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# 1.1: The Pythagorean Theorem

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## Introduction

Right triangles play an integral part in the study of trigonometry. It is from right triangles that the basic definitions of the trigonometric functions are formed. In this chapter we will explore right triangles and their properties. Through this, we will introduce the six basic trig functions and the unit circle.

## Learning Objectives

• Recognize and use the Pythagorean Theorem.
• Recognize basic Pythagorean Triples.
• Use the Distance Formula.

## The Pythagorean Theorem

From Geometry, recall that the Pythagorean Theorem is $a^2 + b^2 = c^2$ where $a$ and $b$ are the legs of a right triangle and $c$ is the hypotenuse. Also, the side opposite the angle is lower case and the angle is upper case. For example, angle $A$ is opposite side $a$.

The Pythagorean Theorem is used to solve for the sides of a right triangle.

Example 1: Use the Pythagorean Theorem to find the missing side.

Solution: $a = 8, \ b = 15$, we need to find the hypotenuse.

$8^2 + 15^2 & = c^2\\64 + 225 & = c^2\\289 & = c^2\\17 & = c$

Notice, we do not include -17 as a solution because a negative number cannot be a side of a triangle.

Example 2: Use the Pythagorean Theorem to find the missing side.

Solution: Use the Pythagorean Theorem to find the missing leg.

$\left ( 5\sqrt{7} \right )^2 + x^2 & = \left ( 5\sqrt{13} \right )^2\\25 \cdot 7 + x^2 & = 25 \cdot 13\\175 + x^2 & = 325\\x^2 & = 150\\x & = 5\sqrt{6}$

## Pythagorean Triples

Pythagorean Triples are sets of whole numbers for which the Pythagorean Theorem holds true. The most well-known triple is 3, 4, 5. This means, that 3 and 4 are the lengths of the legs and 5 is the hypotenuse. The largest length is always the hypotenuse. If we were to multiply any triple by a constant, this new triple would still represent sides of a right triangle. Therefore, 6, 8, 10 and 15, 20, 25, among countless others, would represent sides of a right triangle.

Example 3: Determine if the following lengths are Pythagorean Triples.

a. 7, 24, 25

b. 9, 40, 41

c. 11, 56, 57

Solution: Plug each set of numbers into the Pythagorean Theorem.

a. $7^2 + 24^2 & \overset{\underset{?}{}}{=} 25^2\\49 + 576 & = 625\\625 & = 625$

Yes, 7, 24, 25 is a Pythagorean Triple and sides of a right triangle.

b. $9^2 + 40^2 & \overset{\underset{?}{}}{=} 41^2\\81 + 1600 & =1681\\1681 & =1681$

Yes, 9, 40, 41 is a Pythagorean Triple and sides of a right triangle.

c. $11^2 + 56^2 & \overset{\underset{?}{}}{=} 57^2\\121 + 3136 & = 3249\\3257 & \ne 3249$

No, 11, 56, 57 do not represent the sides of a right triangle.

## Converse of the Pythagorean Theorem

Using the technique from Example 3, we can determine if sets of numbers are acute, right or obtuse triangles. Examples 3a and 3b were both right triangles because the two sides equaled each other and made the Pythagorean Theorem true. However in Example 3c, the two sides were not equal. Because $3257 > 3249$, we can say that 11, 56, and 57 are the sides of an acute triangle. To help you visualize this, think of an equilateral triangle with sides of length $5$. We know that this is an acute triangle. If you plug in $5$ for each number in the Pythagorean Theorem we get $5^2 + 5^2 = 5^2$ and $50 > 25$. Therefore, if $a^2 + b^2 > c^2$, then lengths $a, \ b$, and $c$ make up an acute triangle. Conversely, if $a^2 + b^2 < c^2$, then lengths $a, \ b$, and $c$ make up the sides of an obtuse triangle. It is important to note that the length $c''$ is always the longest.

Example 4: Determine if the following lengths make an acute, right or obtuse triangle.

a. 5, 6, 7

b. 5, 10, 14

c. 12, 35, 37

Solution: Plug in each set of lengths into the Pythagorean Theorem.

a. $5^2 + 6^2 & \ ? \ 7^2\\25 + 36 & \ ? \ 49\\61 & > 49$

Because $61>49$, this is an acute triangle.

b. $5^2 + 10^2 & \ ? \ 14^2\\25+ 100 & \ ? \ 196\\125 & < 196$

Because $125<196$, this is an obtuse triangle.

c. $12^2 + 35^2 & \ ? \ 37^2\\144 + 1225 & \ ? \ 1369\\1369 & = 1369$

Because the two sides are equal, this is a right triangle.

NOTE: All of the lengths in Example 4 represent the lengths of the sides of a triangle. Recall the Triangle Inequality Theorem from geometry which states: The length of a side in a triangle is less than the sum of the other two sides. For example, 4, 7 and 13 cannot be the sides of a triangle because $4+7$ is not greater than 13.

## The Distance Formula

An application of the Pythagorean Theorem is to find the distance between two points. Consider the points (-1, 6) and (5, -3). If we plot them on a grid, they make a diagonal line. Draw a vertical line down from (-1, 6) and a horizontal line to the left of (5, -3) to make a right triangle.

Now we can find the distance between these two points by using the vertical and horizontal distances that we determined from the graph.

$9^2 + (-6)^2 & = d^2\\81 + 36 & = d^2\\117 & = d^2\\\sqrt{117} & = d\\3\sqrt{13} & = d$

Notice, that the $x-$values were subtracted from each other to find the horizontal distance and the $y-$values were subtracted from each other to find the vertical distance. If this process is generalized for two points $(x_1, \ y_1)$ and $(x_2, \ y_2)$, the Distance Formula is derived.

$(x_1-x_2)^2 + (y_1-y_2)^2 = d^2$

This is the Pythagorean Theorem with the vertical and horizontal differences between $(x_1, \ y_1)$ and $(x_2, \ y_2)$. Taking the square root of both sides will solve the right hand side for $d$, the distance.

$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = d$

This is the Distance Formula. The following example shows how to apply the distance formula.

Example 5: Find the difference between the two points.

a. (4, 2) and (-9, 5)

b. (-10, 3) and (0, -15)

Solution: Plug each pair of points into the distance formula.

a. $d & = \sqrt{(4-(-9))^2 + (2-5)^2}\\& = \sqrt{13^2 + (-3)^2}\\& = \sqrt{169+9}\\& = \sqrt{178}$

b. $d & = \sqrt{(-10-0)^2 + (3-(-15))^2}\\& = \sqrt{(-10)^2 + (18)^2}\\& = \sqrt{100+324}\\& = \sqrt{424} = 2\sqrt{106}$

## Points to Consider

• Does the Pythagorean Theorem apply to all real numbers?
• Can a Pythagorean Triple have irrational numbers in the set?
• What is the difference between the Distance Formula and the Pythagorean Theorem?

## Review Questions

Determine if the lengths below represent the sides of a right triangle. If not, state if the triangle is acute or obtuse.

1. 6, 9, 13
2. 9, 10, 11
3. 16, 30, 34
4. 20, 23, 40
5. 11, 16, 29
6. $2\sqrt{6}, \ 6\sqrt{3}, \ 2\sqrt{33}$

Find the missing side of each right triangle below. Leave the answer in simplest radical form.

1. The general formula for a Pythagorean Triple is $n^2-m^2, 2nm, n^2+m^2$ where $n$ and $m$ are natural numbers. Use the Pythagorean Theorem to prove this is true.
2. Find the distance between the pair of points.
1. (5, -6) and (18, 3)
2. $\left ( \sqrt{3}, \ -\sqrt{2} \right )$ and $\left ( -2\sqrt{3}, \ 5\sqrt{2} \right )$

1. $6^2 + 9^2 ? 13^2 \rightarrow 36 + 81 ? 169 \rightarrow 117 < 169$ The triangle is obtuse.
2. $9^2 + 10^2 ? 11^2 \rightarrow 81 + 100 ? 121 \rightarrow 181 > 121$ The triangle is acute.
3. $16^2 + 30^2 ? 34^2 \rightarrow 256 + 900 ? 1156 \rightarrow 1156 = 1156$ This is a right triangle.
4. $20^2 + 23^2 ? 40^2 \rightarrow 400 + 529 ? 1600 \rightarrow 929 < 1600$ The triangle is obtuse.
5. These lengths cannot make up the sides of a triangle. $11+16<29$
6. $\left ( 2\sqrt{6} \right )^2 + \left ( 6\sqrt{3} \right )^2 ? \left ( 2\sqrt{33} \right )^2 \rightarrow (4 \cdot 6) + (36 \cdot 3) ? (4 \cdot 33) \rightarrow 24 + 108 ? 132 \rightarrow 132 = 132$ This is a right triangle.
7. $7^2 + x^2 & = 18^2\\49 + x^2 & = 324\\x^2 & = 275\\x & = \sqrt{275} = 5 \sqrt{11}$
8. $5^2 + \left ( 5\sqrt{3} \right )^2 & = x^2\\25 + (25 \cdot 3) & = x^2\\25 + 75 & = x^2\\100 & = x^2\\10 & = x$
9. Both legs are 11. $11^2 + 11^2 & = x^2\\121 + 121 & = x^2\\242 & = x^2\\\sqrt{242} & = x\\11\sqrt{2} & = x$
10. Plug $n^2-m^2, 2nm, n^2+m^2$ into the Pythagorean Theorem. $(n^2-m^2)^2 + (2nm)^2 & = (n^2+m^2)^2\\n^4-2n^2m^2+m^4+4n^2m^2 & = n^4+2n^2m^2+m^4\\-2n^2m^2 + 4n^2m^2 & = 2n^2m^2\\4n^2m^2 & = 4n^2m^2$
11. (a) (5, -6) and (18, 3) $d & = \sqrt{(5-18)^2 + (-6-3)^2}\\& = \sqrt{(-13)^2 + (-9)^2}\\& = \sqrt{169 + 81}\\& = \sqrt{250}\\& = 5\sqrt{10}$ (b) $\left ( \sqrt{3}, \ -\sqrt{2} \right )$ and $\left ( -2\sqrt{3}, \ 5\sqrt{2} \right )$ $d & = \sqrt{ \left ( \sqrt{3}- \left ( -2\sqrt{3} \right ) \right )^2 + \left ( -\sqrt{2}-5\sqrt{2} \right )^2}\\& = \sqrt{\left ( 3\sqrt{3} \right )^2 + \left ( -6\sqrt{2} \right )^2}\\& = \sqrt{(9 \cdot 3) + (36 \cdot 2)}\\& = \sqrt{27 + 72}\\& = \sqrt{99} = 3\sqrt{11}$

Feb 23, 2012

Aug 21, 2014