Use the Pythagorean Theorem to solve applications involving right triangles
Find the domain of a radical function
This section will discuss applications which use square roots, in particular the Pythagorean Theorem. As always, the following steps will help to translate and solve the problem.
1. Read through the entire problem
2. Organize the information (a picture may be useful)
3. Write the equation
4. Solve the equation
5. Check the answer
Pythagorean Theorem
If we are given the measurements of two sides of a right triangle (A right triangle is one where one of the angles is 90 degrees; indicated with a square in the angle), we can find the measurement of the third side by using the Pythagorean Theorem. The Pythagorean Theorem states that the square of hypotenuse is equal to the sum of the squares of the other two sides.
The Pythagorean Theorem formula is:
[latex]a^2+b^2=c^2[/latex] where c is the hypotenuse; a and b are the legs.
The video below will walk you through the proof and a few examples of using The Pythagorean Theorem, including an application example.
Example 1
Find the length of the hypotenuse of the right triangle pictured below:
Show Answer
Use the Pythagorean Theorem
[latex]c^2=a^2+b^2[/latex]
Substitute the values into the formula. Note, both given sides are the legs, a and b.
[latex]c^2=7^2+24^2[/latex]
Simplify the squares
[latex]c^2=49+576[/latex]
Add
[latex]c^2=625[/latex]
Take the square root of each side
[latex]\sqrt{c^2}=\sqrt{625}[/latex]
[latex]c=25[/latex]
Solution:
The hypotenuse of the triangle is 25 units long.
Example 2
Find the missing length of the leg of the right triangle pictured below:
Show Answer
Use the Pythagorean Theorem
[latex]a^2+b^2=c^2[/latex]
Substitute the values into the formula. Note, this time sides labeled 13 and 5 aren’t both legs of the triangle like in the last example. The side labeled 13 is the hypotenuse so must replace the [latex]c[/latex] in the formula.
The vertical leg of this triangle is 12 units long.
The Pythagorean Theorem can be used in a variety of real-world applications. The next two examples show some simple applications.
Example 3
Given a rectangular picture frame whose opening is 12 inches by 10 inches, how long is the diagonal from one corner of the opening to the other? Give both an exact answer and an approximation to the nearest hundredth.
Show Answer
Use the Pythagorean Theorem
[latex]c^2=a^2+b^2[/latex]
Substitute the values into the formula. Note, both given sides are the legs, a and b.
[latex]c^2=12^2+10^2[/latex]
Simplify the squares
[latex]c^2=144+100[/latex]
Add
[latex]c^2=244[/latex]
Take the square root of each side
[latex]\sqrt{c^2}=\sqrt{244}[/latex]
Simplify square root to list answer in exact form
[latex]c=2\sqrt{61}[/latex]
Type root in calculator to find approximation for the answer
[latex]c≈15.6204...[/latex]
Round to the nearest hundredth
[latex]c≈15.62[/latex]
Solution:
The diagonal from one corner of the opening to the other is of length [latex]2\sqrt{61}[/latex] inches which is approximately 15.62 inches long.
Example 4
A 6–foot ladder is placed against a wall. If the base of the ladder is 3 feet from the wall, how high up the wall will the ladder reach? Give both an exact answer and an approximation to the nearest hundredth.
Show Answer
Since the triangle formed has a right angle we can use the Pythagorean Theorem
[latex]a^2+b^2=c^2[/latex]
Substitute the values into the formula. Note, this time sides labeled 3 and 6 aren’t both legs of the triangle like in the last example. The side 6 ft ladder is the hypotenuse of the triangle so must replace the [latex]c[/latex] in the formula.
As mentioned earlier, the Pythagorean Theorem is used in many applications. In fact, it has played a critical role in real-world applications for centuries! If you are interested in exploring another real-world example, celestial navigation and the mathematics behind it, watch this video for fun.
Domain of Radical Functions
In module 3, we introduced the domain of a function as the set of input values that produce valid outputs. In this module, we have been exploring radical expressions. In the next example, let us examine a radical function and determine if any domain issues arise.
Example 6
Consider the radical function [latex]f(x)=\sqrt{3x-2}[/latex]
A. Compute [latex]f(1)[/latex], [latex]f(6)[/latex], [latex]f(7)[/latex], [latex]f\left(\frac{2}{3}\right)[/latex], and [latex]f(0)[/latex]. Round to three decimal places if needed.
B. Determine if each of the [latex]x[/latex]-values from part A is in the domain of the function.
Show Solution
A. As with any function, we will substitute each value into the function for [latex]x[/latex].
[latex]f(0)=\sqrt{3(0)-2}=\sqrt{0-2}=\sqrt{-2}\hspace{.2in}[/latex] Not a real number
B. The first four inputs produced valid results, but the last input of [latex]0[/latex] did not, as we are reminded that we cannot currently take the even root of a negative number. We conclude that
[latex]1, 6, 7, \frac{2}{3}[/latex] are in the domain
[latex]0[/latex] is not in the domain
The previous example shows us that any input that results in a negative radicand is excluded from the domain. In other words, we must always ensure that the radicand is nonnegative ([latex]\geq 0[/latex]).
Finding the domain of a radical function with an even index
Set the radicand (the expression under the radical) to be [latex]\geq 0[/latex].
Solve the corresponding inequality for the variable.
Write your answer in the appropriate notation.
So now, let us find the domain of the radical function given in the previous example.
Example 7
Find the domain of the radical function [latex]f(x)=\sqrt{3x-2}[/latex]. Give your answer in set-builder notation and interval notation.
Show Solution
We need to determine when the radicand, [latex]3x-2[/latex], is nonnegative. We can do this by setting it to be [latex]\geq 0[/latex] and then solving.
In set-builder notation, we simply write this as [latex]\left\{x|x\geq \frac{2}{3}\right\}[/latex].
On a number line, this corresponds to all numbers to the right of [latex]\frac{2}{3}[/latex], including [latex]\frac{2}{3}[/latex] itself. So, the interval notation is [latex]\left[ \frac{2}{3},\infty\right)[/latex].
It is worth noting that had the function in the previous example been a fourth root, sixth root, or any even root, the process and final answer would not have changed. However, we saw in the beginning of this module that odd roots do not have the same issue with negative radicands. For example, [latex]\sqrt[3]{-8}=-2[/latex]. It follows that odd roots will not have any domain issues.
Domain of a radical function with an odd index
The domain will be all real numbers. In interval notation, this is [latex](-\infty,\infty)[/latex].
Let us conclude the section with one more example to emphasize the difference between these two cases.
Example 8
Find the domain in set-builder notation for each radical function. If the domain is all real numbers, state this.
A. [latex]f(x)=\sqrt{5-x}[/latex]
B. [latex]g(x)=\sqrt[3]{5-x}[/latex]
Show Solution
A. First note that this is an even radical (an index of 2). So, we set the radicand to be [latex]\geq 0[/latex].
