AQA · Mathematics · 8300

GCSE Maths: Intersecting Chord Theorem

Geometry & Measures · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Intersecting Chord Theorem (Corbettmaths Video 368), and every card links back to the exact page it came from. Practice tests generate fresh variations of these questions, so the numbers change every attempt. Source material © Corbettmaths — used with attribution.

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  1. State the intersecting chord theorem for chords AC and BD crossing at Y inside a circle.

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    AY × CY = BY × DY — the products of the two segments of each chord are equal.

  2. Chords AC and BD meet at Y. AY = 12cm, CY = 4cm, DY = 8cm. Find BY.

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    6cm. AY × CY = BY × DY → 12 × 4 = 8 × BY → BY = 48 ÷ 8 = 6.

  3. Chords AC and BD meet at Y. AY = 3cm, CY = 5cm, DY = 6cm. Find BD.

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    8.5cm. 3 × 5 = 6 × BY → BY = 2.5cm, so BD = BY + DY = 2.5 + 6 = 8.5.

  4. Chord BD bisects chord AC at Y. BY = 9cm, DY = 4cm. Find CY.

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    6cm. Bisecting means AY = CY = x, so x² = 9 × 4 = 36 → x = 6.

  5. Chords AB and CD meet at Y. DY = 12cm, CY = 7cm, AB = 20cm, AY > YB. Find AY.

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    14cm. AY × YB = 12 × 7 = 84 and AY + YB = 20; the pair 14 and 6 works, so AY = 14.

  6. Chord AB (105cm) meets diameter CD at P, AP = 80cm, CP:PO = 1:2. Find the circumference.

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    120π ≈ 377cm. With radius r: CP = r/3, PD = 5r/3, so (r/3)(5r/3) = 80 × 25 = 2000 → r² = 3600 → r = 60.

  7. State the intersecting chord theorem when chords CD and EF meet at P outside the circle.

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    PC × PD = PE × PF — each product uses the distances from P to both ends of the same chord.

  8. Chords CD and EF meet at P outside the circle. CD = 11cm, PD = 9cm, PF = 10cm. Find EF.

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    8cm. PC = 20, so 9 × 20 = 10 × PE → PE = 18, and EF = PE − PF = 18 − 10 = 8.

  9. Chords CD and EF meet at external point P. CD = 12cm, PD = 20cm, PF = 16cm, radius 15cm. Find the area of triangle OEF.

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    108cm². 20 × 32 = 16 × PE → PE = 40, so EF = 24. Isosceles triangle OEF with OE = OF = 15 has height √(15² − 12²) = 9, so area = ½ × 24 × 9 = 108.

  10. Chords CD and EF meet at external point P. CD = 12cm, PF = 13cm, EF = 15cm, PD = x. Form and solve the equation for x.

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    x(x + 12) = 13 × 28 → x² + 12x − 364 = 0 → (x + 26)(x − 14) = 0 → x = 14.

  11. When a chord passes through the centre of a circle, its length is the ?, equal to twice the radius.

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    diameter

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