The lengths of three sides of a triangle are 4, 6, and 10. Is this a right triangle?

### Pythagorean Theorem

The **Pythagorean Theorem** states that for right triangles with legs of lengths \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and hypotenuse of length **\begin{align*}c\end{align*}, \begin{align*}a^2+b^2=c^2\end{align*}.**

There are many different proofs of the Pythagorean Theorem. The following picture leads to one of those proofs.

First, you can verify that the quadrilateral in the center is a square. All sides are the same length and each angle must be \begin{align*}90^\circ\end{align*}. The angles must be \begin{align*}90^\circ\end{align*} because the three angles that make a straight angle at each corner of the interior quadrilateral are the same three angles that make up each of the triangles. This relationship is shown with the angle markings in the picture below.

In order to prove the Pythagorean Theorem, find the area of the interior square in two ways. First, find the area directly:

\begin{align*}Area \ of \ square=c^2\end{align*}

Next, find the area as the difference between the area of the large square and the area of the triangles.

\begin{align*}Area \ of \ square=(a+b)^2-4 \left(\frac{1}{2} ab \right)\end{align*}

Since you are referring to the same square each time, those two areas must be equal.

\begin{align*}(a+b)^2-4 \left(\frac{1}{2} ab \right)=c^2\end{align*}

Use algebra to simplify.

\begin{align*}a^2+2ab+b^2-2ab &=c^2 \\ a^2+b^2 &=c^2\end{align*}

The converse of the Pythagorean Theorem is also true. The converse switches the “if” and “then” parts of the theorem. **The converse says that if** \begin{align*}a^2+b^2=c^2\end{align*}, **then the triangle is a right triangle.**

With the Pythagorean Theorem and its converse, you can solve many types of problems. You can:

- Find the missing side of a right triangle when you know the other two sides.
- Determine whether a triangle is right, acute, or obtuse.
- Find the distance between two points.

#### Finding the Length of the Hypotenuse

The two legs of a right triangle have lengths 3 and 4. What is the length of the hypotenuse?

Because it is a right triangle, you can use the Pythagorean Theorem. It doesn't matter whether you assign \begin{align*}a\end{align*} as 3 or \begin{align*}b\end{align*} as 3.

\begin{align*}a^2+b^2 &=c^2 \\ 3^2+4^2 &=c^2 \\ 9+16 &=c^2 \\ 25 &=c^2 \\ \pm 5 &=c \\ \end{align*}

Because length must be positive, the hypotenuse has a length of 5. Side lengths of 3, 4 and 5 are common in geometry. You should remember that they are the lengths of a right triangle. Triples of whole numbers that satisfy the Pythagorean Theorem are called **Pythagorean triples**. “3, 4, 5” is an example of a Pythagorean triple.

#### Classifying Triangles

A triangle has side lengths of 4, 8 and 9. What type of triangle is this?

If the numbers satisfy the Pythagorean Theorem, then it is a right triangle. If \begin{align*}a^2+b^2 > c^2\end{align*}, then \begin{align*}c\end{align*} is shorter than it would be in a right triangle, so the angle opposite it is smaller and it is an acute triangle. If \begin{align*}a^2+b^2 < c^2\end{align*}, then \begin{align*}c\end{align*} is longer than it would be in a right triangle, so the angle opposite it is larger and it is an obtuse triangle.

In this case, \begin{align*}4^2+8^2=80\end{align*}. \begin{align*}9^2=81\end{align*}. \begin{align*}a^2+b^2 < c^2\end{align*}, so the triangle is obtuse.

#### Deriving the Distance Formula

Use the Pythagorean Theorem to derive the distance formula: \begin{align*}d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}.

It can help to draw in a right triangle, where \begin{align*}d\end{align*} is the hypotenuse. Then, find the lengths of the sides in terms of \begin{align*}x_1\end{align*}, \begin{align*}y_1\end{align*}, \begin{align*}x_2\end{align*}, \begin{align*}y_2\end{align*}. Finally, use the Pythagorean Theorem to find \begin{align*}d\end{align*}.

\begin{align*}d^2 &=(x_2-x_1)^2+(y_2-y_1)^2 \\ d &=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}

**Examples **

**Example 1**

Earlier, you were asked if the length of the three sides given is a right triangle.

The lengths of the three sides of the triangle are 4, 6, and 10. \begin{align*}4^2+6^2=52\end{align*}. \begin{align*}10^2=100\end{align*}. \begin{align*}a^2+b^2 < c^2\end{align*} so this is not a right triangle, it is an obtuse triangle.

#### Example 2

Will a multiple of a Pythagorean triple always also be a Pythagorean triple? For example, “6, 8, 10” is a multiple of “3, 4, 5”. Is “6, 8, 10” a Pythagorean triple? Is any multiple of “3, 4, 5” (or any other Pythagorean triple) also a Pythagorean triple?

Yes. Assume “\begin{align*}a\end{align*}, \begin{align*}b\end{align*}, \begin{align*}c\end{align*}” is a Pythagorean triple, so \begin{align*}a^2+b^2=c^2\end{align*}. “\begin{align*}ka\end{align*}, \begin{align*}kb\end{align*}, \begin{align*}kc\end{align*}” where \begin{align*}k\end{align*} is a whole number is a multiple of this Pythagorean triple.

\begin{align*}(ka)^2+(kb)^2 &=k^2a^2+k^2b^2 \\ &=k^2(a^2+b^2) \\ &=k^2c^2 \\ &=(kc)^2 \\ \end{align*}

Since \begin{align*}(ka)^2+(kb)^2=(kc)^2\end{align*}, “\begin{align*}ka\end{align*}, \begin{align*}kb\end{align*}, \begin{align*}kc\end{align*}” is also a Pythagorean triple.

#### Example 3

Find the distance between \begin{align*}(3, -4)\end{align*} and \begin{align*}(-1, 5)\end{align*}.

Use the Pythagorean Theorem or the distance formula.

\begin{align*}d &=\sqrt{(3-(-1))^2+(-4-5)^2} \\ d &=\sqrt{4^2+(-9)^2} \\ d &=\sqrt{16+81} \\ d &=\sqrt{97} \\ d & \approx 9.85\end{align*}

#### Example 4

The length of one leg of a triangle is 5 and the length of the hypotenuse is 8. What is the length of the other leg?

Use the Pythagorean Theorem.

\begin{align*}a^2+5^2 &=8^2 \\ a^2+25 &=64 \\ a^2 &=39 \\ a & \approx 6.24\end{align*}

### Review

Use the Pythagorean Theorem to solve for \begin{align*}x\end{align*} in each right triangle below.

1.

2.

3.

4.

5.

Three side lengths for triangles are given. Determine whether or not each triangle is right, acute, or obtuse.

6. 2, 5, 6

7. 4, 7, 8

8. 6, 8, 10

9. 6, 9, 10

Find the distance between each pair of points.

10. \begin{align*}(2, 5)\end{align*} and \begin{align*}(1, -3)\end{align*}

11. \begin{align*}(-4.5, 2)\end{align*} and \begin{align*}(1.6, 5)\end{align*}

12. \begin{align*}(-3.7, 2.1)\end{align*} and \begin{align*}(-3.2, -1.5)\end{align*}

13. \begin{align*}(-3, -5)\end{align*} and \begin{align*}(5, 6)\end{align*}

14. Find two more Pythagorean triples that are *not* multiples of “3, 4, 5”.

15. Pick any two whole numbers \begin{align*}m\end{align*} and \begin{align*}n\end{align*} with \begin{align*}n > m\end{align*}. Then \begin{align*}n^2-m^2\end{align*}, \begin{align*}2mn\end{align*}, and \begin{align*}n^2+m^2\end{align*} will be a Pythagorean triple. Test this with a few values of \begin{align*}n\end{align*} and \begin{align*}m\end{align*} and then show why this process works using algebra.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 1.7.