# What is the perimeter of a triangle with corners at (6 ,5 ), (8 ,2 ), and (4 ,7 )?

Perimeter = 62

#### Explanation:

Let $\left(6 , 5\right)$ = $A$ , $\left(8 , 2\right)$ = $B$ and $\left(4 , 7\right)$ = $C$
Formula of distance between points is :
sqrt ((x_2-x_1)^2+(y_2-y_1)^2
Find the distance between A and B:
Let $\left(6 , 5\right)$ = $\left({x}_{1} , {y}_{1}\right)$ $\mathmr{and}$ $\left(8 , 2\right)$ = $\left({x}_{2} , {y}_{2}\right)$
$\implies$ $\sqrt{{\left(8 - 6\right)}^{2} + {\left(2 - 5\right)}^{2}}$
$\implies$ $\sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2}}$
$\implies$ $\sqrt{{\left(4 + 9\right)}^{2}}$
$\implies$ $\sqrt{{\left(13\right)}^{2}}$
$\implies$ $13$

Find the distance between B and C:
Let $\left(8 , 2\right)$ = $\left({x}_{1} , {y}_{1}\right)$ $\mathmr{and}$ $\left(4 , 7\right)$ = $\left({x}_{2} , {y}_{2}\right)$
$\implies$ $\sqrt{{\left(4 - 8\right)}^{2} + {\left(7 - 2\right)}^{2}}$
$\implies$ $\sqrt{{\left(- 4\right)}^{2} + {\left(5\right)}^{2}}$
$\implies$ $\sqrt{{\left(16 + 25\right)}^{2}}$
$\implies$ $\sqrt{{\left(41\right)}^{2}}$
$\implies$ $41$

Find the distance between C and A:
Let $\left(4 , 7\right)$ = $\left({x}_{1} , {y}_{1}\right)$ $\mathmr{and}$ $\left(6 , 5\right)$ = $\left({x}_{2} , {y}_{2}\right)$
$\implies$ $\sqrt{{\left(6 - 4\right)}^{2} + {\left(5 - 7\right)}^{2}}$
$\implies$ $\sqrt{{\left(2\right)}^{2} + {\left(- 2\right)}^{2}}$
$\implies$ $\sqrt{{\left(4 + 4\right)}^{2}}$
$\implies$ $\sqrt{{\left(8\right)}^{2}}$
$\implies$ $8$

Perimeter = $A B + B C + C A$
$\implies$ $13 + 41 + 8$
$\implies$ $62$