# What is the perimeter of a triangle with corners at (3 ,4 ), (4 ,8 ), and (8 ,7 )?

Nov 19, 2016

$P = 2 \sqrt{17} + \sqrt{34}$

#### Explanation:

You can determine the perimeter of a triangle (and other shapes) given the coordinates of its vertices by calculating the distance between each vertex and taking the sum of these distances to be the perimeter. This can be done with the distance formula, accounting for two dimensions (x and y).

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

If we plot the coordinates of the triangle's vertices, we get this:

We can imagine that line segments connect these points, giving us a triangle. We can then calculate the distance between the two points which make up each side, and add those distances.

Between (3,4) and (4,8):

$d = \sqrt{{\left(4 - 3\right)}^{2} + {\left(8 - 4\right)}^{2}}$
$d = \sqrt{1 + 16}$
$d = \sqrt{17}$

Note: Because both quantities are squared, it does not matter which point you choose to be point 1 $\left({x}_{1} , {y}_{1}\right)$ and which you decide to be point 2 $\left({x}_{2} , {y}_{2}\right)$.

Between (4,8) and (8,7):

$d = \sqrt{{\left(8 - 4\right)}^{2} + {\left(7 - 8\right)}^{2}}$
$d = \sqrt{16 + 1}$
$d = \sqrt{17}$

Between (8,7) and (3,4):

$d = \sqrt{{\left(3 - 8\right)}^{2} + {\left(4 - 7\right)}^{2}}$
$d = \sqrt{25 + 9}$
$d = \sqrt{34}$

And it looks like we have an isosceles triangle, as two of the sides are the same length.

Adding these lengths together, the perimeter of the triangle is

$P = s i {\mathrm{de}}_{1} + s i {\mathrm{de}}_{2} + s i {\mathrm{de}}_{3}$
$P = \sqrt{17} + \sqrt{17} + \sqrt{34}$
$P = 2 \sqrt{17} + \sqrt{34}$

in exact form. This is ~14.08 in decimal form.