Pythagoras' Theorem in 2D

SPEC G20 — GEOMETRY & MEASURES

Finding sides of right-angled triangles, recognising Pythagorean triples, distance between two points, and exact surd answers — one of the most-used results in GCSE mathematics.

What You'll Learn

Pythagoras' theorem appears on every paper, often inside a longer multi-step problem. The theorem itself is short to state, but the most common error — adding when you should subtract — accounts for nearly three-quarters of the marks lost on these questions.

By the end of this lesson, you will be able to:

State Pythagoras' theorem: a² + b² = c², where c is the hypotenuse

Identify the hypotenuse as the side opposite the right angle (always the longest side)

Find the hypotenuse from two given legs

Find a missing leg by rearranging — subtraction, not addition

Recognise common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) to skip computation

Give exact answers in surd form when the result isn't an integer

Use Pythagoras for the distance between two coordinate points

Identify and correct typical errors in another student's working

Why It's High Priority

Pythagoras is one of the few results you must memorise (it is not on the GCSE formula sheet). It also underpins right-angled trigonometry (Lesson 4.16), the slant height of a cone (Lesson 4.14), 3D Pythagoras at Higher tier (Lesson 4.17), and distance calculations across the spec.

The Theorem & The Hypotenuse

a² + b² = c²

For any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The labels: a and b are the two legs (sides forming the right angle); c is the hypotenuse.

Identify the Hypotenuse First

The hypotenuse is always opposite the right angle — and always the longest side. Label it c before you do anything else. The other two sides (the legs) are a and b; either order works for them.

Right Angle Required

Pythagoras' theorem only works for right-angled triangles. If the triangle has no right angle, you need the sine or cosine rule (Higher tier — Lesson 4.18). Check for the right-angle marker in the diagram before applying the theorem.

Finding the Hypotenuse

If you know both legs and want the hypotenuse, the formula goes the "forward" direction — add the two squares, then take the square root.

c = √(a² + b²)

Step-by-step routine

Identify which side is the hypotenuse (opposite the right angle)

Write a² + b² = c² with the two known legs as a and b

Compute the squares: a² and b²

Add them to get c²

Square-root to get c

Quote with units (cm, m, etc.)

Exam-paper Pattern

AQA papers love this question: a triangle with legs given as integers, asking for the hypotenuse with answer expected in centimetres or as a surd. Examples: legs 3 and 4 → c = 5; legs 5 and 12 → c = 13; legs 7 and 24 → c = 25.

01Worked Example — Find the Hypotenuse

A right-angled triangle has legs of length 5 cm and 12 cm. Find the length of the hypotenuse. (2 marks)

MMethod. The two legs are 5 and 12, so a² + b² = c² with a = 5, b = 12.

CCalculate.

c² = 5² + 12² = 25 + 144 = 169

c = √169 = 13

EEvaluate. The hypotenuse is the longest side — and 13 > 12 > 5 ✓. The triangle has sides 5, 12, 13 — a famous Pythagorean triple. Linear units (cm).

c = 13 cm

Finding a Shorter Side

If you know the hypotenuse and one leg, you must rearrange the theorem. The key change: you subtract, not add.

a = √(c² − b²)

The Most Common Pythagoras Error

Examiners' reports flag this as the dominant Pythagoras error: when finding a leg, students add the two given squares instead of subtracting. The correct approach: hypotenuse² = sum of legs², so to find a leg, subtract the other leg² from hypotenuse².

When to add vs subtract

Finding the hypotenuse (longest side, opposite right angle): ADD the leg² values, then square-root

Finding a leg (shorter side, forming the right angle): SUBTRACT the known leg² from hypotenuse², then square-root

Quick Sanity Check

Whatever you compute, the hypotenuse must end up as the longest side. If your "leg" answer is bigger than the hypotenuse, you've added when you should have subtracted.

02Worked Example — Ladder Against a Wall

A 5 m ladder leans against a vertical wall. The foot of the ladder is 1.4 m from the base of the wall. How far up the wall does the top of the ladder reach? Give your answer to 2 d.p. (3 marks)

MMethod. The ladder is the hypotenuse (5 m, opposite the right angle between wall and ground). The base is one leg (1.4 m). The height up the wall is the other leg, h. Use a² + b² = c² rearranged to find a leg: h² = c² − b². Subtract.

CCalculate.

h² = 5² − 1.4² = 25 − 1.96 = 23.04

h = √23.04 = 4.8 m (exact)

EEvaluate. Sense-check: h = 4.8 m must be less than the ladder length 5 m ✓. Convert to 2 d.p. as required: 4.80 m. The ladder reaches 4.80 m up the wall.

h = 4.80 m

Pythagorean Triples — Recognise & Skip

A Pythagorean triple is a set of three integers a, b, c that satisfy a² + b² = c². Recognising one saves computation time on every paper.

Triplea²b²c²Verify

3, 4, 5916259 + 16 = 25 ✓

5, 12, 132514416925 + 144 = 169 ✓

8, 15, 176422528964 + 225 = 289 ✓

7, 24, 254957662549 + 576 = 625 ✓

Multiples Work Too

Any multiple of a triple is also a triple. So 6, 8, 10 (= 2 × 3-4-5) and 9, 12, 15 (= 3 × 3-4-5) and 10, 24, 26 (= 2 × 5-12-13) all satisfy a² + b² = c². If you spot any of these patterns, you can skip the calculation and write the answer directly.

When triples don't apply

Most exam triangles are not nice triples. If your numbers don't match a triple immediately, just compute a² + b² (or c² − b²) directly — don't waste time hunting for a pattern that isn't there.

Converse — Checking if a Triangle Is Right-Angled

Pythagoras works in both directions: if the three sides of a triangle satisfy a² + b² = c² (with c the longest), the triangle must be right-angled — with the right angle opposite the longest side. So given three sides, you can test whether the triangle is right-angled by squaring all three and checking the equation.

Example: a triangle has sides 9, 40, 41. Test: 9² + 40² = 81 + 1600 = 1681; 41² = 1681. Equal ✓ — so the triangle is right-angled (with the right angle opposite the side of length 41).

Exact Surd Answers

When a² + b² doesn't give a perfect square, the hypotenuse is a surd (irrational). At GCSE, leave it in the exact form √n unless the question asks for a decimal.

Quick demo — legs 4 and 7

c² = 4² + 7² = 16 + 49 = 65

c = √65 cm (exact)

Since 65 = 5 × 13 has no perfect-square factors greater than 1, √65 is already fully simplified — you cannot write it as k√m with smaller m.

Simplifying Surds When Possible

If the radicand has a perfect-square factor, simplify. For example, √50 = √(25 × 2) = 5√2. But √65 (and similarly √34) has no such factor, so it stays as √65. The lesson on surds (1.10 — Surds) covers full simplification.

"Exact" vs "Decimal"

Read the question carefully. "Exact" or "in surd form" means leave as √n (or k√m). "To 2 d.p." or "to 3 s.f." means convert to a decimal. The same calculation answers both — only the final form changes.

03Worked Example — Surd Answer

A right-angled triangle has legs of 3 cm and 5 cm. Find the length of the hypotenuse, giving your answer in exact surd form. (3 marks)

MMethod. Both legs are given (3 and 5), and we want the hypotenuse. Use c² = a² + b² (add direction).

CCalculate.

c² = 3² + 5² = 9 + 25 = 34

c = √34 cm

EEvaluate. 34 = 2 × 17 — no perfect-square factors, so √34 is fully simplified. Sense-check: √34 ≈ 5.83, which is greater than 5 (the longest leg) and consistent with the diagonal-style longest side ✓.

c = √34 cm

Distance Between Two Coordinate Points

Pythagoras applied to coordinate geometry: the straight-line distance between two points is just the hypotenuse of the right triangle formed by their horizontal and vertical separations.

distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

The construction

Given points A(x₁, y₁) and B(x₂, y₂), draw the line from A to B. Then drop a vertical from B and a horizontal from A — they meet at a corner forming a right angle. The right triangle has:

Horizontal leg = (x₂ − x₁) — the difference in x-coordinates

Vertical leg = (y₂ − y₁) — the difference in y-coordinates

Hypotenuse = the distance from A to B

Squaring Removes the Sign

Either order in the subtraction works — (x₂ − x₁) and (x₁ − x₂) give the same value once squared. So you don't need to worry about getting "negative" differences. Just pick one order and stay consistent.

04Worked Example — Distance Between Two Points

Find the distance between the points A(−2, 3) and B(4, 11). (3 marks)

MMethod. Compute Δx = x₂ − x₁ and Δy = y₂ − y₁; these are the legs of the right triangle on the grid. The distance d is the hypotenuse: d = √(Δx² + Δy²).

CCalculate.

Δx = 4 − (−2) = 6

Δy = 11 − 3 = 8

d² = 6² + 8² = 36 + 64 = 100

d = √100 = 10

EEvaluate. The numbers 6, 8, 10 form a Pythagorean triple — it's just the 3-4-5 triple scaled up by 2. Linear units (no specific unit given, so just "10 units").

Distance AB = 10 units

Pythagoras in Other Shapes

Pythagoras only works on right-angled triangles — but you can create a right triangle inside many other shapes by drawing a perpendicular.

Common constructions

Diagonal of a rectangle: the diagonal is the hypotenuse of a right triangle whose legs are the rectangle's sides

Altitude of an isosceles triangle: drop a perpendicular from the apex to the base — it bisects the base, creating two congruent right triangles

Altitude of an equilateral triangle: same as isosceles, but the half-base equals half the side length

Slant height of a cone: right triangle with legs r (radius) and h (perpendicular height) — the hypotenuse is the slant l (Lesson 4.14)

Diagonal of a regular polygon can sometimes be split into right triangles

Strategy — Find the Right Triangle

If the question doesn't have an obvious right-angled triangle, look for ways to construct one. Common moves: drop a perpendicular from a vertex, draw a diagonal, or split a symmetrical shape down its axis. Then label the right triangle's three sides and apply Pythagoras.

05Worked Example — Isosceles Triangle Altitude

An isosceles triangle has a base of 8 cm and two equal sides of 5 cm. Find the perpendicular height (altitude) of the triangle. (3 marks)

MMethod. Drop the altitude from the apex perpendicular to the base. By symmetry, this altitude bisects the base, splitting the isosceles triangle into two congruent right triangles. Each right triangle has hypotenuse 5 cm (the original equal side), one leg of 4 cm (half the base), and the altitude h as the other leg. Apply Pythagoras to find h.

CCalculate.

In the right triangle: 5² = 4² + h²

Rearrange: h² = 25 − 16 = 9

h = √9 = 3 cm

EEvaluate. The three lengths 3, 4, 5 form a Pythagorean triple — that's why the altitude is exactly 3 cm. Sense-check: the altitude must be less than the equal side (5) since it's a leg, not the hypotenuse, of the right triangle ✓.

Altitude h = 3 cm

Spotting Errors in Working

AQA papers regularly include "identify and correct the error" questions. The errors come from a small set of typical mistakes — once you know them, you can spot them quickly.

The four classic Pythagoras errors

Adding when looking for a leg. "Hypotenuse² + leg² = ?" instead of "hypotenuse² − leg²".

Calling the unknown "c" when it isn't the hypotenuse. The letter c is reserved for the hypotenuse — calling a leg "c" then writing a² + b² = c² puts the wrong value in the wrong slot.

Forgetting to square-root at the end. Writing c² = 169 then quoting c = 169.

Squaring sides incorrectly. 12² = 12 + 12 = 24 (a common error — should be 12 × 12 = 144).

Examiners' Reports Say

On Nov24 P1 Q13(b), nearly three quarters of students who lost marks did so by either using subtraction where addition was needed, or by adding squares incorrectly. This is the single biggest source of lost Pythagoras marks.

06Worked Example — Identify and Correct an Error

A right-angled triangle has hypotenuse 17 cm and one leg 8 cm. A student wants to find the other leg and writes:

17² + 8² = c² 289 + 64 = 353 c = √353 ≈ 18.79

(a) Identify two errors in the student's working.

(b) Find the correct length of the missing leg. (3 marks)

MMethod. The triangle has hypotenuse 17 and one leg 8 — we need the other leg. Since we're finding a leg, the formula is rearranged: leg² = hypotenuse² − other leg² (subtract).

CCalculate (find the errors and correct).

(a) Two errors:

Error 1: the student added the squares (17² + 8²). They should have subtracted because the unknown is a leg, not the hypotenuse.

Error 2: the student called the unknown "c". The letter c is reserved for the hypotenuse — the unknown here is a leg, so it should be a (or b).

(b) Correct working:

Let the unknown leg be a.

a² + 8² = 17²

a² = 17² − 8² = 289 − 64 = 225

a = √225 = 15

EEvaluate. Sense-check: the leg 15 must be less than the hypotenuse 17 ✓. The triangle has sides 8, 15, 17 — another famous Pythagorean triple.

Missing leg = 15 cm

Examiner Alerts

Pythagoras marks are mostly lost to a small set of recurring errors:

Adding Instead of Subtracting (Most Common)

What students write: hypotenuse² + leg² = (other leg)² when finding a leg. What to write instead: hypotenuse² − (known leg)² = (unknown leg)². Subtract because the hypotenuse is on its own side of the equation, with the two legs on the other side.

Pythagoras on Non-Right Triangles

What students write: a² + b² = c² for a triangle with all acute angles. What to write instead: Pythagoras only works for right-angled triangles. Look for the small square symbol marking the right angle. For non-right triangles, use the cosine rule (Higher tier — Lesson 4.18).

Forgetting the Square Root

What students write: c² = 25 + 144 = 169 → c = 169. What to write instead: c = √169 = 13. The theorem gives you c² — you must square-root to get c.

Squaring Errors

What students write: 12² = 24 (treating the exponent as multiplication by 2). What to write instead: 12² = 12 × 12 = 144. The exponent 2 means "multiply by itself," not "double".

Mislabelling the Hypotenuse

What students write: calling a leg "c" because it's the longest given length. What to write instead: the hypotenuse is the side opposite the right angle — and only that side is c. If two legs are given without the hypotenuse, neither is c; the answer you compute (a² + b²)^(1/2) IS c.

Lesson Checkpoint

You should now be able to:

The Theorem (memorise — not on the formula sheet)

Pythagoras: a² + b² = c², where c is the hypotenuse

Find hypotenuse: c = √(a² + b²) — ADD

Find a leg: a = √(c² − b²) — SUBTRACT

Distance between (x₁, y₁) and (x₂, y₂): √[(x₂−x₁)² + (y₂−y₁)²]

Pythagorean Triples (memorise the common ones)

3, 4, 5 — and its multiples (6, 8, 10; 9, 12, 15)

5, 12, 13 — and 10, 24, 26

8, 15, 17

7, 24, 25

Discipline

Always identify the hypotenuse first — opposite the right angle, longest side

ADD for hypotenuse; SUBTRACT for a leg

Don't forget the square root at the end

Pythagoras only works for right-angled triangles — check the right-angle marker

Leave answers in exact surd form when "exact" is requested; convert to decimal only when explicitly asked

Up Next

Lesson 4.16 covers Right-Angled Trigonometry — SOH CAH TOA: using sine, cosine and tangent to find sides and angles in right-angled triangles, building directly on the right-triangle setup you've used here.