What is the perimeter of a triangle with corners at (8 ,5 ), (9 ,7 ), and (1 ,4 )?

Jun 30, 2017

$P = \sqrt{5} + 5 \sqrt{2} + \sqrt{73}$

Explanation:

We would find the lengths of the sides and then sum then to get the perimeter of the triangle.

We can find each side by applying the formula of distance between two points:

$A B = \sqrt{{\left({x}_{B} - {x}_{A}\right)}^{2} + {\left({y}_{B} - {y}_{A}\right)}^{2}}$

Then the sides lengths are:

1) $a = \sqrt{{\left(9 - 8\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
2)$b = \sqrt{{\left(1 - 8\right)}^{2} + {\left(4 - 5\right)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2}$
3)$c = \sqrt{{\left(1 - 9\right)}^{2} + {\left(4 - 7\right)}^{2}} = \sqrt{64 + 9} = \sqrt{73}$

and the perimeter is:

$P = \sqrt{5} + 5 \sqrt{2} + \sqrt{73}$