In any right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides.
$$c^2 = a^2 + b^2$$
where $c$ is the hypotenuse and $a$, $b$ are the two shorter sides (legs).
Rearranging to find a leg: $a = \sqrt{c^2 - b^2}$.
The hypotenuse is always:
- Opposite the right angle (marked with a small square)
- The longest side in the triangle
A ladder leans against a wall. The base is 1.8 m from the wall and reaches 4.2 m up the wall. How long is the ladder?
$$c = \sqrt{(1.8)^2 + (4.2)^2} = \sqrt{3.24 + 17.64} = \sqrt{20.88} \approx 4.57 \text{ m}$$
A rectangular paddock is 120 m long and has a diagonal of 150 m. Find the width.
$$w = \sqrt{150^2 - 120^2} = \sqrt{22500 - 14400} = \sqrt{8100} = 90 \text{ m}$$
Pythagoras’ theorem extends to 3D problems by applying it in stages.
Diagonal of a rectangular box with dimensions $l$, $w$, $h$:
$$d = \sqrt{l^2 + w^2 + h^2}$$
Worked Example: A box is 6 cm × 4 cm × 3 cm. Find the space diagonal.
$$d = \sqrt{6^2 + 4^2 + 3^2} = \sqrt{36 + 16 + 9} = \sqrt{61} \approx 7.81 \text{ cm}$$
Alternatively, find the base diagonal first: $\sqrt{36+16} = \sqrt{52}$, then apply Pythagoras again: $\sqrt{52 + 9} = \sqrt{61}$.
Common whole-number solutions worth memorising:
| $(a, b, c)$ | Multiples |
|---|---|
| (3, 4, 5) | (6,8,10), (9,12,15) |
| (5, 12, 13) | (10,24,26) |
| (8, 15, 17) | — |
Recognising triples can save calculation time.
REMEMBER: Pythagoras’ theorem only applies to right-angled triangles. Always confirm the right angle exists before applying the formula.
EXAM TIP: In 3D problems, draw a 2D cross-section showing the right angle. Label all known lengths before applying the theorem. Never try to visualise 3D geometry without a diagram.
