Pythagoras' Theorem

UK Secondary (7-11)

PYTHAG - Calculating Other Side Lengths

Lesson

In a previous lesson, we saw that Pythagorean Theorem states that in a * right-angled triangle *:

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.

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.

We've already been looking at finding side lengths of right-angled triangles here, and in that set, all the answers ended up being whole number values. This meant that the sides of the triangles were forming Pythagorean Triples. That isn't always the case, so now we look at finding the hypotenuse in other cases, we will need to remember how to round to the required number of decimal places.

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.

Question 1

Find the length of unknown side $b$`b` of a right-angled 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).