# What is the perimeter of a triangle with corners at (3 ,7 ), (1 ,3 ), and (4 ,9 )?

May 18, 2017

$= 6 \sqrt{5}$ unit

#### Explanation:

Let say $A \left(3 , 7\right) , B \left(1 , 3\right) \mathmr{and} C \left(4 , 9\right)$
The distance $D$, between 2 point $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is given by

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

therefore,
the distance between A& B $= \sqrt{{\left(3 - 1\right)}^{2} + {\left(7 - 3\right)}^{2}} = \sqrt{20}$

the distance between B&C $= \sqrt{{\left(1 - 4\right)}^{2} + {\left(3 - 9\right)}^{2}} = \sqrt{45}$

the distance between C&A $= \sqrt{{\left(4 - 3\right)}^{2} + {\left(9 - 7\right)}^{2}} = \sqrt{5}$

Total perimenter,
$= \sqrt{20} + \sqrt{45} + \sqrt{5}$

$= \sqrt{4 \cdot 5} + \sqrt{9 \cdot 5} + \sqrt{5}$

$= 2 \sqrt{5} + 3 \sqrt{5} + \sqrt{5}$
$= 6 \sqrt{5}$ unit