What is the perimeter of a triangle with corners at (3 ,4 ), (2 ,7 ), and (4 ,5 )?

Jun 8, 2017

$P = \sqrt{10} + 3 \sqrt{2}$

Explanation:

The perimeter of an object is the sum of the lengths of its sides. To find the lengths of the sides in coordinates, we use a formula that is based off of Pythagoras's Theorem:
$\sqrt{{\left(x {\text{_1 - x""_2)^2+(y""_1 - y}}_{2}\right)}^{2}}$, where
$\left(x {\text{_1,y""_1),(x""_2,y}}_{2}\right)$ are the coordinates. With this formula you will get the exact distance from point A to point B. Since we have three coordinates, we need the exact distance from Point A-B, A-C and B-C. Our formula will now look like:

$\sqrt{{\left(x {\text{_1 - x""_2)^2+(y""_1 - y""_2)^2) + sqrt((x""_1 - x""_3)^2+(y""_1 - y""_3)^2) + sqrt((x""_2 - x""_3)^2+(y""_2 - y}}_{3}\right)}^{2}}$ Where $\left(x {\text{_1,y""_1),(x""_2,y""_2), (x""_3,y}}_{3}\right)$ are the coordinates of
$\left(3 , 4\right) , \left(2 , 7\right) , \left(4 , 5\right)$. Now we substitute the numbers into the formula, and solve.

$\sqrt{{\left(3 - 2\right)}^{2} + {\left(4 - 7\right)}^{2}} + \sqrt{{\left(3 - 4\right)}^{2} + {\left(4 - 5\right)}^{2}} + \sqrt{{\left(2 - 4\right)}^{2} + {\left(7 - 5\right)}^{2}}$
$= \sqrt{{\left(1\right)}^{2} + {\left(- 3\right)}^{2}} + \sqrt{{\left(- 1\right)}^{2} + {\left(- 1\right)}^{2}} + \sqrt{{\left(- 2\right)}^{2} + {\left(2\right)}^{2}}$
$= \sqrt{1 + 9} + \sqrt{1 + 1} + \sqrt{4 + 4}$
$= \sqrt{10} + \sqrt{2} + \sqrt{8}$.

The last part involves the simplification and addition of square roots (sqrt), and if you want further information on how to do that then ask the question and I, or other, will answer. As it is I will just speed through the explanations. Thus,

$= \sqrt{10} + \sqrt{2} + \sqrt{8}$

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

$= \sqrt{10} + 3 \sqrt{2}$

There you have the perimeter of a triangle with coordinates (3,4),(2,7) and (4,5).

I hope I helped!