O-LevelAdditional MathematicsIndices and surds4037/22 · May/June 2015 · Q35 Marks

Do not use a calculator in this question. [Figure 3.1] The diagram shows the right-angled triangle ABC, where AB = (6 + 3√5)cm and angle B = 90°. The area of this triangle is (36+15√5)/2 cm².

(i)Find the length of the side BC in the form (a + b√5)cm, where a and b are integers.[3]
(ii)Find (AC)² in the form (c + d√5)cm², where c and d are integers.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

[M1] ((36+15√5)/(6+3√5)) x ((6-3√5)/(6-3√5)) oe or ((12+5√5)/(2+√5)) x ((2-√5)/(2-√5)) oe
[DM1] (216+90√5-108√5-225) / -9 or (24+10√5-12√5-25) / -1 oe or -(24+10√5)-12√5-25
[A1] 1+2√5 cao allow a = 1 and b = 2
Alternative method for (i):
[M1] 36+15√5 = (6a+15b)+(3a+6b)√5
[DM1] 6a+15b = 36 and 3a+6b = 15
[A1] a = 1 and b = 2 or 1+2√5
[M1] AC² = (6+3√5)² + their (1+2√5)² correct or correct ft expansions, using Pythagoras with (6+3√5) and their BC
[A1] 102 + 40√5 cao ignore attempts to square root after correct answer seen

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Topic: Indices and surds, O-Level Additional Mathematics (syllabus 4037). Structured question, 5 marks.
Source: O-Level Additional Mathematics, May/June 2015 Paper 2 Variant 2, question 3.

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