Question

What is the perimeter of a triangle with vertices of $\left( 1,2 \right),\left( 3,-4 \right)$ and $\left( -4,5 \right)$ ?

Answer 1

Hint: To find the perimeter of the triangle, we have to use the distance formula to find the sides of the triangle. We have to find the distance between $\left( 1,2 \right)\text{ and }\left( 3,-4 \right)$ , $\left( 3,-4 \right)$ and $\left( -4,5 \right)$ and $\left( -4,5 \right)$ and $\left( 1,2 \right)$ . To find the perimeter, we have to add these three distances.

Complete step by step solution:
We have to find the perimeter of the triangle whose vertices are $\left( 1,2 \right),\left( 3,-4 \right)$ and $\left( -4,5 \right)$ . We have to use the distance formula to find the sides of the triangle. We know that distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by
$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Here, let us assume the sides of the triangle as a, b and c.

Therefore, the distance between $\left( 1,2 \right),\left( 3,-4 \right)$ can be found as follows.
$a=\sqrt{{{\left( 3-1 \right)}^{2}}+{{\left( -4-2 \right)}^{2}}}$
Let us simplify the above equation.
$\begin{align}
  & \Rightarrow a=\sqrt{{{2}^{2}}+{{\left( -6 \right)}^{2}}} \\
 & \Rightarrow a=\sqrt{4+36} \\
 & \Rightarrow a=\sqrt{40} \\
\end{align}$
Now, we have to take the square root of 40.
$\Rightarrow a=2\sqrt{10}$
Let us convert the above result into decimals.
$\begin{align}
  & \Rightarrow a=2\times 3.162 \\
 & \Rightarrow a=6.324\text{ units} \\
\end{align}$
Now, we have to find the distance between $\left( 3,-4 \right)$ and $\left( -4,5 \right)$ .
$\Rightarrow b=\sqrt{{{\left( -4-3 \right)}^{2}}+{{\left( 5-\left( -4 \right) \right)}^{2}}}$
Let us simplify the above equation.
$\begin{align}
  & \Rightarrow b=\sqrt{{{\left( -7 \right)}^{2}}+{{\left( 9 \right)}^{2}}} \\
 & \Rightarrow b=\sqrt{49+81} \\
 & \Rightarrow b=\sqrt{130} \\
\end{align}$
Let us take the square root of 130.
$\Rightarrow b=11.402\text{ units}$
Now, we have to find the distance between $\left( -4,5 \right)$ and $\left( 1,2 \right)$ .
$\Rightarrow c=\sqrt{{{\left( 1-\left( -4 \right) \right)}^{2}}+{{\left( 2-5 \right)}^{2}}}$
Let us simplify the above equation.
$\begin{align}
  & \Rightarrow c=\sqrt{{{5}^{2}}+{{\left( -3 \right)}^{2}}} \\
 & \Rightarrow c=\sqrt{25+9} \\
 & \Rightarrow c=\sqrt{34} \\
 & \Rightarrow c=5.831\text{ units} \\
\end{align}$
Now, we have to find the perimeter by adding all the sides.
Perimeter of the triangle $=6.324+11.402+5.831=23.557\text{ units}$
Therefore the perimeter of the given triangle is 23.557 units.

Note: Students must know the distance formula thoroughly to find the sides of any shape when their vertices are given. They must never forget to write the units. To find the side c, we have considered the order of the vertex to be $\left( -4,5 \right)$ and $\left( 1,2 \right)$. We can also write the vertices in the order $\left( 1,2 \right)$ and $\left( -4,5 \right)$. In either order, the distance will be the same.