NZ Level 6 (NZC) Level 1 (NCEA)

PYTHAG - Review

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.

So if two sides of a triangle are known and one side is unknown this relationship can be used to find the length of the unknown side.

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

$c=\sqrt{a^2+b^2}$`c`=√`a`2+`b`2

$b=\sqrt{c^2-a^2}$`b`=√`c`2−`a`2

$a=\sqrt{c^2-b^2}$`a`=√`c`2−`b`2

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 |
$=$= | $10^2+12^2$102+122 |

$c$c |
$=$= | $\sqrt{10^2+12^2}$√102+122 |

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

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**:

$6^2$62 | $=$= | $b^2+4^2$b2+42 |

$b^2$b2 |
$=$= | $6^2-4^2$62−42 |

$b$b |
$=$= | $\sqrt{6^2-4^2}$√62−42 |

$b$b |
$=$= | $4.47$4.47 mm (rounded to $2$2 decimal places) |

Here are some worked examples.

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

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

Give your answer correct to two decimal places.

Find the length of the unknown side in this right-angled triangle, expressing your answer as a decimal approximation to two decimal places.

We can combine this knowledge of Pythagoras, with our existing work on trigonometric ratios.

Remember the Trigonometric Ratios

$\sin\theta=\frac{opposite}{hypotenuse}$`s``i``n``θ`=`o``p``p``o``s``i``t``e``h``y``p``o``t``e``n``u``s``e`

$\cos\theta=\frac{adjacent}{hypotenuse}$`c``o``s``θ`=`a``d``j``a``c``e``n``t``h``y``p``o``t``e``n``u``s``e`

$\tan\theta=\frac{opposite}{adjacent}$`t``a``n``θ`=`o``p``p``o``s``i``t``e``a``d``j``a``c``e``n``t`

We can use these ratios to find unknown side lengths and angles in 2D shapes other than triangles, 3D shapes, as well as real-life situations.

The process is exactly the same. We just need to locate the right-angled triangle, the use to applicable trig ratio to work out the unknown value.

Let's look at some examples so we can see this in action.

Consider a cone with slant height $13$13m and perpendicular height $12$12m.

Find the length of the radius, $r$

`r`, of the base of this cone.Hence, find the length of the diameter of the cone's base.

Find the length of the unknown side, x, in the given trapezium.

Give your answer correct to $2$2 decimal places.

Use trigonometric ratios and Pythagoras’ theorem in two and three dimensions

Apply geometric reasoning in solving problems

Apply right-angled triangles in solving measurement problems