrt{25 + 9}$

${d}_{1} = \sqrt{34}$

Distance Between $\left(6 , 2\right)$ and $\left(5 , 7\right)$:

${d}_{2} = \sqrt{{\left(\textcolor{red}{5} - \textcolor{b l u e}{6}\right)}^{2} + {\left(\textcolor{red}{7} - \textcolor{b l u e}{2}\right)}^{2}}$

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

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

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

Distance Between $\left(5 , 7\right)$ and $\left(1 , 5\right)$:

${d}_{3} = \sqrt{{\left(\textcolor{red}{1} - \textcolor{b l u e}{5}\right)}^{2} + {\left(\textcolor{red}{5} - \textcolor{b l u e}{7}\right)}^{2}}$

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

${d}_{3} = \sqrt{16 + 4}$

${d}_{3} = \sqrt{20}$

${d}_{3} = \sqrt{4 \cdot 5}$

${d}_{3} = \sqrt{4} \cdot \sqrt{5}$

${d}_{3} = 2 \sqrt{5}$

The Perimeter of the Triangel is:

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

$p = \sqrt{34} + \sqrt{26} + 2 \sqrt{5}$

If you need the answer as a single number:

$p = 5.831 + 5.099 + \left(2 \times 2.236\right)$

$p = 5.831 + 5.099 + 4.472$

$p = 15.402$ rounded to the nearest thousandth.