# What is the perimeter of a triangle with corners at (6 ,4 ), (9 ,2 ), and (5 ,1 )?

Feb 10, 2018

Perimeter of triangle ABC $= \textcolor{g r e e n}{10.891}$

#### Explanation:

Using distance formula, let us calculate the lengths of the triangle.

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Given A (6,4), B (9,2), C(5,1)

$\vec{A B} = \sqrt{{\left(9 - 6\right)}^{2} + {\left(2 - 4\right)}^{2}} = 3.6056$

$\vec{B C} = \sqrt{{\left(9 - 5\right)}^{2} + {\left(2 - 1\right)}^{2}} = 4.1231$

$\vec{C A} = \sqrt{{\left(6 - 5\right)}^{2} + {\left(4 - 1\right)}^{2}} = 3.1623$

Perimeter of $\Delta A B C = \vec{A B} + \vec{B C} + \vec{C A}$

$\implies \left(3.6056 + 4.1231 + 3.1623\right) = \textcolor{g r e e n}{10.891}$