The Pythagorean Theorem can be formulated as follows: in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

In the right triangle shown in the image below, we use the first letters of the alphabet to indicate its sides:

$a$ and $b$ are the legs.

$c$ is the hypotenuse.

Using these, we can express the Pythagorean theorem in an algebraic form as follows:

$c²=a²+b²$

We can express the Pythagorean Theorem in a geometric form in the following way, showing that the area of the square ($c$) (square of the hypotenuse) is the sum of the areas of the squares ($a$) and ($b$) (squares of the legs).

## Examples with solutions for Pythagorean Theorem

### Exercise #1

What is the length of the hypotenuse?

### Step-by-Step Solution

We use the Pythagorean theorem

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

We insert the known data:

$3^2+4^2=BC^2$

$9+16=BC^2$

$25=BC^2$

We extract the root:

$\sqrt{25}=BC$

$5=BC$

5

### Exercise #2

Look at the triangle in the diagram. Calculate the length of side AC.

### Step-by-Step Solution

To solve the exercise, we have to use the Pythagorean theorem:

A²+B²=C²

We replace the data we have:

3²+4²=C²

9+16=C²

25=C²

5=C

5 cm

### Exercise #3

Look at the triangle in the diagram. How long is side AB?

### Step-by-Step Solution

To find side AB, we will need to use the Pythagorean theorem.

The Pythagorean theorem allows us to find the third side of a right triangle, if we have the other two sides.

Pythagorean theorem:

$A^2+B^2=C^2$

That is, one side squared plus the second side squared equals the third side squared.

We replace the existing data:

$3^2+2^2=AB^2$

$9+4=AB^2$

$13=AB^2$

We find the root:

$\sqrt{13}=AB$

$\sqrt{13}$ cm

### Exercise #4

Look at the following triangle.

What is the value of X?

### Step-by-Step Solution

It is important to remember: the Pythagorean theorem is only valid for right-angled triangles.

This triangle does not have a right angle, and therefore, the missing side cannot be calculated in this way.

Cannot be solved

### Exercise #5

What is the length of the side marked X?

### Step-by-Step Solution

We use the Pythagorean theorem:

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

$15$

### Exercise #6

Look at the triangle in the diagram. How long is side BC?

### Step-by-Step Solution

To solve the exercise, it is necessary to know the Pythagorean Theorem:

A²+B²=C²

We replace the known data:

2²+B²=7²

4+B²=49

We input into the formula:

B²=49-4

B²=45

We find the root

B=√45

This is the solution. However, we can simplify the root a bit more.

First, let's break it down into prime numbers:

B=√(9*5)

We use the property of roots in multiplication:

B=√9*√5

B=3√5

This is the solution!

$3\sqrt{5}$ cm

### Exercise #7

Given the triangle ABC, find the length BC

### Step-by-Step Solution

To answer this question, we must know the Pythagorean Theorem

The theorem allows us to calculate the sides of a right triangle.

We identify the sides:

ab = a = 5
bc = b = ?

ac = c = 13

We replace the data in the exercise:

5²+?² = 13²

We swap the sections

?²=13²-5²

?²=169-25

?²=144

?=12

12 cm

### Exercise #8

The triangle in the drawing is rectangular and isosceles.

Calculate the length of the legs of the triangle.

### Step-by-Step Solution

We use the Pythagorean theorem as shown below:

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

Since the triangles are isosceles, the theorem can be written as follows:

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

We then insert the known data:

$2AC^2=(8\sqrt{2})^2=64\times2$

Finally we reduce the 2 and extract the root:

$AC=\sqrt{64}=8$

$BC=AC=8$

8 cm

### Exercise #9

Look at the following rectangle:

BC = 8

BD = 17

Calculate the area of the rectangle ABCD.

### Step-by-Step Solution

We will find side DC by using the Pythagorean theorem in triangle DBC:

$BC^2+CD^2=BD^2$

Let's substitute the known data:

$8^2+CD^2=17^2$

$CD^2=289-64=225$

Let's take the square root:

$CD=15$

Now we have the length and width of rectangle ABCD and we'll calculate the area:

$15\times8=120$

120

### Exercise #10

Look at the triangle in the figure.

What is its perimeter?

### Step-by-Step Solution

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:
$\sqrt{58}+7+3=\sqrt{58}+10$

$10+\sqrt{58}$ cm

### Exercise #11

The Egyptians decided to build another pyramid that looks like an isosceles triangle when viewed from the side.

Each side of the pyramid measures 150 m, while the base measures 120 m.

What is the height of the pyramid?

### Step-by-Step Solution

Since the height divides the base into two equal parts, each part will be called X

We begin by calculating X:$120:2=60$

We then are able to calculate the height of the pyramid using the Pythagorean theorem:

$X^2+H^2=150^2$

We insert the corresponding data:

$60^2+h^2=150^2$

Finally we extract the root: $h=\sqrt{150^2-60^2}=\sqrt{22500-3600}=\sqrt{18900}$

$h=30\sqrt{21}$

$30\sqrt{21}$ m

### Exercise #12

Given the triangles in the drawing

What is the length of the side DB?

### Step-by-Step Solution

In this question, we will have to use the Pythagorean theorem twice.

A²+B²=C²

Let's start by finding side CB:

6²+CB²=(2√11)²

36+CB²=4*11

CB²=44-36

CB²=8

CB=√8

We will use the exact same way to find side DB:

2²+DB²=(√8)²

4+CB²=8

CB²=8-4

CB²=4

CB=√4=2

2 cm

### Exercise #13

The rectangle ABCD is shown below.

$BD=25,BC=7$

Calculate the area of the rectangle.

### Step-by-Step Solution

We will use the Pythagorean theorem in order to find the side DC:

$(BC)^2+(DC)^2=(DB)^2$

We begin by inserting the existing data into the theorem:

$7^2+(DC)^2=25^2$

$49+DC^2=625$

$DC^2=625-49=576$

Finally we extract the root:

$DC=\sqrt{576}=24$

168

### Exercise #14

ABCD is a rectangle.

AC = 13

AB = 12

Calculate the length of the side BC.

### Step-by-Step Solution

When writing the name of a polygon, the letters will always be in the order of the sides:

This is a rectangle ABCD:

This is a rectangle ABDC:

Always go in order, and always with the right corner to the one we just mentioned.

5

### Exercise #15

Below is the rectangle ABCD.

O is the intersection point of the diagonals of the rectangle.

BO = 8.5

Calculate the area of the triangle ABD.

### Step-by-Step Solution

According to the given information, we can claim that:

$BD=2BO=8.5\times2=17$

Now let's look at triangle ABD to calculate side AB

$AB^2+AD^2=BD^2$

Let's input the known data:

$AB^2+8^2=17^2$

$AB^2=289-64=225$

We'll take the square root

$AB=15$

Now let's calculate the area of triangle ABD:

$\frac{15\times8}{2}=\frac{120}{2}=60$