# What is the perimeter of a triangle with corners at (3 ,0 ), (5 ,2 ), and (1 ,4 )?

Perimeter $P = 2 \sqrt{2} + 4 \sqrt{5} = 11.772699$

#### Explanation:

Compute perimeter

Let $A \left({x}_{a} , {y}_{a}\right) = A \left(3 , 0\right)$
Let $B \left({x}_{b} , {y}_{b}\right) = B \left(5 , 2\right)$
Let $C \left({x}_{c} , {y}_{c}\right) = C \left(1 , 4\right)$

Perimeter $P = {d}_{a} + {d}_{b} + {d}_{c}$
P=sqrt((x_b-x_c)^2+(y_b-y_c)^2)+sqrt((x_a-x_c)^2+(y_a-y_c)^2)+ sqrt((x_b-x_a)^2+(y_b-y_a)^2)

$P = \sqrt{{\left(5 - 1\right)}^{2} + {\left(2 - 4\right)}^{2}} + \sqrt{{\left(3 - 1\right)}^{2} + {\left(0 - 4\right)}^{2}} + \sqrt{{\left(5 - 3\right)}^{2} + {\left(2 - 0\right)}^{2}}$

$P = \sqrt{{\left(4\right)}^{2} + {\left(- 2\right)}^{2}} + \sqrt{{\left(2\right)}^{2} + {\left(- 4\right)}^{2}} + \sqrt{{2}^{2} + {2}^{2}}$
$P = \sqrt{20} + \sqrt{20} + \sqrt{8}$
$P = 2 \sqrt{2} + 4 \sqrt{5}$
$P = 11.772699$

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