O-LevelMathematicsVectors and transformations4024/22 · May/June 2018 · Q810 Marks
OYC is a triangle. A is a point on OY and B is a point on CY. AB is parallel to OC. AC and OB intersect at X. [Figure: Triangle OYC with AB parallel to OC. X is intersection of AC and OB.]
- (a)Prove that triangle ABX is similar to triangle COX. Give a reason for each statement you make.[3]
- (b)(i)OA = 3a and OC = 6c and CB : BY = 1 : 2 . Find, as simply as possible, in terms of a and/or c, AB,[1]
- (b)(ii)Find, as simply as possible, in terms of a and/or c, CỶ.[2]
- (c)(i)Find, in its simplest form, the ratio OX : XB,[2]
- (c)(ii)Find, in its simplest form, the ratio area of triangle COX : area of triangle ABX,[1]
- (c)(iii)Find, in its simplest form, the ratio area of triangle AYB : area of trapezium OABC.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [3] B1 for two correct pairs of angles
- B1 for correct reason for one pair of angles
- ∠BAX= ∠OCX, alternate [angles]
- ∠ABX= ∠COX, alternate [angles]
- ZAXB = ∠CXO, [vertically] opposite
- [1] 4c
- [2] B1 for answer 9a + kc or ka – 6c (k≠ 0)
- [2] B1 for 3k: 2k, where k is an integer
- [1] FT their 3^2 : their 2^2
- [1] 4:5
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