Using the Pythagorean Theorem
Using the Pythagorean Theorem
We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem. This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.
Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a \(90\text{°}\) angle, marked with a small square in the corner. The side of the triangle opposite the \(90\text{°}\) angle is called the hypotenuse and each of the other sides are called legs.
The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, \({a}^{2}+{b}^{2}={c}^{2},\) where \(a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b\) are the lengths of the legs and \(c\) is the length of the hypotenuse.
Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.
The Pythagorean Theorem
In any right triangle, \({a}^{2}+{b}^{2}={c}^{2}.\)
where a and b are the lengths of the legs, c is the length of the hypotenuse.
To solve exercises that use the Pythagorean Theorem, we will need to find square roots. We have used the notation \(\sqrt{m}\) and the definition:
If \(m={n}^{2},\) then \(\sqrt{m}=n,\) for \(n\ge 0.\)
For example, we found that \(\sqrt{25}\) is 5 because \(25={5}^{2}.\)
Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.
Example
Use the Pythagorean Theorem to find the length of the hypotenuse shown below.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the length of the hypotenuse of the triangle |
| Step 3. Name. Choose a variable to represent it.
|
Let c = the length of the hypotenuse. |
| Step 4. Translate. | |
| Write the appropriate formula. | \({a}^{2}+{b}^{2}={c}^{2}\) |
| Substitute. | \({3}^{2}+{4}^{2}={c}^{2}\) |
| Step 5. Solve the equation. | \(\phantom{\rule{0.4em}{0ex}}9+16={c}^{2}\) |
| Simplify. | \(\phantom{\rule{2em}{0ex}}25={c}^{2}\) |
| Use the definition of square root. | \(\phantom{\rule{1.5em}{0ex}}\sqrt{25}=c\) |
| Simplify. | \(\phantom{\rule{2.5em}{0ex}}5=c\) |
| Step 6. Check. |
|
| Step 7. Answer the question. | The length of the hypotenuse is 5. |
Example
Use the Pythagorean Theorem to find the length of the leg shown below.
Solution
| Step 1. Read the problem. | ||
| Step 2. Identify what you are looking for. | the length of the leg of the triangle | |
| Step 3. Name. Choose a variable to represent it. | Let b = the leg of the triangle. | |
| Lable side b. | ||
| Step 4. Translate | ||
| Write the appropriate formula. | \({a}^{2}+{b}^{2}={c}^{2}\) | |
| Substitute. | \({5}^{2}+{b}^{2}={13}^{2}\) | |
| Step 5. Solve the equation. | \(25+{b}^{2}=169\) | |
| Isolate the variable term. | \(\phantom{\rule{2.1em}{0ex}}{b}^{2}=144\) | |
| Use the definition of square root. | \(\phantom{\rule{2.1em}{0ex}}{b}^{2}=\sqrt{144}\) | |
| Simplify. | \(\phantom{\rule{2.6em}{0ex}}b=12\) | |
| Step 6. Check. |
||
| Step 7. Answer the question. | The length of the leg is 12. |
Example
Kelvin is building a gazebo and wants to brace each corner by placing a \(10\text{″}\) piece of wood diagonally as shown above.
If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.
Solution
\(\begin{array}{cccc}\mathbf{\text{Step 1.}}\phantom{\rule{0.2em}{0ex}}\text{Read the problem.}\hfill & & & \\ \mathbf{\text{Step 2.}}\phantom{\rule{0.2em}{0ex}}\text{Identify what we are looking for.}\hfill & & & \begin{array}{c}\text{the distance from the corner that the}\hfill \\ \text{bracket should be attached}\hfill \end{array}\hfill \\ \mathbf{\text{Step 3.}}\phantom{\rule{0.2em}{0ex}}\text{Name. Choose a variable to represent it.}\hfill & & & \text{Let}\phantom{\rule{0.2em}{0ex}}x=\text{the distance from the corner.}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4.}}\phantom{\rule{0.2em}{0ex}}\text{Translate.}\hfill \\ \text{Write the appropriate formula and substitute.}\hfill \\ \mathbf{\text{Step 5.}}\phantom{\rule{0.2em}{0ex}}\text{Solve the equation.}\hfill \\ \\ \phantom{\rule{2.5em}{0ex}}\text{Isolate the variable.}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Use the definition of square root.}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Simplify. Approximate to the nearest tenth.}\hfill \end{array}\hfill & & & \hfill \begin{array}{}\\ \hfill {a}^{2}+{b}^{2}& =\hfill & {c}^{2}\hfill \\ \hfill {x}^{2}+{x}^{2}& =\hfill & {10}^{2}\hfill \\ \hfill 2{x}^{2}& =\hfill & 100\hfill \\ \hfill {x}^{2}& =\hfill & 50\hfill \\ \hfill x& =\hfill & \sqrt{50}\hfill \\ \hfill x& \approx \hfill & 7.1\hfill \end{array}\hfill \\ \mathbf{\text{Step 6.}}\phantom{\rule{0.2em}{0ex}}\text{Check.}\hfill & & & \\ \begin{array}{ccc}\hfill \phantom{\rule{2.5em}{0ex}}{a}^{2}+{b}^{2}& =\hfill & {c}^{2}\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}{\left(7.1\right)}^{2}+{\left(7.1\right)}^{2}& \approx \hfill & {10}^{2}\phantom{\rule{0.2em}{0ex}}\text{Yes.}\hfill \end{array}\hfill & & & \\ \mathbf{\text{Step 7.}}\phantom{\rule{0.2em}{0ex}}\text{Answer the question.}\hfill & & & \begin{array}{c}\text{Kelvin should fasten each piece of}\hfill \\ \text{wood approximately}\phantom{\rule{0.2em}{0ex}}7.1\text{″}\phantom{\rule{0.2em}{0ex}}\text{from the corner.}\hfill \end{array}\hfill \end{array}\)
This lesson is part of:
Math Models and Geometry II