What is the perimeter of a triangle with corners at (7 ,3 ), (1 ,5 ), and (2 ,4 )?

Jun 7, 2016

Perimeter is $12.84$.

Explanation:

Let us find the sides of triangle formed by $\left(7 , 3\right)$, $\left(1 , 5\right)$ and $\left(2 , 4\right)$. This will be surely distance between pair of points, which is

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

$b = \sqrt{{\left(2 - 1\right)}^{2} + {\left(4 - 5\right)}^{2}} = \sqrt{1 + 1} = \sqrt{2} = 1.4142$ and

$c = \sqrt{{\left(2 - 7\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{25 + 1} = \sqrt{26} = 5.099$

Hence perimeter is $6.3246 + 1.4142 + 5.099 = 12.8378$ or say $12.84$.