O-LevelMathematicsGeometry4024/22 · Oct/Nov 2015 · Q39 Marks
AC is a diameter of the circle, centre O. BCD and OED are straight lines. AC = 6 cm and CD = 3 cm. BẬC = 34°. Explain why BĈA = 56°.
- (a)(i)AC is a diameter of the circle, centre O. BCD and OED are straight lines. AC = 6 cm and CD = 3 cm. BẬC = 34°. Explain why BĈA = 56°.[1]
- (a)(ii)Find CÔD.[2]
- (a)(iii)Find OĈE.[1]
- (b)(i)In the diagram, PS is the bisector of QPR. QPT and QSR are straight lines. RT is parallel to SP. Explain why PT = PR.[2]
- (b)(ii)This diagram shows part of the above diagram. PQ = 12 cm, PR = 5 cm and QR = 8.5 cm. It is given that PQ/PR = QS/SR. Find SR.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.
- [1] Convincing explanation
- [2] B1 for OCD = 124 or triangle COD isosceles soi
- 28
- [1ft] 76
- [2] B1 for a correct pair of equal angles stated
- Convincing explanation
- [3] B1 for 8.5 - SR or 8.5 – QS seen and M1 for 12/5 = (8.5-SR)/SR or 12/5 = QS/(8.5-QS)
- 2.5
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