The diagram shows triangle ABC in which angle B is a right angle. The length of AB is 8 cm and the length of BC is 4 cm. The point D on AB is such that AD = 5cm. The sector DAC is part of a circle with centre D. [Figure 8.1]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.

1. [M1] tan BDC = 4/3 or sin BDC = 4/5 or cos BDC = 3/5 used to find ADC May use cosine rule or CAD = tan⁻¹ (4/8).
2. [A1] BDC = 0.927[3] → ADC = π−0.927[3] [=2.214to2.215] Allow degrees, 126.87, and 0.7048 π or 0.705 π.
3. [M1] Arc AC=5×their2.214 Use of rθ or (θ/360)*2πr Expect 11.07 .
4. [M1] AC = √8² +4² or 2×5×sin1.107 Expect 8.94 .
5. [A1] [Perimeter=11.07+8.94=]20.0 Accept AWRT [20.01, 20.02].
6. [M1] Sector ACD = ½×5²×their 2.214 Use of ½r²θ or (θ/360)*πr². Expect 27.7.
7. [M1] Subtracting the area of ∆ADC= ½×5×4 or ½5² sin their 2.214 or ½x8x4 - ½x3x4 Subtracting the area of ∆ADC, expect – 10.
8. [A1] Shaded area = 27.7-10=17.7 Accept AWRT [17.67, 17.68]. Correct answer cannot come from an angle of 2.215 .