8.1: The Pythagorean Theorem
Learning Objectives
- Review simplifying and reducing radicals.
- Prove and use the Pythagorean Theorem.
- Use the Pythagorean Theorem to derive the distance formula.
Review Queue
- Draw a right scalene triangle.
- Draw an isosceles right triangle.
- List all the factors of 75.
- 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?
Simplifying and Reducing Radicals
In algebra, you learned how to simplify radicals. Let’s review it here.
Example 1: Simplify the radical.
a) \begin{align*}\sqrt{50}\end{align*}
b) \begin{align*}\sqrt{27}\end{align*}
c) \begin{align*}\sqrt{272}\end{align*}
Solution: For each radical, find the square number(s) that are factors.
a) \begin{align*}\sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2}\end{align*}
b) \begin{align*}\sqrt{27} = \sqrt{9 \cdot 3} = 3 \sqrt{3}\end{align*}
c) \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, \begin{align*}2 \sqrt{5} + 3 \sqrt{6}\end{align*}
Example 2: Simplify the radicals.
a) \begin{align*}2 \sqrt{10} + \sqrt{160}\end{align*}
b) \begin{align*}5 \sqrt{6} \cdot 4 \sqrt{18}\end{align*}
c) \begin{align*}\sqrt{8} \cdot 12 \sqrt{2}\end{align*}
d) \begin{align*}\left( 5 \sqrt{2} \right)^2\end{align*}
Solution:
a) Simplify \begin{align*}\sqrt{160}\end{align*}
b) To multiply two radicals, multiply what is under the radicals and what is in front.
\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) \begin{align*}\sqrt{8} \cdot 12 \sqrt{2} = 12 \sqrt{8 \cdot 2} = 12 \sqrt{16} = 12 \cdot 4=48\end{align*}
d) \begin{align*}\left( 5 \sqrt{2} \right )^2 = 5^2 \left( \sqrt{2} \right )^2 = 25 \cdot 2 = 50 \rightarrow\end{align*}
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) \begin{align*}4 \sqrt{6} \div \sqrt{3}\end{align*}
b) \begin{align*}\frac{\sqrt{30}}{\sqrt{8}}\end{align*}
c) \begin{align*}\frac{8 \sqrt{2}}{6 \sqrt{7}}\end{align*}
Solution: Rewrite all division problems like a fraction.
a)
b) \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) \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 “\begin{align*}a\end{align*}
Pythagorean Theorem: Given a right triangle with legs of lengths \begin{align*}a\end{align*}
Investigation 8-1: Proof of the Pythagorean Theorem
Tools Needed: pencil, 2 pieces of graph paper, ruler, scissors, colored pencils (optional)
- 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.
- Cut out the triangle and square and arrange them like the picture on the right.
- This theorem relies on area. Recall that the area of a square is side\begin{align*}^2\end{align*}
2 . In this case, we have three squares with sides 3 in., 4 in., and 5 in. What is the area of each square? - Now, we know that \begin{align*}9 + 16 = 25\end{align*}
9+16=25 , or \begin{align*}3^2+4^2=5^2\end{align*}32+42=52 . 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 \begin{align*}6^2+7^2=8^2\end{align*}
\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 \begin{align*}a = 8\end{align*}
\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 \begin{align*}b\end{align*}.
\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 \begin{align*}d\end{align*}.
\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.
\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 \begin{align*}20^2 + 21^2 = 29^2\end{align*}, then the set is a Pythagorean triple.
\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.
\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 \begin{align*}A(x_1, y_1)\end{align*} and point \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.
Simplify the radicals.
- \begin{align*}2 \sqrt{5} + \sqrt{20}\end{align*}
- \begin{align*}\sqrt{24}\end{align*}
- \begin{align*}\left( 6 \sqrt{3} \right)^2\end{align*}
- \begin{align*}8 \sqrt{8} \cdot \sqrt{10}\end{align*}
- \begin{align*}\left( 2 \sqrt{30} \right )^2\end{align*}
- \begin{align*}\sqrt{320}\end{align*}
- \begin{align*}\frac{4 \sqrt{5}}{\sqrt{6}}\end{align*}
- \begin{align*}\frac{12}{\sqrt{10}}\end{align*}
- \begin{align*}\frac{21 \sqrt{5}}{9 \sqrt{15}}\end{align*}
Find the length of the missing side. Simplify all radicals.
- If the legs of a right triangle are 10 and 24, then the hypotenuse is __________.
- If the sides of a rectangle are 12 and 15, then the diagonal is _____________.
- If the sides of a square are 16, then the diagonal is ____________.
- If the sides of a square are 9, then the diagonal is _____________.
Determine if the following sets of numbers are Pythagorean Triples.
- 12, 35, 37
- 9, 17, 18
- 10, 15, 21
- 11, 60, 61
- 15, 20, 25
- 18, 73, 75
Find the height of each isosceles triangle below. Simplify all radicals.
Find the length between each pair of points.
- (-1, 6) and (7, 2)
- (10, -3) and (-12, -6)
- (1, 3) and (-8, 16)
- What are the length and width of a 42” HDTV? Round your answer to the nearest tenth.
- 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.
- 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.
- If the sides are length \begin{align*}s\end{align*}, what would the height be?
Review Queue Answers
- Factors of 75: 1, 3, 5, 15, 25, 75
- Prime Factorization of 75: \begin{align*}3 \cdot 5 \cdot 5\end{align*}
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