The diagram shows points A and B on a circle with centre O and radius r. The tangents to the circle at A and B meet at T. The shaded region is bounded by the minor arc AB and the lines AT and BT. Angle AOB is 2θ radians. [Figure 8]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1] (i) AT or BT = rtanθ or OT = r/cosθ. May be seen on diagram.
2. [M1] (i) Both formulae, (½r²θ, ½bh or ½absinθ), seen with 2θ used when needed. Area of sector AOB = ½r²(2θ) = r²θ. Area of triangle OAT = ½ × OA × AT = ½r(rtanθ) = ½r²tanθ. Area of kite OATB = 2 × ½r²tanθ = r²tanθ. Area of shaded region = Area of kite OATB - Area of sector AOB = r²tanθ - r²θ. Equating area of sector AOB to shaded region: r²θ = r²tanθ - r²θ.
3. [A1] (i) Divide by r² and rearrange: θ = tanθ - θ, leading to tanθ = 2θ. Fully correct working from a correct statement. Note: ½r²2θ = ½ r²tanθ is a valid statement.
4. [B1] (ii) Length of arc AB = r(2θ). Given 19.2 = 8(2θ), so 2θ = 19.2/8 = 2.4 radians. Thus θ = 1.2 radians. Area of sector AOB = ½r²(2θ) = ½(8)²(2.4) = 32 × 2.4 = 76.8 cm².
5. [B1] (ii) Area of triangle OAT = ½r²tanθ = ½(8)²tan(1.2) = 32 tan(1.2) ≈ 32 × 2.572 = 82.30 cm². Area of kite OATB = 2 × 82.30 = 164.6 cm² (awrt 165).
6. [B1] (ii) Area of shaded region = Area of kite OATB - Area of sector AOB = 164.6 - 76.8 = 87.8 cm² (awrt 87.8 with little or no working can be awarded 3/3. SC Final answers that round to 88 with little or no working can be awarded 2/3.)