Derive and Use Pythagorean Theorem

Pythagoras

The Pythagoreans worshiped their leader and had very strange rules like "Don't urinate towards the sun." They also believed that all numbers were rational.

Rational numbers are numbers that can be written as an integer divided by a nonzero integer. A Pythagorean, Hippasus, discovered that the square root of $2$ wasn't rational.

All numbers are rational he believed. So he wrote $c$ like $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is positive. He applied the theorem of Pythagoras $1^2+1^2=\left(\frac{p}{q}\right)^2=\frac{p^2}{q^2}$.

This gives us $2q^2=p^2$. $2|2q^2$ and therefore $2|p^2$. $p$ is even, because $p^2$ is even.

We know now that $2q^2=4r^2$. Dividing by $2$ gives us $q^2=2r^2$. We apply the same trick again to find that $q=2s$, where $s$ is an integer. So $c=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}$. We have now written $c$ in a new way by halving the numerator and the denominator. We keep repeating this process. The denominator is getting smaller and smaller every time, but it should remain a positive integer forever. Wait a sec, that isn't true!

This was a great discovery, but the Pythagoreans had always believed that all numbers were rational. They were shocked and drowned Hippasus trying to keep it a secret.

The Theorem of Pythagoras

Some terminology used here:

Right-angled triangle: This is a triangle that has one angle that is a right angle. These triangles are also called right triangles.

Hypotenuse: The hypotenuse is the name for the longest side in a right-angled triangle.

Pythagorean Triples

In ancient times, people noticed relationships between the lengths of sides of right-angled triangles. At first they found certain sets of similar right-angled triangles that always had their sides in the same ratio.

One of the simplest examples is the set of similar triangles with their sides in a ratio of $3:4:5$. From experience, people knew that any triangle with its sides in this ratio would be a right-angled triangle.

Similarly, from experience, people discovered a number of other sets of similar right-angled triangles, where the lengths of the sides were in an integer ratio. A 3800 year-old tablet from Babylonian times shows a list of some Pythagorean triples.

These Pythagorean triples (the triple refers to three natural numbers in the ratio) were tremendously useful in calculating lengths.

A simple exercise using a triple:

Remember that early people found Pythagorean Triples by observation and experience. We don't have to do that; instead we can use the Theorem of Pythagoras to calculate triples. (There is a formula for finding them too).

The Theorem of Pythagoras

In the diagram below, a right-angled triangle with sides of length $a$, $b$, and $c$ is given. Note that $c$ is the length of the hypotenuse.

By working with the Pythagorean triples mentioned above, people noticed a pattern:

In any right-angled triangle with sides of length $a$, $b$, and $c$ (where $c$ is length of the hypotenuse), $a^2+b^2=c^2$.

The converse of this is true as well:

If a triangle has sides of length $a$, $b$, and $c$, and $a^2+b^2=c^2$, then the triangle is a right-angled triangle.

"Behold" Proof

There are many different proofs for this theorem. (One book gives something like 370 different proofs for the theorem!)

A beautiful animation demonstrating this proof can be seen online.

The reason that this is sometimes known as the "Behold" proof is that the mathematician Bhaskara included the proof in one of his books on Mathematics. His proof consisted of a diagram and only one word: "Behold!", the idea being that the proof would be obvious from the diagram.

Applications of the Theorem

The most useful application of the Pythagorean Theorem is calculating the side lengths of right-angled triangles when two sides are given.

As a straight-forward worked example (since we already know the answer), let us use a right-angled triangle with a sides of length 3 units and 4 units, and a hypotenuse of unknown length. In order to calculate the length of the hypotenuse, we use the theorem.

Let $a=3$, $b=4$, and let $c$ be the length of the hypotenuse. Then:

$$\begin{align} c^2 & = a^2 + b^2 \qquad\text{(By Thm of Pythagoras)}\\ & = 3^2 + 4^2\\ & = 9 + 16 \\ & = 25\\ \therefore c & = 5\ \text{units} \end{align}$$

If we are given the hypotenuse, but do not know one of the other sides, we can rearrange the equation given by the theorem to help us find the answer.

For example, if we are given $a=20$ and $c=29$, but $b$ is unknown, then: $$\begin{align} c^2 & = a^2 + b^2 \qquad\text{(By Thm of Pythagoras)} \\ b^2 & = c^2 - a^2 \\ & = 29^2 - 20^2\\ & = 441 \\ \therefore b & = 21 \end{align}$$

In each of the following cases, you are given the lengths of two of the sides of a right-angled triangle. Calculate the length of the unknown side:

Welcome to Pythagorean Land

Triangle 1:

Triangle 2:

When we have a right triangle, we often call two of the sides "legs", and one of the sides the "hypotenuse". But rather than just telling you the definitions, I'd like to let you guess what's what first!

Identifying right triangles

In ancient times, people sometimes had to determine whether angles were right, for example, when determining farmland owned by different people:

(Image credit: Wikipedia, uploaded by Lisa.davis)

It turns out that the side lengths in right triangles always share certain coincidences. Let's guess the pattern together!

This right triangle has sides of length 3, 4, and 5.

For fun, let's calculate some things.

Let's try this again, with another right triangle, which happens to have side lengths 5, 12, and 13.

We can calculate: $\text{(short leg)}^2 = 25$, $\text{(long leg)}^2 = 144$, so $\text{(short)}^2 + \text{(long)}^2 = 169$. But $\text{(hypotenuse)}^2 = 169$ too.

We're now ready for the statement of the...

Pythagorean Theorem

In every right triangle, the sum of the squares of the legs equals the square of the hypotenuse!

Proof of Pythagorean Theorem

There are many beautiful proofs. Here's a proof-by-picture:

What? How is that a proof? It just looks like some lines. Well, let's walk through it. :-)

We've started with a right triangle, with legs $a$ and $b$ and hypotenuse $c$, and we constructed a square on its hypotenuse, like this:

Then, we extended line $CA$, and dropped the perpendicular from $M$, like so:

Hooray! We've shown that triangles $ABC$ and $MAF$ are congruent!

Next, let's drop a perpendicular from $P$ to $CF$, and construct lines through $B$ and $M$ that are parallel to $CF$, like so:

Great! Now, let's focus on two squares that appear in the picture:

And, the resulting picture looks like this!

Since the area was preserved after the rotations, we conclude that the area before ($a^2 + b^2$) equals the area after ($c^2$), and so we have proven the Pythagorean theorem:

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

Hooray! :-)

The Pythagorean Theorem is a rule relating the two legs and the hypotenuse of a right triangle.

We see in the diagram, $a$ and $b$ are the legs and $c$ is the hypotenuse:

The Pythagorean Theorem states that $a^2+b^2=c^2$, no matter the right triangle $a,b,c$.

There are many proofs of the Pythagorean Theorem; in fact, the Pythagorean Theorem is probably the most proved mathematical theorem of all time! You can find 103 different proofs here.

For this explanation, there is a proof 'without words.' See if you can figure out why it proves the theorem.

Pythagorean Theorem - Motivation

Let's assume we just finished exploring the SAS congruence theorem. What would be a good part of SAS to fix in search of a relation?

Side lengths are pretty arbitrary - you can choose any positive real number. We fix the angle in between. What would be a good angle to explore?

Many things occur with $90^\circ$ angles... Now what's the relationship?

We want to find a relationship between three variable sides. What can we do to simplify that and get started on a diagram?

We fix one of the sides, without loss of generality $AB$. However, we still have a point free to move around. What options do we have for that point?

Recall that one of the angles is $90^\circ$, and remember your goal is to vary this point to easily find a pattern. Which seems easiest?

Let's assume, without loss of generality, that $\angle ABC=90^\circ$.

Do something!

Now that we only have one thing to vary, $C$, we take note of side length combinations, such as $(AB,\frac{3}{4}AB,\frac{5}{4}AB)$ and $(AB,\frac{12}{5}AB,\frac{13}{5}AB)$.

What would make the side lengths convenient to work with?

If they were integers. You may have noted different side lengths, but if we noted those in the last part, then we get $(3,4,5)$ and $(5,12,13)$. What type of relationship is it?

The result we're about to discover is a pretty primitive result. What did they use geometry for back in the old days?

They used it for farming. What shape would make a usual plot of land?

Square.

What can we do with $(3,4,5)$ and $(5,12,13)$ to experiment, given the direction we just found?

Square them, and we get $(9,16,25)$ and $(25,144,169)$. What's the most basic operation we can perform on these?

Add them. What relationship do you notice?

$9+16=25$ and $25+144=169$. What would this mean for $\triangle ABC$?

$AB^2+BC^2=AC^2$. Let's try to prove it!

What's the easiest side of the equation to deal with?

The right hand side, $AC^2$. Let's draw it in. How?

We have to draw in $AC$ squared...

$AC^2$ is the area of a square with side length $AC$. Which side do we draw it on, though?

We draw a square with side $AC$ on the side opposite $B$ to avoid getting cluttered.

What would the other side of the equation look like geometrically?

What do we want to show?

We want to show that the sum of the areas of the squares with side lengths $AB$ and $BC$ is the area of the square with side length $AC$. Can we prove it directly?

Not easily. Back to this diagram.

Let's consider whether we need each part - the triangle and the square. Why did we put in the square?

To get the right side of the equation, so there's no use in removing it. What would we have without the triangle?

We would be removing the main thing, so the triangle, and thus both parts, are to stay. But what's arbitrary?

Is anything arbitrary from the triangle's perspective? Could we have constructed a square just as special?

No, this square is built on the hypotenuse, the side opposite the right angle. What about from the perspective of the square?

Choose side $AC$ for the triangle is arbitrary. Which side would be a non-arbitrary place to put another triangle?

Due to restrictions with discovery trees in expii, we're just going to go with side $(AC)(CA)$.

Reflectional Symmetry:

Rotational Symmetry:

That's the spirit! Now what is it?

What did we just add?

We need two triangles for the other sides of the squares.

Placement 1:

Placement 2:

Hopefully we added enough... What do we have left to prove?

The other side of the equation. What was the last thing we left off with dealing with that side of the equation?

We left off with this beauty.

What areas do we have left to fill in? Once you thought of an idea to continue, do so and then choose from the available options.

First Approach

Start with the specifications of the square that we want to equate areas with.

What is the square we want to equate areas with?

The beautiful square from earlier.

What is the side length of that square?

$AB+BC$

Where do you find this length in the new figure?

$A(BC)$ and $C(BA)$

Complete the square.

How do we get the desired result?

We want the union of $AB^2$ and $BC^2$ to be $AC^2$, and we added on four triangles. Where are the remaining two triangles?

Since this and our previous "beautiful" square have the same side length, they have the same area. The area of the square from earlier is composed of $AC^2$ along with four triangles congruent to $\triangle ABC$. The area of this square is composed of $AB^2+BC^2$ along with four triangles congruent to $\triangle ABC$. Subtract the four areas of triangles congruent to $\triangle ABC$, and you get that $AB^2+BC^2=AC^2$.

Second Approach

Fill in three triangles congruent to $\triangle ABC$ and hope for the best.

Where do you see something that would be congruent to $\triangle ABC$?

What type of congruence are we looking for?

What inspired this whole thing?

SAS. Now look again.

Two more and we'd be done! How do we orient one of them such that we can use our square from earlier?

What seems like a reasonable way to finish?

Since this and our previous "beautiful" square have the same side length, they have the same area. The area of the square from earlier is composed of $AC^2$ along with four triangles congruent to $\triangle ABC$. The area of this square is composed of $AB^2+BC^2$ along with four triangles congruent to $\triangle ABC$. Subtract the four areas of triangles congruent to $\triangle ABC$, and you get that $AB^2+BC^2=AC^2$.

Vocabulary

Right triangles are one of the most important topics to learn in geometry and are used heavily throughout trigonometry as well.

There are two different ways to classify sides of a right triangle:

Legs: these are the sides of a right triangle that make up the right angle.

Hypotenuse: this is the remaining side of right triangle. This will also always be the longest side of the triangle (You'll see why in a minute)

The lengths of these sides are related to one another by arguably the most important theorem in geometry, known as the $\color{purple}{\text{Pythagorean Theorem}}$.

This $\color{purple}{\text{theorem}}$ states that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. For this to be true, the hypotenuse must always be the longest side length.

Let's try a few problems.

These sets of numbers that satisfy the $\color{purple}{\text{Pythagorean Theorem}}$ are known as Pythagorean triples.

As exhibited in the practice problems, $(6,8,10)$ and $(7, 24, 25)$ are two of the many Pythagorean triples. Some other commonly used triples are $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, and $(20, 21, 29)$.

Notice how $(6,8,10)$ is just $(3,4,5)$ with every term doubled. If you think about this geometrically, you can see why it would be true. By doubling the side lengths, you are creating a similar triangle, which would still be a right triangle, therefore satisfying the $\color{purple}{\text{Pythagorean Theorem}}$.

Pythagorean theorem is a very easy and a very useful theorem. it is used to find the sides of a right angled triangle.if we are given two sides of a right angled triangle, we can easily find the third side by the formula- $a^2 + b^2 =c^2$ where a,b and c represents perpendicular,base and hypotenuse respectively. But have you ever thought why this formula is true?What is the proof of the Pythagorean theorem? Well ! Don't worry! Here is the answer.

Here there are two identical right angled triangles and another right angled triangle OCD. The angle of DOC is 90 degree because if we take AOD as x , BOC will be 90-x. This is because BCO will also be equal to x. As we add 90-x and x we get 90 degree so we can say that DOC is 90 degree. Now we will make an equation showing that the sum of area all the three triangles is equal to the area of the trapezium. So-

$$\frac12\cdot ab +\frac12\cdot ab +\frac12\cdot c^2 \\= \frac12(a+b)\text{sum of parallel sides}\cdot (a+b)\text{height of trapezium}$$ $$\frac{ab}2 + \frac{ab}2 +\frac{c^2}2 =\frac12 (a+b)^2$$ $$ab + \frac{c^2}2 = \frac12 (a^2 + b^2 +2ab)$$

Now we can multiply both the sides by two to get rid of fractions.

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

Now the 2ab on both the side will cut leaving-

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

Hence proved !

Like all triangles, the three angles of a right triangle sum to a total measure of $180$ degrees; however, a right triangle has one right (90-degree) angle.

Right triangles also have another special property known as the Pythagorean Theorem. If $a$ and $b$ are the lengths of the triangle's legs and $c$ is the length of the triangle's hypotenuse, then

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

There are also some special forms of right triangles:

30-60-90 right triangle: the angles have measures of $30$, $60$, and $90$ degrees. The sides have lengths $m$, $m\sqrt{3}$, and $2m$ for a positive value of $m$. Note that this satisfies the Pythagorean Theorem:

$m^2 + (m\sqrt{3})^2 = (2m)^2 \Rightarrow$ $m^2 + 3m^2 = 4m^2$

45-45-90 right triangle: the two non-right angles of the triangle each have measures of $45$ degrees. These triangles are also known as isosceles right triangles. These sides have lengths of $a$, $a$, and $a\sqrt{2}$ for a positive value of $a$. Note that these values also satisfy the Pythagorean Theorem:

$n^2 + n^2 = (n\sqrt{2})^2 \Rightarrow 2n^2 = 2n^2$

Finally, note that besides the right angle, there are two other angles, each of which has a leg of the triangle opposing it. The larger of these two angles is always opposite the longer leg for right triangles.

The Pythagorean Theorem is one of the most basic and useful relationships in all of geometry. Suppose we have a right triangle with sides $a$, $b$, and $c$. Two of the sides must be separated by and angle of $90º$ for this to be a right triangle. Let us say that the two sides in this case are $a$ and $b$. The Pythagorean Theorem states that for any such right triangle:

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

For a simple geometric proof of this concept, let us inscribe a tilted square with side length $c$ inside another larger square.

This figure now gives us four congruent right triangles and an inscribed square, EFGH. Let us call the length of $AE = BF = CG = DH = a$ and $AH = BE = CF = DG = b$, as in the original example. The side length of the square is $c$, which also forms the hypotenuse of the right triangles with leg a and b in this case. Now, the length of one side of the outer square is $a+b$, so we can say that the area of the outer square is $(a+b)^2$. However, another way we can define the area of this outer square is by adding up the areas of the four triangles and the inner square. (See: Area of triangles; Area of Square) This gives us $4(\dfrac{1}{2}ab)+c^2$ as another formula for the area of the larger square. Since the former and the latter are both representing the same area, we can set them equal to each other. This gives us:

$$(a+b)^2=4(\dfrac{1}{2}ab)+c^2$$

Expanding on the left and simplifying on the right:

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

Subtracting $2ab$ from each side, we are left with our original formula for the Pythagorean Theorem:

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

There are numerous other proofs for the Pythagorean Theorem, some more complex than others. This is simply a very basic introduction into the Pythagorean Theorem and how to conceptualize its principle.

Pythagoras once planted a seed.

30 years later, it grew into a tall elm and said "Gee, I'm a tree."

...

...

Don't get it?

"Geometry!"

The Pythagorean theorem is one of the most famous equations you will come across in your early mathematic career. But how is it derived?

Let us briefly explore this theorem and its proof

Pythagorean Theorem

The sum of the squares of the two legs is equal to the square of the hypotenuse, or:

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

But what is the theorem saying?

Colloquially, it says that if you were to build squares on each of the sides of the triangle, the sum of the areas of the squares with side length $a$ (area $a^2$) and side length $b$ (area $b^2$) would be equal to the area of the square made with side length c (area $c^2$)

But why is this true? To answer that we need only consult with Pythagoras' original proof.

Pythagoras' Pythagorean Proof

Pythagoras was a clever man, so his proof is also quite clever. His is a proof by rearrangement or, simply, he moved things around to make the proof easier to visualize. This is certainly allowed, as long as the figure that results, and each of its components, are perfectly equivilant to the original.

Pythagoras' original statement is as follows:

"Let a, b, c denote the legs and the hypotenuse of the given right triangle, and consider the two squares in the accompanying figure, each having a+b as its side. The first square is dissected into six pieces-namely, the two squares on the legs and four right triangles congruent to the given triangle. The second square is dissected into five pieces-namely, the square on the hypotenuse and four right triangles congruent to the given triangle. By subtracting equals from equals, it now follows that the square on the hypotenuse is equal to the sum of the squares on the legs, Q.E.D."- Pythagoras

What Does it Mean?

For some, this still might be hard to visualize, therefore, I encourage all of you to use the image below (never be ashamed to draw things in mathematics, especially geometry, as it is often imperative to understanding the question or proof!)

Pythagoras proposes two rearranged squares. The first is composed of side lengths (a+b). There are six component shapes that comprise this square. Four congruent triangles and two squares, one with an area of $a^2$ and the other with area $b^2$.

In square two, the set up is different. The sides are still (a+b) however this triangle is only made of of five pieces, four congruent triangles and one large square with area $c^2$

Now, since this equation deals with areas, let us find the areas.

The area of the first square can be: $4(1/2)ab+ a^2 + b^2$ or $(a+b)^2$

The area of the second can be: $4(1/2)ab+ c^2$ or $(a+b)^2$

Now what to do? Well, all these should be equivalent so let's set them equal and see what happens.

Clearly, there is a best choice for proving the theorem.

By setting $4(1/2)ab+a^2+b^2$ = $4(1/2)ab+c^2$ we get

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

And that is how we prove the Pythagorean theorem!

In case you're still a little confused, there is a great depiction of the Pythagorean Theorem with water down below!

I hope you've learned to love this equation, because you're going to be using it until you stop doing math:)

There's a really easy geometric proof of the Pythagorean Theorem.

Imagine a square where each side is of length a + b. If you cut an axa square out of one corner and a bxb square out of the opposite corner, there will be two rectangles of area $a \times b$. So $(a+b)^2 = a^2 + 2ab + b^2$, which you already knew.

Now divide the sides of the square into segments of length a, b, a, b, a, b, a, b. Cut off 4 right triangles with sides a and b, and area $ab/2$ and let's call the hypotenuse of those triangles c. The center part is a square of area $c \times c$. (We know it's a square, because where its corner touches the side of the bigger square, there are three angles and those three angles sum to 180 degrees. The other two angles sum to 90 degrees.) So we can see that the area of the big square is $c \times c + 4 a b/2$.

Putting the two parts together, $a^2 + b^2 = c^2$.

The Pythagorean Theorem can be useful in solving problems pertaining to related rates. In this box we will demonstrate a generic derivation that may be helpful for using the Pythagorean Theorem in related rates problems.

Related rates problems ALWAYS involve differentiating with respect to time. We assume both legs and the hypotenuse have lengths that are functions of time.

$A(t)^2 + B(t)^2 = C(t)^2$

Now we differentiate the whole equation, noting the necessity of use of the Chain Rule:

$2A(t)\frac{dA}{dt} + 2B(t)\frac{dB}{dt} = 2C(t)\frac{dC}{dt} \Rightarrow$ $A(t)\frac{dA}{dt} + B(t)\frac{dB}{dt} = C(t)\frac{dC}{dt}$

That's a lot of information you need, but luckily most of it will be given to you. Furthermore, it may be said (or interpreted) that one side of the triangle has a constant length. For instance, if the hypotenuse has a constant length, then the equation becomes

$A(t)\frac{dA}{dt} + B(t)\frac{dB}{dt} = 0$

The final thing to note is the domain restriction that arises when working with triangles. We must have the following hold:

$A(t) + B(t) > C(t)$

Otherwise we violate the triangle inequality.

The Pythagorean Theorem

The Pythagorean Theorem is a theorem which tells us that for a right triangle with legs $a$ and $b$ along with a hypotenuse $c$, the sum of the squares of the legs is equal to the square of the hypotenuse. In mathematical terms, this means that $a^2 + b^2 = c^2$.

This theorem is arguably one of the most used theorems in geometry. In general, right triangles along with their respective properties, theorems, and formulas are all very prominent in geometry. But before we decide to have fun with this theorem, we should be able to prove it/understand why it works.

Garfield's Proof of the Pythagorean Theorem

There are many proofs of the Pythagorean Theorem. In this lesson, we will examine two ways to prove it (one that is simple, and another that requires a bit more knowledge in geometry). The first proof is credited to James A. Garfield (though it is not written in the way he wrote it in his time), a former president of the United States of America, who discovered it before his presidency while he was serving as a member of the House of Representatives.

To better understand the proof, refer to the diagram below.

Proof: Recall that the area of a trapezoid is $$A_{\text{trapezoid}}=\frac{1}{2} \; (\text{sum of bases})(\text{height}).$$ By referring to the above diagram, we can rewrite the equation in terms of $a, b,$ and $c$ to get $$A_{\text{trapezoid}} =\frac{1}{2} \; (a+b)(a+b) = \frac{1}{2} \; (a^2 + 2ab + b^2) = ab + \frac{c^2}{2}$$ by rearranging the terms, we have $$\frac{1}{2} \; (a^2 + b^2 + 2ab) = \frac{c^2}{2} + ab \\ \frac{a^2}{2} + \frac{b^2}{2} + ab = \frac{c^2}{2} + ab \\ \frac{a^2}{2} + \frac{b^2}{2} = \frac{c^2}{2}$$ to conclude we multiply both sides by $2$ to get our desired result. $\Box$

Now let's look at a different proof of this theorem which will require the use of a different theorem.

A Lovely Application of Pappou-Clairaut

Before we go on to proving the Pythagorean Theorem, let us first learn about another theorem: Pappou-Clairaut.

Theorem (Pappou-Clairaut): Consider parallelograms BB'A'A and CC'A''A which are in the exterior of $\triangle ABC$, where $D$ is a point common to lines C'A'' and $B'A'$, while $E$ is the point of intersection of lines $BC$ and $AP$. Furthermore, consider a straight line segment $DA = EN$ that is in the exterior of $\triangle ABC$ and lies on the extension of segment $AE$.

Let B''C''CB be a parallelogram s.t. the point $N$ lies on side B''C''. Then P_{B''C''CB} = P_{CC'A''A} + P_{B'BAA'}where $P$ is the area enclosed by the corresponding quadrilateral.

We will use this theorem to prove the Pythagorean Theorem. To better understand the proof, refer to the diagram below.

Proof: We begin by noticing that since $AC = AH$ and $AE = AB = AB = HT$, it must be true that $\triangle HTC$ is equal to $\triangle ABC$. From this, we get $$BC=BK=AT$$ which tells us that $\angle MAB = \angle C = \angle A_1$ where $M$ lies on side $BC$ s.t. it is the point of intersection of lines $AT$ and $BC$. It follows that $$\angle B + \angle A_2 = \angle B + \angle C = \frac{\pi}{2}.$$ This gives us $BC \perp AT \Rightarrow BK \parallel AT.$ Now let us recall a well-known property of squares, that they are a type of parallelogram. Hence by using Pappou-Clairaut, we get $$[ABDE] + [ACZH] = [BKLC]$$ which is equivalent to saying $AB^2 + AC^2 = BC^2$. $\Box$

Through these two proofs we have come to understand why (and how) the Pythagorean Theorem works. Now, let's try applying it to some problems!

Answer: Let $a = 3, b = 4,$ and $c$ be the hypotenuse. By using the Pythagorean Theorem, we find that $3^2 + 4^2 = c^2$ thus $c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Remark: As you solve more and more problems by using the Pythagorean Theorem, you will find that, since we are using exponents, the length of the side you are solving for won't always be an integer! The combination of $3\text{-}4\text{-}5$ is actually one of the members of a family of combinations called Pythagorean Triples which are combinations of integers where the Pythagorean Theorem gives nice integers as your side-lengths

Answer: Let $a$ and $b$ be the unknown leg lengths and $c$ be the hypotenuse. We know that $a^2 + b^2 = (4\sqrt{5})^2 = 80$ and, since $a$ and $b$ are integers, we know $a^2$ and $b^2$ are perfect square integers. The only perfect square integers which add up to $80$ are $4^2 = 16$ and $8^2 = 64$. Thus, our answer is $4 + 8 = 12$.

Conclusion

Thank you for taking the time to go through this lesson! I hope you enjoyed it and learned a lot about the Pythagorean Theorem. Keep exploring new concepts and solve fun problems!

Part One

Section 1 A

The Pythagorean Theorem

Let's say that you have a wooden triangle that you're making a ruler out of. A right triangle, to be specific. The legs are $12$ inches and $5$ inches. The third leg, which is called the hypotenuse, is labeled "$c$ inches."

Section 1 B

The Pythagorean Theorem: The Formula

Look at a square like this:

The Pythagorean Theorem States that $a^2+b^2=c^2$

Section 2 A The Proof

The Pythagorean theorem is $a^2+b^2=c^2$ and only applies to right triangles where $c$ is the hypotenuse and $a$ and $b$ are the legs.

What do you notice about the triangles outside of the small square? Is there any similarity between one triangle and the next?

Look at the shaded triangle. Did you notice that all of the triangles are the same?

So far we have that,

Area of Inner Square = IS = $c^2$

Area of Outer Square= OS = $\left(a+b\right)^2$

Area of 1 Triangle=AT = 0.25 • OS

Let's solve for AT. We use substitution to find the area: 0.25•(OS)= 0.25•$\left(a+b\right)^2$ , minus the small square.