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

Feb 11, 2018

$4 + 2 \sqrt{10} + 2 \sqrt{2} \cong 13.15$

#### Explanation:

Well, perimeter is simply the sum of the sides for any 2D shape.

We have three sides in our triangle: from $\left(3 , 3\right)$ to $\left(7 , 3\right)$; from $\left(3 , 3\right)$ to $\left(9 , 5\right)$; and from $\left(7 , 3\right)$ to $\left(9 , 5\right)$.

The lengths of each are found by Pythagoras' theorem, using the difference between the $x$ and the $y$ coordinates for a pair of points. .

For the first:

${l}_{1} = \sqrt{{\left(7 - 3\right)}^{2} + {\left(3 - 3\right)}^{2}} = 4$

For the second:

${l}_{2} = \sqrt{{\left(9 - 3\right)}^{2} + {\left(5 - 3\right)}^{2}} = \sqrt{40} = 2 \sqrt{10} \cong 6.32$

And for the final one:

${l}_{3} = \sqrt{{\left(9 - 7\right)}^{2} + {\left(5 - 3\right)}^{2}} = \sqrt{8} = 2 \sqrt{2} \cong 2.83$

so the perimeter is going to be

$P = {l}_{1} + {l}_{2} + {l}_{3} = 4 + 6.32 + 2.83 = 13.15$

or in surd form,

$4 + 2 \sqrt{10} + 2 \sqrt{2}$