As you'll see, the Pythagorean theorem applies only to right triangles.

In this tutorial, we'll look at a theorem that lies at the core of our study of triangles, the Pythagorean theorem.

As you'll see, the Pythagorean theorem applies only to right triangles.

So let's start by taking a quick look at the anatomy of a right triangle.

A triangle with one of its internal angles equal to $\hspace{0.2em} 90 \degree \hspace{0.2em}$ (a right angle) is known as a right triangle or a right-angled triangle.

In a right triangle, the side opposite to the right angle is the largest side and is known as the hypotenuse.

The other two sides - those adjacent to the right angle - are known as legs.

The Pythagorean theorem gives us the relation between the three sides of a right triangle. It is named after Pythagoras, the great Greek polymath (although the theorem was known to different civilizations for centuries before Pythagoras).

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

One way to visualize the Pythagorean theorem is as follows.

If we construct squares using each side of a right triangle, as shown in the figure above, the area of the largest square (that on the hypotenuse) would be equal to the sum of the areas of the other two squares (those on the legs).

You will see this idea in use when we talk about the proofs of the Pythagorean theorem.

As far as proofs are concerned, it's difficult to beat the Pythagorean theorem. It has literally hundreds of proofs. Of course, we won't be discussing all of them in this tutorial. But let's look at a couple of them.

We start with a right triangle and a square shown in the figure below.

Next, we make four copies of the triangle and arrange them in the square in two different ways. Again, as shown in the figure below.

The yellow region in the first arrangement $\hspace{0.2em} (YR_1) \hspace{0.2em}$ consists of two squares. The top one has a side length of a (smaller side of the triangle) and so its area would be $\hspace{0.2em} a^2 \hspace{0.2em}$. For the bottom square, the side length is b (longer side of the triangle) and so its area would be $\hspace{0.2em} b^2 \hspace{0.2em}$.

$\text{Area}(R_1) = a^2 + b^2$

In the second arrangement, the yellow region $\hspace{0.2em} (YR_2) \hspace{0.2em}$ is a square with a side length of c (the hypotenuse of our right triangle), so its area would be $\hspace{0.2em} c^2 \hspace{0.2em}$.

$\text{Area}(R_2) = c^2$

Also, it's clear that the yellow regions visible in the two arrangements have the same area. In both cases, it's the area of the square (the one we started with) minus the area of the four triangles.

$\begin{align*} \text{Area}(R_1) &= \text{Area}(R_2) \\[1em] a^2 + b^2 &= c^2 \end{align*}$

For this proof, we will work with the area of the yellow square in the figure below.

Well, one way to get the area of any square is to take the square of its side. So -

But for the area of the yellow square, we can also subtract the area of the 4 right triangles from the area of the larger square.

From (1) and (2), we have,

$a^2 + b^2 = c^2$

The third and last proof in this tutorial uses the idea of similar triangles. Feel free to skip this one if you are not familiar with the concept.

In the figure below, we have a right triangle ABC. And CD is a perpendicular drawn from C to the hypotenuse AB.

To begin with, the three triangles - △ ABC, △ ACD, and △ CBD - are similar to one another by AA similarity.

Now, since △ ABC ~ △ ACD,

$\frac{AC}{AB} = \frac{AD}{AC}$

[Corresponding sides of similar triangles are in proportion.]

Rearranging the above equation, we have

$AC^2 = AB \cdot AD \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1)$

Similarly, since △ ABC ~ △ CBD,

$\begin{align*} &\frac{BC}{AB} = \frac{DB}{BC} \\[1.3em] &BC^2 = AB \cdot DB \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2) \end{align*}$

Adding (1) and (2)

$AC^2 + BC^2 = {\color{Red} AB} \cdot AD + {\color{Red} AB} \cdot DB$

Factoring out AB on the right side

$\begin{align*} AC^2 + BC^2 &= {\color{Red} AB} \cdot (AD + DB) \\[1em] AC^2 + BC^2 &= {\color{Red} AB} \cdot AB \\[1em] AC^2 + BC^2 &= AB^2 \end{align*}$

And that's it. We have proved the Pythagorean theorem.

Alright. Now that we know what the Pythagorean theorem is, let's see how we can use it to solve problems.

Example

Find the length of AC in the figure below.

Solution

The triangle is a right triangle. So we can use the Pythagorean theorem.

Applying the theorem, we have -

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

Substituting the values for AB and BC,

$\begin{align*} AC^2 &= 6^2 + 8^2 \\[1em] AC^2 &= 100 \end{align*}$

And now we just need to take the square root on both sides.

$AC = 10$

Example

Find the length of BC in the figure below.

Solution

Again, we have a right triangle and so, again, using the Pythagorean theorem -

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

Substituting the values for AB and AC, and solving for BC, we have -

$\begin{align*} 3^2 &= 1^2 + BC^2 \\[1em] BC^2 &= 3^2 - 1^2 \\[1em] BC^2 &= 8 \\[1em] BC &= 2 \sqrt{2} \end{align*}$

A Pythagorean triple consists of three positive integers such that the sum of the squares of the two smaller integers is equal to the square of the third integer.

So, if a, b, and c make a Pythagorean triple (and a < b < c), then -

$a^2 + b^2 = c^2$

No wonder they have that name!

Here are some examples of Pythagorean triples.

For every Pythagorean triple (a, b, c), there exists a right triangle with side lengths a, b, and c.

And that brings us to the end of this tutorial on the Pythagorean theorem. Until next time.

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