# What is the perimeter of a triangle with corners at (7 ,3 ), (9 ,5 ), and (3 ,9 )?

Mar 15, 2018

$\sqrt{8} + 4 \sqrt{13}$

#### Explanation:

The easiest way to calculate the perimeter of this triangle is to find the distance between each of the points, and then add them together.

We will use the distance formula to find the distance between these three sets of points (which make up the 3 sides of the triangle):

$\text{(7,3) and (9,5) = side 1}$
$\text{(9,5) and (3,9) = side 2}$
$\text{(3,9) and (7,3) = side 3}$

The distance formula is as follows:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Using the first points of $\text{side 1}$, we can find the first distance. Remember that points are in the form $\left(x , y\right)$. We'll call $\left(7 , 3\right)$ point 1 and $\left(9 , 5\right)$ point 2.

Point 1: $\left(7 , 3\right)$ so ${x}_{1} = 7$ and ${y}_{1} = 3$
Point 2: $\left(9 , 5\right)$ so ${x}_{2} = 9$ and ${y}_{2} = 5$

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$d = \sqrt{{\left(9 - 7\right)}^{2} + {\left(5 - 3\right)}^{2}}$
$d = \sqrt{{\left(2\right)}^{2} + {\left(2\right)}^{2}}$
$d = \sqrt{4 + 4}$
$d = \sqrt{8}$

$\text{side 1" = sqrt8}$

To avoid making this answer too long, I'm not going to show the whole process of finding the length of the next sides. You may do them on your own, using the first side as a guide.

Using the same formula, I found that:

$\text{side 2} = \sqrt{52} \rightarrow 2 \sqrt{13}$
$\text{side 3" = sqrt52 rarr 2sqrt13}$

Now add the radicals together as far as you can:
$\text{side 1" + "side 2" + "side 3" = "perimeter}$

$\sqrt{8} + 2 \sqrt{13} + 2 \sqrt{13} = \text{perimeter}$

$\sqrt{8} + 4 \sqrt{13}$ is as simple as we can get the equation. That is our perimeter.