# 9.3: The Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Big Play

“You should have seen it!” Miguel exclaimed at breakfast one morning. “The bottom of the ninth, two outs and a man on second, then the man on second runs for home and the second baseman catches the ball and throws it home!”

“Then what happened?” Carmen, Miguel’s sister asked, her mouth wide open in anticipation.

“We got the guy out and ended up winning the game!” Miguel grinned from ear to ear.

The whole family continued eating for a few minutes.

“That must have been quite a moment son,” Miguel’s Dad added, drinking his coffee.

“And a long way to throw the ball,” Carmen commented.

Miguel started to think about this last statement. It was a long way to throw the ball. Baseball players did it all the time, but that throw was so accurate. It went right to the catcher! Miguel began wondering about that distance. How far had the second baseman thrown the ball?

While Miguel thinks about this, you can learn how to figure it out. Triangles, squares and the Pythagorean Theorem are all part of this solution. Pay attention because you will see this problem again at the end of the lesson.

What You Will Learn

By the end of this lesson you will learn the following skills.

• Recognize the Pythagorean Theorem.
• Use the Pythagorean Theorem to find the length of the hypotenuse, approximating as necessary.
• Use the Pythagorean Theorem to find the length of a leg, approximating as necessary.
• Solve real-world problems involving the application of the Pythagorean Theorem.

Teaching Time

I. Recognize the Pythagorean Theorem

In the last two lessons, we have been working with square roots. Notice that the key word in the word square root is “square”. This lesson introduces you to a theorem called “The Pythagorean Theorem.” This is a very useful theorem about right triangles, but first let’s see how the Pythagorean Theorem connects to our work with squares.

Here is a square. Notice that this square has been divided in half by a diagonal. When we divide a square in half, we can see that it forms two right triangles.

Think about the work that we have just finished with square roots. We know that we can take the area of a square and find the square root of it and that will give us the length of one of the sides.

Well, we can use this information to figure out the length of the diagonal. The length of the diagonal has a relationship with the length of the sides of a right triangle.

Before we get into all of that, let’s look at the parts of right triangle.

You can see that the sides of the right triangle are labeled \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*}. Sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are called the legs of the triangle.

Side \begin{align*}c\end{align*} is called the hypotenuse of a right triangle and it is the longest side of the triangle.

Now let’s look at how the Pythagorean Theorem connects all of these pieces together.

What is the Pythagorean Theorem?

The Pythagorean Theorem is a theorem or rule for working with a mathematical principle. It states that if we take the square of one of the legs of a right triangle and add it to the square of one of the other legs of a right triangle that we will end up with the square of the length of the hypotenuse.

\begin{align*}a^2 + b^2=c^2 \end{align*}

If you think a little broader, you can see that if we square the sides lengths of square, that we will have the total units or area of that square. In the same instance, if we square the sides of a right triangle and add those products together, then we end up with the length of the hypotenuse squared.

When we have right triangle, we are going to be using the Pythagorean Theorem to help us to figure out the side lengths of the triangle.

II. Use the Pythagorean Theorem to Find the Length of the Hypotenuse, Approximating as Necessary

The Pythagorean Theorem makes the most sense when it is used in practice. Without examples, it can be tricky to get a grasp on how cool this theorem really is. We can use the Pythagorean Theorem in problem solving.

Let’s look at an example to help make this clearer.

Example

Here we have a right triangle. The legs have been labeled with 3 and 4. The hypotenuse is labeled \begin{align*}c\end{align*}.

We want to figure out the length of the hypotenuse. We can do this by using the formula for the Pythagorean Theorem.

\begin{align*}a^2+b^2=c^2\end{align*}

Let’s substitute the values that we know into the formula.

\begin{align*}3^2+4^2=c^2\end{align*}

Now we can find the products of the sides and add them together.

\begin{align*}9 + 16 &= c^2\\ 25 &= c^2\end{align*}

You can solve for \begin{align*}c\end{align*} by using what you have learned about solving equations. We need to get \begin{align*}c\end{align*} alone to solve for its value. Right now, \begin{align*}c\end{align*} is squared. The inverse of squaring is to find the square root. Therefore, if we take the square root of both sides of the equation we will find the value of \begin{align*}c\end{align*}.

\begin{align*}\sqrt{25} &= \bcancel{\sqrt{c^2}}\\ 5&=c \end{align*}

The length of side \begin{align*}c\end{align*} is 5.

This triangle is called a 3, 4, 5 triangle and is one of the building blocks of the Pythagorean Theorem.

We can check our work by substituting these values back into the formula to see if both sides of the equation balance.

\begin{align*}3^2+4^2&=5^2 \\ 9 + 16 &= 25\\ 25 &= 25\end{align*}

Our work checks out.

Anytime you know the lengths of the legs of a right triangle, you can use the Pythagorean Theorem to find the length of the hypotenuse.

Example

Find the length of the hypotenuse in this triangle.

We can use the Pythagorean Theorem to find the length of \begin{align*}c\end{align*}.

\begin{align*}6^2+8^2&=c^2\\ 36 + 64 &= c^2\\ 100 &= c^2\\ \sqrt{100}&=\bcancel{\sqrt{c^2}}\\ 10&=c\end{align*}

The value of the hypotenuse is 10.

Do you notice anything interesting about the square roots of the last two hypotenuses?

If you notice, they were perfect squares. Sometimes, we will have a perfect square as the hypotenuse, but other times we won’t. When this happens, you will need to find a decimal approximation for the square root.

Let’s see how this works.

Example

What is the length of the hypotenuse of this triangle?

Begin by entering the known values into the formula for the Pythagorean Theorem.

\begin{align*}7^2+10^2&=c^2 \\ 49 + 100 &= c^2\\ 149 &= c^2\\ \sqrt{149}&=\bcancel{\sqrt{c^2}}\end{align*}

Up until this point, everything has been perfect, but now we need to find the square root of 149. It is not a perfect square. You can use your calculator to find the square root of this number.

\begin{align*}\sqrt{149} = 12.20655...\end{align*}

We can round it to 12.2 as our approximate answer for \begin{align*}c\end{align*}.

Approximate answers are a bit tricky!! Be sure to apply the rules for rounding numbers: round up if the next number is 5 or greater, keep the current value if the next number is 4 or less.

9E. Lesson Exercises

Find the length of the hypotenuse given each triangle.

1. A right triangle with legs of 9 and 12.
2. A right triangle with legs of 12 and 16.
3. A right triangle with legs of 3 and 6.

III. Use the Pythagorean Theorem to Find the Length of a Leg, Approximating as Necessary

You just learned how to find the length of the hypotenuse using the Pythagorean Theorem. What about if you have been given the length of one of the legs and the hypotenuse? Can we use the Pythagorean Theorem to find the length of the missing leg?

Sure we can. Let’s look at how this can be done.

Example

You have already worked on this triangle in the last section, and you may even remember the lengths of the sides. However, let’s use it as an example to see how the Pythagorean Theorem can help us solve for a missing leg, not just a missing hypotenuse.

First, fill the given values into the formula.

\begin{align*}3^2+b^2=5^2\end{align*}

Here we know the values of \begin{align*}a\end{align*} and \begin{align*}c\end{align*} so we can fill those in, but \begin{align*}b\end{align*} is a mystery so it remains an unknown variable.

\begin{align*}9 + b^2=25\end{align*}

Now we want to solve this equation by getting the variable alone. To do this, we subtract 9 from both sides of the equation.

\begin{align*}9 - 9 + b^2 &= 25-9\\ b^2&=16 \end{align*}

Next we take the square root of both sides.

\begin{align*}\bcancel{\sqrt{b^2}} & = \sqrt{16}\\ b &=4\end{align*}

Sometimes you won’t be working with perfect squares. When this happens, you will need to approximate the length of a leg just as we approximated the length of the hypotenuse in the last section.

Example

First, take the given side lengths and substitute them into the formula.

\begin{align*}4^2+b^2&=12^2\\ 16 + b^2&=144\end{align*}

Next, we subtract 16 from both sides of the equation.

\begin{align*}16 - 16 + b^2&=144 - 16\\ b^2&=128\end{align*}

Now we take the square root of both sides of the equation.

\begin{align*}\bcancel{\sqrt{b^2}} &= \sqrt{128}\\ b &= 11.3\end{align*}

9F. Lesson Exercises

Use the Pythagorean Theorem to find each missing side length. You may round to the nearest tenth as necessary.

1. A right triangle with \begin{align*}a, \ b = 6\end{align*} and \begin{align*}c = 13\end{align*}
2. A right triangle with \begin{align*}a = 8, \ b,\end{align*} and \begin{align*}c = 12\end{align*}
3. A right triangle with \begin{align*}a = 6, \ b,\end{align*} and \begin{align*}c = 10\end{align*}

IV. Solve Real-World Problems Involving the Application of the Pythagorean Theorem

Now that we are familiar with the different ways we can use the Pythagorean Theorem, let’s practice using it in some real-world situations. When you are given a word problem, it may help to draw a picture and fill in the information you have been given. This will help you keep track of what you know and what you need to find. Most importantly, it will help you see which side of a triangle is the hypotenuse and which sides are the legs so that you can put the numbers you know into the appropriate places in the formula. Let’s give it a try.

Example

Claire wants to hang a banner from the sill of a second-story window in her house. She needs to find a ladder that, when rested against the outside wall of her house, will be long enough to reach the second-story window. If the window is 16 feet above the ground and Claire places the foot of the ladder 12 feet from the wall, how long will the ladder need to be?

First things first: let's draw a picture so that we understand this situation better. Claire will rest a ladder against the wall of her house to reach the second-story window. This forms a triangle, so we might draw something like this.

The wall side is 16 feet and the ground side is 12 feet.

We need to solve for the length of the ladder side. Is this a leg or the hypotenuse? It is the hypotenuse, so we will use the Pythagorean Theorem to solve for \begin{align*}c\end{align*}. Let’s fill in the information and solve.

\begin{align*}a^2 + b^2 &= c^2\\ 12^2 + 16^2 & = c^2\\ 144 + 256 &= c^2\\ 400 & = c^2\\ \sqrt{400} &= \sqrt{c^2}\\ 20 &= c\end{align*}

The square root of 400 is 20 because \begin{align*}20 \times 20\end{align*} equals 400.

Claire needs a ladder 20 feet long in order to reach the window.

Now let’s apply what we have learned in solving the problem from the introduction.

## Real–Life Example Completed

The Big Play

Here is the original problem once again. Reread it and underline any important information.

“You should have seen it!” Miguel exclaimed at breakfast one morning. “The bottom of the ninth, two outs and a man on second, then the man on second runs for home and the second baseman catches the ball and throws it home!”

“Then what happened?” Carmen, Miguel’s sister asked, her mouth wide open in anticipation.

“We got the guy out and ended up winning the game!” Miguel grinned from ear to ear.

The whole family continued eating for a few minutes.

“That must have been quite a moment son,” Miguel’s Dad added, drinking his coffee.

“And a long way to throw the ball,” Carmen commented.

Miguel started to think about this last statement. It was a long way to throw the ball. Baseball players did it all the time, but that throw was so accurate. It went right to the catcher! Miguel began wondering about that distance. How far had the second baseman thrown the ball?

To solve this problem, we need to think about the distance from second base to home plate. Look at this diagram.

Now you can see that the distance from second base to home plate divides this square infield into two triangles. The legs of the triangles are the distance from home to \begin{align*}1^{st}\end{align*} and from \begin{align*}1^{st}\end{align*} to second.

We know that the distance for each leg is 90 feet. We can use this information and the Pythagorean Theorem to figure out the distance from \begin{align*}2^{nd}\end{align*} base to home plate. This is the hypotenuse of the triangle.

\begin{align*}a^2+b^2&=c^2 \\ 90^2 + 90^2 &= c^2 \\ 8100 + 8100 &= c^2 \\ 16,200 &= c^2 \\ \sqrt{16200} &= \bcancel{\sqrt{c^2}}\\ 127.279 \ feet &= c\end{align*}

We can say that the distance is approximately 127.28 or a little more that \begin{align*}127 \frac{1}{4}\end{align*} feet. That is how far the \begin{align*}2^{nd}\end{align*} baseman threw the ball.

## Vocabulary

Here are the vocabulary words found in this lesson.

Pythagorean Theorem
the theorem that states that the square of leg \begin{align*}a\end{align*} of a right triangle plus the square of leg \begin{align*}b\end{align*} of a right triangle is equal to the square of the hypotenuse side \begin{align*}c\end{align*} of the same right triangle.
Legs of the Right Triangle
legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the two shorter sides of a right triangle.
Hypotenuse
side \begin{align*}c\end{align*}, the longest side of a right triangle. It forms the diagonal in a square too.

## Technology Integration

Other Videos:

1. http://www.mathplayground.com/mv_pythagorean_theorem.html – This is a Brightstorm video on the Pythagorean Theorem and on how to apply the Pythagorean Theorem.

Resources

You may also want to check out this fictional book “What’s Your Angle Pythagoras” by Julie Ellis and Phyllis Hornung – this is a fun way to understand the Pythagorean Theorem.

## Time to Practice

Directions: Use the Pythagorean Theorem to find the length of side \begin{align*}c\end{align*}, the hypotenuse. You may round to the nearest tenth when necessary.

1. \begin{align*}a = 6, \ b = 8, \ c = ?\end{align*}

2. \begin{align*}a = 9, \ b = 12, \ c = ?\end{align*}

3. \begin{align*}a = 12, \ b = 16, \ c = ?\end{align*}

4. \begin{align*}a = 15, \ b = 20, \ c = ?\end{align*}

5. \begin{align*}a = 18, \ b = 24, \ c = ?\end{align*}

6. \begin{align*}a = 24, \ b = 32, \ c = ?\end{align*}

7. \begin{align*}a = 5, \ b = 7, \ c = ?\end{align*}

8. \begin{align*}a = 9, \ b = 11, \ c = ?\end{align*}

9. \begin{align*}a = 10, \ b = 12, \ c = ?\end{align*}

10. \begin{align*}a = 12, \ b = 8, \ c = ?\end{align*}

Directions: Use the Pythagorean Theorem to find the length of each missing leg. You may round to the nearest tenth when necessary.

11. \begin{align*}a = 6, \ b = ?, \ c = 12\end{align*}

12. \begin{align*}a = 9, \ b = ?, \ c = 15\end{align*}

13. \begin{align*}a = 4, \ b = ?, \ c = 5\end{align*}

14. \begin{align*}a = 9, \ b = ?, \ c = 18\end{align*}

15. \begin{align*}a = 15, \ b = ?, \ c = 25\end{align*}

16. \begin{align*}a = ?, \ b = 10, \ c = 12\end{align*}

17. \begin{align*}a = ?, \ b = 11, \ c = 14\end{align*}

18. \begin{align*}a = ?, \ b = 13, \ c = 15\end{align*}

Directions: Use problem solving to draw a picture and solve each problem.

19. Joanna laid a plank of wood down to make a ramp so that she could roll a wheelbarrow over a low wall in her garden. The wall is 1.5 meters tall, and the plank of wood touches the ground 2 meters from the wall. How long is the wooden plank?

20. Chris rode his bike 4 miles west and then 3 miles south. What is the shortest distance he can ride back to the point where he started?

21. Naomi is cutting triangular patches to make a quilt. Each has a diagonal side of 14.5 inches and a short side of 5.5 inches. What is the length of the third side of each triangular patch?

22. The distance around the entire baseball diamond at Seth’s school is 80 feet. How far is it from first base across to third base?

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