# 9.7: Pythagorean Theorem and its Converse

**At Grade**Created by: CK-12

**Practice**Pythagorean Theorem and its Converse

“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 Concept.**

### Guidance

In the last few Concepts, we have been working with square roots. Notice that the key word in the word square root is “square”. This lesson introduced 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.**

**The Pythagorean Theorem makes the most sense when it is used in practice.**

**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.**

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.

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 select the one that makes the most sense. Since the number after the two is a zero in this answer, it made sense to leave it as 12.2. Sometimes, it will make more sense to round. You have to think it through!!*

Find the length of the hypotenuse given each triangle.

#### Example A

**A right triangle with legs of 9 and 12.**

**Solution: \begin{align*}c = 15\end{align*}**

#### Example B

**A right triangle with legs of 12 and 16.**

**Solution: \begin{align*}c = 20\end{align*}**

#### Example C

**A right triangle with legs of 3 and 6.**

**Solution: \begin{align*}c = 6.7\end{align*}**

Here is the original problem once again.

“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 in this Concept.

- 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.

### Guided Practice

Here is one for you to try on your own.

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?

**Answer**

First things first: we need to 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.**

### Video Review

Here is a video for review.

This is a James Sousa video on the Pythagorean Theorem.

### 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*}

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

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

13. \begin{align*}a = 15, \ b = 7, \ c = ?\end{align*}

14. \begin{align*}a = 14, \ b = 3, \ c = ?\end{align*}

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

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

### Image Attributions

Here you'll learn to recognize and use the Pythagorean Theorem.