O-LevelAdditional MathematicsIndices and surds4037/12 · May/June 2024 · Q59 Marks

DO NOT USE A CALCULATOR IN THIS QUESTION. In this question, all lengths are in centimetres. The diagram shows the trapezium ABCD. The lengths of AB, BC and CD are 8√7-7, √7+2 and 9√7–9 respectively. The line BC is perpendicular to the lines AB and CD. [Figure 5.1]

(a)Find the perimeter of the trapezium, giving your answer in its simplest form.[3]
(b)Find the area of the trapezium, giving your answer in the form p√7+q, where p and q are rational numbers.[3]
(c)Find cotDBC, giving your answer in the form r√7+s, where r and s are simplified rational numbers.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[B1] AD² = (√7−2)² + (√7+2)² soi, oe
[B1] AD² =11–4√7+11+4√7 or AD = √11–4√7+11+4√7 This is the minimum detail that is acceptable for this mark
[B1] Perimeter = 18√7+ √22-14 isw Allow if previous B mark not awarded because of lack of detail.
[B1] 1/2 (8√7-7+9√7-9)(√7+2) oe
[B1] 1/2 (119+18√7-32) This is the minimum detail that is acceptable
[B1] 9√7+87/2 oe isw Allow if previous B mark not awarded because of lack of detail. Must be 2 separate terms.
[B1] cot θ = (√7+2)/(9(√7-1)) or 1/(9(3-√7)) oe May have factors of 9 multiplied out
[M1] (√7+2)(√7+1)/(9(√7-1)(√7+1)) or (3+√7)/(9(3-√7)(3+√7)) For rationalisation of their cot θ
[A1] √7/18 + 1/6

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