# What is the perimeter of a triangle with corners at (1 ,4 ), (6 ,7 ), and (4 ,5 )?

Mar 11, 2016

Perimeter is $\left(5.831 + 2.828 + 3.162\right) = 11.821$

#### Explanation:

Let us find the distance between each pair of points A$\left(1 , 4\right)$, B$\left(6 , 7\right)$ and C$\left(4 , 5\right)$.

AB=$c$=$\sqrt{{\left(6 - 1\right)}^{2} + {\left(7 - 4\right)}^{2}} = \sqrt{25 + 9} = \sqrt{34} = 5.831$

BC=$a$=$\sqrt{{\left(6 - 4\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2.828$

AC=$b$=$\sqrt{{\left(4 - 1\right)}^{2} + {\left(5 - 4\right)}^{2}} = \sqrt{9 + 1} = \sqrt{10} = 3.162$

Hence perimeter is $\left(5.831 + 2.828 + 3.162\right) = 11.821$