# Pythagorean Theorem

## Introduction

### Triangle Relationships

In previous lessons, you explored some patterns in triangles. For instance, all the angles inside a triangle add up to 180 degrees. It is also the case that similar triangles have equivalent corresponding angles and corresponding side lengths that are proportional to each other in the same ratio. Right-angle triangles are more orderly, which gives rise to more relationships. You will explore the relationship between the side lengths of right triangles, in the following lesson.

### Conventions

The relationship you will be studying is usually called the Pythagorean Theorem (or minor variations thereof), in English and "Western" countries. Sometimes the slang "Pythagoras" is used instead. In China, it is called the Gougu Theorem. Historically, India has called this relationship the Baudhayana Theorem, but has adopted the English naming convention more recently.

There are many conventions for identifying the side lengths of a right-angle triangle when using the Pythagorean Theorem. In Chinese, the hypotenuse is called 弦/Xian and the other two sides are called 勾/Gou and 股/Gu. In English, it is common to use the letters $a$ , $b$ , and $c$ to identify the side lengths. People often switch between using the $c$ and the $a$ for the hypotenuse and then use the remaining two letters for the other two sides. Identifying the hypotenuse is the most important part. The other two sides are interchangeable. This is because you can simply flip and rotate the triangle, without changing it. This website will choose to follow (wherever possible) the convention that sets the letter $c$ as the value of the length of the hypotenuse. The letters $a$ and $b$ will be used for the other two sides.

### An Experimental Approach

 $a$ $b$ $c$ 3 4 5 5 12 13 8 15 17 7 24 25

At this point, an ambitious student could begin drawing right-angle triangles and looking for a pattern in the side lengths. The pattern is fairly complex, and the numbers are really messy unless very specific choices are made. To avoid unnecessary frustration, a table with a few sets of side lengths is provided. They are special numbers called Pythagorean Triples. Their special property is that they are all whole numbers and can be used as the lengths of a right-angle triangle (ie. they satisfy the relationship you are looking for). If you'd like to sketch them (to check that they work) feel free. Now please take a moment and try to find the pattern that relates these triples.

It is worthwhile to note that each Pythagorean Triple has a twin Pythagorean Triple, in which the $a$ and $b$ values have been swapped. If you took $a$ , $b$ , and $c$ to be valued 3, 4, and 5; then another valid triple would be 4, 3, and 5.

It is also valuable to notice that similar triangles sometimes also work. For example, if you were to multiply all the values of a Pythagorean Triple by a single positive whole number, the new values would also form a new Pythagorean Triple. Explicitly, if you took $a$ , $b$ , and $c$ to be valued 3, 4, and 5; then if you were to multiply each of these values by 2 to get 6, 8, and 10; you would find that 6, 8, and 10 make another Pythagorean Triple.

Feel free to take a break if you need it. The pattern is particularly unexpected, so there is no shame in moving on to the next section, once you've made a strong effort.

## Finding the Pythagorean Theorem: A Visual Approach

Now that you have tried to find the relationship, take a look at this app. Leave the action button (beside the full-screen button) and hint buttons alone for now. You can use the full-screen button and slider if you like.

What do you see? What does the slider do?

Hint 1
There is a large square, with two squares and two rectangles inside it. The rectangles can also be squares, if the slider is set to the initial position. Each rectangle is further divided into two gray triangles. When the slider is moved, the rectangles and triangles change size and shape. Do you notice anything else about the triangles?
Hint 2
All the triangles are identical right triangles, which have been rotated. What does changing the triangles with the slider achieve?
Hint 3
The slider allows you to consider right triangles where the two non-90-degree angles can take on any valid value. Do you notice anything else about the diagram? Remember we are trying to find a relationship for all right-angle triangles.
Hint 4
There are no sizes indicated, therefore the diagram can be scaled to any size you choose. Taking this hint and the previous hint together, what can you conclude?
Hint 5
The diagram can represent a right triangle with any valid set of angles and any valid side lengths. That means that any possible right triangle can be represented with this diagram. Now it's time to start trying to find the relationship. Is there anything that jumps out at you?
Hint 6
Feel free to use the action button going forward. What do you see?
Hint 7
The large square remains the same size as does each of the four triangles, but the triangles are arranged so that there is only one inner square. Does this help you find the relationship?
Hint 8
Try using the first hint button. What do you see?
Hint 9
The sides of the triangles have been labeled red, green, and blue. The hypotenuse is red. Play with the action button and the slider. Does this help you find the relationship?
Hint 10
Try using the second hint button. What do you see? Don't forget to play with the slider and the action button.
Hint 11
Depending on which value the action button is set to, you will see either one (red) square or two smaller squares (one blue and one green). Does this help you find the relationship?
Hint 12
Notice that the area of the large square and the area of each of the four triangles does not change. What does this suggest?
Hint 13
It suggests that, when the action button is pressed, the area inside the large square, not covered by the triangles, remains constant. Does this help you find the relationship?
Hint 14
Try using the final hint button. What do you see?
Hint 15
The app seems to suggest that the green square plus the blue square is equal to the red square. Considering Hint 13, this suggests that the area of the green square plus the area of the blue square is equal to the area of the red square. Does this help you find the relationship?
Hint 16
Following the convention, label the lengths of the sides of the triangles with the letters $a$ , $b$ , and $c$ . The green side will be labeled $a$ , the blue side will be labeled $b$ , and the red side will be labeled $c$ . Does this help you find the relationship? Think back to the Pythagorean Triples exercise in the previous section. If you did not find the relationship in the previous section try that exercise again now.
Hint 17
What are the areas of each of the colored squares? Give the values in terms of $a$ , $b$ , or $c$ .
Hint 18
The area of the green square is $a^{2}$ , the area of the blue square is $b^{2}$ , and the area of the red square is $c^{2}$ . This is because their respective side lengths are $a$ , $b$ , and $c$ . Does this help you find the relationship?
Hint 19
Since the area of the green square plus the area of the blue square is equal to the area of the red square, simply replace the green blue and red squares with $a^{2}$ , $b^{2}$ , and $c^{2}$ . This gives the Pythagorean Theorem $a^{2}+b^{2}=c^{2}$ .

## Finding the Pythagorean Theorem: An Algebraic Approach Here is another way to justify the Pythagorean Theorem. It is my favorite. It requires more algebra, but has the benefit that it is more direct. It does not require an elaborate construction and, therefore (in my opinion), is more likely to be understood (or even reinvented) by a first-time learner. Have a look at the diagram provided and see if you can justify the Pythagorean Theorem in a new way.

Hint 1
What do you see?
Hint 2
There is a large right triangle labeled $\triangle ABC$ . Its sides are labeled $a$ , $b$ , and $c$ according to the convention used for the Pythagorean Theorem on this website. If one considers the side with length $c$ to be the base of the triangle, then a dotted line (an altitude) labeled with an $h$ is showing the height. There are two smaller right triangles, which are only partially labeled. The labels correspond to the large triangle, with the exception of the heights. The height of the triangle on the left is labeled $h_{L}$ and the height of the triangle on the right is labeled $h_{R}$ . There are two arrows pointing from opposite sides of the large triangle towards the smaller triangles. What do you make of this?
Hint 3
Taking the labels and the arrows together seems to suggest that if the large triangle was cut along the dotted line, it would split into the two smaller triangles. What does this mean?
Hint 4
This means that if you were to add the area of the two smaller triangles together you would get the area of the bigger triangle $A_{L}+A_{R}=A$ . It also has implications for the angles of the smaller triangles. What are these implications?
Hint 5
The triangle on the left shares $\angle A$ with the large triangle. The triangle on the right shares $\angle B$ with the large triangle. The remaining angle in each of the smaller triangles must sum to 90 degrees, because they were made by splitting the 90-degree angle in the large triangle. Can you find the values of the remaining angles in the small triangles?
Hint 6
Recall that the interior angles of a triangle must sum to 180 degrees. Can you find the value of the angles now?
Hint 7
Note, the angles of the large triangle sum to 180 degrees giving $\angle A+\angle B+90^{\circ }=180^{\circ }$ . What about the smaller triangles?
Hint 8
For the triangle on the left we have $?+\angle B+90^{\circ }=180^{\circ }$ and for the triangle on the right we have $\angle A+?+90^{\circ }=180^{\circ }$ . What angles are missing?
Hint 9
It turns out all three triangles have the same three angles. This means that they are all similar. Now rewrite the area equivalence $A_{L}+A_{R}=A$ substituting in the equation for the area of each of the triangles.
Hint 10
You should end up with the equation ${\frac {1}{2}}h_{L}a+{\frac {1}{2}}h_{R}b={\frac {1}{2}}hc$ . There are many things to notice about this equation, but there are two things of particular importance. Can you spot them?
Hint 11
One of the things to notice is a way to simplify the equation. Can you spot it?
Hint 12
The one halfs can be collected ${\frac {1}{2}}(h_{L}a+h_{R}b)={\frac {1}{2}}hc$ . It can be further simplified though can you spot it?
Hint 13
The one halfs cancel $h_{L}a+h_{R}b=hc$ . Do you notice the other interesting thing about this equation?
Hint 14
It is starting to look a lot like the Pythagorean Theorem. It has a term with an $a$ plus a term with a $b$ equal to a term with a $c$ . Can you find a way to deal with the heights?
Hint 15
The triangles were all found to be similar. Could you use this to your advantage?
Hint 16
There are relationships (ratios) between the measurements of similar triangles. Can you use this to your advantage?
Hint 17
Consider these ratios, found using the similar triangles relationship. ${\frac {h_{L}}{h}}={\frac {a}{c}}$ and ${\frac {h_{R}}{h}}={\frac {b}{c}}$ . How might you use these ratios to simplify the equation found in Hint 13?
Hint 18
Perhaps it would be helpful to manipulate the ratios a bit?
Hint 19
The ratios can be rewritten $h_{L}=h{\frac {a}{c}}$ and $h_{R}=h{\frac {b}{c}}$ . Does this help you simplify the equation?
Hint 20
Yes it does. Now the $h_{L}$ and $h_{R}$ terms can be replaced, this shrinks the number of variables from 6 to 4, and 3 of the remaining 4 are the variables that you want to have in the final equation. What does the new simplified equation look like?
Hint 21
The simplified equation is $h{\frac {a}{c}}a+h{\frac {b}{c}}b=hc$ . How can this equation be further simplified? There are three simplifications that need to happen.
Hint 22
One simplification is the elimination of the remaining height. How would you do this?
Hint 23
You do this by collecting the heights on the left-hand side $h({\frac {a}{c}}a+{\frac {b}{c}}b)=hc$ and then canceling ${\frac {a}{c}}a+{\frac {b}{c}}b=c$ . How else might you simplify this equation?
Hint 24
Another simplification, would be to collect the variable $c$ on the left-hand side ${\frac {1}{c}}(a*a+b*b)=c$ and then multiply both sides by $c$ which gives $a*a+b*b=c*c$ . What final simplification do you think will finish off the derivation?
Hint 25
Finally use the definition of squaring $a^{2}+b^{2}=c^{2}$ . There you have it. The Pythagorean Theorem, derived again.

## Other Derivations

The Pythagorean Theorem can be derived in many ways. I invite you to investigate other approaches. There are many good videos on YouTube and an article on Wikipedia. Feel free to Google other derivations as well.

## Conclusion

The Pythagorean Theorem is an important result in trigonometry. It lays some of the groundwork for learning about the distance formula, vectors, polar coordinates, and even the Euclidean Metric.