Geometry (B): READ: Pythagorean Theorem Reading

The Pythagorean Theorem

Learning Objectives

Identify and employ the Pythagorean Theorem when working with right triangles.
Identify common Pythagorean triples.
Use the Pythagorean Theorem to find the area of isosceles triangles.
Use the Pythagorean Theorem to derive the distance formula on a coordinate grid.

Introduction

The triangle below is a right triangle.

The sides labeled $a$ and $b$ are called the legs of the triangle and they meet at the right angle. The third side, labeled $c$ is called the hypotenuse. The hypotenuse is opposite the right angle. The hypotenuse of a right triangle is also the longest side.

The Pythagorean Theorem states that the length of the hypotenuse squared will equal the sum of the squares of the lengths of the two legs. In the triangle above, the sum of the squares of the legs is $a^2 + b^2$ and the square of the hypotenuse is $c^2$ .

The Pythagorean Theorem: Given a right triangle with legs whose lengths are $a$ and $b$ and a hypotenuse of length $c$ ,

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

Be careful when using this theorem—you must make sure that the legs are labeled $a$ and $b$ and the hypotenuse is labeled $c$ to use this equation. A more accurate way to write the Pythagorean Theorem is:

$(\text{leg}_1)^2 + (\text{leg}_2)^2 = \text{hypotenuse}^2$

Example 1

Use the side lengths of the following triangle to test the Pythagorean Theorem.

The legs of the triangle above are $3\;\mathrm{inches}$ and $4\;\mathrm{inches}$ . The hypotenuse is $5\;\mathrm{inches}$ . So, $a = 3 , b = 4$ , and $c = 5$ . We can substitute these values into the formula for the Pythagorean Theorem to verify that the relationship works:

$a^2+b^2 &= c^2\\ 3^2+4^2 &= 5^2\\ 9+16 &= 25\\ 25 &=25$

Since both sides of the equation equal $25$ , the equation is true. Therefore, the Pythagorean Theorem worked on this right triangle.

Proof of the Pythagorean Theorem

There are many ways to prove the Pythagorean Theorem. One of the most straightforward ways is to use similar triangles. Start with a right triangle and construct an altitude from the right angle to the opposite sides. In the figure below, we can see the following relationships:

Proof.

Given: $\triangle WXY$ as shown in the figure below
Prove: $a^{2}+b^{2}=c^{2}$

: First we start with a triangle similarity statement:

$\triangle{WXY} \sim \triangle{WZX} \sim \triangle{XZY}$

: Now, using similar triangles, we can set up the following proportion:
: These are all true by the $AA$ triangle similarity postulate.

$\frac{d}{a} &= \frac{a}{c} \\ a^2 &= dc$

: and

$\frac{e}{b} &= \frac{b}{c} \\ b^2 &= ec$

: Putting these equations together by using substitution,

$a^2 + b^2 = dc + ec$

: factoring the right hand side,

$a^2 + b^2 = c(d+e)$

: but notice $d + e = c$ , so this becomes

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

$a^2 + b^2 = c^2. \blacklozenge$

We have finished proving the Pythagorean Theorem. There are hundreds of other ways to prove the Pythagorean Theorem and one of those alternative proofs is in the exercises for this section.

Making Use of the Pythagorean Theorem

As you know from algebra, if you have one unknown variable in an equation, you can solve to find its value. Therefore, if you know the lengths of two out of three sides in a right triangle, you can use the Pythagorean Theorem to find the length of the missing side, whether it is a leg or a hypotenuse. Be careful to use inverse operations properly and avoid careless mistakes.

Example 2

What is the length of $b$ in the triangle below?

Use the Pythagorean Theorem to find the length of the missing leg, $b$ . Set up the equation $a^2+b^2=c^2$ , letting $a=6$ and $b=10$ . Be sure to simplify the exponents and roots carefully, remember to use inverse operations to solve the equation, and always keep both sides of the equation balanced.

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

In algebra you learned that $\sqrt{x^2}=\pm x$ because, for example, $(5)^2=(-5)^2=25$ . However, in this case (and in much of geometry), we are only interested in the positive solution to $b=\sqrt{64}$ because geometric lengths are positive. So, in example 2, we can disregard the solution $b=-8$ , and our final answer is $b=8\;\mathrm{inches}$ .

Example 3

Find the length of the missing side in the triangle below.

Use the Pythagorean Theorem to set up an equation and solve for the missing side. Let $a = 5$ and $b = 12$ .

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

So, the length of the missing side is $13\;\mathrm{centimeters}$ .

Using Pythagorean Triples

In example 1, the sides of the triangle were $3 , 4$ , and $5$ . This combination of numbers is referred to as a Pythagorean triple. A Pythagorean triple is three numbers that make the Pythagorean Theorem true and they are integers (whole numbers with no decimal or fraction part). Throughout this chapter, you will use other Pythagorean triples as well. For instance, the triangle in example 2 is proportional to the same ratio of $3:4:5$ . If you divide the lengths of the triangle in example 2 $(6, 8, 10)$ by two, you find the same proportion— $3:4:5$ . Whenever you find a Pythagorean triple, you can apply those ratios with greater factors as well. Finally, take note of the side lengths of the triangle in example 3— $5:12:13$ . This, too, is a Pythagorean triple. You can extrapolate that this ratio, multiplied by greater factors, will also yield numbers that satisfy the Pythagorean Theorem.

There are infinitely many Pythagorean triples, but a few of the most common ones and their multiples are:

Triple	$\times 2$	$\times 3$	$\times 4$
$3-4-5$	$6-8-10$	$9-12-15$	$12-16-20$
$5-12-13$	$10-24-26$	$15-36-39$	$20-48-52$
$7-24-25$	$14-48-50$	$21-72-75$	$28-96-100$
$8-15-17$	$16-30-34$	$24-45-51$	$32-60-68$

The Distance Formula

You have already learned that you can use the Pythagorean Theorem to understand different types of right triangles, find missing lengths, and identify Pythagorean triples. You can also apply the Pythagorean Theorem to a coordinate grid and learn how to use it to find distances between points.

Example 5

Look at the points on the grid below.

Find the length of the segment connecting $(1,5)$ and $(5,2)$ .

The question asks you to identify the length of the segment. Because the segment is not parallel to either axis, it is difficult to measure given the coordinate grid. However, it is possible to think of this segment as the hypotenuse of a right triangle. Draw a vertical line at $x = 1$ and a horizontal line at $y = 2$ and find the point of intersection. This point represents the third vertex in the right triangle.

You can easily count the lengths of the legs of this triangle on the grid. The vertical leg extends from $(1,2)$ to $(1,5)$ , so it is $|5-2|=|3|=3\;\mathrm{units}$ long. The horizontal leg extends from $(1,2)$ to $(5,2)$ , so it is $|5-1|=|4|=4\;\mathrm{units}$ long. Use the Pythagorean Theorem with these values for the lengths of each leg to find the length of the hypotenuse.

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

The segment connecting $(1,5)$ and $(5,2)$ is $5\;\mathrm{units}$ long.

Mathematicians have simplified this process and created a formula that uses these steps to find the distance between any two points in the coordinate plane. If you use the distance formula, you don’t have to draw the extra lines.

Distance Formula: Give points $(x_1,y_1)$ and $(x_2,y_2)$ , the length of the segment connecting those two points is $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Example 6

Use the distance formula $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find the distance between the points $(1,5)$ and $(5,2)$ on a coordinate grid.

You already know from example 1 that the distance will be $5\;\mathrm{units}$ , but you can practice using the distance formula to make sure it works. In this formula, substitute $1$ for $x_1 , 5$ for $y_1 , 5$ for $x_2$ , and $2$ for $y_2$ because $(1,5)$ and $(5,2)$ are the two points in question.

$D &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\ &= \sqrt{(5 - 1)^2 + (2 - 5)^2}\\ &= \sqrt{(4)^2 + (-3)^2}\\ &= \sqrt{16 + 9}\\ &= \sqrt{25}\\ &= 5$

Now you see that no matter which method you use to solve this problem, the distance between $(1,5)$ and $(5,2)$ on a coordinate grid is $5\;\mathrm{units}$ .

Lesson Summary

In this lesson, we explored how to work with different radical expressions both in theory and in practical situations. Specifically, we have learned:

How to identify and employ the Pythagorean Theorem when working with right triangles.
How to identify common Pythagorean triples.
How to use the Pythagorean Theorem to find the area of isosceles triangles.
How to use the Pythagorean Theorem to derive the distance formula on a coordinate grid.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to apply the Pythagorean Theorem to mathematical situations.

Points to Consider

Now that you have learned the Pythagorean Theorem, there are countless ways to apply it. Could you use the Pythagorean Theorem to prove that a triangle contained a right angle if you did not have an accurate diagram?

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

What is the distance between $(-5,-5)$ and $(-2,-1)$ ?
Do the numbers $12$ , $16$ , and $20$ make a Pythagorean triple?
What is the length of $p$ in the triangle below?
Do the numbers $13 , 26$ , and $35$ make a Pythagorean triple?
What is the distance between $(1,9)$ and $(9,4)$ ?
What is the length of $m$ in the triangle below?
What is the distance between $(-3,7)$ and $(6,5)$ ?
An alternative proof of the Pythagorean Theorem uses the area of a square. The diagram below shows a square with side lengths $a + b$ , and an inner square with side lengths $c$ . Use the diagram below to prove $a^2 + b^2 = c^2.$
Hint: Find the area of the inner square in two ways: once directly, and once by finding the area of the larger square and subtracting the area of each triangle.

Review Answers

$5$
yes
$17\;\mathrm{inches}$
no
$\sqrt{89}$
$15\;\mathrm{inches}$
$60\;\mathrm{square yards}$
Proof. The plan is, we will find the area of the green square in two ways. Since those two areas must be equal, we can set those areas equal to each other.
For the inner square (in green), we can directly compute the area: $\text{Area of inner square} = c^2$ .

Now, the area of the large, outer square is $(a+b)^2$ . Don’t forget to multiply this binomial carefully!

$\text{area} &= (a+b)^2\\ &=(a+b)(a+b)\\ &=a^2+2ab+b^2$

The area of each small right triangle (in yellow) is

$\text{area} = \frac{1}{2} ab$ .

Since there are four of those right triangles, we have the combined area

$4\left(\frac {1}{2}ab \right) = 2ab$

Finally, subtract the area of the four yellow triangles from the area of the larger square, and we are left with

$\text{large square}-\text{four triangles} & = \text{area of inner square}\\ a^2+ 2ab + b^2 -2ab & = a^2 + b^2$

Putting together the two different ways for finding the area of the inner square, we have $a^2 + b^2 = c^2.$

Last modified: Tuesday, June 29, 2010, 10:17 AM