What is the perimeter of a triangle with corners at (7 ,3 ), (1 ,5 ), and (2 ,1 )?

Nov 13, 2016

Perimeter of the triangle is $15.8329$

Explanation:

As the corners are $A \left(7 , 3\right)$, $B \left(1 , 5\right)$ and $C \left(2 , 1\right)$, the three sides of a triangle are

$a = \sqrt{{\left(2 - 1\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{1 + 16} = \sqrt{17} = 4.1231$

$b = \sqrt{{\left(7 - 2\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{25 + 4} = \sqrt{29} = 5.3852$

$c = \sqrt{{\left(7 - 1\right)}^{2} + {\left(3 - 5\right)}^{2}} = \sqrt{36 + 4} = \sqrt{40} = 6.3246$

and perimeter is $4.1231 + 5.3852 + 6.3246 = 15.8329$