(a) [Figure 1] In Fig. 1, OAB is a sector of a circle with centre O and radius r. AX is the tangent at A to the arc AB and angle BAX = α. (b) [Figure 2] In Fig. 2, ABC is an equilateral triangle of side 4 cm. The lines AX, BX and CX are tangents to the equal circular arcs AB, BC and CA. Use the results in part (a) to find the area of the shaded region, giving your answer in terms of π and √3.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2016). The mark or mark type is shown in brackets.

1. [M1] Angle between tangent AX and radius OA is 90°. In triangle OAX, angle OAX = 90°. Angle OAB = 90° - α. Triangle OAB is isosceles (OA=OB=r), so angle OBA = OAB = 90° - α (Allow use of 90° or 180°)
2. [A1] Angle AOB = 180° - 2(90° - α) = 180° - 180° + 2α = 2α AG (Or other valid reasoning)
3. [B2,1,0] Area of sector = 1/2 r²(2α). Area of triangle OAB = 1/2 r² sin(2α). Shaded segment area = 1/2 r²(2α) - 1/2 r² sin(2α) oe (SCB1 for reversed subtraction)
4. [B1] For Fig. 2, consider one arc, say arc AB. The triangle AOB formed by the radii and chord AB. The line AX is tangent at A. The angle BAX is internal to the triangle ABC, so it is 60°/2 = 30° = π/6 rad. Thus α = π/6. (From (a)(i), AOB = 2α = π/3). The radius r = 4 cm. (Use of α = π/6, r = 4)
5. [B1] Correct identification of α and r.
6. [M1] Area of 1 segment = 1/2 r²(2α) - 1/2 r² sin(2α) = 1/2 (4²)(π/3) - 1/2 (4²) sin(π/3) (Ft their (a)(ii), α,r)
7. [B1] = 8π/3 - 8(√3/2) = 8π/3 - 4√3 (per segment)
8. [M1] Area of equilateral triangle ABC T = 1/2 (4²)(sin(π/3)) = 1/2 (16)(√3/2) = 4√3. (OR AXB = π/3, T = 4 tan(π/6))
9. [M1] Total shaded area = Area of equilateral triangle - 3 × (Area of 1 segment) = 16√3 - 3(8π/3 - 4√3) (OR 3(T/3 - S) = 3(4√3/3 - (8π/3 - 4√3)))
10. [A1] = 16√3 - 8π cao