Perimeter of a Triangle

Find the perimeter of the triangle whose vertices are located at A(–3, 1), B(2, 1) and C(2, –1).

Solution:

In order to find the length of the perimeter, we need to find the length of the respective sides and we can do so by using the distance formula for all the three vertices.

For the first side AB:

$$\begin{gathered} AB=\sqrt{{\left(-3-2\right)}^2+{\left(1-1\right)}^2} \\ =5 \end{gathered}$$

For the second side, BC:

$$\begin{gathered} BC=\sqrt{{\left(2-2\right)}^2+{\left(1+1\right)}^2} \\ =2 \end{gathered}$$

And for the third side, AC:

$$\begin{gathered} AC=\sqrt{{\left(-3-2\right)}^2+{\left(1+1\right)}^2} \\ =\sqrt{29} \end{gathered}$$

And now the perimeter can be calculated by adding all these sides:

$$\begin{gathered} P=AB+BC+AC \\ P=5+2+\sqrt{29}=7+\sqrt{29} \end{gathered}$$

Hence the perimeter of the triangle whose vertices are A(–3, 1), B(2, 1) and C(2, –1) is $7+\sqrt{29}$ units.

Find the perimeter of a triangle whose two sides are of length 4 and 5 units, and the angle between those sides is $\frac{\pi }{3}$ radian.

Solution:

Let us label the two sides as a and b respectively, so that a = 4 and b = 5 , and the angle be C, we need to find the length of the remaining side in order to find the perimeter.

Using the cosine law here:

$$\begin{gathered} c=\sqrt{a^2+b^2-2ab\cos C} \\ {c}^2=16+25-2.4.5·\frac{1}{2} \\ c=\sqrt{41-20} \\ c=\sqrt{21} \end{gathered}$$

Therefore, the perimeter is given by:

$$\begin{gathered} P=a+b+c \\ P=4+5+\sqrt{21} \\ P=9+\sqrt{21} \end{gathered}$$

Thus the perimeter of the given triangle is $9+\sqrt{21}$ units.