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# 7.4: Using the Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

## Introduction

Now that the students in Mrs. Richardson’s class understand that they will need to use the Pythagorean Theorem, they are left with the question of which ladder to use. Mrs. Richardson has told them that the height of the shed is 23 feet.

They have the choice of two different ladders. One ladder is 20 feet long and one ladder is 25 feet long.

“If we choose the 20 foot ladder and it is about 4 feet from the shed, how high up will it go?” Aran asked taking out a piece of paper and a pencil.

“I don’t know. We also have to think about the 25 foot ladder. If it is also 4 feet from the shed, how high up will it reach?” Amy said. “We need the Pythagorean Theorem for this one,” Aran said.

Both students took out a piece of paper and a pencil and began to work.

Learning to use the Pythagorean Theorem in real – world problems is important because of its many uses. Pay attention to this lesson and you will be able to figure out how high each ladder will reach by the end of the lesson.

What You Will Learn

In this lesson, you will how to execute the following skills.

• Use the Pythagorean Theorem to measure indirectly.
• Use the Pythagorean Theorem to find the perimeter and area of right triangles with a missing dimension.
• Use the converse of the Pythagorean Theorem to identify Pythagorean triples.
• Solve real – world problems involving applications of the Pythagorean Theorem.

Teaching Time

I. Use the Pythagorean Theorem to Measure Indirectly

Now that you are familiar with the Pythagorean Theorem, there are many different ways you can use it. Math questions will often require you to use the theorem even if it isn’t mentioned specifically in the text. Whenever you look at a shape, think about whether or not a right angle is present. If there is a right triangle in the shape, you may need to use the Pythagorean Theorem.

Write this statement down in your notebook.

There are many situations in which you may need to use the Pythagorean Theorem on objects that don’t even appear (at first) to be right triangles. Often, you will find that measurement involves a figure that looks like a right triangle. You can use the Pythagorean Theorem in these situations. This skill is called indirect measurement, and it is important to be comfortable using it in many situations. Indirect measurement allows you to figure out lengths or distances that would be difficult by using logic and mathematical knowledge. The example below requires indirect measurement to solve the problem.

Example

The diagram below shows the distance on roads to get from point $A$ to point $B$. If you bike, however, you can travel in a straight line between those two points. What is the shortest possible distance between points $A$ and $B$?

If you look at the image above, you see a rectangle, not a triangle. So, at first, you are unlikely to notice that you must use the Pythagorean Theorem to solve the problem. The question asks for the shortest distance between points $A$ and $B$. You know from your geometry studies that the shortest distance between two points is always a straight line, so this distance will be a diagonal in the rectangle.

Now that the diagonal is drawn in, the triangle is more noticeable. In fact, this triangle is a 3:4:5 triangle, so you can quickly see that the hypotenuse will be 5 miles. Use the Pythagorean Theorem to confirm your answer.

$a^2 + b^2 & = c^2\\3^2 + 4^2 & = c^2\\(3 \times 3) + (4 \times 4) & = c^2\\9 + 16 & = c^2\\25 & = c^2\\\sqrt{25} & = \sqrt{c^2}\\5 & = c$

Five miles is the correct answer. If you bike between points $A$ and $B$ on the map, and you can go in a straight line, the distance will be 5 miles.

Working with the Pythagorean Theorem in this way requires you to be a detective of sorts. When you see a figure, you can think about what the characteristics of the figure are. This can help you. A big clue in the last example is that a rectangle has $90^\circ$ angles. When you see a figure with $90^\circ$ angles, you will know that you can form right triangles in that figure.

II. Use the Pythagorean Theorem to Find the Perimeter and Area of Right Triangles with a Missing Dimension

You should already be familiar with the terms perimeter and area. The perimeter of an object is the distance around the outside of it. So, for a triangle, it would be the sum of the two legs and the hypotenuse. The area of an object is the amount of space it occupies, or how many square it would cover on a grid. The area formula for a triangle is $A = \frac{1}{2} bh$ where $b$ is the length of the base of the triangle, and $h$ is the height. In a right triangle, the product $bh$ can always be found by multiplying the two legs together.

Sometimes, you won’t know all of the dimensions of a right triangle. In order to find the perimeter of a right triangle, you have to know all three sides. In order to find the area of a right triangle, you have to know the base and the height. If you don’t know all of the dimensions that you need to solve a problem, you can use the Pythagorean Theorem to help you in problem solving.

Let’s look at an example.

Example

What is the perimeter and area of the triangle below?

The first thing to notice is that there is a missing leg of this triangle. Before we can do anything, we have to figure out the length of the missing side. This is a right triangle, so the first step in completing this problem is using the Pythagorean Theorem to identify the length of the missing leg.

$a^2 + b^2 = c^2$, where $a = 5$ and $c = 13$

$5^2 + b^2 & = 13^2\\25 + b^2 & = 169\\25 + b^2 - 25 & = 169 - 25\\b^2 & = 144\\\sqrt{b^2} & = \sqrt{144}\\b & = 12$

The length of the missing side is 12 inches.

Now, to find the perimeter $(P)$ of the triangle, add up the lengths of the three sides.

$P & = leg + leg + hypotenuse\\P & = 5 + 12 +13\\P & = 30$

The perimeter of the triangle is 30 inches.

To find the area, use the area formula shown above. Remember that in this triangle, $b=5$ and $h=12$. The height is the other leg because it is a right triangle.

$A & = \frac{1}{2}bh\\A & = \frac{1}{2} (5)(12)\\A & =\frac{1}{2} (60)\\A & = 30$

The area of the triangle is 30 square inches. Remember to use square units when measuring area.

This triangle is unique in that the numerical value for both the perimeter and the area are the same. Remember however, that the true values are different because the units are different.

III. Use the Converse of the Pythagorean Theorem to Identify Pythagorean Triples

In the previous lesson, you practiced using the converse of the Pythagorean Theorem. That is, if three numbers make the Pythagorean Theorem true, they can represent the three sides of a right triangle. We can use the converse to determine Pythagorean triples too.

An important time-saving technique is to identify and remember as many Pythagorean triples as you can. You learned in the previous lesson that 3:4:5 is a Pythagorean triple, and that any multiple would also be a Pythagorean triple. If you multiply each number by 2, you get 6:8:10. If you multiply each number by 3, you get 9:12:15, and so on. Each of these combinations of numbers makes the Pythagorean Theorem true, and if you can commit them to memory, it will save you a great deal of calculation time when solving problems.

Example

Is 5:12:13 a Pythagorean triple?

These numbers may look familiar, but you may not know off the top of your head whether or not they comprise a Pythagorean triple. Use the theorem to evaluate whether they are correct.

$a^2 + b^2 = c^2$, where $a = 5, b = 12,$ and $c = 13$

$5^2 + 12^2 & = 13^2\\25 + 144 & = 169\\169 & = 169$

So, 5:12:13 is a true Pythagorean triple.

How can we find more triples?

Now that you know the ratio of all Pythagorean triples originate at 3:4:5, if you multiply each value in the triple by the same number, you will find more triples. For example, if you multiply each number by 2, you get the triple 10:24:26. If you multiply each number by 3, you get 15:36:39. Each of these sets of numbers are Pythagorean triples and can be used to solve problems of many different types.

Let’s look at using the Pythagorean Theorem when problem solving.

IV. Solve Real – World Problems Involving Applications of the Pythagorean Theorem

Pythagorean Theorem concepts will show up in real-world problems just like any other mathematical concept. Remember to solve problems carefully. Look for Pythagorean triples that can save you time in calculations.

Example

Bethany’s mom made her a sandwich with a length of 8 cm and a height of 6 cm. If she cuts it in half diagonally as shown, what is the length of the cut edge?

This question is asking you to find the hypotenuse of a triangle with legs 6 cm and 8 cm. You may recognize those numbers as belonging to the Pythagorean triple 6:8:10, so the answer is probably 10 cm. You can check your answer using the Pythagorean Theorem.

$a^2 + b^2 = c^2$, where $a = 6$ and $b = 8$

$6^2 + 8^2 & = c^2\\36 + 64 & = c^2\\100 & = c^2\\\sqrt{100} & = \sqrt{c^2}\\10 & = c$

The length of the cut side of the sandwich is 10 cm.

Example

Lena had a piece of poster board that measured 12 inches by 16 inches. She cuts the poster board in half diagonally and wants to know the perimeter of one piece. What is the perimeter of half of Lena’s board?

To find the perimeter, you must find the length of the missing side (the hypotenuse). You may recognize 12 and 16 as being multiples of 3 and 4 (greater by a factor of 4) and could conclude that the hypotenuse will be four times five, or 20. However, you can solve this using the Pythagorean Theorem even if you don’t notice the Pythagorean triple.

$a^2 + b^2 = c^2$, where $a = 12$ and $b = 16$

$12^2 + 16^2 & = c^2\\144 + 256 & = c^2\\400 & = c^2\\\sqrt{400} = & \sqrt{c^2}\\20 & = c$

The missing side is 20 inches.

Don’t stop here, though! You have to find the perimeter of the triangle, so add the three sides together.

$P & = leg + leg + hypotenuse\\P & = 12 + 16 + 20\\P & = 48$

The perimeter of half of Lena’s poster board is 48 inches.

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

## Real-Life Example Completed

Here is the problem from the introduction. Reread it and then solve for the heights of each ladder.

Now that the students in Mrs. Richardson’s class understand that they will need to use the Pythagorean Theorem, they are left with the question of which ladder to use. Mrs. Richardson has told them that the height of the shed is 23 feet.

They have the choice of two different ladders. One ladder is 20 feet long and one ladder is 25 feet long.

“If we choose the 20 foot ladder and it is about 4 feet from the shed, how high up will it go?” Aran asked taking out a piece of paper and a pencil.

“I don’t know. We also have to think about the 25 foot ladder. If it is also 4 feet from the shed, how high up will it reach?” Amy said.

“We need the Pythagorean Theorem for this one,” Aran said.

Both students took out a piece of paper and a pencil and began to work.

Solution to Real – Life Example

First, let’s start with the ladder that is 20 feet long. We use the ladder length as the $c$ in the Pythagorean Theorem. It is the long side of the triangle. The height that the ladder will reach on the shed is what we are looking to solve, we will call that $a$ and that is our unknown value. We know that the ladder is four feet from the shed, and so this is our $b$ value.

$a^2+b^2&=c^2 \\a^2+4^2 & =20^2 \\a^2+16&=400 \\a^2&=400-16=384 \\a&=19.59 \ feet$

Now we can look at the ladder that is 25 feet long.

$a^2+b^2&=c^2 \\a^2+4^2&=25^2 \\a^2+16&=625 \\a^2&=625-16 \\a&=24.5 \ feet$

If the shed is 23 feet high, then the students should use the ladder that is 25 feet long so that they can paint all the way to the top of the shed.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Indirect Measurement
using geometric properties to figure out distances and lengths that would otherwise be challenging to measure.
Perimeter
the distance around a figure.
Area
the measure of the space inside a figure.
Pythagorean Theorem
the formula for figuring out the side lengths of a right triangle - $a^2+b^2=c^2$
Pythagorean triple
different forms of the ratio 3:4:5 which represent the side lengths of a right triangle.

## Time to Practice

Directions: Identify whether or not each set of measurements indicates a Pythagorean Triple.

1. 3, 4, 5
2. 6, 8, 12
3. 6, 8, 10
4. 15, 20, 25
5. 5, 9, 14
6. 9, 12, 15
7. 18, 24, 30
8. 1.5, 2, 4
9. 1.5, 2, 2.5
10. 21, 28, 35

Directions: Find the missing side length of each right triangle by using the Pythagorean Theorem. You may round to the nearest tenth when necessary.

1. $a = 6, b = 10, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
2. $a = 5, b = 7, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
3. $a = 7, b = 9, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
4. $a = 6, b = 8, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
5. $a = 9, b = 12, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
6. $a = 10, b = 14, c = \underline{\;\;\;\;\;\;\;\;\;\;}$
7. $a = 6, b = \underline{\;\;\;\;\;\;\;\;\;\;}, c = 10$
8. $a = \underline{\;\;\;\;\;\;\;\;\;\;}, b = 12, c = 15$
9. $a = 15, b = \underline{\;\;\;\;\;\;\;\;\;\;}, c = 25$
10. $a = \underline{\;\;\;\;\;\;\;\;\;\;}, b = 32, c = 40$

1. What is the area of a right triangle that has one leg of 6 yards and a hypotenuse of 10 yards?
2. A television is measured by the length of the diagonal from one corner to another. If the screen is 8 inches by 15 inches, what is the length of the diagonal?
3. Do the numbers 15, 20 and 25 comprise a Pythagorean triple?
4. What is the perimeter of a right triangle with a hypotenuse of 30 inches and a leg of 18 inches?

Jan 14, 2013

Jan 13, 2015