POP

G4.9 Pythagoras' Theorem

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01 — Explanation

What does Pythagoras' theorem say, and when can I use it?

In any right-angled triangle, $a^2+b^2=c^2$ , where $c$ is the hypotenuse — the longest side, always found directly opposite the $90°$ angle. This only works for right-angled triangles; applying it to an acute or obtuse triangle is a fundamental error, not just an arithmetic one.

Exam Tip

Before doing anything else, locate the right angle and label the side opposite it as $c$ . Getting this one identification right determines whether you add or subtract in the next step — it's the single most important decision in the whole calculation.

02 — Explanation

How do I find the hypotenuse?

Square both known legs, add them, then square-root the total:

c = \sqrt{a^2+b^2}

A frequent slip is stopping at $c^2$ and forgetting the final square-root step — always check your answer is a length, not a length-squared.

03 — Explanation

How do I find one of the shorter sides?

Rearrange to subtract instead of add — and the hypotenuse-squared must always come first in the subtraction:

a = \sqrt{c^2-b^2}

04 — Explanation

How do I use Pythagoras to check if a triangle has a right angle?

This is the converse: label the longest side $c$ , and test whether $a^2+b^2=c^2$ holds exactly. If it does, the triangle is right-angled; if not, it isn't.

05 — Explanation

How do I handle two Pythagoras steps in a row, like a cuboid's space diagonal?

Decompose the 3D shape into two separate 2D right-angled triangles. First find the diagonal of the base rectangle using its length and width as the two legs. Then use that diagonal, together with the cuboid's height, as the two legs of a second right-angled triangle to find the overall space diagonal:

\text{space diagonal} = \sqrt{(\text{base diagonal})^2+h^2}

The same "two-triangle" thinking applies to classic contextual problems — a ladder against a wall, the diagonal across a rectangular field, or the diagonal of a phone screen — the key skill is spotting which two lengths form the legs of the hidden right-angled triangle.

06 — Explanation

Why does precision matter so much in multi-step problems?

If you round an intermediate answer too early, the error compounds through every later step and can push your final answer outside the examiner's accepted range. Keep the full calculator value (or leave it as a surd, e.g. $\sqrt{72}$ ) until the very last step, and round only the final answer — usually to 3 significant figures unless told otherwise.

Exam Tip

Always check units match before calculating (don't mix cm and m), and show your substitution explicitly (e.g. $c^2=6^2+6^2$ ) — this alone secures the method mark even if a later arithmetic slip costs the accuracy mark.

07 — Explanation

How does Pythagoras combine with other topics?

It's frequently the first step in a longer problem — the length you calculate becomes the input for a perimeter, area, or trigonometry calculation that follows. It's also the geometric basis of the coordinate distance formula:

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

08 — Worked example

Worked example

A rectangular field is 48 m long and 20 m wide. Dana walks diagonally across it instead of along two edges. How much shorter is her route, to 1 d.p.?

The diagonal is the hypotenuse of a right triangle with legs 48 m and 20 m: $c=\sqrt{48^2+20^2}=\sqrt{2304+400}=\sqrt{2704}=52$ m.
The edge route is $48+20=68$ m.
Difference: $68-52=16.0$ m shorter.