O-LevelMathematicsGeometry4024/22 · May/June 2020 · Q99 Marks
[Figure 9.1] EMF and GLNH are parallel lines. LM = LN and GÎM = 126°. Find FMN. Give a reason for each step of your working.
- (a)[Figure 9.1] EMF and GLNH are parallel lines. LM = LN and GÎM = 126°. Find FMN. Give a reason for each step of your working.[4]
- (b)[Figure 9.2] A, B, C and D are points on the circumference of a circle, centre O. BD and AC intersect at E and BC is a diameter of the circle. ACD = x° and DỘC = y°. Find an expression, in terms of x and/or y, for
- (b)(i)DBC,[1]
- (b)(ii)ABD,[1]
- (b)(iii)AED,[2]
- (b)(iv)BDA.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [4] (a) 63
- [4] B1 MÎN = 54 angles on a straight line or LMN + MNL = 126 exterior angle of a triangle
- [4] B1 MÑL = 63 or NML = 63 angles in isosceles (triangle)
- [4] B1 MNL = FMN alternate angles or GÎM = LMF alternate angles or EML = 54 co-interior with GLM (or alternate with MLN ) and angles on a straight line
- [1] (b)(i) y/2 oe
- [1] (b)(ii) x
- [2] (b)(iii) 90+x oe
- [2] FT 90+their (a)(ii)
- [2] B1 for BÂC = 90° or BDC =90°
- [1] (b)(iv) 90-x-y/2 oe
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