Teorema de Pitágoras

(Video) Math Life Hack - How to Use the Pythagorean Theorem to Cut Corners

This video, created by the Expii team, talks about the pythagorean theorem in a real life situation!

$\large{\text{Summary}}$

What is the Pythagorean Theorem?

The Pythagorean Theorem was discovered by Pythagoras around $500$ BCE when Pythagoras was running late and saw his bus pulling away. Struck by the sudden understanding of geometry and the universe, he realized he could cut through an alley (doing sick parkour along the way) and arrive before the bus.

He later realized this worked because of a simple geometric shape: right triangles.

After much calculation he discovered that when you square the length of each side of the triangle, they are related by the following formula: $$a^2+b^2=c^2$$

Okay, and?

This simple relation was a game changer. As you saw in the video, it's pretty effective at catching buses. Let's take a closer look at why that is.

In the video, we are given the diagram:

There are two options to catch the bus at the next stop (finish):

1. Run along the outside of the triangle: going down then toward the finish (the same way the bus is going).
2. Run along the diagonal of the triangle.

We know the lengths of the outside of the triangle. If we were to run this path, we would run down a distance of $3$ and then would run toward the finish for a distance of $4$. To get the total distance we would travel, just add them up.

So, the distance would be $7$ this way. Is that shorter or longer than taking the path along the diagonal? Hmm, if only there was a theorem to figure the length of the diagonal...oh wait PYTHAGOREAN THEOREM!

Okay so we have the Pythagorean Theorem. $$a^2+b^2=c^2$$ Now, we need to fill it in with the information we know. The $a$ and $b$ were the outer lengths of the triangle.

Yup! Another nice thing about the Pythagorean Theorem is it doesn't matter which outer length we set as $a$ and which we set as $b$. For now, let's let $a=3$ and $b=4$. So, just plug these into our formula. $$\begin{align} a^2+b^2&=c^2\\ \color{red}{3}^2+\color{red}{4}^2&=c^2 \end{align}$$

Right, $c$ is just a question mark in the problem above. To make it clear, I'm going to change it to a question mark but you can also leave it as $c$. $$\begin{align} \color{red}{3}^2+\color{red}{4}^2&=c^2\\ \color{red}{3}^2+\color{red}{4}^2&=\,\color{red}{?}^2\\ \end{align}$$

Next, we need to square everything. Since we can't really square a question mark, just leave that as is. $$\begin{align} \color{red}{3}^2+\color{red}{4}^2&=\,\color{red}{?}^2\\ 9+16&=\,?^2 \end{align}$$

Now, simplify the left hand side by adding the numbers together. $$\begin{align} 9+16&=\,?^2\\ 25&=\,?^2 \end{align}$$

We're almost there. One final step. We can either ask ourselves, "What number, when squared, equals $25$?" or we can take the square root of both sides of the equation. Let's try that. $$\begin{align} 25&=\,?^2\\ \color{red}{\sqrt{\color{black}{25}}}&=\color{red}{\sqrt{\color{black}{?^2}}} \end{align}$$

The square root of $25$ is $5$ and the square root of a number squared is just the number. So: $$\begin{align} \color{red}{\sqrt{\color{black}{25}}}&=\color{red}{\sqrt{\color{black}{?^2}}}\\ 5&=\,? \end{align}$$ The length of the diagonal is just $5$.

Using the Pythagorean Theorem, you can quickly find unknown lengths for right triangles. Who knows, it just might help you catch your next bus.

Try it Yourself!

(Video) The Pythagorean Theorem

by PatrickJMT

In this video by PatrickJMT, he explains the Pythagorean Theorem and goes over some problems that you might encounter with it.

$\large{\text{Summary}}$

The Pythagorean Theorem is by far one of the most well known theorems in all of math. Its states that for any right triangle: $$a^2+b^2=c^2$$

Using the Pythagorean Theorem

One way to identify a right triangle is by using the Pythagorean Theorem. The Pythagorean Theorem shows the relationship among the lengths of the sides of a right triangle.

$c$ is the longest side, called the hypotenuse, and its the side that is across from the right angle. $a$ and $b$ are the lengths of the other two sides, called legs.

If you plug in the lengths of each side of the triangle into the Pythagorean equation and get a TRUE statement, then the triangle is a right triangle.

Note: In order for $a^2+b^2=c^2$ to be true, the side length $c$ must be the largest side length. So in order to test whether a triangle is right, you only need to plug in values where $c$ is the largest among $a,b,c$.

$$3^2+4^2\overset{?}{=}5^2$$ $$9+16\overset{?}{=}25$$ $$25=25$$

This produces a true statement ($25=25$), which tells us that our triangle is in fact a right triangle.

The order of $3$ and $4$ (as $a$ and $b$) doesn't really matter. However, it is important that we checked the Pythagorean rule by plugging in $c=5$, the largest among $3,4,5$. If we had plugged in $c=4$ or $c=3$, we would've gotten something like $5^2+3^2\neq 4^2$ or $4^2+5^2\neq 3^2$ instead. The test is only useful when $c$ is the largest side.

$$4^2+4^2\overset{?}{=}6^2$$ $$16+16\overset{?}{=}36$$ $$32\neq 36$$

We can tell that this triangle is not a right triangle because the result of using the Pythagorean equation produces a FALSE statement, even though we plugged in the largest side length, $6$, for $c$.

You can also use the Pythagorean Theorem to calculate the length of one of the sides of a known right triangle. (This only works if the triangle is a right triangle, so be careful when you use it).

$$2^2+b^2=5^2$$ $$4+b^2=25$$ $$b^2=21$$ $$b=\sqrt{21}$$

The Pythagorean Theorem

Let's start with what a theorem is. A theorem is a statement that's been proven to be true from other things that are either known or accepted to be true.

Pythagoras was this guy a long time ago who was insanely smart and was able to prove this theorem.

The important thing to know is that the Pythagorean Theorem describes an important geometric relationship between all three sides of a right triangle.

The Pythagorean Theorem

For a right triangle with legs $a$ and $b$ and hypotenuse $c$ $$a^2+b^2=c^2$$

It's important to note that the Pythagorean Theorem only applies to right triangles. A right triangle is any triangle that has a right angle, an angle that equals $90^{\circ}$. Right triangles are usually denoted with a square around their right angle.

The right angle also helps you identify an important side of a right triangle, the hypotenuse. The hypotenuse is always the longest side of a right triangle and it is always located opposite the right angle. It is denoted by the letter $c$.

The sides that touch, or form, the right angle are called the legs of the triangle. It doesn't matter which we call $a$ and which we call $b$ as log as we keep track of it and stay consistent.

3,4,5 Triangle

Let's look at a special right triangle in order to make sense of this theorem. This special right triangle is called a $3$,$4$,$5$ Triangle because its sides have the relative lengths of $3$, $4$, and $5$.

Let's see how the Pythagorean Theorem works for this. $$a^2+b^2=c^2$$

Our first side, $a$, has a length of $3$. The Pythagorean Theorem tells us to square this. $$3^2=3*3=9$$

Our second side, $b$ has a length of $4$. The Pythagorean Theorem tells us to square this. $$4^2=4*4=16$$

It says to add these two values together now. $$9+16=25$$

According to the Pythagorean Theorem this will be the same number as when we square the hypotenuse, $c$, which is $5$. $$5^2=5*5=25$$

As you get see we got $25$ both times! So $a^2+b^2=c^2$!

That's cool, but why should I care?

The reason the Pythagorean Theorem is so important is that it's a useful tool to help us use information we do know to find out information we don't know.

If you have a right triangle and only know how long two of its sides are, the Pythagorean Theorem tells you how to find the length of the third, unknown, side.

For example, suppose we had this triangle:

We could easily find the length of this unknown side using the Pythagorean Theorem. $$\begin{align} a^2+b^2&=c^2\\ \color{red}{2}^2+\color{red}{3}^2&=c^2\\ 4+9&=c^2\\ 13&=c^2\\ \sqrt{13}&=\sqrt{c^2}\\ \sqrt{13}&=c \end{align}$$

Let's look at another example where we know only one of the legs and the hypotenuse.

Use the Pythagorean Theorem to find the length. $$\begin{align} a^2+b^2&=c^2\\ \color{red}{4}^2+b^2&=\color{red}{6}^2\\ 16+b^2&=36\\ b^2&=36-16\\ b^2&=20\\ \sqrt{b^2}&=\sqrt{20}\\ b&=2\sqrt5 \end{align}$$

Test a Triangle

We can also use the Pythagorean Theorem to test if a triangle is a right triangle. For example, lets look at this triangle:

The Pythagorean Theorem can tell us if it's a right triangle as long as we know all three sides. Since $4$ is the longest side, we can plug that into $c$ and the others into $a$ and $b$ to test it. $$\begin{align} a^2+b^2≟c^2\\ \color{red}{3}^2+\color{red}{3}^2≟\color{red}{4}^2\\ 9+9≟16\\ 18\ne16 \end{align}$$ Since these two sides are not equal, we can conclude that the triangle is NOT a right triangle.

(Video) The Pythagorean Theorem

by Mahalo Math

Mahalo Math has a good video explaining what the Pythagorean theorem is and why it is useful.

The Pythagorean theorem only applies to right triangles and is written like $$a^2+b^2=c^2$$

You can call either side $a$ and $b$ but $c$ must be the hypotenuse which is the longest side across from the right angle.

You can use the pythagorean theorem to find the length of an unknown side of a right triangle.

(Video) Word Problem Using the Pythagorean Theorem

by PatrickJMT

Patrick JMT has some great Pythagorean theorem word problems. The Pythagorean theorem is one of your most useful tools in mathematics so being able to identify when to use it will help you immensely.

$\large{\text{Summary}}$

The Pythagorean theorem states $a^2+b^2=c^2$ when $a$ and $b$ are the legs and $c$ is the hypotenuse.

Example $1$:

If you have an equilateral triangle with height $8$, find the length of each side.

We draw the right triangle such that the one leg is $8$ and the other is $\frac{1}{2}x$ and $c$ is $x$. We set up the equation: $8^2+(\frac{1}{2}x)^2=x^2$

Now, we simplify. $$\begin{align} 8^2+(\frac{1}{2}x)^2&=x^2\\ 64+(\frac{1}{4}x)&=x^2\\ 64&=x^2-\frac{1}{4}x^2\\ 64&=x^2-\frac{1}{4}x^2\\ 64&=\frac{3}{4}x^2\\ \frac{4}{3}\times 64&=x^2\\ \frac{256}{3}=x^2\\ \sqrt{\frac{256}{3}}&=\sqrt{x^2}\\ \frac{16\sqrt{3}}{3}&=x \end{align}$$

Example $2$:

You have two cyclists that start from the same direction. They are going the same speed and one is going east and one is going north. Find the speed of each in minutes per hour if after two hours they are $17\sqrt{2}$ miles apart.

We draw the triangle such that they hypotenuse is $17\sqrt{2}$ and each side is $x$. So, $x^2+x^2=(17\sqrt{2})^2$. We simplify this to get $17$ which we then divide by two to find the speed of each of the cyclists. $\frac{17}{2}$ minutes per hour is how fast each of the cyclists are going.

(Videos) Pythagorean Theorem

by tecmath

This video by tecmath works through a Pythagorean theorem example problem.

$\large{\text{Summary}}$

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

We are given a quadrilateral. We can make a triangle and a square out of the quadrilateral. We then know the sides are $10$ and $3$. We can plug in those numbers to find $x$. $$10^2+3^2=x^2$$

This is the same as $$\sqrt{109}=x$$ This is equal to $10.44$

Feel free to watch the other example in the video for more practice!

(Videos) Quadratic Equations and the Pythagorean Theorem

by PatrickJMT

Patrick JMT has a bunch of more complicated problems that use both the Pythagorean theorem and the Quadratic Formula together. Here is an example below.

$\large{\text{Summary}}$

In his example, he tries to solve for $x$ for a triangle with sides $5(x-1)$ and $x$, and a hypotenuse of $6$ We need to use the pythagorean theorem to set up the equation $$[5(x-1)]^2 + (x)^2 = 6^2$$ Then, distributing the squares, we get an equation of: $$25[(x-1)(x-1)] + x^2 = 36$$ $$25[x^2 – 2x + 1] + x^2 = 36$$ $$25x^2 – 50x + 25 + x^2 = 36$$ Adding like terms will give us: $$26x^2 -50x + 25 = 36$$ Now we want the equation on the left to equal $0$, so we subtract $36$ from both sides to get $$26x^2 – 50x – 11 = 0$$ Using the quadratic formula with values of $a = 26, b = -50,$ and $c = -11$, we can find the values for $x$ $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$x = \frac{-(-50)\pm\sqrt{(-50)^2-4(26)(-11)}}{2(26)}$$ $$x=\frac{50 \pm \sqrt{2500 +1144}}{52}$$ $$x = \frac{50 \pm \sqrt{3644}}{52}$$ Note that in the video, he reduced $\sqrt{3644}$ to $60.37$. You can either leave it as a radical, or you can also reduce it to a decimal.

So, what you get in the end is a lower limit and an upper limit. For our problem, the lower limit is negative. However, given the nature of the problem, which is finding the length of a triangle side, we do not want negative numbers. For sides of shapes, or any measurements at all, you cannot have negative numbers. So, our $x$ value will be $x = \frac {50 + \sqrt{3644}}{52}$.

(Video) Pythagorean Theorem

by yaymath

Here's a video by yaymath explaining the Pythagorean theorem and showing you how you can use it to see if a triangle is right triangle or not.

$\large{\text{Summary}}$

A right triangle is a triangle with a right angle in it.

The Pythagorean Theorem states that given any right triangle: $$a^2+b^2=c^2$$ You can also prove that any triangle is right triangle by plugging in the lengths of its sides and seeing if the Pythagorean Theorem holds true.

When solving problems involving right triangles the steps are usually:

1. Figure out the $a$, $b$, and $c$ of the triangle and what's missing.
2. Use the formula $a^2+b^2=c^2$.
3. Plug in the sides that you know.
4. Solve for the side that you don't know.

(Video) The Pythagorean Theorem

by mathman1024

Here's a video by mathman1024 explaining the Pythagorean Theorem.

$\large{\text{Summary}}$

The Pythagorean Theorem is very simple. It states: $$a^2+b^2=c^2$$ Cool! But what do those letters mean? Well, they're all the sides of right triangle.

Here $c$ is the hypotenuse of the triangle. The hypotenuse is always the longest side of the triangle. It will also always be on the opposite side of the right angle. The $a$ and $b$ denote the legs of the triangle.

The first problem he looks at is:

Where we want to use the Pythagorean Theorem to find $x$. $$\begin{align} &a^2+b^2=c^2\\ &(3)^2+(x)^2=(5)^2\\ &9+x^2=25\\ &9\color{red}{-9}+x^2=25\color{red}{-9}\\ &x^2=16\\ &x=\pm4 \end{align}$$ Since we can't have a negative length we can toss $x=-4$ out the window. That means $x=4$.

This problem was actually an example of a Pythagorean Triples. Pythagorean Triples are integer solutions to the Pythagorean Theorem. In the last problem, $3^2+4^2=5^2$ so the triple is $3,4,5$. Remember that the larger number is always the hypotenuses length. These are useful to remember because they can let you solve a problem quickly. Some other common ones are: $$5,12,13\\ 7,24,25\\ 8,15,17\\ 9,40,41$$ Here's a list of a bunch of Pythagorean Triples.