# What is the perimeter of a triangle with corners at (7 ,8 ), (8 ,3 ), and (4 ,5 )?

Perimeter $P = 13.8138$

#### Explanation:

Perimeter $P = a + b + c$
Let $A \left({x}_{a} , {y}_{a}\right) = \left(7 , 8\right)$
Let $B \left({x}_{b} , {y}_{b}\right) = \left(8 , 3\right)$
Let $C \left({x}_{c} , {y}_{c}\right) = \left(4 , 5\right)$

Let a be distance from B to C
Let b be distance from A to C
Let c be distance from A to B

$a = \sqrt{{\left({x}_{b} - {x}_{c}\right)}^{2} + {\left({y}_{b} - {y}_{c}\right)}^{2}} = 2 \sqrt{5}$
$b = \sqrt{{\left({x}_{a} - {x}_{c}\right)}^{2} + {\left({y}_{a} - {y}_{c}\right)}^{2}} = 3 \sqrt{2}$
$c = \sqrt{{\left({x}_{a} - {x}_{b}\right)}^{2} + {\left({y}_{a} - {y}_{b}\right)}^{2}} = \sqrt{26}$

$P = 13.8138$

God bless....I hope the explanation is useful.