# Perimeter of a Trapezoid - Examples, Exercises and Solutions

## Perimeter of a trapezoid

The trapezoid is a quadrilateral defined as having 2 parallel opposite sides. The calculation of the perimeter of the trapezoid is solved using a very simple formula that we will see below: all sides are added together. This type of questions can appear in tests of the first and second level in the first years of high school and also in final exams of level 3, 4 and 5 for the graduation of the secondary cycle.

## Examples with solutions for Perimeter of a Trapezoid

### Exercise #1

Look at the trapezoid in the diagram.

What is its perimeter?

### Step-by-Step Solution

To calculate the perimeter, we'll add up all the sides of the trapezoid:

7+10+7+12 =

36

And that's the solution!

36

### Exercise #2

What is the perimeter of the trapezoid in the figure?

### Step-by-Step Solution

To find the perimeter we will add all the sides:

$4+5+9+6=9+9+6=18+6=24$

24

### Exercise #3

Look at the trapezoid in the figure.

The long base is 1.5 times longer than the short base.

Find the perimeter of the trapezoid.

### Step-by-Step Solution

First, we calculate the long base from the existing data:

Multiply the short base by 1.5:

$5\times1.5=7.5$

Now we will add up all the sides to find the perimeter:

$2+5+3+7.5=7+3+7.5=10+7.5=17.5$

17.5

### Exercise #4

Given an isosceles trapezoid, calculate its perimeter

### Step-by-Step Solution

Since this is an isosceles trapezoid, and the two legs are equal, we can claim that:

$AB=CD=6$

Now let's add all the sides together to find the perimeter

$6+6+10+12=$

$12+22=34$

34

### Exercise #5

The perimeter of the trapezoid in the diagram is 25 cm. Calculate the missing side.

### Step-by-Step Solution

We replace the data in the formula to find the perimeter:

$25=4+7+11+x$

$25=22+x$

$25-22=x$

$3=x$

$3$ cm

### Exercise #6

Since X=3

Calculate the perimeter of the trapezoid

### Step-by-Step Solution

To calculate the perimeter, we add up all the sides:

$10+x+(6+x)+(x+1)$

Now, given that x equals 3, we substitute in the appropriate places:

$10+3+(6+3)+(3+1)=$

$10+3+9+4=$

$13+13=26$

26

### Exercise #7

Shown below is the isosceles trapezoid ABCD.

Given in cm:
BC = 7

Height of the trapezoid (h) = 5

Perimeter of the trapezoid (P) = 34

Calculate the area of the trapezoid.

### Step-by-Step Solution

Since ABCD is a trapezoid, one can determine that:

$AD=BC=7$

Thus the formula to find the area will be

$S_{ABCD}=\frac{(AB+DC)\times h}{2}$

Since we are given the perimeter of the trapezoid, we can find$AB+DC$

$P_{ABCD}=7+AB+7+DC$

$34=14+AB+DC$

$34-14=AB+DC$

$20=AB+DC$

Now we will place the data we obtained into the formula in order to calculate the area of the trapezoid:

$S=\frac{20\times5}{2}=\frac{100}{2}=50$

50

### Exercise #8

ABCD is an isosceles trapezoid.

AB = 3

CD = 6

The area of the trapezoid is 9 cm².

What is the perimeter of the trapezoid?

### Step-by-Step Solution

We can find the height BE by calculating the trapezoidal area formula:

$S=\frac{(AB+CD)}{2}\times h$

We replace the known data: $9=\frac{(3+6)}{2}\times BE$

We multiply by 2 to get rid of the fraction:

$9\times2=9\times BE$

$18=9BE$

We divide the two sections by 9:

$\frac{18}{9}=\frac{9BE}{9}$

$2=BE$

If we draw the height from A to CD we get a rectangle and two congruent triangles. That is:

$AF=BE=2$

$AB=FE=3$

$ED=CF=1.5$

Now we can find one of the legs through the Pythagorean theorem.

We focus on triangle BED:

$BE^2+ED^2=BD^2$

We replace the known data:

$2^2+1.5^2=BD^2$

$4+2.25=DB^2$

$6.25=DB^2$

We extract the root:

$\sqrt{6.25}=DB$

$2.5=DB$

Now that we have found DB, it can be argued that:

$AC=BD=2.5$

We calculate the perimeter of the trapezoid:$6+3+2.5+2.5=$

$9+5=14$

14

### Exercise #9

ABC is an isosceles triangle.

AD is the height of triangle ABC.

AF = 5

AB = 17
AG = 3

What is the perimeter of the trapezoid EFBC?

### Step-by-Step Solution

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We perform the same process with the side DB of the triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts so:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

62

### Exercise #10

What can be said about the two trapezoids in the diagram?

### Step-by-Step Solution

We calculate the perimeter of the left trapezoid:

$P=10+12+x+y$

$P=22+x+y$

We calculate the perimeterof the right trapezoid:

$P=x+5+17+\frac{y}{3}+\frac{2y}{3}$

$P=x+22+\frac{2y+y}{3}$

$P=x+22+\frac{3y}{3}$

$P=x+22+y$

It can be seen that the two perimeters are identical to each other.

Their perimeters are identical.

### Exercise #11

Given the trapezoids in the drawing.

Are they the same trapezoid?

### Step-by-Step Solution

We calculate the perimeter of the left trapezoid:

$P=6+10+7+\frac{5}{2}x+5$

$P=28+\frac{5}{2}x$

We calculate the perimeter of the right trapezoid:

$P=7+x+x+16+\frac{x}{2}+5$

$P=2\frac{1}{2}x+28$

$P=\frac{5}{2}x+28$

The perimeters of the two trapezoids are equal to each other.

No, but their perimeter is identical.

### Exercise #12

Look at the trapezoid in the figure.

Calculate its perimeter.

24.2

### Exercise #13

What is the perimeter of the trapezoid in the figure?

16

### Exercise #14

Given the trapezoid:

What is its perimeter?

32

### Exercise #15

AB = 5

CD = 7

AC = 4

BD = 4

Calculate the perimeter of the rectangle.