O-LevelMathematicsGeometry4024/21 · May/June 2017 · Q1012 Marks
(a) [Figure: Circle with points A, B, C, D on circumference, O at centre, angles 2y°, 120°, x°, y° shown] A, B, C and D lie on the circumference of a circle, centre O. AÔB = 120°, ADB = x°, OBD = y° and DÂO = 2y° . (b) [Figure: Sector OAB with angle 120° and radius r+4] Sector OAB has radius (r + 4) cm and its area is the same as the area of a circle of radius r cm.
- (a)(i)Find x, giving a reason for your answer.[2]
- (a)(ii)Find DÂB.[3]
- (a)(iii)Find BCD.[1]
- (b)(i)Show that r² – 4r – 8 = 0 .[3]
- (b)(ii)Calculate r.[3]
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Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [2] 60° angle at centre is twice angle at circumference
- [1] B1 for either correct
- [3] 70°
- [1] B2 for y = 20
- [1] B1 for OAB = 30 or OBA = 30 or 240
- [1] 110°
- [1] FT 180-(a)(ii) provided not negative answer
- [3] 120/360 π(r + 4)² = πr² leading to r² +8r + 16 = 3r² leading to r² – 4r – 8 = 0 without error
- [1] B1 for 120/360 π(r +4)²
- [1] M1 for forming equation and expanding (r + 4)²
- [3] r = 5.46 to 5.47
- [2] B2 for (-(-4) ± √(-4)²-4×1×-8) / 2 oe
- [1] B1 for (-(-4) ± √p) / (2×1) oe or (q ± √(-4)²-4×1×-8) / r
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