AQA · Mathematics · 8300

GCSE Maths: Intersecting Secant Theorem

Geometry & Measures · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Intersecting Secant 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 chords theorem for chords AC and BD meeting 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 intersect at Y. AY = 12 cm, DY = 8 cm, CY = 4 cm. Find BY.

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    6 cm. AY × CY = BY × DY, so 12 × 4 = 8 × BY, giving BY = 48 ÷ 8 = 6.

  3. Chords AC and BD intersect at Y. AY = 3 cm, CY = 5 cm, DY = 6 cm. Find the length of BD.

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

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

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    6 cm. Since AY = CY = x, x² = 9 × 4 = 36, so x = 6.

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

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    14 cm. Let AY = x: x(20 − x) = 84, so x² − 20x + 84 = 0, (x − 6)(x − 14) = 0; AY > YB gives x = 14.

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

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    120π ≈ 377 cm. PB = 25, so CP × PD = 80 × 25 = 2000. With CP = k, PO = 2k, r = 3k, PD = 5k: 5k² = 2000, k = 20, r = 60, circumference = 2π × 60.

  7. State the intersecting secant theorem for chords CD and EF extended to meet at external point P.

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    PC × PD = PE × PF — for each secant, the distance to the near point times the distance to the far point is the same.

  8. Secants from external point P: CD = 11 cm, PD = 9 cm, PF = 10 cm. Find EF.

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    8 cm. PC = 9 + 11 = 20, so 9 × 20 = 180 = 10 × PE, giving PE = 18 and EF = 18 − 10 = 8.

  9. Secants from external P: CD = 12 cm, PD = 20 cm, PF = 16 cm, radius OE = 15 cm. Find the area of triangle OEF.

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    108 cm². 20 × 32 = 640 = 16 × PE, so PE = 40 and EF = 24. Triangle OEF is isosceles (OE = OF = 15); height = √(15² − 12²) = 9; area = ½ × 24 × 9.

  10. Secants from external P: CD = 12 cm, PF = 13 cm, EF = 15 cm, PD = x. Which quadratic does x satisfy?

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    x² + 12x − 364 = 0. PE = 13 + 15 = 28, so x(x + 12) = 13 × 28 = 364.

  11. Solve x² + 12x − 364 = 0 for a positive length x.

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    x = 14. Discriminant = 144 + 1456 = 1600, so x = (−12 + 40) ÷ 2 = 14 (the negative root is rejected).

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