What if you had a 52" High Definition Television (52" being the length of the diagonal of the rectangular viewing area)? High Definition Televisions (HDTVs) have sides in the ratio of 16:9. What is the length and width of a 52” HDTV? What is the length and width of an HDTV with a \begin{align*}y''\end{align*}

### Applications of the Pythagorean Theorem

There are many different applications of the Pythagorean Theorem. Three applications are explored below.

##### Find the Height of an Isosceles Triangle

One way to use The Pythagorean Theorem is to identify the heights in isosceles triangles so you can calculate the area. The area of a triangle is \begin{align*}\frac{1}{2} \ bh\end{align*}

If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height.

##### Prove the Distance Formula

Another application of the Pythagorean Theorem is the Distance Formula.

First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to find these distances.

Now that we have a right triangle, we can use the Pythagorean Theorem to find \begin{align*}d\end{align*}

**Distance Formula:** The distance \begin{align*}A(x_1, y_1)\end{align*}

##### Determine if a Triangle is Acute, Obtuse, or Right

We can extend the converse of the Pythagorean Theorem to determine if a triangle has an obtuse angle or is acute. We know that if the sum of the squares of the two smaller sides equals the square of the larger side, then the triangle is right. We can also interpret the outcome if the sum of the squares of the smaller sides does not equal the square of the third.

**Theorem:** (1) If the sum of the squares of the two shorter sides in a right triangle is ** greater** than the square of the longest side, then the triangle is

**. (2) If the sum of the squares of the two shorter sides in a right triangle is**

*acute***than the square of the longest side, then the triangle is**

*less***.**

*obtuse*In other words: The sides of a triangle are \begin{align*}a, b\end{align*}

If \begin{align*}a^2 + b^2 > c^2\end{align*}

If \begin{align*}a^2 + b^2 = c^2\end{align*}

If \begin{align*}a^2 + b^2 < c^2\end{align*}

**Proof of Part 1:**

Given: In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*}, where \begin{align*}c\end{align*} is the longest side.

In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle.

Prove: \begin{align*}\triangle ABC\end{align*} is an acute triangle. (all angles are less than \begin{align*}90^\circ\end{align*})

Statement |
Reason |
---|---|

1. In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*}, and \begin{align*}c\end{align*} is the longest side. In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle. | Given |

2. \begin{align*}a^2 + b^2 = h^2\end{align*} | Pythagorean Theorem |

3. \begin{align*}c^2 < h^2\end{align*} | Transitive PoE |

4. \begin{align*}c < h\end{align*} | Take the square root of both sides |

5. \begin{align*}\angle C\end{align*} is the largest angle in \begin{align*}\triangle ABC\end{align*}. | The largest angle is opposite the longest side. |

6. \begin{align*}m \angle N = 90^\circ\end{align*} | Definition of a right angle |

7. \begin{align*}m \angle C < m \angle N\end{align*} | SSS Inequality Theorem |

8. \begin{align*}m \angle C < 90^\circ\end{align*} | Transitive PoE |

9. \begin{align*}\angle C\end{align*} is an acute angle. | Definition of an acute angle |

10. \begin{align*}\triangle ABC\end{align*} is an acute triangle. | If the largest angle is less than \begin{align*}90^\circ\end{align*}, then all the angles are less than \begin{align*}90^\circ\end{align*}. |

#### Finding the Area of an Isosceles Triangle

What is the area of the isosceles triangle?

First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, find the length of the altitude using the Pythagorean Theorem.

\begin{align*}7^2 + h^2 & = 9^2\\ 49 + h^2 & = 81\\ h^2 & = 32\\ h & = \sqrt{32} = 4 \sqrt{2}\end{align*}

Now, use \begin{align*}h\end{align*} and \begin{align*}b\end{align*} in the formula for the area of a triangle.

\begin{align*}A = \frac{1}{2} \ bh = \frac{1}{2} (14) \left (4 \sqrt{2} \right ) = 28 \sqrt{2} \ units^2\end{align*}

#### Calculating the Distance between Two Points

Find the distance between (1, 5) and (5, 2).

Make \begin{align*}A(1, 5)\end{align*} and \begin{align*}B(5, 2)\end{align*}. Plug into the distance formula.

\begin{align*}d & = \sqrt{(1 - 5)^2 + (5 - 2)^2}\\ & = \sqrt{(-4)^2 + (3)^2}\\ & = \sqrt{16 + 9 } = \sqrt{25} = 5\end{align*}

You might recall that the distance formula was presented as \begin{align*}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\end{align*}, with the first and second points switched. It does not matter which point is first as long as \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are both first in each parenthesis. In Example 7, we could have switched \begin{align*}A\end{align*} and \begin{align*}B\end{align*} and would still get the same answer.

\begin{align*}d & = \sqrt{(5 - 1)^2 + (2 - 5)^2}\\ & = \sqrt{(4)^2 + (-3)^2}\\ & = \sqrt{16 + 9} = \sqrt{25} = 5\end{align*}

Also, just like the lengths of the sides of a triangle, distances are always positive.

#### Classifying Triangles

Determine if the following triangles are acute, right or obtuse.

Set the shorter sides in each triangle equal to \begin{align*} a \end{align*} and \begin{align*} b \end{align*} and the longest side equal to \begin{align*}c\end{align*}.

a)

\begin{align*}6^2 + (3 \sqrt{5})^2 \ &? \ 8^2\\ 36 + 45 \ &? \ 64\\ 81 &> 64\end{align*}

The triangle is acute

b)

\begin{align*}15^2 + 14^2 \ &? \ 21^2\\
225 + 196 \ &? \ 441\\
421 &< 441\end{align*}

The triangle is obtuse.

#### TV Problem Revisited

To find the length and width of a 52” HDTV, plug in the ratios and 52 into the Pythagorean Theorem. We know that the sides are going to be a multiple of 16 and 9, which we will call \begin{align*}n\end{align*}.

\begin{align*}(16n)^2 + (9n)^2 & = 52^2\\ 256n^2 + 81n^2 & = 2704\\ 337n^2 & = 2704\\ n^2 & = 8.024\\ n & = 2.83\end{align*}

Therefore, the dimensions of the TV are \begin{align*}16(2.83'')\end{align*} by \begin{align*}9(2.833'')\end{align*}, or \begin{align*}45.3''\end{align*} by \begin{align*}25.5''\end{align*}. If the diagonal is \begin{align*}y''\end{align*} long, it would be \begin{align*}n \sqrt{337}''\end{align*} long. The extended ratio is \begin{align*}9 : 16 : \sqrt{337}\end{align*}.

### Examples

#### Example 1

Graph \begin{align*}A(-4, 1), B(3, 8)\end{align*}, and \begin{align*}C(9, 6)\end{align*}. Determine if \begin{align*}\triangle ABC\end{align*} is acute, obtuse, or right.

This looks like an obtuse triangle, but we need proof to draw the correct conclusion. Use the distance formula to find the length of each side.

\begin{align*}AB & = \sqrt{(-4-3)^2 + (1 - 8)^2} = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2}\\ BC & = \sqrt{(3 - 9)^2 + (8 - 6)^2} = \sqrt{36 + 4 } = \sqrt{40} = 2 \sqrt{10}\\ AC & = \sqrt{(-4-9)^2 + (1-6)^2} = \sqrt{169 + 25} = \sqrt{194}\end{align*}

Now, let’s plug these lengths into the Pythagorean Theorem.

\begin{align*}\left ( \sqrt{98} \right )^2 + \left ( \sqrt{40} \right )^2 & \ ? \ \left ( \sqrt{194} \right )^2\\ 98 + 40 & \ ? \ 194\\ 138 & < 194\end{align*}

\begin{align*}\triangle ABC\end{align*} is an obtuse triangle.

#### Example 2

Do the lengths 7, 8, 9 make a triangle that is right, acute, or obtuse?

Acute because \begin{align*}7^2 + 8^2>9^2\end{align*}.

#### Example 3

Do the lengths 14, 48, 50 make a triangle that is right, acute, or obtuse?

Right because \begin{align*}14^2+48^2=50^2\end{align*}

### Review

Find the area of each triangle below. Round your answers to the nearest tenth.

Find the length between each pair of points.

- (-1, 6) and (7, 2)
- (10, -3) and (-12, -6)
- (1, 3) and (-8, 16)
- What are the length and width of a \begin{align*}42''\end{align*} HDTV? Round your answer to the nearest tenth.
- Standard definition TVs have a length and width ratio of 4:3. What are the length and width of a \begin{align*}42''\end{align*} Standard definition TV? Round your answer to the nearest tenth.
An equilateral triangle is an isosceles triangle. If all the sides of an equilateral triangle are \begin{align*}s\end{align*}, find the area, using the technique learned in this section. Leave your answer in simplest radical form.*Challenge*- Find the area of an equilateral triangle with sides of length 8.

- The two
*shorter*sides of a triangle are 9 and 12.- What would be the length of the third side to make the triangle a right triangle?
- What is a possible length of the third side to make the triangle acute?
- What is a possible length of the third side to make the triangle obtuse?

- The two
*longer*sides of a triangle are 24 and 25.- What would be the length of the third side to make the triangle a right triangle?
- What is a possible length of the third side to make the triangle acute?
- What is a possible length of the third side to make the triangle obtuse?

- The lengths of the sides of a triangle are \begin{align*}8x, 15x,\end{align*} and \begin{align*}17x\end{align*}. Determine if the triangle is acute, right, or obtuse.

Determine if the following triangles are acute, right or obtuse.

- 5, 12, 15
- 13, 84, 85
- 20, 20, 24
- 35, 40, 51
- 39, 80, 89
- 20, 21, 38
- 48, 55, 76

Graph each set of points and determine if \begin{align*}\triangle ABC\end{align*} is acute, right, or obtuse.

- \begin{align*}A(3, -5), B(-5, -8), C(-2, 7)\end{align*}
- \begin{align*}A(5, 3), B(2, -7), C(-1, 5)\end{align*}
Explain the two different ways you can show that a triangle in the coordinate plane is a right triangle.*Writing*

### Review (Answers)

To view the Review answers, open this PDF file and look for section 8.2.