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# 8.1: The Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Prove and use the Pythagorean Theorem.
• 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.

## Review Queue

1. Draw a right scalene triangle.
2. Draw an isosceles right triangle.
1. \begin{align*}\sqrt{50}\end{align*}
2. \begin{align*}\sqrt{27}\end{align*}
3. \begin{align*}\sqrt{272}\end{align*}
4. Perform the indicated operations on the following numbers. Simplify all radicals.
1. \begin{align*}2 \sqrt{10} + \sqrt{160}\end{align*}
2. \begin{align*}5 \sqrt{6} \cdot 4 \sqrt{18}\end{align*}
3. \begin{align*}\sqrt{8} \cdot 12 \sqrt{2}\end{align*}

Know What? All televisions dimensions refer to the diagonal of the rectangular viewing area. Therefore, for a 52” TV, 52” is the length of the diagonal. High Definition Televisions (HDTVs) have sides in the ratio of 16:9. What is the length and width of a 52” HDTV? What is the length and width of an HDTV with a \begin{align*}y”\end{align*} long diagonal?

## The Pythagorean Theorem

We have used the Pythagorean Theorem already in this text, but we have never proved it. Recall that the sides of a right triangle are called legs (the sides of the right angle) and the side opposite the right angle is the hypotenuse. For the Pythagorean Theorem, the legs are “\begin{align*}a\end{align*}” and “\begin{align*}b\end{align*}” and the hypotenuse is “\begin{align*}c\end{align*}”.

Pythagorean Theorem: Given a right triangle with legs of lengths \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and a hypotenuse of length \begin{align*}c\end{align*}, then \begin{align*}a^2 + b^2 = c^2\end{align*}.

There are several proofs of the Pythagorean Theorem. We will provide one proof within the text and two others in the review exercises.

Investigation 8-1: Proof of the Pythagorean Theorem

Tools Needed: pencil, 2 pieces of graph paper, ruler, scissors, colored pencils (optional)

1. On the graph paper, draw a 3 in. square, a 4 in. square, a 5 in square and a right triangle with legs of 3 and 4 inches.
2. Cut out the triangle and square and arrange them like the picture on the right.
3. This theorem relies on area. Recall from a previous math class, that the area of a square is length times width. But, because the sides are the same you can rewrite this formula as \begin{align*}A_{square} = length \times width = side \times side = side^2\end{align*}. So, the Pythagorean Theorem can be interpreted as \begin{align*}(square \ with \ side \ a)^2 + (square \ with \ side \ b)^2 = (square \ with \ side \ c)^2\end{align*}. In this Investigation, the sides are 3, 4 and 5 inches. What is the area of each square?
4. Now, we know that \begin{align*}9 + 16 = 25\end{align*}, or \begin{align*}3^2 + 4^2 = 5^2\end{align*}. Cut the smaller squares to fit into the larger square, thus proving the areas are equal.

## Another Proof of the Pythagorean Theorem

This proof is “more formal,” meaning that we will use letters, \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} to represent the sides of the right triangle. In this particular proof, we will take four right triangles, with legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and hypotenuse \begin{align*}c\end{align*} and make the areas equal.

For two animated proofs, go to http://www.mathsisfun.com/pythagoras.html and scroll down to “And You Can Prove the Theorem Yourself.”

## Using the Pythagorean Theorem

The Pythagorean Theorem can be used to find a missing side of any right triangle, to prove that three given lengths can form a right triangle, to find Pythagorean Triples, to derive the Distance Formula, and to find the area of an isosceles triangle. Here are several examples. Simplify all radicals.

Example 1: Do 6, 7, and 8 make the sides of a right triangle?

Solution: Plug in the three numbers into the Pythagorean Theorem. The largest length will always be the hypotenuse. \begin{align*}6^2 + 7^2 = 36 + 49 = 85 \neq 8^2\end{align*}. Therefore, these lengths do not make up the sides of a right triangle.

Example 2: Find the length of the hypotenuse of the triangle below.

Solution: Let’s use the Pythagorean Theorem. Set \begin{align*}a\end{align*} and \begin{align*}b\end{align*} equal to 8 and 15 and solve for \begin{align*}c\end{align*}, the hypotenuse.

\begin{align*}8^2 + 15^2 & = c^2\\ 64 + 225 & = c^2\\ 289 & = c^2 \qquad \quad Take \ the \ square \ root \ of \ both \ sides.\\ 17 & = c\end{align*}

When you take the square root of an equation, usually the answer is +17 or -17. Because we are looking for length, we only use the positive answer. Length is never negative.

Example 3: Find the missing side of the right triangle below.

Solution: Here, we are given the hypotenuse and a leg. Let’s solve for \begin{align*}b\end{align*}.

\begin{align*}7^2 + b^2 & = 14^2\\ 49 + b^2 & = 196\\ b^2 & = 147\\ b & = \sqrt{147} = \sqrt{7 \cdot 7 \cdot 3} =7 \sqrt{3}\end{align*}

Example 4: What is the diagonal of a rectangle with sides 10 and \begin{align*}16 \sqrt{5}\end{align*}?

Solution: For any square and rectangle, you can use the Pythagorean Theorem to find the length of a diagonal. Plug in the sides to find \begin{align*}d\end{align*}.

\begin{align*}10^2 + \left ( 16 \sqrt{5} \right )^2 & = d^2\\ 100 + 1280 & = d^2\\ 1380 & = d^2\\ d & = \sqrt{1380} = 2 \sqrt{345}\end{align*}

## Pythagorean Triples

In Example 2, the sides of the triangle were 8, 15, and 17. This combination of numbers is referred to as a Pythagorean triple.

Pythagorean Triple: A set of three whole numbers that makes the Pythagorean Theorem true.

The most frequently used Pythagorean triple is 3, 4, 5, as in Investigation 8-1. Any multiple of a Pythagorean triple is also considered a triple because it would still be three whole numbers. Therefore, 6, 8, 10 and 9, 12, 15 are also sides of a right triangle. Other Pythagorean triples are:

\begin{align*}3, 4, 5 \qquad 5, 12, 13 \qquad 7, 24, 25 \qquad 8, 15, 17\end{align*}

There are infinitely many Pythagorean triples. To see if a set of numbers makes a triple, plug them into the Pythagorean Theorem.

Example 5: Is 20, 21, 29 a Pythagorean triple?

Solution: If \begin{align*}20^2 + 21^2\end{align*} is equal to \begin{align*}29^2\end{align*}, then the set is a triple.

\begin{align*}20^2 + 21^2 & = 400 + 441 = 841\\ 29^2 & = 841\end{align*}

Therefore, 20, 21, and 29 is a Pythagorean triple.

## Area of an Isosceles Triangle

There are many different applications of the Pythagorean Theorem. One way to use The Pythagorean Theorem is to identify the heights in isosceles triangles so you can calculate the area. The area of a triangle is \begin{align*}\frac{1}{2} \ bh\end{align*}, where \begin{align*}b\end{align*} is the base and \begin{align*}h\end{align*} is the height (or altitude).

If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height.

Example 6: What is the area of the isosceles triangle?

Solution: First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, find the length of the altitude using the Pythagorean Theorem.

\begin{align*}7^2 + h^2 & = 9^2\\ 49 + h^2 & = 81\\ h^2 & = 32\\ h & = \sqrt{32} = 4 \sqrt{2}\end{align*}

Now, use \begin{align*}h\end{align*} and \begin{align*}b\end{align*} in the formula for the area of a triangle.

\begin{align*}A = \frac{1}{2} \ bh = \frac{1}{2} (14) \left (4 \sqrt{2} \right ) = 28 \sqrt{2} \ units^2\end{align*}

## The Distance Formula

Another application of the Pythagorean Theorem is the Distance Formula. We have already been using the Distance Formula in this text, but we can prove it here.

First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to find these distances.

Now that we have a right triangle, we can use the Pythagorean Theorem to find \begin{align*}d\end{align*}.

Distance Formula: The distance \begin{align*}A(x_1, y_1)\end{align*} and \begin{align*}B(x_2, y_2)\end{align*} is \begin{align*}d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\end{align*} .

Example 7: Find the distance between (1, 5) and (5, 2).

Solution: Make \begin{align*}A(1, 5)\end{align*} and \begin{align*}B(5, 2)\end{align*}. Plug into the distance formula.

\begin{align*}d & = \sqrt{(1 - 5)^2 + (5 - 2)^2}\\ & = \sqrt{(-4)^2 + (3)^2}\\ & = \sqrt{16 + 9 } = \sqrt{25} = 5\end{align*}

You might recall that the distance formula was presented as \begin{align*}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\end{align*}, with the first and second points switched. It does not matter which point is first as long as \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are both first in each parenthesis. In Example 7, we could have switched \begin{align*}A\end{align*} and \begin{align*}B\end{align*} and would still get the same answer.

\begin{align*}d & = \sqrt{(5 - 1)^2 + (2 - 5)^2}\\ & = \sqrt{(4)^2 + (-3)^2}\\ & = \sqrt{16 + 9} = \sqrt{25} = 5\end{align*}

Also, just like the lengths of the sides of a triangle, distances are always positive.

Know What? Revisited To find the length and width of a 52” HDTV, plug in the ratios and 52 into the Pythagorean Theorem. We know that the sides are going to be a multiple of 16 and 9, which we will call \begin{align*}n\end{align*}.

\begin{align*}(16n)^2 + (9n)^2 & = 52^2\\ 256n^2 + 81n^2 & = 2704\\ 337n^2 & = 2704\\ n^2 & = 8.024\\ n & = 2.83\end{align*}

Therefore, the dimensions of the TV are \begin{align*}16(2.83”)\end{align*} by \begin{align*}9(2.833”)\end{align*}, or \begin{align*}45.3”\end{align*} by \begin{align*}25.5”\end{align*}. If the diagonal is \begin{align*}y”\end{align*} long, it would be \begin{align*}n \sqrt{337}”\end{align*} long. The extended ratio is \begin{align*}9 : 16 : \sqrt{337}\end{align*}.

## Review Questions

Find the length of the missing side. Simplify all radicals.

1. If the legs of a right triangle are 10 and 24, then the hypotenuse is _____________.
2. If the sides of a rectangle are 12 and 15, then the diagonal is _____________.
3. If the legs of a right triangle are \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, then the hypotenuse is ____________.
4. If the sides of a square are 9, then the diagonal is _____________.

Determine if the following sets of numbers are Pythagorean Triples.

1. 12, 35, 37
2. 9, 17, 18
3. 10, 15, 21
4. 11, 60, 61
5. 15, 20, 25
6. 18, 73, 75

Find the area of each triangle below. Simplify all radicals.

Find the length between each pair of points.

1. (-1, 6) and (7, 2)
2. (10, -3) and (-12, -6)
3. (1, 3) and (-8, 16)
4. What are the length and width of a \begin{align*}42”\end{align*} HDTV? Round your answer to the nearest tenth.
5. Standard definition TVs have a length and width ratio of 4:3. What are the length and width of a \begin{align*}42”\end{align*} Standard definition TV? Round your answer to the nearest tenth.
6. Challenge An equilateral triangle is an isosceles triangle. If all the sides of an equilateral triangle are \begin{align*}s\end{align*}, find the area, using the technique learned in this section. Leave your answer in simplest radical form.
7. Find the area of an equilateral triangle with sides of length 8.

Pythagorean Theorem Proofs

The first proof below is similar to the one done earlier in this lesson. Use the picture below to answer the following questions.

1. Find the area of the square with sides \begin{align*}(a + b)\end{align*}.
2. Find the sum of the areas of the square with sides \begin{align*}c\end{align*} and the right triangles with legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.
3. The areas found in the previous two problems should be the same value. Set the expressions equal to each other and simplify to get the Pythagorean Theorem.

Major General James A. Garfield (and former President of the U.S) is credited with deriving this next proof of the Pythagorean Theorem using a trapezoid.

1. Find the area of the trapezoid using the trapezoid area formula: \begin{align*}A = \frac{1}{2} (b_1 + b_2)h\end{align*}
2. Find the sum of the areas of the three right triangles in the diagram.
3. The areas found in the previous two problems should be the same value. Set the expressions equal to each other and simplify to get the Pythagorean Theorem.

1. \begin{align*}\sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2}\end{align*}
2. \begin{align*}\sqrt{27} = \sqrt{9 \cdot 3} = 3 \sqrt{3}\end{align*}
3. \begin{align*}\sqrt{272} = \sqrt{16 \cdot 17} = 4 \sqrt{17}\end{align*}
1. \begin{align*}2 \sqrt{10} + \sqrt{160} = 2 \sqrt{10} + 4 \sqrt{10} = 4 \sqrt{10}\end{align*}
2. \begin{align*}5 \sqrt{6} \cdot 4 \sqrt{18} = 5 \sqrt{6} \cdot 12 \sqrt{2} = 60 \sqrt{12} = 120 \sqrt{3}\end{align*}
3. \begin{align*}\sqrt{8} \cdot 12 \sqrt{2} = 12 \sqrt{16} = 12 \cdot 4 = 48\end{align*}

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