What if you knew the lengths of two sides of a right triangle but not the third? How could you find the length of this missing side? After completing this Concept, you'll be able to use the Pythagorean Theorem to solve problems like this one where variables are involved.

### Guidance

In the previous concept, we learned about the Pythagorean theorem and how to use it to find the hypotenuse. In this concept, we will learn how to use the Pythagorean theorem to find any side of a right triangle.

#### Example A

*Determine the value of the missing side. You may assume that each triangle is a right triangle.*

**Solution**

Apply the Pythagorean Theorem.

\begin{align*}a^2+b^2 &= c^2\\ x^2+15^2 &= 21^2\\ x^2+225 &= 441\\ x^2 &= 216 \Rightarrow \\ x & =\sqrt{216}=6 \sqrt{6}\end{align*}

#### Example B

*Determine the values of the missing sides. You may assume that each triangle is a right triangle.*

**Solution**

Apply the Pythagorean Theorem.

\begin{align*}a^2+b^2 &= c^2\\ 18^2+15^2 &= z^2\\ 324+225 &= z\\ z^2 &= 549 \Rightarrow \\ z & =\sqrt{549}=3 \sqrt{61}\end{align*}

#### Example C

*One leg of a right triangle is 5 units longer than the other leg. The hypotenuse is one unit longer than twice the size of the short leg. Find the dimensions of the triangle.*

**Solution**

Let \begin{align*}x =\end{align*} length of the short leg.

Then \begin{align*}x + 5 =\end{align*} length of the long leg

And \begin{align*}2x + 1 =\end{align*} length of the hypotenuse.

The sides of the triangle must satisfy the Pythagorean Theorem.

\begin{align*}\text{Therefore:} && x^2+(x+5)^2& =(2x+1)^2\\ \text{Eliminate the parentheses:} && x^2+x^2+10x+25& =4x^2+4x+1\\ \text{Move all terms to the right hand side of the equation:} && 0& =2x^2-6x-24\\ \text{Divide all terms by} \ 2: && 0& =x^2-3x-12\\ \text{Solve using the quadratic formula:} && x& =\frac{3 \pm \sqrt{9+48}}{2}=\frac{3 \pm \sqrt{57}}{2}\\ && x& =\underline{\underline{5.27}} \ \text{or} \ x=-2.27\end{align*}

The negative solution doesn’t make sense when we are looking for a physical distance, so we can discard it. Using the positive solution, we get: **\begin{align*}\text{short leg} = 5.27, \text{long leg} = 10.27\end{align*} and \begin{align*}\text{hypotenuse} = 11.54\end{align*}.**

Watch this video for help with the Examples above.

CK-12 Foundation: The Pythagorean Theorem with Variables

### Guided Practice

*Determine the values of the missing sides. You may assume that each triangle is a right triangle.*

**Solution**

Apply the Pythagorean Theorem. \begin{align*}a^2+b^2 &= c^2\\ y^2+3^2 &= 7^2\\ y^2+9 &= 49\\ y^2 &= 40 \Rightarrow\\ y & =\sqrt{40}=2 \sqrt{10}\end{align*}

### Explore More

Find the missing length of each right triangle.

- \begin{align*}a = 12, b = 16, c = ?\end{align*}
- \begin{align*}a = ?, b = 20, c = 30\end{align*}
- \begin{align*}a = 4, b = ?, c = 11\end{align*}
- \begin{align*}a = 12, b = ?, c = 37\end{align*}
- One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle.
- One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle.
- Two sides of a right triangle are 5 units and 8 units respectively. Those sides could be the legs, or they could be one leg and the hypotenuse. What are the possible lengths of the third side?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 11.9.