The diagram shows a circle with centre A and radius r passing through points B, C and D. A larger circle of radius s has centre C and passes through B and D. The length BD is also s. [Figure X.X]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [M1] Forms a correct trigonometric or geometric equation relating r and s (e.g., cosine rule s² = r²+r²-2r²cos(∠BAC), or considering a right-angled triangle).
2. [A1] Shows s = √3r (AG - convincing proof with correct values of trig ratios if used).
3. [B1] Angle BAD = 2π/3 (or 120°) (seen or implied).
4. [B1] Area of sector DAB = (1/2)r²(2π/3) = πr²/3.
5. [B1] Angle BCD = π/3 (or 60°) (angle subtended by chord BD at centre C of larger circle).
6. [DB1] Area of sector BCD = (1/2)s²(π/3) = (1/2)(3r²)(π/3) = πr²/2 (dependent on previous B1).
7. [B1] Area of triangle ABD = (1/2)r²sin(2π/3) = (1/2)r²(√3/2) = √3r²/4.
8. [DB1] Area of triangle BCD = (1/2)s²sin(π/3) = (1/2)(3r²)(√3/2) = 3√3r²/4 (dependent on B1 for angle BCD).
9. [B1] Area of shaded region = Area of sector DAB - Area of triangle ABD + Area of triangle BCD - Area of sector BCD = (πr²/3 - √3r²/4) + (3√3r²/4 - πr²/2) = (√3/2 - π/6)r². (a = √3/2, b = -1/6).