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*}**\begin{align*}c\end{align*} c, \begin{align*}a^2+b^2=c^2\end{align*}a2+b2=c2.**

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*}

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.

**Solve the following problems**

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*}

\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.

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*}

In this case, \begin{align*}4^2+8^2=80\end{align*}

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.

### Answers for Review Problems

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