# What is the perimeter of a triangle with corners at (3 ,6 ), (1 ,8 ), and (8 ,7 )?

Apr 26, 2018

$7 \sqrt{2} + \sqrt{26}$

#### Explanation:

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ is the distance formula. If you simply plug in all the points one by one by replace ${x}_{1}$ and ${x}_{2}$ as well as ${y}_{1}$ and ${y}_{2}$ with three pairs of points you can solve each single one.

So the first pair between (3,6) and (1,8) is $\sqrt{{\left(1 - 3\right)}^{2} + {\left(8 - 6\right)}^{2}}$ is $\sqrt{8}$.

Second is the pair of points (1,8) and (8,7). Plug this in once again for $\sqrt{{\left(8 - 1\right)}^{2} + {\left(7 - 8\right)}^{2}}$ and simply to $\sqrt{50}$.

Finally for the third pair of points, (8,7) and (3,6) you plug it in for $\sqrt{{\left(3 - 8\right)}^{2} + {\left(6 - 7\right)}^{2}}$. Simply to $\sqrt{26}$.

Now simply the radicals as much as you can. $\sqrt{8}$ can go to $2 \sqrt{2}$, $\sqrt{50}$ can simplify to $5 \sqrt{2}$, and $\sqrt{26}$ cannot be simplified.

Now you just add like terms and thats it so in the end you get $2 \sqrt{2} + 5 \sqrt{2} + \sqrt{26}$. The answer is $7 \sqrt{2} + \sqrt{26}$