The Pythagorean Theorem

Ontario 09 Academic (MPM1D)

Calculating side lengths using Pythagorean Theorem

Lesson

Pythagorean Theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written algebraically.

$a^2+b^2=c^2$`a`2+`b`2=`c`2

where $c$`c` represents the length of the hypotenuse and $a$`a`, $b$`b` are the two shorter sides. To see why this is true you can check out the lesson here.

We can use the formula to find any side if we know the lengths of the two others.

Remember!

$a^2+b^2$a2+b2 |
$=$= | $c^2$c2 |

other side lengths | hypotenuse |

The value $c$`c` is used to represent the hypotenuse which is the longest side of the triangle. The other two lengths are $a$`a`, $b$`b`.

Use the letters provided to you in the questions, if no letters are provided you can use $a$`a` and $b$`b` for either of the sides.

When you have to round to a required number of decimal places, look to the digit in the following place value column. If that value is 5 or higher, we round up. If that value is less than 5 we round down. This lesson will help you with rounding decimals if you need a refresher.

Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $3$3 cm and $4$4 cm.

Think: Here we want to find $c$`c`, and are given $a$`a` and $b$`b`.

**Do**: We will substitute the values we know in the formula and then solve to find $c$`c`.

$c^2$c2 |
$=$= | $3^2+4^2$32+42 | fill in the values for $a$a and $b$b |

$c^2$c2 |
$=$= | $9+16$9+16 | evaluate the squares |

$c^2$c2 |
$=$= | $25$25 | add the numbers together |

$c$c |
$=$= | $\sqrt{25}$√25 | take square root of both sides |

$c$c |
$=$= | $5$5 cm |

**Reflect**: Becuase all the lengths for the sides of the triangle are whole numbers, the numbers form a Pythagorean Triple, sometimes called a Pythagorean triple, you can read more about the triples here.

Find the length of the hypotenuse, $c$`c` in this triangle.

Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $10$10 cm and $12$12 cm.

**Think**: Here we want to find $c$`c`, and are given $a$`a` and $b$`b`.

**Do**:

$c^2$c2 |
$=$= | $a^2+b^2$a2+b2 |
start with the formula |

$c^2$c2 |
$=$= | $10^2+12^2$102+122 | fill in the values for $a$a and $b$b |

$c^2$c2 |
$=$= | $100+144$100+144 | evaluate the squares |

$c^2$c2 |
$=$= | $244$244 | add the $100$100 and $144$144 together |

$c$c |
$=$= | $\sqrt{244}$√244 | take the square root of both sides |

$c$c |
$=$= | $15.62$15.62 cm | rounded to $2$2 decimal places |

If we need to find one of the shorter side lengths ($a$`a` or $b$`b`), using the formula we will have 1 extra step of rearranging to consider.

We can rearrange the equation in any way to make the unknown side the subject.

Question 4

Find the length of unknown side $b$`b` of a right triangle whose hypotenuse is $10$10 mm and one other side is $6$6 mm.

**Think**: Here we want to find $b$`b`, the length of a shorter side.

**Do**:

$c^2$c2 |
$=$= | $a^2+b^2$a2+b2 |
start with the formula |

$10^2$102 | $=$= | $6^2+b^2$62+b2 |
fill in the values we know |

$b^2$b2 |
$=$= | $10^2-6^2$102−62 | rearrange to get the $b^2$b2 on its own |

$b^2$b2 |
$=$= | $100-36$100−36 | evaluate the right-hand side |

$b^2$b2 |
$=$= | $64$64 | |

$b$b |
$=$= | $8$8 | take the square root of both sides |

**Reflect**: Again, we can see that because all the lengths for the sides of the triangle are whole numbers, the numbers form a Pythagorean Triple, sometimes called a Pythagorean triple, you can read more about the triples here.

Calculate the value of $b$`b` in the triangle below.

Question 6

Find the length of unknown side $b$`b` of a right triangle whose hypotenuse is $6$6 mm and one other side is $4$4 mm.

**Think**: Here we want to find $b$`b`, the length of a shorter side.

**Do**:

$c^2$c2 |
$=$= | $a^2+b^2$a2+b2 |
start with the formula |

$6^2$62 | $=$= | $4^2+b^2$42+b2 |
fill in the values we know |

$b^2$b2 |
$=$= | $6^2-4^2$62−42 | rearrange to get the $b^2$b2 on its own |

$b^2$b2 |
$=$= | $36-16$36−16 | evaluate the right-hand side |

$b^2$b2 |
$=$= | $20$20 | |

$b$b |
$=$= | $\sqrt{20}$√20 | take the square root of both sides |

$b$b |
$=$= | $4.47$4.47 mm | round to the required number of decimal places |

**Reflect**: This gives us a triangle with hypotenuse $6$6 mm and side lengths $4$4 and $4.47$4.47 mm. It's good to check at the end that you haven't ended up with a side length longer than the hypotenuse, as the hypotenuse has to be the longest length.

Calculate the value of $b$`b` in the triangle below.

Give your answer correct to two decimal places.

Consider the triangle below. One of the side lengths is unknown.

Find the length of the unknown side in this triangle, rounding your answer to two decimal places (if necessary).

Solve problems using the Pythagorean Theorem, as required in applications