Perimeter of a Triangle

Find the perimeter of a triangle whose sides are of length a = 4, b = 5 and c = 7 units.

Solution:

The perimeter is given by the following:

$$\begin{gathered} P=a+b+c \\ =4+5+7 \\ =16units \end{gathered}$$

Find the perimeter of an isosceles triangle whose two sides are of length 4 units each and the third side is 5 units.

Solution:

The perimeter of an isosceles triangle is given by:

$P=2a+b$.

where a = 4 and b = 5 are given:

$$\begin{gathered} P=2\times 4+5 \\ =8+5 \\ =13 \end{gathered}$$

Hence, the perimeter of this triangle is 13 units.

Find the perimeter of an equilateral triangle whose length of each side is 3 units.

Solution:

We can use the formula to find the perimeter of an equilateral triangle:

$$\begin{gathered} P=3a \\ =3\times 3 \\ =9 \end{gathered}$$

Hence the perimeter of the triangle is 9 units.

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.