O-LevelMathematicsGeometry4024/22 · Oct/Nov 2014 · Q26 Marks
[Figure 2.1] A, B, C and D are points on the circumference of the circle and AC is a diameter. AFBE and DCE are straight lines. DF is perpendicular to AE and CDF = 67°. [Figure 2.2] In the triangle PQR, the bisectors of PQR and PÂRQ intersect at Y. The straight line XYZ is parallel to QR. Prove that the perimeter of triangle PXZ = PQ+ PR.
- (a)(i)Find AÊD.[1]
- (a)(ii)Find CBE, giving a reason for your answer.[1]
- (a)(iii)Explain why DF is parallel to CB.[1]
- (b)Prove that the perimeter of triangle PXZ = PQ+ PR.[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 2014). The mark or mark type is shown in brackets.
- [1] 23
- [1] 90 with reason
- [1] Parallel lines established
- [3] Convincing argument
- [3] This must have e.g. XQ = XY justified. If there is no justification, then MAX B2 from B1 for XQ = XY oe And B1 for relating this to the perimeter of PXZ Or B1 for equal (alternate or bisected) angles
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