# What is the perimeter of a triangle with corners at (7 ,3 ), (4 ,5 ), and (3 ,1 )?

May 11, 2016

While not difficult, this question is tedious and involves the same formula 3 times. Perimeter = $\sqrt{13} + \sqrt{20} + \sqrt{17} = 12.2$

#### Explanation:

length of a line = distance between 2 points = $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$\left(7 , 3\right) \mathmr{and} \left(4 , 5\right) \Rightarrow$ $\sqrt{{\left(7 - 4\right)}^{2} + {\left(3 - 5\right)}^{2}} = \sqrt{9 + 4}$

$\left(7 , 3\right) \mathmr{and} \left(3 , 1\right) \Rightarrow$ $\sqrt{{\left(7 - 3\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{16 + 4}$

$\left(4 , 5\right) \mathmr{and} \left(3 , 1\right) \Rightarrow$ $\sqrt{{\left(4 - 3\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{1 + 16}$

Perimeter = $\sqrt{13} + \sqrt{20} + \sqrt{17} = 12.2$