# What is the perimeter of a triangle with corners at (8 ,5 ), (9 ,1 ), and (3 ,4 )?

Jul 27, 2016

Perimeter is $15.930$

#### Explanation:

As perimeter is sum of all the sides, let us find all the sides of triangle formed by $\left(8 , 5\right)$, $\left(9 , 1\right)$ and $\left(3 , 4\right)$. This will be surely distance between pair of points, (which is given by $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$. Hence the three sides are:

$a = \sqrt{{\left(9 - 8\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{1 + 16} = \sqrt{17} = 4.123$

$b = \sqrt{{\left(3 - 9\right)}^{2} + {\left(4 - 1\right)}^{2}} = \sqrt{36 + 9} = \sqrt{45} = 6.708$ and

$c = \sqrt{{\left(3 - 8\right)}^{2} + {\left(4 - 5\right)}^{2}} = \sqrt{25 + 1} = \sqrt{26} = 5.099$

Hence perimeter is $4.123 + 6.708 + 5.099 = 15.930$