# Perimeter of a Triangle - Examples, Exercises and Solutions

### Perimeter: calculating the perimeter of a triangle

To calculate the perimeter of a triangle, all you have to do is add its three sides. If you have all the necessary information, you can solve such a problem in a matter of seconds, for example:

#### Formulafor the perimeter of a triangle:

P = Side 1 + Side 2 + Side 3.

If you give us a triangle whose sides have the following measurements:

$AB = 5$

$BC = 8$

$CA = 6$

In this case, the perimeter of the triangle, which is the sum of the $3$ sides will be equal to. $19$

## examples with solutions for perimeter of a triangle

### Exercise #1

Look at the triangle below:

What is the perimeter of the triangle?

### Step-by-Step Solution

The perimeter of the triangle is equal to the sum of all sides together, therefore:

$6+8+10=14+10=24$

24

### Exercise #2

Given the triangle:

What is its perimeter?

### Step-by-Step Solution

The perimeter of a triangle is equal to the sum of all its sides together:

$11+7+13=11+20=31$

31

### Exercise #3

Given an equilateral triangle:

What is its perimeter?

### Step-by-Step Solution

Since the triangle is equilateral, that is, all sides are equal to each other.

The perimeter of the triangle is equal to the sum of all sides together, the perimeter of the triangle in the drawing is equal to:

$5+5+5=15$

15

### Exercise #4

Below is an equilateral triangle:

If the perimeter of the triangle is 33 cm, then what is the value of X?

### Step-by-Step Solution

We know that in an equilateral triangle all sides are equal.

Therefore, if we know that one side is equal to X, then all sides are equal to X.

We know that the perimeter of the triangle is 33.

The perimeter of the triangle is equal to the sum of the sides together.

We replace the data:

$x+x+x=33$

$3x=33$

We divide the two sections by 3:

$\frac{3x}{3}=\frac{33}{3}$

$x=11$

11

### Exercise #5

Look at the isosceles triangle below:

What is its perimeter?

### Step-by-Step Solution

Since we are referring to an isosceles triangle, the two legs are equal to each other.

In the drawing, they give us the base which is equal to 4 and one side is equal to 6, therefore the other side is also equal to 6.

The perimeter of the triangle is equal to the sum of the sides and therefore:

$6+6+4=12+4=16$

16

### Exercise #6

Given the isosceles triangle,

What is its perimeter?

### Step-by-Step Solution

Since the triangle is isosceles, that means its two legs are equal to each other.

Therefore, the base is 7 and the other two sides are 12.

The perimeter of a triangle is equal to the sum of all sides together:

$12+12+7=24+7=31$

31

### Exercise #7

Look at the isosceles triangle below:

The perimeter of the triangle is 50.

What is the value of X?

### Step-by-Step Solution

Since we know the triangle is isosceles, the other side will also be equal to X

Now we can replace the data to calculate X.

The perimeter of the triangle is equal to:

$x+x+5.6=50$

$2x=50-5.6$

$2x=44.4$

We divide both sides by 2:

$\frac{2x}{2}=\frac{44.4}{2}$

$x=22.2$

22.2

### Exercise #8

Look at the following triangle:

The perimeter of the triangle is 17.

What is the value of X?

### Step-by-Step Solution

We know that the perimeter of a triangle is equal to the sum of all sides together, so we replace the data:

$3x+2x+3.5x=17$

$8.5x=17$

Divide the two sections by 8.5:

$\frac{8.5x}{8.5}=\frac{17}{8.5}$

$x=2$

2

### Exercise #9

The triangle ABC has a perimeter measuring 42 cm.

AC = 15

AB = 13

Calculate the area of the triangle.

### Step-by-Step Solution

Given that the perimeter of triangle ABC is 42.

We will use this data to find side CB:

$13+15+CB=42$

$CB+28=42$

$CB=42-28=14$

Now we can calculate the area of triangle ABC:

$\frac{AD\times BC}{2}=\frac{12\times14}{2}=\frac{168}{2}=84$

84 cm²

### Exercise #10

Below is the right triangle ABD, which has a perimeter of 36 cm.

AB = 15

AC = 13

DC = 5

CB = 4

Work out the area of the triangle.

### Step-by-Step Solution

According to the data:

$BD=4+5=9$

Now that we are given the perimeter of triangle ABD we can find the missing side AD:

$AD+15+9=36$

$AD+24=36$

$AD=36-24=12$

Now we can calculate the area of triangle ABD:

$\frac{AD\times BD}{2}=\frac{12\times9}{2}=\frac{108}{2}=54$

54 cm²

### Exercise #11

Look at the triangle in the figure.

What is its perimeter?

### Step-by-Step Solution

To find the perimeter of a triangle, first we need to find all its sides.

Given two sides and only the perimeter remains to be found.

We can use the Pythagorean Theorem
$AB^2+BC^2=AC^2$
We replace all the known data:

$AC^2=7^2+3^2$
$AC^2=49+9=58$
We extract the square root:

$AC=\sqrt{58}$
Now that we have all the sides, we can add them up and thus find the perimeter:
$\sqrt{58}+7+3=\sqrt{58}+10$

$10+\sqrt{58}$ cm

### Exercise #12

The perimeter of the triangle ABD shown below is 36 cm.

Given in cm:

AB = 15

AC = 13

DC = 5

CB = 4

Calculate the area of triangle ADC.

### Step-by-Step Solution

We use the data of the triangle's perimeter, with the help of which we will first find the side AD by calculating the sum of all the sides of the triangle:

$AD+9+15=36$

$AD+24=36$

$AD=36-24=12$

Now that we know that AD is equal to 12, we will note that AD is also the height from BD since it forms a 90-degree angle.

If AD is the height from BD, it is also the height from DC.

Now we calculate the area of the triangle ADC:

$\frac{AD\times DC}{2}$

$\frac{12\times5}{2}=\frac{60}{2}=30$

30 cm²

### Exercise #13

Look at the triangle in the figure.

What is the length of the hypotenuse given that its perimeter is $12+4\sqrt{5}$ cm?

### Step-by-Step Solution

We calculate the perimeter of the triangle:

$12+4\sqrt{5}=4+AC+BC$

As we want to find the hypotenuse (BC), we isolate it:

$12+4\sqrt{5}-4-AC=BC$

$BC=8+4\sqrt{5}-AC$

Then calculate AC using the Pythagorean theorem:

$AB^2+AC^2=BC^2$

$4^2+AC^2=(8+4\sqrt{5}-AC)^2$

$16+AC^2=(8+4\sqrt{5})^2-2\times AC(8+4\sqrt{5})+AC^2$

We then simplify the two:$AC^2$

$16=8^2+2\times8\times4\sqrt{5}+(4\sqrt{5})^2-2\times8\times AC-2AC4\sqrt{5}$

$16=64+64\sqrt{5}+16\times5-16AC-8\sqrt{5}AC$

$16AC+8\sqrt{5}AC=64+64\sqrt{5}+16\times5-16$

$AC(16+8\sqrt{5})=128+64\sqrt{5}$

$AC=\frac{128+64\sqrt{5}}{16+8\sqrt{5}}=\frac{8(16+8\sqrt{5})}{16+8\sqrt{5}}$

We simplify to obtain:

$AC=8$

Now we can replace AC with the value we found for BC:

$BC=8+4\sqrt{5}-AC$

$BC=8+4\sqrt{5}-8=4\sqrt{5}$

$4\sqrt{5}$ cm

### Exercise #14

Find the perimeter of the triangle ABC

12

### Exercise #15

Find the perimeter of the triangle ABC