# What is the perimeter of a triangle with corners at (1 ,5 ), (6 , 2 ), and (2 ,7 )?

Perimeter $P = \sqrt{41} + \sqrt{5} + \sqrt{34}$
Perimeter $P = 14.47014411 \text{ }$units

#### Explanation:

let the points be $A \left(1 , 5\right)$, $B \left(6 , 2\right)$, $C \left(2 , 7\right)$

compute lengths of sides a, b , c

$a = \sqrt{{\left({x}_{B} - {x}_{C}\right)}^{2} + {\left({y}_{B} - {y}_{C}\right)}^{2}}$
$a = \sqrt{{\left(6 - 2\right)}^{2} + {\left(2 - 7\right)}^{2}}$
$a = \sqrt{{\left(4\right)}^{2} + {\left(- 5\right)}^{2}}$
$a = \sqrt{16 + 25}$
$a = \sqrt{41}$

$b = \sqrt{{\left({x}_{A} - {x}_{C}\right)}^{2} + {\left({y}_{A} - {y}_{C}\right)}^{2}}$
$b = \sqrt{{\left(1 - 2\right)}^{2} + {\left(5 - 7\right)}^{2}}$
$b = \sqrt{{\left(- 1\right)}^{2} + {\left(- 2\right)}^{2}}$
b=sqrt((1+4)
$b = \sqrt{5}$

$c = \sqrt{{\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}}$
$c = \sqrt{{\left(1 - 6\right)}^{2} + {\left(5 - 2\right)}^{2}}$
$c = \sqrt{{\left(- 5\right)}^{2} + {\left(3\right)}^{2}}$
$c = \sqrt{25 + 9}$
$c = \sqrt{34}$

Perimeter $P = \sqrt{41} + \sqrt{5} + \sqrt{34}$
Perimeter $P = 14.47014411 \text{ }$units