# What is the perimeter of a triangle with corners at (5 ,2 ), (9 ,7 ), and (1 ,4 )?

May 15, 2018

The perimeter is 19.419, or $\left(\sqrt{41} + \sqrt{73} + 2 \sqrt{5}\right)$

#### Explanation:

Let's call each of the corners/vertices $A$, $B$, and $C$.

$A = \left(5 , 2\right)$
$B = \left(9 , 7\right)$
$C = \left(1 , 4\right)$

The Perimeter $P$, can be described as:

$P = \overline{A B} + \overline{A C} + \overline{B C}$

Where the value with the bar over it is the length of the triangle's side.

We can calculate the individual lengths by treating the $x$ and $y$ displacements as right triangles, and using The Pythagorean Theorem to calculate the length:

If $c = \sqrt{{a}^{2} + {b}^{2}}$, and:

$a = \Delta x = \left({x}_{2} - {x}_{1}\right)$
$b = \Delta y = \left({y}_{2} - {y}_{1}\right)$
$c = \text{side length}$

We can re-write a general equation for any length as:

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

Now, let's do the math for one leg, $\overline{A B}$:

$\overline{A B} = \sqrt{{\left({x}_{B} - {x}_{A}\right)}^{2} + {\left({y}_{B} - {y}_{A}\right)}^{2}}$

$\overline{A B} = \sqrt{{\left(9 - 5\right)}^{2} + {\left(7 - 2\right)}^{2}} = \sqrt{{4}^{2} + {5}^{2}}$

$\overline{A B} = \sqrt{16 + 25}$

$\overline{A B} = \sqrt{41} = 6.403$

Repeating the process for $\overline{B C}$ and $\overline{A C}$:

$\overline{B C} = \sqrt{73} = 8.544$

$\overline{A C} = \sqrt{20} = 2 \sqrt{5} = 4.472$

Now we can calculate the perimeter, $P$:

$P = \sqrt{41} + \sqrt{73} + 2 \sqrt{5}$

$P = 6.403 + 8.544 + 4.472$

color(green)(P=19.419