Pythagorean Theorem - Examples, Exercises and Solutions

To find the length of the hypotenuse BC in a right-angled triangle where AB and AC are the other two sides, we use the Pythagorean theorem: $c^2 = a^2 + b^2$.

Here, $a = 6 \text{ cm}$ and $b = 8 \text{ cm}$.

Plugging the values into the Pythagorean theorem, we get:

$c^2 = 6^2 + 8^2$.

Calculating further:

$c^2 = 36 + 64$

$c^2 = 100$.

Taking the square root of both sides gives:

$c = 10 \text{ cm}$.

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

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.

You can read all about the theorem here.

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:

$\sqrt{13}=AB$

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:

$\sqrt{25}=BC$

$5=BC$

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

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.

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!

We use the Pythagorean theorem:

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

Let's focus on triangle BCD in order to find side DC

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$6^2+DC^2=10^2$

$DC^2=100-36=64$

Let's take the square root:

$DC=8$

Since in a rectangle each pair of opposite sides are equal to each other, we can state that:

$DC=AB=8$

$BC=AD=6$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$8+6+8+6=16+12=28$

Let's focus on triangle BCD in order to find side BC

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$BC^2+24^2=25^2$

$BC^2=625-576=49$

Let's take the square root:

$BC=7$

Since in a rectangle, each pair of opposite sides are equal to each other, we can state that:

$DC=AB=24$

$BC=AD=7$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$24+7+24+7=14+48=62$

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:

$AC=\sqrt{64}=8$

$BC=AC=8$

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

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:

$15\times8=120$

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

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: $h=\sqrt{150^2-60^2}=\sqrt{22500-3600}=\sqrt{18900}$

$h=30\sqrt{21}$