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# 9.8: Solving Equations Using the Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12
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Practice Solving Equations Using the Pythagorean Theorem

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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.

32+b2=52\begin{align*}3^2+b^2=5^2\end{align*}

Here we know the values of a\begin{align*}a\end{align*} and c\begin{align*}c\end{align*} so we can fill those in. B\begin{align*}B\end{align*} is a mystery so it remains an unknown variable.

9+b2=25\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.

99+b2b2=259=16\begin{align*}9 - 9 + b^2 &= 25-9\\ 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*}

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*}

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*} and \begin{align*}c = 13\end{align*}

Solution: \begin{align*}a = 11.5\end{align*}

#### Example B

A right triangle with \begin{align*}a = 8, \ b,\end{align*} and \begin{align*}c = 12\end{align*}

Solution: \begin{align*}8.9\end{align*}

#### Example C

A right triangle with \begin{align*}a = 6, \ b,\end{align*} and \begin{align*}c = 10\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*} of a right triangle plus the square of leg \begin{align*}b\end{align*} of a right triangle is equal to the square of the hypotenuse side \begin{align*}c\end{align*} of the same right triangle.
Legs of the Right Triangle
legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the two shorter sides of a right triangle.
Hypotenuse
side \begin{align*}c\end{align*}, 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*}

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*}

### Video Review

Here is a video for review.

### 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.

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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