# What is the perimeter of a triangle with corners at (7 ,5 ), (1 ,2 ), and (4 ,8 )?

Jun 8, 2017

Perimeter of the triangle is $17.66 \left(2 \mathrm{dp}\right)$ unit.

#### Explanation:

The vertices of the triangle are $A \left(7 , 5\right) , B \left(1 , 2\right) \mathmr{and} C \left(4 , 8\right)$. The mesurement of sides are

$A B = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{{\left(7 - 1\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{45} = 3 \sqrt{5}$

$B C = \sqrt{{\left(1 - 4\right)}^{2} + {\left(2 - 8\right)}^{2}} = \sqrt{45} = 3 \sqrt{5}$

$C A = \sqrt{{\left(4 - 7\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{18} = 3 \sqrt{2}$

Perimeter of the triangle is $P = A B + B C + C A = 3 \sqrt{5} + 3 \sqrt{5} + 3 \sqrt{2}$ or

$P = 6 \sqrt{5} + 3 \sqrt{2} \approx 17.66 \left(2 \mathrm{dp}\right)$ unit

Perimeter of the triangle is $17.66 \left(2 \mathrm{dp}\right)$ unit [Ans]

Jun 9, 2017

$17.66 \text{ units}$

#### Explanation:

To find the perimeter, you need to find the distance of each side and add them up right?

So first you need to figure out the length of each side, and you can do this by plotting the corner points and assigning vectors to each side.
When you plot the points, it should look something like this:

To find a vector $\vec{P Q}$ between two points $P \left({x}_{1} , {y}_{1}\right) \text{ and } Q \left({x}_{2} , {y}_{2}\right)$ You can use the equation $\vec{P Q} = \left[\begin{matrix}{x}_{2} - {x}_{1} \\ {y}_{2} - {y}_{1}\end{matrix}\right]$

Hence to calculate the vectors between $A \left(7 , 5\right) , B \left(1 , 2\right) \text{ and } C \left(4 , 8\right)$, the below calculations can be conducted:
$\vec{A B} = \left[\begin{matrix}1 - 7 \\ 2 - 5\end{matrix}\right] = \left[\begin{matrix}- 6 \\ - 3\end{matrix}\right]$

$\vec{B C} = \left[\begin{matrix}4 - 1 \\ 8 - 2\end{matrix}\right] = \left[\begin{matrix}3 \\ 6\end{matrix}\right]$

$\vec{A C} = \left[\begin{matrix}4 - 7 \\ 8 - 5\end{matrix}\right] = \left[\begin{matrix}- 3 \\ 3\end{matrix}\right]$

Then the length of each vector needs to be calculated. To calculate the length of vector $\vec{M N} = \left[\begin{matrix}x \\ y\end{matrix}\right]$

You can use:
$\text{distance of } \vec{M N} = | \vec{M N} | = \sqrt{{x}^{2} + {y}^{2}}$

Hence when we are calculating the distances of the vectors of the triangle:

$| \vec{A B} | = \sqrt{{\left(- 6\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{36 + 9} = \sqrt{45} \text{ units}$

$| \vec{B C} | = \sqrt{{\left(3\right)}^{2} + {\left(6\right)}^{2}} = \sqrt{9 + 36} = \sqrt{45} \text{ units}$

$| \vec{A C} | = \sqrt{{\left(- 3\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{9 + 9} = \sqrt{18} \text{ units}$

Therefore the total perimeter is
$| \vec{A B} | + | \vec{B C} | + | \vec{A C} | = \sqrt{45} + \sqrt{45} + \sqrt{18} = 6 \sqrt{5} + 3 \sqrt{2} \approx 17.66 \text{ units}$