# What is the perimeter of a triangle with corners at (7 ,6 ), (4 ,5 ), and (3 ,1 )?

Aug 10, 2017

$P e r i m e t e r = \sqrt{10} + \sqrt{17} + \sqrt{41}$

#### Explanation:

Equation for finding distance between $2$ coordinates $=$ $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Let,
$\left(7 , 6\right) =$point $A$

$\left(4 , 5\right) =$point $B$

$\left(3 , 1\right) =$point $C$

Distance between $A B$

${x}_{1} = 7 , {x}_{2} = 4 , {y}_{1} = 6 , {y}_{2} = 5$

$A B = \sqrt{{\left(4 - 7\right)}^{2} + {\left(5 - 6\right)}^{2}}$

=>sqrt((-3)^2+(-1)^2

$\implies \sqrt{9 + 1}$

$\implies \sqrt{10}$

$A B = \sqrt{10}$

Distance between $B C$

${x}_{1} = 4 , {x}_{2} = 3 , {y}_{1} = 5 , {y}_{2} = 1$

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

=>sqrt((-1)^2+(-4)^2

$\implies \sqrt{1 + 16}$

$\sqrt{17}$

$B C = \sqrt{17}$

Distance between $A C$

${x}_{1} = 7 , {x}_{2} = 3 , {y}_{1} = 6 , {y}_{2} = 1$

$A C = \sqrt{{\left(3 - 7\right)}^{2} + {\left(1 - 6\right)}^{2}}$

=>sqrt((-4)^2+(-5)^2

$\implies \sqrt{16 + 25}$

$\implies \sqrt{41}$

$A C = \sqrt{41}$

Perimeter of a triangle is the sum of its $3$ sides, that $A B + B C + A C .$

$P e r i m e t e r = \sqrt{10} + \sqrt{17} + \sqrt{41}$