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8.1: The Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Review simplifying and reducing radicals.
• Prove and use the Pythagorean Theorem.
• Use the Pythagorean Theorem to derive the distance formula.

Review Queue

1. Draw a right scalene triangle.
2. Draw an isosceles right triangle.
3. List all the factors of 75.
4. Write the prime factorization of 75.

Know What? For a 52” TV, 52” is the length of the diagonal. High Definition Televisions (HDTVs) have sides in a ratio of 16:9. What are the length and width of a 52” HDTV?

In algebra, you learned how to simplify radicals. Let’s review it here.

a) 50\begin{align*}\sqrt{50}\end{align*}

b) 27\begin{align*}\sqrt{27}\end{align*}

c) 272\begin{align*}\sqrt{272}\end{align*}

Solution: For each radical, find the square number(s) that are factors.

a) 50=252=52\begin{align*}\sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2}\end{align*}

b) 27=93=33\begin{align*}\sqrt{27} = \sqrt{9 \cdot 3} = 3 \sqrt{3}\end{align*}

c) 272=1617=417\begin{align*}\sqrt{272} = \sqrt{16 \cdot 17} = 4 \sqrt{17}\end{align*}

When adding radicals, you can only combine radicals with the same number underneath it. For example, 25+36\begin{align*}2 \sqrt{5} + 3 \sqrt{6}\end{align*} cannot be combined, because 5 and 6 are not the same number.

a) 210+160\begin{align*}2 \sqrt{10} + \sqrt{160}\end{align*}

b) 56418\begin{align*}5 \sqrt{6} \cdot 4 \sqrt{18}\end{align*}

c) 8122\begin{align*}\sqrt{8} \cdot 12 \sqrt{2}\end{align*}

d) (52)2\begin{align*}\left( 5 \sqrt{2} \right)^2\end{align*}

Solution:

a) Simplify 160\begin{align*}\sqrt{160}\end{align*} before adding: 210+160=210+1610=210+410=610\begin{align*}2 \sqrt{10} + \sqrt{160} = 2 \sqrt{10} + \sqrt{16 \cdot 10} = 2 \sqrt{10} + 4 \sqrt{10} = 6 \sqrt{10}\end{align*}

b) To multiply two radicals, multiply what is under the radicals and what is in front.

56418=54618=20108=20363=2063=1203\begin{align*}5 \sqrt{6} \cdot 4 \sqrt{18} = 5 \cdot 4 \sqrt{6 \cdot 18} = 20 \sqrt{108} = 20 \sqrt{36 \cdot 3} = 20 \cdot 6 \sqrt{3} = 120 \sqrt{3}\end{align*}

c) 8122=1282=1216=124=48\begin{align*}\sqrt{8} \cdot 12 \sqrt{2} = 12 \sqrt{8 \cdot 2} = 12 \sqrt{16} = 12 \cdot 4=48\end{align*}

d) (52)2=52(2)2=252=50\begin{align*}\left( 5 \sqrt{2} \right )^2 = 5^2 \left( \sqrt{2} \right )^2 = 25 \cdot 2 = 50 \rightarrow\end{align*} the \begin{align*}\sqrt{}\end{align*} and the 2\begin{align*}^2\end{align*} cancel each other out

Lastly, to divide radicals, you need to simplify the denominator, which means multiplying the top and bottom of the fraction by the radical in the denominator.

Example 3: Divide and simplify the radicals.

a) 46÷3\begin{align*}4 \sqrt{6} \div \sqrt{3}\end{align*}

b) 308\begin{align*}\frac{\sqrt{30}}{\sqrt{8}}\end{align*}

c) 8267\begin{align*}\frac{8 \sqrt{2}}{6 \sqrt{7}}\end{align*}

Solution: Rewrite all division problems like a fraction.

a)

b) 30888=24064=16158=4158=152\begin{align*}\frac{\sqrt{30}}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{\sqrt{240}}{\sqrt{64}} = \frac{\sqrt{16 \cdot 15}}{8} = \frac{4 \sqrt{15}}{8} = \frac{\sqrt{15}}{2}\end{align*}

c) 826777=81467=41437=41421\begin{align*}\frac{8 \sqrt{2}}{6 \sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{8 \sqrt{14}}{6 \cdot 7} = \frac{4 \sqrt{14}}{3 \cdot 7} = \frac{4 \sqrt{14}}{21}\end{align*}

Notice, we do not really “divide” radicals, but get them out of the denominator of a fraction.

The Pythagorean Theorem

We have used the Pythagorean Theorem already in this text, but have not proved it. Recall that the sides of a right triangle are the legs (the sides of the right angle) and the hypotenuse (the side opposite the right angle). For the Pythagorean Theorem, the legs are “a\begin{align*}a\end{align*}” and “b\begin{align*}b\end{align*}” and the hypotenuse is “c\begin{align*}c\end{align*}”.

Pythagorean Theorem: Given a right triangle with legs of lengths a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} and a\begin{align*}a\end{align*} hypotenuse of length c\begin{align*}c\end{align*}, then a2+b2=c2\begin{align*}a^2+b^2=c^2\end{align*}.

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 in. and 4 in.
2. Cut out the triangle and square and arrange them like the picture on the right.
3. This theorem relies on area. Recall that the area of a square is side2\begin{align*}^2\end{align*}. In this case, we have three squares with sides 3 in., 4 in., and 5 in. What is the area of each square?
4. Now, we know that 9+16=25\begin{align*}9 + 16 = 25\end{align*}, or 32+42=52\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.

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

Using the Pythagorean Theorem

Here are several examples of the Pythagorean Theorem in action.

Example 4: Do 6, 7, and 8 make the sides of a right triangle?

Solution: Plug in the three numbers to the Pythagorean Theorem. The largest length will always be the hypotenuse. If 62+72=82\begin{align*}6^2+7^2=8^2\end{align*}, then they are the sides of a right triangle.

62+7282=36+49=85=648564, so the lengths are not the sides of a right triangle.\begin{align*}6^2 + 7^2 &= 36 + 49=85\\ 8^2 &= 64 && 85 \neq 64, \ \text{so the lengths are not the sides of a right triangle.}\end{align*}

Example 5: Find the length of the hypotenuse.

Solution: Use the Pythagorean Theorem. Set a=8\begin{align*}a = 8\end{align*} and b=15\begin{align*}b = 15\end{align*}. Solve for c\begin{align*}c\end{align*}.

82+15264+22528917=c2=c2=c2=cTake the square root of both sides.\begin{align*}8^2 + 15^2 &= c^2\\ 64 + 225 &= c^2\\ 289 &= c^2 && Take \ the \ square \ root \ of \ both \ sides.\\ 17 &= c\end{align*}

When you take the square root of an equation, the answer is 17 or -17. Length is never negative, which makes 17 the answer.

Example 6: Find the missing side of the right triangle below.

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

72+b249+b2b2b=142=196=147=147=493=73\begin{align*}7^2 + b^2 &= 14 ^2\\ 49 + b^2 &= 196\\ b^2 &= 147\\ b &= \sqrt{147} = \sqrt{49 \cdot 3} = 7 \sqrt{3}\end{align*}

Example 7: What is the diagonal of a rectangle with sides 10 and 16?

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 d\begin{align*}d\end{align*}.

102+162100+256356d=d2=d2=d2=356=28918.87\begin{align*}10^2 + 16^2 &= d^2\\ 100 + 256 &= d^2\\ 356 &= d^2\\ d &= \sqrt{356} = 2 \sqrt{89} \approx 18.87\end{align*}

Pythagorean Triples

In Example 5, the sides of the triangle were 8, 15, and 17. This combination of numbers is called a Pythagorean triple.

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

3,4,55,12,137,24,258,15,179,12,1510,24,26\begin{align*}3, 4, 5 && 5, 12, 13 && 7, 24, 25 && 8, 15, 17 && 9, 12, 15 && 10, 24, 26\end{align*}

Any multiple of a Pythagorean triple is also considered a triple because it would still be three whole numbers. Multiplying 3, 4, 5 by 2 gives 6, 8, 10, which is another triple. To see if a set of numbers makes a triple, plug them into the Pythagorean Theorem.

Example 8: Is 20, 21, 29 a Pythagorean triple?

Solution: If 202+212=292\begin{align*}20^2 + 21^2 = 29^2\end{align*}, then the set is a Pythagorean triple.

202+212292=400+441=841=841\begin{align*}20^2 + 21^2 &= 400+441=841\\ 29^2 &= 841\end{align*}

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

Height of an Isosceles Triangle

One way to use The Pythagorean Theorem is to find the height of an isosceles triangle.

Example 9: What is the height of the isosceles triangle?

Solution: Draw the altitude from the vertex between the congruent sides, which bisect the base.

72+h249+h2h2h=92=81=32=32=162=42\begin{align*}7^2 + h^2 &= 9^2\\ 49 + h^2 &= 81\\ h^2 &= 32\\ h &= \sqrt{32} = \sqrt{16 \cdot 2} = 4 \sqrt{2}\end{align*}

The Distance Formula

Another application of the Pythagorean Theorem is the Distance Formula. We will prove it here.

Let’s start with point A(x1,y1)\begin{align*}A(x_1, y_1)\end{align*} and point B(x2,y2)\begin{align*}B(x_2, y_2)\end{align*}, to the left. We will call the distance between \begin{align*}A\end{align*} and \begin{align*}B, d\end{align*}.

Draw the vertical and horizontal lengths to make a right triangle.

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

\begin{align*}d^2 &= (x_1-x_2)^2 + (y_1-y_2)^2\\ d &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\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 10: 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*}

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*}

The dimensions of the TV are \begin{align*}16(2.83”) \times 9(2.83”),\end{align*} or \begin{align*}45.3” \times 25.5”\end{align*}.

Review Questions

• Questions 1-9 are similar to Examples 1-3.
• Questions 10-15 are similar to Example 5 and 6.
• Questions 16-19 are similar to Example 7.
• Questions 20-25 are similar to Example 8.
• Questions 26-28 are similar to Example 9.
• Questions 29-31 are similar to Example 10.
• Questions 32 and 33 are similar to the Know What?
• Question 34 and 35 are a challenge and similar to Example 9.

1. \begin{align*}2 \sqrt{5} + \sqrt{20}\end{align*}
2. \begin{align*}\sqrt{24}\end{align*}
3. \begin{align*}\left( 6 \sqrt{3} \right)^2\end{align*}
4. \begin{align*}8 \sqrt{8} \cdot \sqrt{10}\end{align*}
5. \begin{align*}\left( 2 \sqrt{30} \right )^2\end{align*}
6. \begin{align*}\sqrt{320}\end{align*}
7. \begin{align*}\frac{4 \sqrt{5}}{\sqrt{6}}\end{align*}
8. \begin{align*}\frac{12}{\sqrt{10}}\end{align*}
9. \begin{align*}\frac{21 \sqrt{5}}{9 \sqrt{15}}\end{align*}

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 sides of a square are 16, then the diagonal 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 height of each isosceles 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 42” 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 42” 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 8, find the height. Leave your answer in simplest radical form.
7. If the sides are length \begin{align*}s\end{align*}, what would the height be?

1. Factors of 75: 1, 3, 5, 15, 25, 75
2. Prime Factorization of 75: \begin{align*}3 \cdot 5 \cdot 5\end{align*}

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