Perimeter of a Triangle - Examples, Exercises and Solutions

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

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

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

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

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:

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

$x=11$

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:

$5+5+5=15$

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:

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

Due to the fact that the the triangle is isosceles, its two legs are equal to one another.

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

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

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

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:

$AD+15+9=36$

$AD+24=36$

$AD=36-24=12$

Thus we can calculate the area of triangle ABD:

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

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:

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

$x=2$

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:

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

$x=22.2$

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:

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

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

Two sides have already been given leaving only one remaining side to find.

We can use the Pythagorean Theorem.
$AB^2+BC^2=AC^2$
We insert all of 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 of the sides, we can add them up and thus find the perimeter:
$\sqrt{58}+7+3=\sqrt{58}+10$

Using the given data of the triangle's perimeter 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 are able to deduce 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 can calculate the area of the triangle ADC:

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

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

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:

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

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