Pythagoras — the Hypotenuse

Section 1 of 2

The Right-Angled Triangle

The sides

A right angle is $90^\circ$ (a perpendicular, $\perp$). The hypotenuse ($H$) is the longest side, opposite the right angle. The other two are the opposite ($O$) and adjacent ($A$).

Pythagoras’ Theorem

$H^2 = O^2 + A^2$ — the hypotenuse squared equals the sum of the squares of the other two sides.

Section 2 of 2

Finding the Hypotenuse

Worked example — the 3, 4, 5 triangle

The two short sides are $3$ and $4$. Find the hypotenuse $x$.

$x^2 = 3^2 + 4^2 = 9 + 16 = 25$

$x = \sqrt{25} = 5$

$x = 5$

Worked example — a surd answer

The two short sides are $1$ and $1$. Find the hypotenuse $x$.

$x^2 = 1^2 + 1^2 = 2$

$x = \sqrt{2}$

Worked example — short sides are surds

The two short sides are $\sqrt{3}$ and $\sqrt{7}$. Find the hypotenuse $x$.

$x^2 = (\sqrt{3})^2 + (\sqrt{7})^2 = 3 + 7 = 10$

$x = \sqrt{10}$

You try

Short sides $5$ and $12$. Find the hypotenuse $x$.

$x^2 = 5^2 + 12^2$.

$x^2 = 25 + 144 = 169 \Rightarrow x = 13$

$x = 13$

You try

Short sides $9$ and $3$. Find the hypotenuse $x$.

$x^2 = 9^2 + 3^2$.

$x^2 = 81 + 9 = 90 \Rightarrow x = \sqrt{90} = 3\sqrt{10}$

$x = 3\sqrt{10}$

You try

Short sides $5$ and $\sqrt{3}$. Find the hypotenuse $x$.

$x^2 = 5^2 + (\sqrt{3})^2$.

$x^2 = 25 + 3 = 28 \Rightarrow x = \sqrt{28} = 2\sqrt{7}$

$x = 2\sqrt{7}$

That’s Class 5.

Pythagoras to find the hypotenuse. Class 6: finding a shorter side and two-step problems.