In any right-angled triangle, a2 + b2 = c2. Here's how to use it to find a missing side — and the one thing you must check first.
Also known as: Pythagorean theorem · Pythagorean triples · right triangle
Pythagoras' theorem was introduced in Grade 8; in Grade 9 you revise it and use it inside bigger figures — two triangles joined together, quadrilaterals and boxes. It is one of the most useful results in all of maths. It says that in a right-angled triangle, the square of the longest side equals the sum of the squares of the other two sides.
Here c is the hypotenuse — the side opposite the right angle, and always the longest. a and b are the two shorter sides.
1Finding the hypotenuse
- 1Write the theorem: c2 = a2 + b2.You're looking for the longest side, so c is the unknown.
- 2Substitute: c2 = 32 + 42 = 9 + 16 = 25.Square each short side and add.
- 3Square-root both sides: c = √25 = 5.Undo the square to get the length.
2Finding a shorter side
If the hypotenuse is known and you want a shorter side, rearrange: a2 = c2 − b2. Subtract instead of add.
Only add when finding the hypotenuse. To find a shorter side you must subtract the squares. Mixing this up gives an impossible triangle (or a maths error).
- 1The unknown side is NOT the hypotenuse, so subtract: a2 = 132 − 52.The hypotenuse is the longest side, so its square is the biggest.
- 2a2 = 169 − 25 = 144.
- 3a = √144 = 12 cm.Check: 52 + 122 = 25 + 144 = 169 = 132 ✓
3Is the triangle right-angled? (the converse)
The theorem also works backwards. If the square of the longest side equals the sum of the squares of the other two sides, the triangle is right-angled, and the right angle is opposite the longest side. If the two totals are different, it is not right-angled.
- 1The longest side is AC = 17, so AC2 = 289.Only the longest side can be the hypotenuse.
- 2AB2 + BC2 = 64 + 225 = 289.Add the squares of the other two sides.
- 3AB2 + BC2 = AC2, so △ABC is right-angled at B, opposite AC.Reason: converse of Pythagoras.
4Two right-angled triangles in one figure
Grade 9 figures often join two right-angled triangles along a shared side. Work in one triangle at a time, and use what you find in the first triangle in the second.
- 1BC = BD + DC = 3 + 5 = 8 cm.The bottom side is shared by both triangles.
- 2In △ABC: AC2 = AB2 + BC2 = 36 + 64 = 100, so AC = 10 cm.Pythagoras in the big triangle (right angle at B).
- 3In △ABD: AD2 = AB2 + BD2 = 36 + 9 = 45, so AD = √45 ≈ 6,71 cm.The same right angle at B belongs to the small triangle too.
5Using it twice: the diagonal of a box
- 1Base diagonal: AC2 = AB2 + BC2 = 62 + 32 = 45.The floor is a rectangle, so AC is the hypotenuse of right-angled △ABC. Keep AC2 = 45 — there is no need to round AC.
- 2CG stands straight up from C, so △ACG is right-angled at C.A vertical edge is perpendicular to the floor.
- 3AG2 = AC2 + CG2 = 45 + 22 = 49, so AG = 7 cm.Pythagoras a second time, in the upright triangle.
Pythagoras only works in a right-angled triangle. Always check there's a 90° angle before using it.
This theorem returns in Grade 10–12 analytical geometry (the distance formula is Pythagoras in disguise) and trigonometry, so it's well worth mastering now.
Practice exercises
Work each one out, then click to reveal the answer.
- 1A right triangle has legs 6 and 8. Find the hypotenuse.
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c2 = 36 + 64 = 100 → c = 10 - 2Legs of 5 and 12 — find the hypotenuse.
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c2 = 25 + 144 = 169 → c = 13 - 3Hypotenuse 10, one leg 6. Find the other leg.
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b2 = 100 − 36 = 64 → b = 8 - 4Is a triangle with sides 3, 4, 6 right-angled?
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32 + 42 = 25 but 62 = 36 — not equal, so no
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