IGCSE Maths 0580 Pythagoras’ Theorem Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Trigonometry · 6.1

Every pythagoras’ theorem question from 9 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 6.1. 9 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/23/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Trigonometry past paper questions.

0580 Pythagoras’ Theorem past paper questions: Core (Papers 1 and 3)

12 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q26

The diagram shows a right-angled triangle with shorter sides 11 cm and 18 cm and hypotenuse x cm. Calculate the value of x.

0580/31 May/June 2026 Q26 – right-angled triangle with sides 11 cm, 18 cm and x cm11 cm18 cmx cmNOT TOSCALE
Show solution
  1. x is the hypotenuse (opposite the right angle), so use Pythagoras: x² = 11² + 18².
  2. x² = 121 + 324 = 445.
  3. x = √445 = 21.09… = 21.1
Answer 21.1
22 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q17(b)

A compound shape is made from two rectangles. The straight line from X to Y is the hypotenuse of a right-angled triangle with sides 3 cm and 5 cm. Calculate the length of the straight line from X to Y.

0580/32 May/June 2026 Q17 – L-shaped compound shape with sides 12 cm, 6 cm, 9 cm and 7 cm, and points X and Y12 cm6 cm9 cm7 cmXYNOT TOSCALE
Show solution
  1. XY is the hypotenuse of a right-angled triangle with shorter sides 3 cm and 5 cm.
  2. Pythagoras: XY² = 3² + 5² = 9 + 25 = 34.
  3. XY = √34 = 5.830… = 5.83 cm
Answer 5.83 cm
33 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q25

A right-angled triangle has base 21 cm. The area of this triangle is 147 cm². Calculate the length of the longest side.

0580/32 February/March 2026 Q25 – right-angled triangle with base 21 cm21 cmNOT TOSCALE
Show solution
  1. Area of a triangle = 1⁄2 × base × height, so 1⁄2 × 21 × h = 147.
  2. 10.5h = 147, so h = 147 ÷ 10.5 = 14 cm.
  3. The longest side is the hypotenuse. Use Pythagoras: √(21² + 14²) = √(441 + 196) = √637.
  4. √637 = 25.23… = 25.2 cm
Answer 25.2 cm
44 marksCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q23

A right-angled trapezium has parallel sides 6 cm and 14 cm and perpendicular height 15 cm. Calculate the perimeter of this trapezium.

0580/32 October/November 2025 Q23 – right-angled trapezium with parallel sides 6 cm and 14 cm and height 15 cm6 cm14 cm15 cmNOT TOSCALE
Show solution
  1. Draw a line down from the end of the 6 cm side to split off a right-angled triangle.
  2. The triangle has height 15 cm and base 14 − 6 = 8 cm. The slanted side is its hypotenuse.
  3. Slant² = 15² + 8² = 225 + 64 = 289, so the slant = √289 = 17 cm.
  4. Perimeter = 14 + 15 + 6 + 17 = 52 cm
Answer 52 cm
55 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q16

An isosceles triangle has base 4 cm and perpendicular height 5 cm. Each sloping side is x cm.

0580/33 October/November 2025 Q16 – isosceles triangle with base 4 cm, height 5 cm and sloping side x cm5 cm4 cmx cmNOT TOSCALE

(a) Calculate the value of x.

(b) The triangle is one face of a square-based pyramid. Each triangular face is the same. On the 1 cm² grid, draw a net of this pyramid.

Show solution
Part (a)
  1. The height cuts the isosceles triangle into two right-angled triangles. Each has base 4 ÷ 2 = 2 cm and height 5 cm.
  2. The sloping side x is the hypotenuse: x² = 5² + 2² = 25 + 4 = 29.
  3. x = √29 = 5.385… = 5.39
Answer 5.39
Part (b)
  1. A square-based pyramid has 1 square face and 4 triangular faces.
  2. Draw a 4 cm by 4 cm square with an isosceles triangle (base 4 cm, height 5 cm) on each of its four sides
0580/33 October/November 2025 Q16(b) solution – net of the square-based pyramid
Answer a 4 cm by 4 cm square with an isosceles triangle (base 4 cm, height 5 cm) on each of its four sides

0580 Pythagoras’ Theorem past paper questions: Extended (Papers 2 and 4)

62 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q13(a)

A, B and C are points on a circle with centre O. AOC is a straight line and M is the midpoint of AB. AB = 6 cm and OM = 4 cm. Work out OA.

0580/23 May/June 2026 Q13 – circle centre O with diameter AC, chord AB and midpoint MOABCMNOT TOSCALE
Show solution
  1. The line from the centre to the midpoint of a chord is perpendicular to the chord, so angle AMO = 90°.
  2. M is the midpoint of AB, so AM = 6 ÷ 2 = 3 cm.
  3. Use Pythagoras in triangle AMO: OA² = AM² + OM² = 3² + 4² = 9 + 16 = 25.
  4. OA = √25 = 5 cm
Answer 5 cm
73 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q7(a)

The point A is on the top of a vertical building of height h m. The horizontal distance from the base of the building to B is 22 m. AB = 25 m. Find the value of h.

0580/43 May/June 2026 Q7 – vertical building with A at the top, B on the ground 22 m away, AB = 25 m22 m25 mh mBANOT TOSCALE
Show solution
  1. The building, the ground and AB make a right-angled triangle. AB = 25 m is the hypotenuse.
  2. Pythagoras: h² + 22² = 25².
  3. h² = 625 − 484 = 141.
  4. h = √141 = 11.87… = 11.9
  5. Tip: To find a shorter side, subtract the squares. To find the hypotenuse, add them.
Answer 11.9
83 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q9

A right-angled trapezium has parallel sides 21 cm and 15 cm, and the perpendicular side is 8 cm. Work out the perimeter of the trapezium.

0580/22 February/March 2026 Q9 – right-angled trapezium with parallel sides 21 cm and 15 cm and height 8 cm21 cm15 cm8 cmNOT TOSCALE
Show solution
  1. Draw a line from the end of the shorter parallel side down to the longer side. This makes a right-angled triangle.
  2. Its sides are 8 cm (the height) and 21 − 15 = 6 cm (the extra length of the longer side).
  3. The sloping side is the hypotenuse: √(8² + 6²) = √(64 + 36) = √100 = 10 cm.
  4. Perimeter = 21 + 15 + 8 + 10 = 54 cm
Answer 54 cm
94 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q10

Triangle ABC is right-angled at A. AB = 8 cm and BC = 11 cm. Angle ABC = x°.

0580/23 October/November 2025 Q10 – right-angled triangle ABC with AB = 8 cm, BC = 11 cm and angle x° at Bx°8 cm11 cmABCNOT TOSCALE

(a) Work out the exact length of AC. (b) cos x = k. Write down the value of k.

Show solution
Part (a)
  1. The right angle is at A, so BC is the hypotenuse (the longest side).
  2. Pythagoras: AC² + AB² = BC², so AC² = 11² − 8² = 121 − 64 = 57.
  3. Exact length: AC = √57 cm
Answer √57 cm
Part (b)
  1. From angle x at B: AB is adjacent and BC is the hypotenuse.
  2. cos x = adjacent⁄hypotenuse = 8⁄11, so k = 8⁄11
Answer k = 8⁄11

Question papers used on this page

All pythagoras’ theorem questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q13(a)
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q26
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q17(b)
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q7(a)
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q9
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q25
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q10
0580/32/O/N/250580_w25_qp_32Oct/Nov 2025 · Paper 3 (Core)Q23
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q16

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