# What is the perimeter of a triangle with corners at (7 ,2 ), (8 ,3 ), and (4 ,5 )?

Apr 22, 2016

Perimeter is $10.1288$.

#### Explanation:

The formula for distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Hence length of three sides will be given by

1. $\sqrt{{\left(8 - 7\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{1 + 1} = \sqrt{2} = 1.4142$
2. $\sqrt{{\left(4 - 8\right)}^{2} + {\left(5 - 3\right)}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5} = 2 \times 2.236 = 4.472$
3. $\sqrt{{\left(4 - 7\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} = 4.2426$

Hence perimeter is $1.4142 + 4.472 + 4.2426 = 10.1288$