9.8: Solving Equations Using the Pythagorean Theorem
Have you ever visited different ball parks? Look at this dilemma.
Miguel visited a ball park in a different town. At this baseball diamond, the distance from the pitcher's mound to home plate measured 60 feet. The distance from first to home is 90 feet. What is the distance from the pitcher's mound to first plate?
These distances form a right triangle in a baseball diamond. You are missing the dimensions of one of the legs of the triangle.
Do you know how to find the missing distance?
This Concept will show you how to use the Pythagorean Theorem to figure this out.
Guidance
You just learned how to find the length of the hypotenuse using the Pythagorean Theorem. What about if you have been given the length of one of the legs and the hypotenuse?
Can we use the Pythagorean Theorem to find the length of the missing leg?
Sure we can. Let’s look at how this can be done.
You have already worked on this triangle in the last section, and you may even remember the lengths of the sides. However, let’s use it as an example to see how the Pythagorean Theorem can help us solve this one.
First, fill the given values into the formula.
\begin{align*}3^2+b^2=5^2\end{align*}
Here we know the values of \begin{align*}a\end{align*}
\begin{align*}9 + b^2=25\end{align*}
Now we want to solve this equation by getting the variable alone. To do this, we subtract 9 from both sides of the equation.
\begin{align*}9  9 + b^2 &= 259\\
b^2&=16 \end{align*}
Next we take the square root of both sides.
\begin{align*}\bcancel{\sqrt{b^2}} & = \sqrt{16}\\
b &=4\end{align*}
Our answer is 4.
Sometimes you won’t be working with perfect squares. When this happens, you will need to approximate the length of a leg just as we approximated the length of the hypotenuse in the last Concept.
First, take the given side lengths and substitute them into the formula.
\begin{align*}4^2+b^2&=12^2\\
16 + b^2&=144\end{align*}
Next, we subtract 16 from both sides of the equation.
\begin{align*}16  16 + b^2&=144  16\\
b^2&=128\end{align*}
Now we take the square root of both sides of the equation.
\begin{align*}\bcancel{\sqrt{b^2}} &= \sqrt{128}\\
b &= 11.3\end{align*}
This is our answer.
Use the Pythagorean Theorem to find each missing side length. You may round to the nearest tenth as necessary.
Example A
A right triangle with \begin{align*}a, \ b = 6\end{align*}
Solution: \begin{align*}a = 11.5\end{align*}
Example B
A right triangle with \begin{align*}a = 8, \ b,\end{align*}
Solution: \begin{align*}8.9\end{align*}
Example C
A right triangle with \begin{align*}a = 6, \ b,\end{align*}
Solution:\begin{align*}b = 8\end{align*}
Here is the original problem once again.
Miguel visited a ball park in a different town. At this baseball diamond, the distance from the pitcher's mound to home plate measured 60 feet. The distance from first to home is 90 feet. What is the distance from the pitcher's mound to first plate?
These distances form a right triangle in a baseball diamond. You are missing the dimensions of one of the legs of the triangle.
Do you know how to find the missing distance?
First, let's write an equation using the Pythagorean Theorem. We'll call the missing leg, \begin{align*}b\end{align*}
\begin{align*}60^2 + b^2 = 90^2\end{align*}
Now we can solve.
\begin{align*}3600 + b^2 = 8100\end{align*}
\begin{align*}b^2 = 8100  3600\end{align*}
\begin{align*}b^2 = 4500\end{align*}
\begin{align*}\sqrt{b^2} = \sqrt{4500}\end{align*}
\begin{align*}b = 67.08 = 67.1\end{align*}
The distance is 67.1 feet.
Vocabulary
Here are the vocabulary words found in this Concept.
 Pythagorean Theorem

the theorem that states that the square of leg \begin{align*}a\end{align*}
a of a right triangle plus the square of leg \begin{align*}b\end{align*}b of a right triangle is equal to the square of the hypotenuse side \begin{align*}c\end{align*}c of the same right triangle.
 Legs of the Right Triangle

legs \begin{align*}a\end{align*}
a and \begin{align*}b\end{align*}b are the two shorter sides of a right triangle.
 Hypotenuse

side \begin{align*}c\end{align*}
c , the longest side of a right triangle. It forms the diagonal in a square too.
Guided Practice
Here is one for you to try on your own.
Find the length of leg b.
\begin{align*}a = 12, b = ?, c = 13\end{align*}
Answer
First use the formula for the Pythagorean Theorem to solve for b.
\begin{align*}a^2 + b^2 =c^2\end{align*}
\begin{align*}12^2 + b^2 = 13^2\end{align*}
\begin{align*}144 + b^2 = 169\end{align*}
\begin{align*}b^2 = 25\end{align*}
\begin{align*}b = 5\end{align*}
This is our answer.
Video Review
Here is a video for review.
This is a James Sousa video on the Pythagorean Theorem.
Practice
Directions: Use the Pythagorean Theorem to find the length of each missing leg. You may round to the nearest tenth when necessary.
1. \begin{align*}a = 6, \ b = ?, \ c = 12\end{align*}
2. \begin{align*}a = 9, \ b = ?, \ c = 15\end{align*}
3. \begin{align*}a = 4, \ b = ?, \ c = 5\end{align*}
4. \begin{align*}a = 9, \ b = ?, \ c = 18\end{align*}
5. \begin{align*}a = 15, \ b = ?, \ c = 25\end{align*}
6. \begin{align*}a = ?, \ b = 10, \ c = 12\end{align*}
7. \begin{align*}a = ?, \ b = 11, \ c = 14\end{align*}
8. \begin{align*}a = ?, \ b = 13, \ c = 15\end{align*}
Directions: Use problem solving to write an equation using the Pythagorean Theorem and solve each problem.
Joanna laid a plank of wood down to make a ramp so that she could roll a wheelbarrow over a low wall in her garden. The wall is 1.5 meters tall, and the plank of wood touches the ground 2 meters from the wall. How long is the wooden plank?
9. Write the equation.
10. Solve for the answer.
Chris rode his bike 4 miles west and then 3 miles south. What is the shortest distance he can ride back to the point where he started?
11. Write the equation.
12. Solve the problem.
Naomi is cutting triangular patches to make a quilt. Each has a diagonal side of 14.5 inches and a short side of 5.5 inches. What is the length of the third side of each triangular patch?
13. Write the equation.
14. Solve the problem.
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Here you'll learn to use the Pythagorean Theorem to find the length of a leg.