# 11.4: The Pythagorean Theorem and Its Converse

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the Pythagorean Theorem.
• Use the converse of the Pythagorean Theorem.
• Solve real-world problems using the Pythagorean Theorem and its converse.

## Introduction

The Pythagorean Theorem is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the hypotenuse and the sides of the triangle adjacent to the 90 degree angle are called the legs.

If we let \begin{align*}a\end{align*} and \begin{align*}b\end{align*} represent the legs of the right triangle and \begin{align*}c\end{align*} represent the hypotenuse, then the Pythagorean Theorem can be stated as:

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

That is,

\begin{align*}(\text{leg}_1)^2 + (\text{leg}_2)^2=(\text{hypotenuse})^2\end{align*}

Or using the labels given in the triangle to the right

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

This theorem is very useful because if we know the lengths of the legs of a right triangle, we can find the length of the hypotenuse. Conversely, if we know the length of the hypotenuse and the length of a leg, we can calculate the length of the missing leg of the triangle. When you use the Pythagorean Theorem, it does not matter which leg you call a and which leg you call \begin{align*}b\end{align*}, but the hypotenuse is always called \begin{align*}c\end{align*}.

Although nowadays we use the Pythagorean Theorem as a statement about the relationship between distances and lengths, originally the theorem made a statement about areas. If we build squares on each side of a right triangle, the Pythagorean Theorem says that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares formed by the legs of the triangle.

## Use the Pythagorean Theorem and Its Converse

The Pythagorean Theorem can be used to verify that a triangle is a right triangle. If you can show that the three sides of a triangle make the equation \begin{align*}(\text{leg}_1)^2 + (\text{leg}_2)^2=(\text{hypotenuse})^2\end{align*} true, then you know that the triangle is a right triangle. This is called the Converse of the Pythagorean Theorem.

Note: When you use the Converse of the Pythagorean Theorem, you must make sure that you substitute the correct values for the legs and the hypotenuse. One way to check is that the hypotenuse must be the longest side. The other two sides are the legs and the order in which you use them is not important.

Example 1

Determine if a triangle with sides 5, 12 and 13 is a right triangle.

Solution

The triangle is right if its sides satisfy the Pythagorean Theorem.

First of all, the longest side would have to be the hypotenuse so we designate \begin{align*}c=13\end{align*}.

We then designate the shorter sides as \begin{align*}a=5\end{align*} and \begin{align*}b=12\end{align*}.

We plug these values into the Pythagorean Theorem.

\begin{align*}a^2 + b^2 &=c^2 \Rightarrow 5^2 + 12^2=c^2 \\ 25 + 144 &=169=c^2 \Rightarrow 169=169\end{align*}

The sides of the triangle satisfy the Pythagorean Theorem, thus the triangle is a right triangle.

Example 2

Determine if a triangle with sides \begin{align*} \sqrt{10}, \sqrt{15}\end{align*} and 5 is a right triangle.

Solution

We designate the hypotenuse \begin{align*}c=5\end{align*} because this is the longest side.

We designate the shorter sides as \begin{align*}a=\sqrt{10}\end{align*} and \begin{align*}b=\sqrt{15}\end{align*}.

We plug these values into the Pythagorean Theorem.

\begin{align*} a^2+b^2 &=c^2 \Rightarrow \left(\sqrt{10}\right)^2 + \left(\sqrt{15}\right)^2=c^2 \\ 10 + 15 &=25=(5)^2\end{align*}

The sides of the triangle satisfy the Pythagorean Theorem, thus the triangle is a right triangle.

Pythagorean Theorem can also be used to find the missing hypotenuse of a right triangle if we know the legs of the triangle.

Example 3

In a right triangle one leg has length 4 and the other has length 3. Find the length of the hypotenuse.

Solution

\begin{align*}\text{Start with the Pythagorean Theorem.}&&a^2 + b^2 &=c^2\\ \text{Plug in the known values of the legs. }&&3^2 + 4^2 &=c^2\\ \text{Simplify.}&& 9 + 16 &=c^2 \\ && 25 &=c^2 \\ \text{Take the square root of both sides. }&&c &=5\end{align*}

## Use the Pythagorean Theorem with Variables

Example 4

Determine the values of the missing sides. You may assume that each triangle is a right triangle.

a)

b)

c)

Solution

Apply the Pythagorean Theorem.

a)

\begin{align*}a^2 + b^2 &=c^2 \\ x^2 + 15^2 &=21^2 \\ x^2 + 225 &=441 \\ x^2 &=216 \Rightarrow \\ x &=\sqrt{216}=6\sqrt{6}\end{align*}

b)

\begin{align*} a^2+b^2 &=c^2 \\ y^2+3^2 &=7^2 \\ y^2+ 9 &=49 \\ y^2 &=40 \Rightarrow \\ y & =\sqrt{40}=2\sqrt{10}\end{align*}

c)

\begin{align*} a^2+b^2 & =c^2 \\ 18^2+15^2 & =z^2 \\ 324+225 & =z^2 \\ z^2 &=216 \Rightarrow \\ x & =\sqrt{549}=3 \sqrt{61}\end{align*}

Example 5

One leg of a right triangle is 5 more than the other leg. The hypotenuse is one more than twice the size of the short leg. Find the dimensions of the triangle.

Solution

Let \begin{align*}x =\end{align*} length of the short leg.

Then, \begin{align*}x + 5 =\end{align*} length of the long leg

And, \begin{align*}2x + 1 =\end{align*} length of the hypotenuse.

The sides of the triangle must satisfy the Pythagorean Theorem,

\begin{align*}\text{Therefore }&&x^2 + (x + 5)^2 &=(2x+ 1)\\ \text{Eliminate the parenthesis. }&&x^2 + x^2 + 10x + 25 &=4x^2 + 4x + 1\\ \text{Move all terms to the right hand side of the equation.}&&0 &=2x^2 - 6x - 24\\ \text{Divide all terms by}\ 2.&&0 &=x^2 - 3x - 12\\ \text{Solve using the quadratic formula. }&& x & =\frac{3\pm \sqrt{9+48}}{2}=\frac{3\pm \sqrt 57}{2}\\ && x & \approx 5.27\ \text{or}\ x \approx -2.27\end{align*}

We can discard the negative solution since it does not make sense in the geometric context of this problem. Hence, we use \begin{align*}x=5.27\end{align*} and we get \begin{align*}\text{short - leg}=5.27, \text{long - leg}=10.27\end{align*} and \begin{align*}\text{hypotenuse}=11.54.\end{align*}

## Solve Real-World Problems Using the Pythagorean Theorem and Its Converse

The Pythagorean Theorem and its converse have many applications for finding lengths and distances.

Example 6

Maria has a rectangular cookie sheet that measures \begin{align*}10 \ inches \times 14 \ inches\end{align*}. Find the length of the diagonal of the cookie sheet.

Solution

1. Draw a sketch.

2. Define variables.

Let \begin{align*}c =\end{align*} length of the diagonal.

3. Write a formula. Use the Pythagorean Theorem

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

4. Solve the equation.

\begin{align*}10^2 + 14^2 &=c^2 \\ 100 + 196 &=c^2 \\ c^2 &=296\Rightarrow c=\sqrt{296}\Rightarrow c=4\sqrt{74}\ \text{or}\ c \approx 17.2\ inches\end{align*}

5. Check

\begin{align*}10^2 + 14^2=100 + 196\end{align*} and \begin{align*}c^2=17.2^2 \approx 296. \end{align*}

The solution checks out.

Example 7

Find the area of the shaded region in the following diagram.

Solution:

1. Diagram

Draw the diagonal of the square on the figure.

Notice that the diagonal of the square is also the diameter of the circle.

2. Define variables

Let \begin{align*}c =\end{align*} diameter of the circle.

3. Write the formula

Use the Pythagorean Theorem: \begin{align*}a^2 +b^2=c^2\end{align*}

4. Solve the equation:

\begin{align*}2^2 + 2^2 &=c^2 \\ 4 + 4 &=c^2 \\ c^2 &=8\Rightarrow c=\sqrt{8}\Rightarrow c=2\sqrt{2}\end{align*}

The diameter of the circle is \begin{align*} 2\sqrt{2}\end{align*}. Therefore, the radius is \begin{align*} r=\sqrt{2}\end{align*}.

Area of a circle is \begin{align*} A=\pi r^2 =\pi \left(\sqrt{2}\right)^2=2\pi\end{align*}.

Area of the shaded region is therefore \begin{align*}2 \pi - 4 \approx 2.28\end{align*}

Example 8

In a right triangle, one leg is twice as long as the other and the perimeter is 28. What are the measures of the sides of the triangle?

Solution

1. Make a sketch. Let’s draw a right triangle.

2. Define variables.

Let: \begin{align*}a =\end{align*} length of the short leg

\begin{align*}2a =\end{align*} length of the long leg

\begin{align*}c =\end{align*} length of the hypotenuse

3. Write formulas.

The sides of the triangle are related in two different ways.

1. The perimeter is 28, \begin{align*}a+2a+c=28\Rightarrow 3a+c=28 \end{align*}

2. This a right triangle, so the measures of the sides must satisfy the Pythagorean Theorem.

\begin{align*} a^2+(2a)^2 &=c^2\Rightarrow a^2+4a^2=c^2\Rightarrow 5a^2=c^2\\ & \text{or}\\ c &=a\sqrt{5}\approx 2.236a\end{align*}

4. Solve the equation

Use the value of \begin{align*}c\end{align*} we just obtained to plug into the perimeter equation \begin{align*}3a + c= 28\end{align*}.

\begin{align*} 3a+2.236a=28\Rightarrow 5.236a=28\Rightarrow a=5.35 \end{align*}

The short leg is:s \begin{align*}a \approx 5.35.\end{align*}

The long leg is: \begin{align*}2a \approx 10.70.\end{align*}

The hypotenuse is: \begin{align*}c \approx 11.95.\end{align*}

5. Check The legs of the triangle should satisfy the Pythagorean Theorem

\begin{align*}a^2 + b^2=5.35^2 + 10.70^2=143.1, c^2=11.95^2=142.80 \end{align*}

The results are approximately the same.

The perimeter of the triangle should be 28.

\begin{align*}a + b + c=5.35 + 10.70 + 11.95=28 \end{align*}

Example 9

Mike is loading a moving van by walking up a ramp. The ramp is 10 feet long and the bed of the van is 2.5 feet above the ground. How far does the ramp extend past the back of the van?

Solution

1. Make a sketch.

2. Define Variables.

Let \begin{align*}x =\end{align*} how far the ramp extends past the back of the van.

3. Write a formula. Use the Pythagorean Theorem:

\begin{align*}x^2 + 2.5^2=10^2\end{align*}

4. Solve the equation.

\begin{align*}x^2 + 6.25 &=100\\ x^2 &=93.5\\ x &=\sqrt{93.5}\approx 9.7 \ ft\end{align*}

5. Check. Plug the result in the Pythagorean Theorem.

\begin{align*}9.7^2+2.5^2=94.09+6.25=100.36 \approx 100.\end{align*}

The ramp is 10 feet long.

## Review Questions

Verify that each triangle is a right triangle.

1. \begin{align*}a=12, b=9, c=15\end{align*}
2. \begin{align*}a=6, b=6, c=6 \sqrt{2}\end{align*}
3. \begin{align*}a=8, b=8 \sqrt{3}, c=16\end{align*}

Find the missing length of each right triangle.

1. \begin{align*}a=12, b=16, c =?\end{align*}
2. \begin{align*}a =?, b=20, c=30\end{align*}
3. \begin{align*}a=4, b =?, c=11\end{align*}
4. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle.
5. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle.
6. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate?
7. Emanuel has a cardboard box that measures \begin{align*}20 \ cm \times 10 \ cm \times 8 \ cm \ (\text{length} \times \text{width} \times \text{height})\end{align*}. What is the length of the diagonal from a bottom corner to the opposite top corner?
8. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house?
9. Find the area of the triangle if area of a triangle is defined as \begin{align*}A=\frac{1}{2} \text{base} \times \text{height}\end{align*}.
10. Instead of walking along the two sides of a rectangular field, Mario decided to cut across the diagonal. He saves a distance that is half of the long side of the field. Find the length of the long side of the field given that the short side is 123 feet.
11. Marcus sails due north and Sandra sails due east from the same starting point. In two hours, Marcus’ boat is 35 miles from the starting point and Sandra’s boat is 28 miles from the starting point. How far are the boats from each other?
12. Determine the area of the circle.

1. \begin{align*}12^2 + 9^2 =225\!\\ 15^2 =225\end{align*}
2. \begin{align*}6^2 + 6^2 =72\!\\ \left(6\sqrt{2}\right)^2 =72\end{align*}
3. \begin{align*}8^2+\left(8\sqrt{3}\right)^2 =256\!\\ 16^2 =256\end{align*}
4. \begin{align*}c=20\end{align*}
5. \begin{align*}a=10\sqrt{5}\end{align*}
6. \begin{align*}b=\sqrt{105}\end{align*}
7. \begin{align*}c=\sqrt{130}\end{align*}
8. \begin{align*}a=28\end{align*}
9. \begin{align*}b=12\sqrt{3}\end{align*}
10. 12, 16, 20
11. 3.62, 10.24, 10.86
12. 127.3 ft
13. 23.75 cm
14. 8 feet
15. 32.24
16. 164 feet
17. 44.82 miles
18. 83.25

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