Question

Find the perimeter of the triangles whose vertices have the following coordinates $(3,10),(5,2),(14,12)$.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let the given vertices be $A=(3,10)$, $B=(5,2)$ and $C=(14,12)$

We first find the distance between $A=(3,10)$ and $B=(5,2)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(5-3)}^2+{(2-10)}^2}=\sqrt{2^2+{(-8)}^2}=\sqrt{4+64}=\sqrt{68}$

Similarly, the distance between $B=(5,2)$ and $C=(14,12)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(14-5)}^2+{(12-2)}^2}=\sqrt{9^2+{10}^2}=\sqrt{81+100}=\sqrt{181}$

Now, the distance between $C=(14,12)$ and $A=(3,10)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(14-3)}^2+{(12-10)}^2}=\sqrt{{11}^2+2^2}=\sqrt{121+4}=\sqrt{125}$

Since the perimeter $P$ of a triangle $ABC$ is $AB+BC+CA$, therefore,

$P=\sqrt{68}+\sqrt{181}+\sqrt{125}$

Hence, the perimeter of the triangle is $\sqrt{68}+\sqrt{181}+\sqrt{125}$ units.